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IB Physics · Concept clusters

IB Physics concept clusters

Original, past-paper-style synoptic questions. Each one is built around a single setup and needs ideas from several subchapters at once, so you practise connecting topics the way long exam questions expect. 48 questions · 1,088 marks · two per subchapter · every part with a worked solution · All courses

Theme

Subchapter

Level

Skill

Theme A · Space, time and motion

● home subchapter · ○ linked subchapter · last column: links to other themes

Setup	A.1	A.2	A.3	A.4	A.5	Other
A delivery drone: climb, hover and parcel drop	●	○	○			—
A big-air ski jump: take-off, flight and landing	●	○	○	○		—
Weighing yourself in a moving lift	○	●	○			—
Curling: friction on ice and a collision between stones	○	●	○			—
An e-bike climbing a hill: power, forces and efficiency	○	○	●			B.5
Pumped-hydro storage: storing a grid's surplus energy	○		●			B.5 B.2
A flywheel energy-recovery system on a city bus	○	○	○	●		—
A bascule bridge: counterweight, torque and opening the deck	○	○	○	●		—
A fast interstellar probe sending laser pulses home	○				●	C.2
A kaon beam in an accelerator: where to place the detector	○				●	E.3

A.1 Kinematics

+ Add to test✓ In your testCC-IBP-A1-122 marks~25 min

A delivery drone: climb, hover and parcel drop

A.1 KinematicsA.2 Forces and MomentumA.3 Work, Energy and PowerSL

A delivery drone carrying a parcel has a total mass of 3.0 kg. The drone takes off vertically from rest on level ground. Figure 1 shows how its vertical velocity v varies with time t during the climb, until it hovers.

Air resistance on the drone is negligible. Take g = 9.8 m s⁻².

Figure 1: velocity–time graph for the vertical climb of the drone

(a)

Determine the acceleration of the drone during the first 4.0 s.

GRAPHNotes

Show solution

acceleration = gradient of the v–t graph = 3.0 ÷ 4.0	✓ 1	Award this mark for a correct gradient expression from the graph.
0.75 m s⁻²	✓ 1	Accept 0.8 m s⁻². Unit required.

Answer: 0.75 m s⁻²

(2)

(b)

Show that the drone hovers at a height of 45 m above the ground.

GRAPHNotes

Show solution

displacement = area under the graph	✓ 1	Must be stated or clearly implied by the working.
½ × 4.0 × 3.0 + 11 × 3.0 + ½ × 4.0 × 3.0 = 6.0 + 33 + 6.0 = 45 m	✓ 1	Accept the trapezium: ½(11 + 19) × 3.0 = 45 m. The full substitution must be seen; 45 m alone scores 0.

Answer: 45 m

(2)

(c)

Calculate the upward force exerted by the air on the drone (the thrust) during the first 4.0 s.

CALCNotes

Show solution

F − mg = ma so F = 3.0 × (9.8 + 0.75)	✓ 1	Allow ECF from (a).
32 N	✓ 1	Accept 31.7 N or 31.65 N. 29 N (weight only) scores 0.

Answer: 32 N

(2)

(d)(i)

The drone climbs at constant velocity between t = 4.0 s and t = 15 s.

Outline why the thrust is equal to the weight of the drone during this time.

REASONNotes

Show solution

velocity is constant so acceleration is zero, hence the resultant force is zero (Newton's first law)

✓ 1

Reference to zero acceleration or zero resultant force is needed; "the forces balance" alone is not enough.

(1)

(d)(ii)

Calculate the useful power output of the drone's motors during this time.

CALCNotes

Show solution

P = Fv = (3.0 × 9.8) × 3.0	✓ 1	Thrust = weight = 29.4 N must be used. Allow use of ΔE_p/Δt = 3.0 × 9.8 × 33 ÷ 11.
88 W	✓ 1	Accept 88.2 W. Using 32 N from (c) gives 95 W and scores 1 mark.

Answer: 88 W

(2)

(e)(i)

The drone then flies horizontally at a constant speed of 8.0 m s⁻¹, 45 m above level ground, and releases the parcel. Air resistance on the parcel is negligible.

Show that the parcel takes about 3 s to reach the ground.

DERIVENotes

Show solution

initial vertical velocity is zero, so 45 = ½ × 9.8 × t²	✓ 1	Recognition that the vertical motion starts from rest must be clear.
t = 3.03 s	✓ 1	Answer to at least 2 s.f. (3.0 s or 3.03 s); 3 s alone is not enough.

Answer: 3.03 s

(2)

(e)(ii)

Calculate the horizontal distance between the point of release and the point where the parcel lands.

CALCNotes

Show solution

8.0 × 3.03 = 24 m

✓ 1

Accept 24.2 m. Allow 24 m from t = 3.0 s.

Answer: 24 m

(1)

(e)(iii)

Determine the speed of the parcel as it reaches the ground.

MULTINotes

Show solution

vertical component = 9.8 × 3.03 = 29.7 m s⁻¹ or ½v² = ½ × 8.0² + 9.8 × 45	✓ 1	Energy method: mass cancels; ΔE_p converts to extra E_k.
v = √(8.0² + 29.7²) = 31 m s⁻¹	✓ 1	Accept 30.8 m s⁻¹. 29.7 m s⁻¹ (vertical component only) scores 1 mark.

Answer: 31 m s⁻¹

(2)

(f)

On the axes in Figure 2, sketch graphs to show how the vertical component of the parcel's velocity varies with time from release until it lands, taking downward as positive:

when air resistance is negligible (label this line N)
when air resistance is not negligible (label this line R).

Figure 2: axes for the vertical component of the parcel's velocity (downward positive)

SKETCHNotes

Show solution

N: straight line from the origin ending at about 30 m s⁻¹ at about 3.0 s	✓ 1	Allow ECF from (e)(i) and (e)(iii) for the end point.
R: starts at the origin with the same initial gradient as N, then its gradient decreases	✓ 1	A curve that starts below N with a smaller initial gradient scores 0 for this mark.
R: always below N and ends at a lower vertical velocity at a later time than N	✓ 1	Both features needed.

(3)

(g)

Explain why, when air resistance is not negligible, the parcel lands closer to the point directly below the point of release.

EXPLAINNotes

Show solution

air resistance acts opposite to the velocity, so it has a horizontal component opposite to the horizontal motion	✓ 1	Accept "drag acts backwards along the path".
the horizontal velocity decreases during the fall, so the parcel covers less than 8.0 m horizontally each second	✓ 1	Do not accept "the parcel falls more slowly" alone.

(2)

(h)

On another delivery, the drone is moving horizontally at 8.0 m s⁻¹ and also climbing vertically at 2.0 m s⁻¹ when it releases the parcel at a height of 45 m. Air resistance is negligible.

Determine the horizontal distance travelled by the parcel before it lands.

MULTINotes

Show solution

initial vertical velocity of the parcel = 2.0 m s⁻¹ upward: −45 = 2.0t − 4.9t²	✓ 1	Sign convention must be consistent. Using u = 0 scores 0 for the part.
t = 3.24 s	✓ 1	Accept 3.2 s. Allow the two-stage method: 0.20 s rising 0.20 m, then fall from 45.2 m in 3.04 s.
horizontal distance = 8.0 × 3.24 = 26 m	✓ 1	Accept 25.9 m. Allow ECF from an incorrect time.

Answer: 26 m

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
acceleration = gradient of the v–t graph = 3.0 ÷ 4.0	✓ 1	Award this mark for a correct gradient expression from the graph.
0.75 m s⁻²	✓ 1	Accept 0.8 m s⁻². Unit required.
Part (b)
displacement = area under the graph	✓ 1	Must be stated or clearly implied by the working.
½ × 4.0 × 3.0 + 11 × 3.0 + ½ × 4.0 × 3.0 = 6.0 + 33 + 6.0 = 45 m	✓ 1	Accept the trapezium: ½(11 + 19) × 3.0 = 45 m. The full substitution must be seen; 45 m alone scores 0.
Part (c)
F − mg = ma so F = 3.0 × (9.8 + 0.75)	✓ 1	Allow ECF from (a).
32 N	✓ 1	Accept 31.7 N or 31.65 N. 29 N (weight only) scores 0.
Part (d)(i)
velocity is constant so acceleration is zero, hence the resultant force is zero (Newton's first law)	✓ 1	Reference to zero acceleration or zero resultant force is needed; "the forces balance" alone is not enough.
Part (d)(ii)
P = Fv = (3.0 × 9.8) × 3.0	✓ 1	Thrust = weight = 29.4 N must be used. Allow use of ΔE_p/Δt = 3.0 × 9.8 × 33 ÷ 11.
88 W	✓ 1	Accept 88.2 W. Using 32 N from (c) gives 95 W and scores 1 mark.
Part (e)(i)
initial vertical velocity is zero, so 45 = ½ × 9.8 × t²	✓ 1	Recognition that the vertical motion starts from rest must be clear.
t = 3.03 s	✓ 1	Answer to at least 2 s.f. (3.0 s or 3.03 s); 3 s alone is not enough.
Part (e)(ii)
8.0 × 3.03 = 24 m	✓ 1	Accept 24.2 m. Allow 24 m from t = 3.0 s.
Part (e)(iii)
vertical component = 9.8 × 3.03 = 29.7 m s⁻¹ or ½v² = ½ × 8.0² + 9.8 × 45	✓ 1	Energy method: mass cancels; ΔE_p converts to extra E_k.
v = √(8.0² + 29.7²) = 31 m s⁻¹	✓ 1	Accept 30.8 m s⁻¹. 29.7 m s⁻¹ (vertical component only) scores 1 mark.
Part (f)
N: straight line from the origin ending at about 30 m s⁻¹ at about 3.0 s	✓ 1	Allow ECF from (e)(i) and (e)(iii) for the end point.
R: starts at the origin with the same initial gradient as N, then its gradient decreases	✓ 1	A curve that starts below N with a smaller initial gradient scores 0 for this mark.
R: always below N and ends at a lower vertical velocity at a later time than N	✓ 1	Both features needed.
Part (g)
air resistance acts opposite to the velocity, so it has a horizontal component opposite to the horizontal motion	✓ 1	Accept "drag acts backwards along the path".
the horizontal velocity decreases during the fall, so the parcel covers less than 8.0 m horizontally each second	✓ 1	Do not accept "the parcel falls more slowly" alone.
Part (h)
initial vertical velocity of the parcel = 2.0 m s⁻¹ upward: −45 = 2.0t − 4.9t²	✓ 1	Sign convention must be consistent. Using u = 0 scores 0 for the part.
t = 3.24 s	✓ 1	Accept 3.2 s. Allow the two-stage method: 0.20 s rising 0.20 m, then fall from 45.2 m in 3.04 s.
horizontal distance = 8.0 × 3.24 = 26 m	✓ 1	Accept 25.9 m. Allow ECF from an incorrect time.

Answers: (a) 0.75 m s⁻² · (b) 45 m · (c) 32 N · (d)(ii) 88 W · (e)(i) 3.03 s · (e)(ii) 24 m · (e)(iii) 31 m s⁻¹ · (h) 26 m

Common error (b) Reading the hover height from the v-axis. The height is the area under a v–t graph; the graph's value is the speed at that instant.

Exam tip (e)(iii) When only the speed at a height is asked for and air resistance is negligible, energy conservation gives it in one line: v² = u² + 2gh, whatever the launch direction.

More A.1 questions in the question bank

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A big-air ski jump: take-off, flight and landing

A.1 KinematicsA.2 Forces and MomentumA.3 Work, Energy and PowerA.4 Rigid Body MechanicsSL

A freestyle skier of mass 72 kg leaves the lip of a snow kicker with a speed of 15.0 m s⁻¹ at 30° above the horizontal. The lip is 2.0 m above a horizontal landing deck, as shown in Figure 1. The skier can be treated as a point, and air resistance is negligible during the flight. Take g = 9.8 m s⁻².

Figure 1: the kicker, the deck and the skier's path (skier treated as a point)

(a)

Calculate the horizontal and vertical components of the skier's velocity at take-off.

CALCNotes

Show solution

horizontal: 15.0 cos 30° = 13.0 m s⁻¹	✓ 1	Accept 12.99 m s⁻¹.
vertical: 15.0 sin 30° = 7.5 m s⁻¹	✓ 1	Components swapped scores 0 for both marks.

Answer: 13.0 m s⁻¹ horizontal; 7.5 m s⁻¹ vertical

(2)

(b)

Determine the maximum height of the skier above the deck.

CALCNotes

Show solution

height above the lip: 0 = 7.5² − 2 × 9.8 × h, h = 2.87 m	✓ 1	Allow ECF from (a).
height above the deck = 2.87 + 2.0 = 4.9 m	✓ 1	Accept 4.87 m. 2.9 m (above the lip) scores 1 mark.

Answer: 4.9 m

(2)

(c)

Show that the horizontal distance from the lip to the landing point is about 23 m.

DERIVENotes

Show solution

vertical motion, upward positive: −2.0 = 7.5t − 4.9t²	✓ 1	Alternative: 0.77 s to the top, then a fall of 4.87 m from rest taking 1.00 s.
t = 1.76 s	✓ 1	Ignore the negative root if seen.
x = 13.0 × 1.76 = 22.9 m	✓ 1	Final answer to at least 3 s.f., not just 23 m.

Answer: 22.9 m

(3)

(d)

On the axes in Figure 2, sketch graphs to show how the horizontal component and the vertical component of the skier's velocity vary with time from take-off until the skier lands. Label the graphs H and V.

Figure 2: axes for the velocity components of the skier (upward and forward positive)

SKETCHNotes

Show solution

H: horizontal line at 13 m s⁻¹ for the whole flight	✓ 1	Allow ECF from (a).
V: straight line starting at 7.5 m s⁻¹ and crossing zero at about 0.77 s	✓ 1	The line must be straight with a negative gradient (−9.8 m s⁻²).
V ends at about −9.8 m s⁻¹ at about 1.76 s; both graphs end at the same time	✓ 1	Allow ECF from (c) for the end time.

(3)

(e)

The skier starts from rest at the top of the in-run, which is 14.0 m vertically above the lip.

Determine the percentage of the gravitational potential energy lost by the skier on the in-run that is dissipated by friction and air resistance.

MULTINotes

Show solution

ΔE_p = 72 × 9.8 × 14.0 = 9.88 × 10³ J	✓ 1	Accept 9878 J.
E_k at the lip = ½ × 72 × 15.0² = 8.10 × 10³ J	✓ 1
(9878 − 8100) ÷ 9878 × 100 = 18 %	✓ 1	Accept 18.0 %. 82 % (fraction transferred to E_k) scores 2 marks.

Answer: 18 %

(3)

(f)

On landing, the skier's vertical velocity is reduced to zero in 0.40 s while the horizontal velocity is unchanged.

Determine the average vertical force exerted by the deck on the skier during the landing.

MULTINotes

Show solution

vertical speed at landing = 7.5 − 9.8 × 1.76 = −9.77 m s⁻¹, so Δp = 72 × 9.77 = 703 kg m s⁻¹	✓ 1	Allow ECF from (c). Energy method for the vertical speed accepted: √(7.5² + 2 × 9.8 × 2.0).
resultant force = 703 ÷ 0.40 = 1.76 × 10³ N	✓ 1
deck force = 1.76 × 10³ + 72 × 9.8 = 2.5 × 10³ N	✓ 1	Accept 2.46 kN. 1.8 kN (weight omitted) scores 2 marks.

Answer: 2.5 × 10³ N

(3)

(g)

Explain why skiers bend their knees as they land.

EXPLAINNotes

Show solution

bending the knees increases the time taken to stop the vertical motion	✓ 1	Accept "increases the stopping distance" only if linked to time.
the change in momentum is the same, so the average force (rate of change of momentum) is smaller	✓ 1	Must refer to the same momentum change or impulse.

(2)

(h)

On a later jump the skier leaves the lip at the same speed of 15.0 m s⁻¹ but at 40° above the horizontal.

Determine the new horizontal distance from the lip to the landing point.

MULTINotes

Show solution

components: 11.5 m s⁻¹ horizontal and 9.64 m s⁻¹ vertical	✓ 1
−2.0 = 9.64t − 4.9t², t = 2.16 s	✓ 1
x = 11.5 × 2.16 = 25 m	✓ 1	Accept 24.8 m. Allow ECF from an incorrect time.

Answer: 25 m

(3)

(i)

During the flight the skier spins about a vertical axis through the body. The resultant external torque on the skier about this axis is negligible.

Explain why pulling the arms in towards the body increases the rate of spin.

HL onlyEXPLAINNotes

Show solution

no external torque, so angular momentum L = Iω is conserved	✓ 1	Reference to conservation of angular momentum is needed.
mass moves closer to the axis, so the moment of inertia decreases and the angular velocity increases	✓ 1	Both steps needed for the mark.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
horizontal: 15.0 cos 30° = 13.0 m s⁻¹	✓ 1	Accept 12.99 m s⁻¹.
vertical: 15.0 sin 30° = 7.5 m s⁻¹	✓ 1	Components swapped scores 0 for both marks.
Part (b)
height above the lip: 0 = 7.5² − 2 × 9.8 × h, h = 2.87 m	✓ 1	Allow ECF from (a).
height above the deck = 2.87 + 2.0 = 4.9 m	✓ 1	Accept 4.87 m. 2.9 m (above the lip) scores 1 mark.
Part (c)
vertical motion, upward positive: −2.0 = 7.5t − 4.9t²	✓ 1	Alternative: 0.77 s to the top, then a fall of 4.87 m from rest taking 1.00 s.
t = 1.76 s	✓ 1	Ignore the negative root if seen.
x = 13.0 × 1.76 = 22.9 m	✓ 1	Final answer to at least 3 s.f., not just 23 m.
Part (d)
H: horizontal line at 13 m s⁻¹ for the whole flight	✓ 1	Allow ECF from (a).
V: straight line starting at 7.5 m s⁻¹ and crossing zero at about 0.77 s	✓ 1	The line must be straight with a negative gradient (−9.8 m s⁻²).
V ends at about −9.8 m s⁻¹ at about 1.76 s; both graphs end at the same time	✓ 1	Allow ECF from (c) for the end time.
Part (e)
ΔE_p = 72 × 9.8 × 14.0 = 9.88 × 10³ J	✓ 1	Accept 9878 J.
E_k at the lip = ½ × 72 × 15.0² = 8.10 × 10³ J	✓ 1
(9878 − 8100) ÷ 9878 × 100 = 18 %	✓ 1	Accept 18.0 %. 82 % (fraction transferred to E_k) scores 2 marks.
Part (f)
vertical speed at landing = 7.5 − 9.8 × 1.76 = −9.77 m s⁻¹, so Δp = 72 × 9.77 = 703 kg m s⁻¹	✓ 1	Allow ECF from (c). Energy method for the vertical speed accepted: √(7.5² + 2 × 9.8 × 2.0).
resultant force = 703 ÷ 0.40 = 1.76 × 10³ N	✓ 1
deck force = 1.76 × 10³ + 72 × 9.8 = 2.5 × 10³ N	✓ 1	Accept 2.46 kN. 1.8 kN (weight omitted) scores 2 marks.
Part (g)
bending the knees increases the time taken to stop the vertical motion	✓ 1	Accept "increases the stopping distance" only if linked to time.
the change in momentum is the same, so the average force (rate of change of momentum) is smaller	✓ 1	Must refer to the same momentum change or impulse.
Part (h)
components: 11.5 m s⁻¹ horizontal and 9.64 m s⁻¹ vertical	✓ 1
−2.0 = 9.64t − 4.9t², t = 2.16 s	✓ 1
x = 11.5 × 2.16 = 25 m	✓ 1	Accept 24.8 m. Allow ECF from an incorrect time.
Part (i)
no external torque, so angular momentum L = Iω is conserved	✓ 1	Reference to conservation of angular momentum is needed.
mass moves closer to the axis, so the moment of inertia decreases and the angular velocity increases	✓ 1	Both steps needed for the mark.

Answers: (a) 13.0 m s⁻¹ horizontal; 7.5 m s⁻¹ vertical · (b) 4.9 m · (c) 22.9 m · (e) 18 % · (f) 2.5 × 10³ N · (h) 25 m

Common error (c) Using a time of flight of 2u sin θ / g. That result only holds when the projectile lands at its launch height; here it lands 2.0 m lower, so the quadratic must be solved.

More A.1 questions in the question bank

A.2 Forces and Momentum

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Weighing yourself in a moving lift

A.2 Forces and MomentumA.1 KinematicsA.3 Work, Energy and PowerSL

A student of mass 55.0 kg stands on a set of bathroom scales on the floor of a lift. The scales display the force they exert on the student, in newtons. The lift starts from rest at the ground floor and travels upwards to the top floor, where it stops.

An app on a phone in the lift records the vertical acceleration a of the lift. Figure 1 shows how a varies with time t during the trip. Take g = 9.8 m s⁻².

Figure 1: acceleration–time graph recorded by the phone during the trip (upward positive)

(a)(i)

The scales exert an upward normal force on the student. State the force that forms a Newton's third law pair with this force, including its direction.

RECALLNotes

Show solution

the (downward) normal force exerted by the student on the scales

✓ 1

Both the body exerting the force and the body it acts on are needed. "The student's weight" scores 0.

(1)

(a)(ii)

When the lift is at rest, the weight of the student and the normal force on the student are equal in magnitude and opposite in direction. Explain why they are not a Newton's third law pair.

EXPLAINNotes

Show solution

both forces act on the same body (the student)	✓ 1	A third-law pair acts on two different bodies.
they are different types of force: one is gravitational, the other is a contact (normal) force	✓ 1	Accept: the partner of the weight is the gravitational pull of the student on the Earth.

(2)

(b)

Using Figure 1, calculate the reading on the scales during each stage of the trip:

0 to 2.0 s
2.0 s to 14.0 s
14.0 s to 16.5 s.

DATANotes

Show solution

0 to 2.0 s: F_N = 55.0 × (9.8 + 1.0) = 594 N	✓ 1	Award the first mark for F_N − mg = ma applied to any stage if the value is wrong.
2.0 s to 14.0 s: F_N = 55.0 × 9.8 = 539 N	✓ 1
14.0 s to 16.5 s: F_N = 55.0 × (9.8 − 0.80) = 495 N	✓ 1	Allow ECF for a misread acceleration, once only.

Answer: 594 N; 539 N; 495 N

(3)

(c)

On the axes in Figure 2, sketch the velocity–time graph for the lift from t = 0 to t = 18 s.

Figure 2: axes for the velocity–time graph of the lift

SKETCHNotes

Show solution

straight line from the origin to 2.0 m s⁻¹ at 2.0 s	✓ 1	Change in velocity = area under the a–t graph: 1.0 × 2.0 = 2.0 m s⁻¹.
horizontal line at 2.0 m s⁻¹ from 2.0 s to 14.0 s	✓ 1
straight line from 2.0 m s⁻¹ at 14.0 s down to zero at 16.5 s, then zero to 18 s	✓ 1	0.80 × 2.5 = 2.0 m s⁻¹ is removed, so the lift is at rest from 16.5 s.

(3)

(d)

Determine the total height travelled by the lift.

MULTINotes

Show solution

area under the v–t graph: ½ × 2.0 × 2.0 + 12.0 × 2.0 + ½ × 2.5 × 2.0	✓ 1	Allow ECF from the graph in (c).
= 2.0 + 24 + 2.5 = 28.5 m	✓ 1	Accept 28 m or 29 m.

Answer: 28.5 m

(2)

(e)

The mass of the lift car, not including the student, is 850 kg. A single cable pulls the lift car vertically upwards. Resistive forces are negligible.

Show that the tension in the cable during the first 2.0 s is about 9.8 kN.

DERIVENotes

Show solution

total mass accelerated = 850 + 55.0 = 905 kg	✓ 1	Using 850 kg only gives 9.2 kN and scores 0.
T = 905 × (9.8 + 1.0) = 9774 N	✓ 1	At least 3 s.f. required: 9770 N or 9774 N.

Answer: 9.77 kN

(2)

(f)(i)

Calculate the useful power output of the motor while the lift moves at constant speed.

CALCNotes

Show solution

at constant speed T = 905 × 9.8 = 8869 N	✓ 1	Using 9774 N from (e) scores 0 for this mark.
P = Tv = 8869 × 2.0 = 1.8 × 10⁴ W	✓ 1	Accept 17.7 kW. Allow ECF for v from (c).

Answer: 1.8 × 10⁴ W

(2)

(f)(ii)

The electrical power input to the motor at this time is 22 kW. Calculate the efficiency of the motor.

CALCNotes

Show solution

17.7 ÷ 22 = 0.81 (81 %)

✓ 1

Allow ECF from (f)(i). Accept 0.80 to 0.82 (0.82 from 1.8 × 10⁴ W) or 80 % to 82 %.

Answer: 0.81

(1)

(g)

In the real installation the cable passes over a pulley wheel driven by the motor, and its other end is attached to a counterweight of mass 600 kg, which moves down as the lift car moves up.

Calculate the useful power output of the motor while the lift car moves upwards at constant speed. Assume the pulley wheel has negligible mass and friction.

MULTINotes

Show solution

the motor must supply only the difference between the tensions: (905 − 600) × 9.8 = 2989 N	✓ 1	Accept an energy argument: rate of gain of E_p of the car minus rate of loss of E_p of the counterweight.
P = 2989 × 2.0 = 6.0 × 10³ W	✓ 1	Accept 5.98 kW.

Answer: 6.0 × 10³ W

(2)

(h)

During a safety test with a dummy load on the scales, the cable is cut and the lift car is allowed to fall freely for a short time.

Predict the reading on the scales during the free fall. Explain your answer.

PREDICTNotes

Show solution

reading = 0 (zero)	✓ 1
the dummy, the scales and the lift all accelerate downwards at g	✓ 1	Accept: F_N = m(g + a) with a = −g.
the weight alone provides the resultant force needed, so no normal (contact) force acts between the dummy and the scales	✓ 1	The weight itself has not become zero; an answer stating "there is no gravity" loses this mark.

Answer: 0 N

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
the (downward) normal force exerted by the student on the scales	✓ 1	Both the body exerting the force and the body it acts on are needed. "The student's weight" scores 0.
Part (a)(ii)
both forces act on the same body (the student)	✓ 1	A third-law pair acts on two different bodies.
they are different types of force: one is gravitational, the other is a contact (normal) force	✓ 1	Accept: the partner of the weight is the gravitational pull of the student on the Earth.
Part (b)
0 to 2.0 s: F_N = 55.0 × (9.8 + 1.0) = 594 N	✓ 1	Award the first mark for F_N − mg = ma applied to any stage if the value is wrong.
2.0 s to 14.0 s: F_N = 55.0 × 9.8 = 539 N	✓ 1
14.0 s to 16.5 s: F_N = 55.0 × (9.8 − 0.80) = 495 N	✓ 1	Allow ECF for a misread acceleration, once only.
Part (c)
straight line from the origin to 2.0 m s⁻¹ at 2.0 s	✓ 1	Change in velocity = area under the a–t graph: 1.0 × 2.0 = 2.0 m s⁻¹.
horizontal line at 2.0 m s⁻¹ from 2.0 s to 14.0 s	✓ 1
straight line from 2.0 m s⁻¹ at 14.0 s down to zero at 16.5 s, then zero to 18 s	✓ 1	0.80 × 2.5 = 2.0 m s⁻¹ is removed, so the lift is at rest from 16.5 s.
Part (d)
area under the v–t graph: ½ × 2.0 × 2.0 + 12.0 × 2.0 + ½ × 2.5 × 2.0	✓ 1	Allow ECF from the graph in (c).
= 2.0 + 24 + 2.5 = 28.5 m	✓ 1	Accept 28 m or 29 m.
Part (e)
total mass accelerated = 850 + 55.0 = 905 kg	✓ 1	Using 850 kg only gives 9.2 kN and scores 0.
T = 905 × (9.8 + 1.0) = 9774 N	✓ 1	At least 3 s.f. required: 9770 N or 9774 N.
Part (f)(i)
at constant speed T = 905 × 9.8 = 8869 N	✓ 1	Using 9774 N from (e) scores 0 for this mark.
P = Tv = 8869 × 2.0 = 1.8 × 10⁴ W	✓ 1	Accept 17.7 kW. Allow ECF for v from (c).
Part (f)(ii)
17.7 ÷ 22 = 0.81 (81 %)	✓ 1	Allow ECF from (f)(i). Accept 0.80 to 0.82 (0.82 from 1.8 × 10⁴ W) or 80 % to 82 %.
Part (g)
the motor must supply only the difference between the tensions: (905 − 600) × 9.8 = 2989 N	✓ 1	Accept an energy argument: rate of gain of E_p of the car minus rate of loss of E_p of the counterweight.
P = 2989 × 2.0 = 6.0 × 10³ W	✓ 1	Accept 5.98 kW.
Part (h)
reading = 0 (zero)	✓ 1
the dummy, the scales and the lift all accelerate downwards at g	✓ 1	Accept: F_N = m(g + a) with a = −g.
the weight alone provides the resultant force needed, so no normal (contact) force acts between the dummy and the scales	✓ 1	The weight itself has not become zero; an answer stating "there is no gravity" loses this mark.

Answers: (b) 594 N; 539 N; 495 N · (d) 28.5 m · (e) 9.77 kN · (f)(i) 1.8 × 10⁴ W · (f)(ii) 0.81 · (g) 6.0 × 10³ W · (h) 0 N

Exam tip (b) Write Newton's second law for the student with upward positive: F_N − mg = ma. A negative a while moving up simply means the lift is slowing down.

Common error (f)(i) Putting the resultant force into P = Fv. At constant speed the resultant force is zero; use the force the motor's cable actually exerts.

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Curling: friction on ice and a collision between stones

A.2 Forces and MomentumA.1 KinematicsA.3 Work, Energy and PowerSL

In a game of curling, stone A of mass 19.0 kg is released with a speed of 2.4 m s⁻¹ along a straight line on horizontal ice. It slides 28.0 m in 14.0 s and then strikes a stationary stone B of the same mass head-on (Figure 1). The friction force on a sliding stone is constant, and air resistance is negligible. Take g = 9.8 m s⁻².

Figure 1: plan view of the ice (not to scale)

(a)

Show that the speed of stone A just before it strikes stone B is 1.6 m s⁻¹.

DERIVENotes

Show solution

uniform deceleration, so s = ½(u + v)t: 28.0 = ½(2.4 + v) × 14.0	✓ 1	Accept: average speed = 28.0 ÷ 14.0 = 2.0 m s⁻¹ = (2.4 + v)/2.
v = 4.0 − 2.4 = 1.6 m s⁻¹	✓ 1	The rearrangement must be seen; the given answer alone scores 0.

Answer: 1.6 m s⁻¹

(2)

(b)(i)

Calculate the magnitude of the deceleration of stone A.

CALCNotes

Show solution

(2.4 − 1.6) ÷ 14.0 = 0.057 m s⁻²

✓ 1

Accept 0.0571 m s⁻².

Answer: 0.057 m s⁻²

(1)

(b)(ii)

Determine the coefficient of dynamic friction between the stone and the ice.

CALCNotes

Show solution

friction is the resultant force: μ_dmg = ma, so μ_d = a/g	✓ 1	F_N = mg on horizontal ice.
μ_d = 0.0571 ÷ 9.8 = 0.0058	✓ 1	Accept 0.0058 to 0.006. No unit. Allow ECF from (b)(i).

Answer: 0.0058

(2)

(c)

Immediately after the collision, stone B moves forward with a speed of 1.4 m s⁻¹. Friction during the brief collision is negligible.

Calculate the velocity of stone A immediately after the collision.

CALCNotes

Show solution

19.0 × 1.6 = 19.0 × v_A + 19.0 × 1.4	✓ 1	Conservation of momentum must be applied.
v_A = 0.20 m s⁻¹ in the original direction	✓ 1	Direction needed. Accept 0.2 m s⁻¹ forward.

Answer: 0.20 m s⁻¹ forward

(2)

(d)

Show that the collision is not perfectly elastic, and calculate the kinetic energy transferred to other forms.

DERIVENotes

Show solution

E_k before = ½ × 19.0 × 1.6² = 24.3 J	✓ 1
E_k after = ½ × 19.0 × (0.20² + 1.4²) = 19.0 J, which is less, so kinetic energy is not conserved	✓ 1	The comparison with the conclusion is needed. Allow ECF from (c).
energy transferred = 24.3 − 19.0 = 5.3 J	✓ 1	Accept 5.32 J.

Answer: 5.3 J

(3)

(e)(i)

Figure 2 shows how the force exerted on stone B by stone A varies with time during the contact, which lasts 2.0 ms.

State what the area under the graph represents, and deduce its value.

Figure 2: force exerted on stone B by stone A during the contact

GRAPHNotes

Show solution

the impulse on stone B, equal to its change of momentum	✓ 1	Accept "change in momentum of B".
19.0 × 1.4 = 27 N s	✓ 1	Accept 26.6 N s or kg m s⁻¹.

Answer: 27 N s

(2)

(e)(ii)

By treating the graph as a triangle, estimate the peak force exerted on stone B.

GRAPHNotes

Show solution

½ × F_peak × 2.0 × 10⁻³ = 26.6	✓ 1	Time must be converted to seconds.
F_peak = 2.7 × 10⁴ N	✓ 1	Accept 26.6 kN. Allow ECF from (e)(i). 1.3 × 10⁴ N (average force) scores 1 mark.

Answer: 2.7 × 10⁴ N

(2)

(e)(iii)

State the size of the peak force exerted on stone A by stone B during the contact.

RECALLNotes

Show solution

the same as in (e)(ii), 2.7 × 10⁴ N, by Newton's third law (in the opposite direction)

✓ 1

Allow ECF from (e)(ii). Reference to the third law or "equal and opposite" needed.

Answer: 2.7 × 10⁴ N

(1)

(f)

Determine the distance stone B slides after the collision before it stops.

MULTINotes

Show solution

same deceleration of 0.0571 m s⁻², since μ_dmg/m does not depend on mass: 0 = 1.4² − 2 × 0.0571 × s	✓ 1	Allow ECF from (b).
s = 17 m	✓ 1	Accept 17.2 m.

Answer: 17 m

(2)

(g)

Players sweep the ice in front of a moving stone. This warms the ice surface and reduces the coefficient of dynamic friction.

Explain the effect of sweeping on the distance travelled by a stone released at a given speed.

EXPLAINNotes

Show solution

smaller friction force (μ_dmg), so a smaller deceleration	✓ 1	Accept: less work is done against friction per metre.
the stone travels further before stopping	✓ 1	Answer must be linked to the reasoning; "travels further" alone scores 0.

(2)

(h)

In another end, stone A strikes a stationary stone C of mass 21.0 kg head-on at 1.6 m s⁻¹. A student predicts that, if this collision were perfectly elastic, stone C would move forward at 1.52 m s⁻¹ and stone A would move backward at 0.080 m s⁻¹.

Verify that this prediction is consistent with both conservation of momentum and conservation of kinetic energy.

REASONNotes

Show solution

momentum before = 19.0 × 1.6 = 30.4 kg m s⁻¹	✓ 1
momentum after = 21.0 × 1.52 − 19.0 × 0.080 = 31.92 − 1.52 = 30.4 kg m s⁻¹	✓ 1	The negative sign for A's velocity must be used.
E_k after = ½ × 21.0 × 1.52² + ½ × 19.0 × 0.080² = 24.26 + 0.06 = 24.3 J = E_k before	✓ 1	Comparison with 24.3 J from (d) needed.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
uniform deceleration, so s = ½(u + v)t: 28.0 = ½(2.4 + v) × 14.0	✓ 1	Accept: average speed = 28.0 ÷ 14.0 = 2.0 m s⁻¹ = (2.4 + v)/2.
v = 4.0 − 2.4 = 1.6 m s⁻¹	✓ 1	The rearrangement must be seen; the given answer alone scores 0.
Part (b)(i)
(2.4 − 1.6) ÷ 14.0 = 0.057 m s⁻²	✓ 1	Accept 0.0571 m s⁻².
Part (b)(ii)
friction is the resultant force: μ_dmg = ma, so μ_d = a/g	✓ 1	F_N = mg on horizontal ice.
μ_d = 0.0571 ÷ 9.8 = 0.0058	✓ 1	Accept 0.0058 to 0.006. No unit. Allow ECF from (b)(i).
Part (c)
19.0 × 1.6 = 19.0 × v_A + 19.0 × 1.4	✓ 1	Conservation of momentum must be applied.
v_A = 0.20 m s⁻¹ in the original direction	✓ 1	Direction needed. Accept 0.2 m s⁻¹ forward.
Part (d)
E_k before = ½ × 19.0 × 1.6² = 24.3 J	✓ 1
E_k after = ½ × 19.0 × (0.20² + 1.4²) = 19.0 J, which is less, so kinetic energy is not conserved	✓ 1	The comparison with the conclusion is needed. Allow ECF from (c).
energy transferred = 24.3 − 19.0 = 5.3 J	✓ 1	Accept 5.32 J.
Part (e)(i)
the impulse on stone B, equal to its change of momentum	✓ 1	Accept "change in momentum of B".
19.0 × 1.4 = 27 N s	✓ 1	Accept 26.6 N s or kg m s⁻¹.
Part (e)(ii)
½ × F_peak × 2.0 × 10⁻³ = 26.6	✓ 1	Time must be converted to seconds.
F_peak = 2.7 × 10⁴ N	✓ 1	Accept 26.6 kN. Allow ECF from (e)(i). 1.3 × 10⁴ N (average force) scores 1 mark.
Part (e)(iii)
the same as in (e)(ii), 2.7 × 10⁴ N, by Newton's third law (in the opposite direction)	✓ 1	Allow ECF from (e)(ii). Reference to the third law or "equal and opposite" needed.
Part (f)
same deceleration of 0.0571 m s⁻², since μ_dmg/m does not depend on mass: 0 = 1.4² − 2 × 0.0571 × s	✓ 1	Allow ECF from (b).
s = 17 m	✓ 1	Accept 17.2 m.
Part (g)
smaller friction force (μ_dmg), so a smaller deceleration	✓ 1	Accept: less work is done against friction per metre.
the stone travels further before stopping	✓ 1	Answer must be linked to the reasoning; "travels further" alone scores 0.
Part (h)
momentum before = 19.0 × 1.6 = 30.4 kg m s⁻¹	✓ 1
momentum after = 21.0 × 1.52 − 19.0 × 0.080 = 31.92 − 1.52 = 30.4 kg m s⁻¹	✓ 1	The negative sign for A's velocity must be used.
E_k after = ½ × 21.0 × 1.52² + ½ × 19.0 × 0.080² = 24.26 + 0.06 = 24.3 J = E_k before	✓ 1	Comparison with 24.3 J from (d) needed.

Answers: (a) 1.6 m s⁻¹ · (b)(i) 0.057 m s⁻² · (b)(ii) 0.0058 · (c) 0.20 m s⁻¹ forward · (d) 5.3 J · (e)(i) 27 N s · (e)(ii) 2.7 × 10⁴ N · (e)(iii) 2.7 × 10⁴ N · (f) 17 m

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A.3 Work, Energy and Power

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An e-bike climbing a hill: power, forces and efficiency

A.3 Work, Energy and PowerA.1 KinematicsA.2 Forces and MomentumB.5 Current and CircuitsSL

A rider and her electrically assisted bicycle (e-bike) have a total mass of 95 kg. The rider can pedal, and an electric motor powered by a 36 V battery adds extra driving force.

Later in the journey the rider climbs a straight hill at a constant speed of 5.5 m s⁻¹ (Figure 1). The road rises 6.0 m for every 100 m travelled along it, so the angle θ between the road and the horizontal has sin θ = 0.060. At this speed the total resistive force (air resistance and rolling resistance) on the rider and e-bike is 16 N.

Figure 1: the rider and e-bike on the hill (not to scale)

(a)(i)

Before reaching the hill, the e-bike starts from rest on a level road and accelerates uniformly to 5.5 m s⁻¹ over a distance of 11 m.

Calculate the time taken.

CALCNotes

Show solution

s = (u + v)t/2 or 11 = (0 + 5.5)t/2	✓ 1	Or find a = 5.5²/(2 × 11) = 1.375 m s⁻² then t = v/a.
t = 4.0 s	✓ 1

Answer: 4.0 s

(2)

(a)(ii)

Calculate the increase in kinetic energy of the rider and e-bike during this acceleration.

CALCNotes

Show solution

E_k = ½ × 95 × 5.5² = 1.4 × 10³ J

✓ 1

Accept 1440 J or 1437 J.

Answer: 1.4 × 10³ J

(1)

(b)

State why the resultant force on the rider and e-bike is zero while they climb the hill.

REASONNotes

Show solution

the velocity is constant (constant speed in a straight line), so the acceleration is zero and so the resultant force is zero (Newton's first/second law)

✓ 1

Constant speed alone is not enough without the straight line / constant direction, or a reference to zero acceleration.

(1)

(c)

Show that the forward driving force on the e-bike is about 72 N.

DERIVENotes

Show solution

component of weight down the slope = mg sin θ = 95 × 9.8 × 0.060 = 55.9 N	✓ 1
driving force = 55.9 + 16 = 71.9 N	✓ 1	Final answer must be given to at least 3 s.f. (71.9 N), or the full substitution must be seen.

Answer: 71.9 N

(2)

(d)(i)

Calculate the mechanical power developed by the driving force.

CALCNotes

Show solution

P = Fv = 71.9 × 5.5 = 395 W

✓ 1

Accept 396 W from 72 N. Allow 4.0 × 10² W.

Answer: 395 W

(1)

(d)(ii)

Calculate the rate at which the gravitational potential energy of the rider and e-bike increases. Hence determine the percentage of the power in (d)(i) that is used to increase the gravitational potential energy.

MULTINotes

Show solution

rate = mg(Δh/Δt) = mgv sin θ = 95 × 9.8 × 5.5 × 0.060 = 307 W	✓ 1	Accept 55.9 × 5.5.
307/395 = 0.78 so 78 %	✓ 1	Allow ECF from (d)(i). The rest (16 × 5.5 = 88 W) is used against the resistive forces.

Answer: 307 W; 78 %

(2)

(e)(i)

On the hill the rider supplies a mechanical power of 115 W by pedalling, and the motor supplies the rest of the power in (d)(i). The efficiency of the motor is 80 %.

Calculate the current in the battery.

MULTINotes

Show solution

motor output = 395 − 115 = 280 W and electrical input = 280/0.80 = 350 W	✓ 1	Allow ECF from (d)(i).
I = P/V = 350/36 = 9.7 A	✓ 1	Using 280 W gives 7.8 A: award the first mark only if 280 W is clearly found.

Answer: 9.7 A

(2)

(e)(ii)

The fully charged battery stores 500 Wh (0.50 kWh) of energy.

Determine the greatest distance the rider could travel up a road of this gradient at 5.5 m s⁻¹ before the battery is empty.

MULTINotes

Show solution

E = 500 × 3600 = 1.8 × 10⁶ J and t = 1.8 × 10⁶/350 = 5.1 × 10³ s	✓ 1	Or t = 500/350 = 1.43 h. Allow ECF from (e)(i).
distance = 5.5 × 5.1 × 10³ = 2.8 × 10⁴ m (28 km)	✓ 1

Answer: 2.8 × 10⁴ m (28 km)

(2)

(f)

On a level road the total resistive force on the rider and e-bike is proportional to v². Figure 2 shows how the power needed to overcome the resistive forces varies with speed v on a level road.

The rider stops pedalling and the motor delivers a mechanical power of 280 W. Use Figure 2 to determine the maximum speed of the e-bike on a level road.

Figure 2: power needed to overcome the resistive forces on a level road against speed

GRAPHNotes

Show solution

at maximum speed all the motor's output power is used against the resistive forces (driving force = resistive force), so read the graph at 280 W	✓ 1
v = 8.1 m s⁻¹	✓ 1	Accept 8.0–8.2 m s⁻¹.

Answer: 8.1 m s⁻¹

(2)

(g)

The range of the e-bike on a level road is the distance it can travel on one full battery. Explain why the range falls sharply as the steady cruising speed increases.

EXPLAINNotes

Show solution

at constant speed the energy needed per metre travelled equals the work done against the resistive force (W = Fs)	✓ 1	Or: power needed ∝ v³ but the time taken per metre ∝ 1/v.
F ∝ v² so the energy per metre ∝ v²: doubling the speed needs four times the energy per metre, so the range is quartered	✓ 1	Accept a correct numerical illustration from Figure 2.

(2)

(h)(i)

Near the top, the hill becomes steeper: the road now rises 10 m for every 100 m travelled along it. The total mechanical power delivered by the rider and motor stays at the value in (d)(i).

Estimate the new constant speed, assuming that the resistive force is negligible.

CALCNotes

Show solution

driving force = mg sin θ = 95 × 9.8 × 0.10 = 93.1 N	✓ 1
v = P/F = 395/93.1 = 4.2 m s⁻¹	✓ 1	Allow ECF from (d)(i).

Answer: 4.2 m s⁻¹

(2)

(h)(ii)

Suggest, with a reason, whether the actual speed is greater or smaller than your estimate in (h)(i).

PREDICTNotes

Show solution

smaller: the resistive force is not zero, so the driving force needed is larger than 93.1 N and, for the same power, v = P/F is smaller

✓ 1

Accept: some of the power is used against the resistive force, so less is available to raise the rider. (Including the resistive force gives about 3.6–3.9 m s⁻¹, depending on how it varies with speed.)

(1)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
s = (u + v)t/2 or 11 = (0 + 5.5)t/2	✓ 1	Or find a = 5.5²/(2 × 11) = 1.375 m s⁻² then t = v/a.
t = 4.0 s	✓ 1
Part (a)(ii)
E_k = ½ × 95 × 5.5² = 1.4 × 10³ J	✓ 1	Accept 1440 J or 1437 J.
Part (b)
the velocity is constant (constant speed in a straight line), so the acceleration is zero and so the resultant force is zero (Newton's first/second law)	✓ 1	Constant speed alone is not enough without the straight line / constant direction, or a reference to zero acceleration.
Part (c)
component of weight down the slope = mg sin θ = 95 × 9.8 × 0.060 = 55.9 N	✓ 1
driving force = 55.9 + 16 = 71.9 N	✓ 1	Final answer must be given to at least 3 s.f. (71.9 N), or the full substitution must be seen.
Part (d)(i)
P = Fv = 71.9 × 5.5 = 395 W	✓ 1	Accept 396 W from 72 N. Allow 4.0 × 10² W.
Part (d)(ii)
rate = mg(Δh/Δt) = mgv sin θ = 95 × 9.8 × 5.5 × 0.060 = 307 W	✓ 1	Accept 55.9 × 5.5.
307/395 = 0.78 so 78 %	✓ 1	Allow ECF from (d)(i). The rest (16 × 5.5 = 88 W) is used against the resistive forces.
Part (e)(i)
motor output = 395 − 115 = 280 W and electrical input = 280/0.80 = 350 W	✓ 1	Allow ECF from (d)(i).
I = P/V = 350/36 = 9.7 A	✓ 1	Using 280 W gives 7.8 A: award the first mark only if 280 W is clearly found.
Part (e)(ii)
E = 500 × 3600 = 1.8 × 10⁶ J and t = 1.8 × 10⁶/350 = 5.1 × 10³ s	✓ 1	Or t = 500/350 = 1.43 h. Allow ECF from (e)(i).
distance = 5.5 × 5.1 × 10³ = 2.8 × 10⁴ m (28 km)	✓ 1
Part (f)
at maximum speed all the motor's output power is used against the resistive forces (driving force = resistive force), so read the graph at 280 W	✓ 1
v = 8.1 m s⁻¹	✓ 1	Accept 8.0–8.2 m s⁻¹.
Part (g)
at constant speed the energy needed per metre travelled equals the work done against the resistive force (W = Fs)	✓ 1	Or: power needed ∝ v³ but the time taken per metre ∝ 1/v.
F ∝ v² so the energy per metre ∝ v²: doubling the speed needs four times the energy per metre, so the range is quartered	✓ 1	Accept a correct numerical illustration from Figure 2.
Part (h)(i)
driving force = mg sin θ = 95 × 9.8 × 0.10 = 93.1 N	✓ 1
v = P/F = 395/93.1 = 4.2 m s⁻¹	✓ 1	Allow ECF from (d)(i).
Part (h)(ii)
smaller: the resistive force is not zero, so the driving force needed is larger than 93.1 N and, for the same power, v = P/F is smaller	✓ 1	Accept: some of the power is used against the resistive force, so less is available to raise the rider. (Including the resistive force gives about 3.6–3.9 m s⁻¹, depending on how it varies with speed.)

Answers: (a)(i) 4.0 s · (a)(ii) 1.4 × 10³ J · (c) 71.9 N · (d)(i) 395 W · (d)(ii) 307 W; 78 % · (e)(i) 9.7 A · (e)(ii) 2.8 × 10⁴ m (28 km) · (f) 8.1 m s⁻¹ · (h)(i) 4.2 m s⁻¹

Common error (c) Forgetting the component of the weight along the slope and using only the 16 N resistive force. On a hill at constant speed the driving force must balance mg sin θ as well.

Exam tip (e)(ii) The watt-hour is a unit of energy: 1 Wh = 3600 J. Convert to joules (or keep the time in hours throughout) before dividing by a power in watts.

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Pumped-hydro storage: storing a grid's surplus energy

A.3 Work, Energy and PowerA.1 KinematicsB.5 Current and CircuitsB.2 Greenhouse EffectSL

A pumped-hydro storage station has an upper reservoir on a hill and a lower lake (Figure 1). At night, when there is surplus electrical energy, reversible pump–turbines pump water from the lake up to the reservoir. In the evening, when demand is high, the water flows back down through the same machines, which then act as turbines driving generators.

The water surface of the upper reservoir is 320 m above the surface of the lake, where the pump–turbines are. When full, the upper reservoir holds 4.0 × 10⁹ kg of water. Treat the height difference as constant at 320 m unless you are told otherwise.

Density of water = 1000 kg m⁻³

Figure 1: the pumped-hydro station (not to scale)

(a)(i)

Calculate the gravitational potential energy stored by the water in the full upper reservoir, relative to the lake.

CALCNotes

Show solution

ΔE_p = mgΔh = 4.0 × 10⁹ × 9.8 × 320 = 1.3 × 10¹³ J

✓ 1

Accept 1.25 × 10¹³ J.

Answer: 1.3 × 10¹³ J

(1)

(a)(ii)

Show that, if no energy were dissipated, water starting from rest at the reservoir surface would reach the pump–turbines with a speed of about 80 m s⁻¹.

DERIVENotes

Show solution

mgΔh = ½mv² or v² = u² + 2as with u = 0 and a = g	✓ 1	Either an energy or an equation-of-motion method.
v = √(2 × 9.8 × 320) = 79.2 m s⁻¹	✓ 1	Answer to at least 3 s.f., or the full substitution, must be seen.

Answer: 79.2 m s⁻¹

(2)

(b)

At full output the station supplies 450 MW of electrical power to the grid. The efficiency of the turbines and generators together is 88 %.

Calculate the volume of water that flows through the turbines each second.

MULTINotes

Show solution

rate of loss of E_p = 450/0.88 = 511 MW	✓ 1
mass per second = 5.11 × 10⁸/(9.8 × 320) = 1.63 × 10⁵ kg s⁻¹, so volume per second = 163 m³ s⁻¹	✓ 1	Accept 160 m³ s⁻¹. Using 450 MW without the efficiency gives 143 m³ s⁻¹: award the second mark only.

Answer: 163 m³ s⁻¹

(2)

(c)

When pumping, the efficiency of the motors and pumps together is 85 %. The round-trip efficiency is the electrical energy returned to the grid divided by the electrical energy taken from the grid to pump the water up.

Show that the round-trip efficiency is about 75 %.

DERIVENotes

Show solution

energy stored = 0.85 × input, and energy returned = 0.88 × energy stored	✓ 1	Or a numerical chain, e.g. 100 MJ in → 85 MJ stored → 74.8 MJ out.
round-trip efficiency = 0.85 × 0.88 = 0.748 (74.8 %)	✓ 1	Answer to 3 s.f. must be seen.

Answer: 0.748 (74.8 %)

(2)

(d)

Sketch a Sankey diagram for one complete pumping and generating cycle, starting with 100 units of electrical energy taken from the grid. Label each arrow with the form of energy and its number of units.

SKETCHNotes

Show solution

arrows labelled: input 100 (electrical); 15 dissipated (thermal) in pumping; 10 dissipated (thermal) in generating; 75 electrical output	✓ 1	Accept 10.2 and 74.8. An intermediate stage of 85 units of gravitational potential energy may be shown.
arrow widths proportional to the values, with the input width equal to the sum of the output widths and the losses branching off	✓ 1

(2)

(e)(i)

Figure 2 shows the electrical power exchanged between the station and the grid during one day. Negative values show power taken from the grid for pumping; positive values show power supplied to the grid.

Determine, in J, the electrical energy taken from the grid during the day.

Figure 2: electrical power supplied to the grid (positive) and taken from the grid (negative) during one day

GRAPHNotes

Show solution

area under the graph for pumping = 300 MW × 6.0 h = 1800 MW h	✓ 1	Or 3.0 × 10⁸ W × 2.16 × 10⁴ s.
= 6.5 × 10¹² J	✓ 1	Accept 6.48 × 10¹² J.

Answer: 6.5 × 10¹² J

(2)

(e)(ii)

Determine the electrical energy supplied to the grid in the evening. Hence comment on whether Figure 2 is consistent with your answer to (c).

GRAPHNotes

Show solution

area of the trapezium = ½ × (4.0 + 2.0) h × 450 MW = 1350 MW h = 4.9 × 10¹² J	✓ 1	Accept 4.86 × 10¹² J.
ratio = 4.86/6.48 = 0.75, equal to the round-trip efficiency, so the graph is consistent	✓ 1	Allow ECF from (e)(i) and (c).

Answer: 4.9 × 10¹² J; consistent

(2)

(f)(i)

The 450 MW output is transmitted to a city at a potential difference of 400 kV.

Calculate the current in the transmission line.

CALCNotes

Show solution

I = P/V = 4.5 × 10⁸/4.0 × 10⁵ = 1.1 × 10³ A

✓ 1

Accept 1125 A.

Answer: 1.1 × 10³ A

(1)

(f)(ii)

The total resistance of the transmission line is 2.0 Ω.

Determine the fraction of the transmitted power that is dissipated in the line.

CALCNotes

Show solution

P = I²R = 1125² × 2.0 = 2.5 × 10⁶ W	✓ 1	Allow ECF from (f)(i). Using V²/R with 400 kV scores 0 for this row.
fraction = 2.53/450 = 5.6 × 10⁻³ (0.56 %)	✓ 1

Answer: 5.6 × 10⁻³ (0.56 %)

(2)

(g)(i)

Calculate the gravitational potential energy stored per cubic metre of water in the upper reservoir. Compare it with the energy density of petrol, 3.4 × 10¹⁰ J m⁻³.

CALCNotes

Show solution

1000 × 9.8 × 320 = 3.1 × 10⁶ J m⁻³	✓ 1
petrol stores about 10⁴ times more energy per unit volume	✓ 1	Accept 1.1 × 10⁴. Allow ECF.

Answer: 3.1 × 10⁶ J m⁻³; about 10⁴ times less than petrol

(2)

(g)(ii)

Many countries are replacing fossil-fuel power stations with wind and solar farms. Explain how pumped storage helps this change to reduce the enhanced greenhouse effect.

EXPLAINNotes

Show solution

wind and solar output varies and does not match demand: surplus energy can be stored and returned to the grid when demand is high	✓ 1
so fewer fossil-fuel power stations are needed to meet peak demand, reducing emissions of CO₂, a greenhouse gas whose increasing concentration causes the enhanced greenhouse effect	✓ 1	Reference to reduced CO₂ (or greenhouse-gas) emissions is needed.

(2)

(h)

Towards the end of an evening of generating, the water surface in the upper reservoir has fallen by 20 m. The volume flow rate and the efficiency are the same as in (b).

Calculate the electrical output power now.

CALCNotes

Show solution

height difference is now 300 m (output ∝ Δh for the same flow rate)	✓ 1
P = 1.63 × 10⁵ × 9.8 × 300 × 0.88 = 4.2 × 10⁸ W (422 MW)	✓ 1	Or 450 × 300/320. Allow ECF from (b).

Answer: 4.2 × 10⁸ W

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
ΔE_p = mgΔh = 4.0 × 10⁹ × 9.8 × 320 = 1.3 × 10¹³ J	✓ 1	Accept 1.25 × 10¹³ J.
Part (a)(ii)
mgΔh = ½mv² or v² = u² + 2as with u = 0 and a = g	✓ 1	Either an energy or an equation-of-motion method.
v = √(2 × 9.8 × 320) = 79.2 m s⁻¹	✓ 1	Answer to at least 3 s.f., or the full substitution, must be seen.
Part (b)
rate of loss of E_p = 450/0.88 = 511 MW	✓ 1
mass per second = 5.11 × 10⁸/(9.8 × 320) = 1.63 × 10⁵ kg s⁻¹, so volume per second = 163 m³ s⁻¹	✓ 1	Accept 160 m³ s⁻¹. Using 450 MW without the efficiency gives 143 m³ s⁻¹: award the second mark only.
Part (c)
energy stored = 0.85 × input, and energy returned = 0.88 × energy stored	✓ 1	Or a numerical chain, e.g. 100 MJ in → 85 MJ stored → 74.8 MJ out.
round-trip efficiency = 0.85 × 0.88 = 0.748 (74.8 %)	✓ 1	Answer to 3 s.f. must be seen.
Part (d)
arrows labelled: input 100 (electrical); 15 dissipated (thermal) in pumping; 10 dissipated (thermal) in generating; 75 electrical output	✓ 1	Accept 10.2 and 74.8. An intermediate stage of 85 units of gravitational potential energy may be shown.
arrow widths proportional to the values, with the input width equal to the sum of the output widths and the losses branching off	✓ 1
Part (e)(i)
area under the graph for pumping = 300 MW × 6.0 h = 1800 MW h	✓ 1	Or 3.0 × 10⁸ W × 2.16 × 10⁴ s.
= 6.5 × 10¹² J	✓ 1	Accept 6.48 × 10¹² J.
Part (e)(ii)
area of the trapezium = ½ × (4.0 + 2.0) h × 450 MW = 1350 MW h = 4.9 × 10¹² J	✓ 1	Accept 4.86 × 10¹² J.
ratio = 4.86/6.48 = 0.75, equal to the round-trip efficiency, so the graph is consistent	✓ 1	Allow ECF from (e)(i) and (c).
Part (f)(i)
I = P/V = 4.5 × 10⁸/4.0 × 10⁵ = 1.1 × 10³ A	✓ 1	Accept 1125 A.
Part (f)(ii)
P = I²R = 1125² × 2.0 = 2.5 × 10⁶ W	✓ 1	Allow ECF from (f)(i). Using V²/R with 400 kV scores 0 for this row.
fraction = 2.53/450 = 5.6 × 10⁻³ (0.56 %)	✓ 1
Part (g)(i)
1000 × 9.8 × 320 = 3.1 × 10⁶ J m⁻³	✓ 1
petrol stores about 10⁴ times more energy per unit volume	✓ 1	Accept 1.1 × 10⁴. Allow ECF.
Part (g)(ii)
wind and solar output varies and does not match demand: surplus energy can be stored and returned to the grid when demand is high	✓ 1
so fewer fossil-fuel power stations are needed to meet peak demand, reducing emissions of CO₂, a greenhouse gas whose increasing concentration causes the enhanced greenhouse effect	✓ 1	Reference to reduced CO₂ (or greenhouse-gas) emissions is needed.
Part (h)
height difference is now 300 m (output ∝ Δh for the same flow rate)	✓ 1
P = 1.63 × 10⁵ × 9.8 × 300 × 0.88 = 4.2 × 10⁸ W (422 MW)	✓ 1	Or 450 × 300/320. Allow ECF from (b).

Answers: (a)(i) 1.3 × 10¹³ J · (a)(ii) 79.2 m s⁻¹ · (b) 163 m³ s⁻¹ · (c) 0.748 (74.8 %) · (e)(i) 6.5 × 10¹² J · (e)(ii) 4.9 × 10¹² J; consistent · (f)(i) 1.1 × 10³ A · (f)(ii) 5.6 × 10⁻³ (0.56 %) · (g)(i) 3.1 × 10⁶ J m⁻³; about 10⁴ times less than petrol · (h) 4.2 × 10⁸ W

More A.3 questions in the question bank

A.4 Rigid Body Mechanics

+ Add to test✓ In your testCC-IBP-A4-120 marks~25 min

A flywheel energy-recovery system on a city bus

A.4 Rigid Body MechanicsA.3 Work, Energy and PowerA.2 Forces and MomentumA.1 KinematicsHL

A city bus has a flywheel energy-recovery system. When the bus brakes, its wheels drive a steel flywheel through a transmission, so that some of the kinetic energy of the bus is stored as rotational kinetic energy of the flywheel. When the bus pulls away again, the flywheel drives the wheels.

The flywheel (Figure 1) is a uniform solid disc of mass 120 kg and radius 0.25 m. Its moment of inertia about its axis is I = ½MR². Its maximum safe angular velocity is 1500 rad s⁻¹. Friction in the flywheel bearings is negligible.

The mass of the bus, including its passengers, is 1.2 × 10⁴ kg.

Figure 1: the flywheel (not to scale)

(a)

Show that the moment of inertia of the flywheel is about 3.8 kg m².

HL onlyCALCNotes

Show solution

I = ½ × 120 × 0.25² = 3.75 kg m²

✓ 1

Substitution and an answer to at least 3 s.f. must be seen.

Answer: 3.75 kg m²

(1)

(b)

Calculate the maximum rotational kinetic energy that the flywheel can store.

HL onlyCALCNotes

Show solution

E_k = ½Iω² = ½ × 3.75 × 1500² = 4.2 × 10⁶ J

✓ 1

Accept 4.22 × 10⁶ J; allow 4.3 × 10⁶ J from 3.8 kg m².

Answer: 4.2 × 10⁶ J

(1)

(c)

The bus travels at 14 m s⁻¹. Calculate the translational kinetic energy of the bus and compare it with your answer to (b).

HL onlyCALCNotes

Show solution

E_k = ½ × 1.2 × 10⁴ × 14² = 1.2 × 10⁶ J	✓ 1	Accept 1.18 × 10⁶ J.
the flywheel can store about 3.6 times as much (so it has enough capacity for a stop from this speed)	✓ 1	Allow ECF from (b). Accept a ratio in the range 3.5–3.7.

Answer: 1.2 × 10⁶ J; the flywheel can store about 3.6 times as much

(2)

(d)(i)

Figure 2 shows how the angular velocity ω of the flywheel varies with time t during one braking event. The bus slows uniformly from 14 m s⁻¹ at t = 0 and comes to rest at t = 6.0 s.

Determine the angular acceleration of the flywheel during braking.

Figure 2: angular velocity of the flywheel during one braking event (the bus comes to rest at t = 6.0 s)

HL onlyGRAPHNotes

Show solution

gradient = (800 − 400)/6.0 = 67 rad s⁻²

✓ 1

Accept 66.7 rad s⁻².

Answer: 67 rad s⁻²

(1)

(d)(ii)

Determine the number of revolutions made by the flywheel between t = 0 and t = 6.0 s.

HL onlyGRAPHNotes

Show solution

angular displacement = area under the graph = ½ × (400 + 800) × 6.0 = 3600 rad	✓ 1	Or Δθ = (ω_i + ω_f)t/2.
number of revolutions = 3600/2π = 573	✓ 1	Accept 570. Not 3600 (radians not converted).

Answer: 573 revolutions

(2)

(d)(iii)

Calculate the average resultant torque on the flywheel during braking.

HL onlyCALCNotes

Show solution

τ = Iα = 3.75 × 66.7 = 250 N m

✓ 1

Allow ECF from (d)(i). Accept 2.5 × 10² N m.

Answer: 250 N m

(1)

(e)

Show that about 77 % of the kinetic energy of the bus is transferred to the flywheel during this braking event.

HL onlyDERIVENotes

Show solution

gain in rotational E_k = ½ × 3.75 × (800² − 400²) = 9.0 × 10⁵ J	✓ 1	Not ½I(800 − 400)².
9.0 × 10⁵/1.176 × 10⁶ = 0.765 (76.5 %)	✓ 1	Answer to 3 s.f. must be seen.

Answer: 0.765 (76.5 %)

(2)

(f)(i)

Calculate the distance travelled by the bus while it brakes.

CALCNotes

Show solution

s = (u + v)t/2 = (14 + 0)/2 × 6.0 = 42 m

✓ 1

Answer: 42 m

(1)

(f)(ii)

Determine the magnitude of the average resultant force on the bus during braking. Verify your answer using the work done on the bus.

MULTINotes

Show solution

F = Δp/Δt = 1.2 × 10⁴ × 14/6.0 = 2.8 × 10⁴ N	✓ 1	Or F = ma with a = 14/6.0 = 2.33 m s⁻².
Fs = 2.8 × 10⁴ × 42 = 1.18 × 10⁶ J, equal to the loss of kinetic energy of the bus	✓ 1	Allow ECF from (c) and (f)(i).

Answer: 2.8 × 10⁴ N

(2)

(g)(i)

The engineers consider replacing the disc with a thin ring (a rim-loaded flywheel) of the same mass and radius. Its moment of inertia is I = MR².

Explain why the ring stores more energy than the disc at the same angular velocity.

HL onlyEXPLAINNotes

Show solution

in the ring all the mass is at the greatest distance R from the axis, so its moment of inertia is larger (twice that of the disc)	✓ 1
rotational E_k = ½Iω² is proportional to I at the same ω, so the ring stores twice the energy	✓ 1

(2)

(g)(ii)

The ring flywheel absorbs the same energy as in (e) during the same braking event, again starting at 400 rad s⁻¹.

Calculate its angular velocity at the end of braking.

HL onlyCALCNotes

Show solution

I = 120 × 0.25² = 7.5 kg m² and ½ × 7.5 × (ω² − 400²) = 9.0 × 10⁵	✓ 1	Allow ECF from (e).
ω = 632 rad s⁻¹	✓ 1	Accept 630 rad s⁻¹.

Answer: 632 rad s⁻¹

(2)

(h)

The bus, now fitted again with the original disc flywheel, pulls away from rest on a level road. The flywheel slows from 800 rad s⁻¹ to 400 rad s⁻¹ in 10 s, and 80 % of the energy it releases becomes kinetic energy of the bus. Resistive forces on the bus can be ignored.

Determine the speed of the bus after 10 s and the average useful power delivered to the bus.

HL onlyMULTINotes

Show solution

energy released = 9.0 × 10⁵ J, so energy to the bus = 0.80 × 9.0 × 10⁵ = 7.2 × 10⁵ J	✓ 1	Allow ECF from (e).
½ × 1.2 × 10⁴ × v² = 7.2 × 10⁵ so v = 11 m s⁻¹	✓ 1	Accept 10.9 m s⁻¹.
P = 7.2 × 10⁵/10 = 7.2 × 10⁴ W	✓ 1

Answer: 11 m s⁻¹; 7.2 × 10⁴ W

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
I = ½ × 120 × 0.25² = 3.75 kg m²	✓ 1	Substitution and an answer to at least 3 s.f. must be seen.
Part (b)
E_k = ½Iω² = ½ × 3.75 × 1500² = 4.2 × 10⁶ J	✓ 1	Accept 4.22 × 10⁶ J; allow 4.3 × 10⁶ J from 3.8 kg m².
Part (c)
E_k = ½ × 1.2 × 10⁴ × 14² = 1.2 × 10⁶ J	✓ 1	Accept 1.18 × 10⁶ J.
the flywheel can store about 3.6 times as much (so it has enough capacity for a stop from this speed)	✓ 1	Allow ECF from (b). Accept a ratio in the range 3.5–3.7.
Part (d)(i)
gradient = (800 − 400)/6.0 = 67 rad s⁻²	✓ 1	Accept 66.7 rad s⁻².
Part (d)(ii)
angular displacement = area under the graph = ½ × (400 + 800) × 6.0 = 3600 rad	✓ 1	Or Δθ = (ω_i + ω_f)t/2.
number of revolutions = 3600/2π = 573	✓ 1	Accept 570. Not 3600 (radians not converted).
Part (d)(iii)
τ = Iα = 3.75 × 66.7 = 250 N m	✓ 1	Allow ECF from (d)(i). Accept 2.5 × 10² N m.
Part (e)
gain in rotational E_k = ½ × 3.75 × (800² − 400²) = 9.0 × 10⁵ J	✓ 1	Not ½I(800 − 400)².
9.0 × 10⁵/1.176 × 10⁶ = 0.765 (76.5 %)	✓ 1	Answer to 3 s.f. must be seen.
Part (f)(i)
s = (u + v)t/2 = (14 + 0)/2 × 6.0 = 42 m	✓ 1
Part (f)(ii)
F = Δp/Δt = 1.2 × 10⁴ × 14/6.0 = 2.8 × 10⁴ N	✓ 1	Or F = ma with a = 14/6.0 = 2.33 m s⁻².
Fs = 2.8 × 10⁴ × 42 = 1.18 × 10⁶ J, equal to the loss of kinetic energy of the bus	✓ 1	Allow ECF from (c) and (f)(i).
Part (g)(i)
in the ring all the mass is at the greatest distance R from the axis, so its moment of inertia is larger (twice that of the disc)	✓ 1
rotational E_k = ½Iω² is proportional to I at the same ω, so the ring stores twice the energy	✓ 1
Part (g)(ii)
I = 120 × 0.25² = 7.5 kg m² and ½ × 7.5 × (ω² − 400²) = 9.0 × 10⁵	✓ 1	Allow ECF from (e).
ω = 632 rad s⁻¹	✓ 1	Accept 630 rad s⁻¹.
Part (h)
energy released = 9.0 × 10⁵ J, so energy to the bus = 0.80 × 9.0 × 10⁵ = 7.2 × 10⁵ J	✓ 1	Allow ECF from (e).
½ × 1.2 × 10⁴ × v² = 7.2 × 10⁵ so v = 11 m s⁻¹	✓ 1	Accept 10.9 m s⁻¹.
P = 7.2 × 10⁵/10 = 7.2 × 10⁴ W	✓ 1

Answers: (a) 3.75 kg m² · (b) 4.2 × 10⁶ J · (c) 1.2 × 10⁶ J; the flywheel can store about 3.6 times as much · (d)(i) 67 rad s⁻² · (d)(ii) 573 revolutions · (d)(iii) 250 N m · (e) 0.765 (76.5 %) · (f)(i) 42 m · (f)(ii) 2.8 × 10⁴ N · (g)(ii) 632 rad s⁻¹ · (h) 11 m s⁻¹; 7.2 × 10⁴ W

Common error (e) Using ½I(Δω)² for the energy gained. The change in energy is ½Iω_f² − ½Iω_i², which is not the same.

More A.4 questions in the question bank

+ Add to test✓ In your testCC-IBP-A4-223 marks~30 min

A bascule bridge: counterweight, torque and opening the deck

A.4 Rigid Body MechanicsA.2 Forces and MomentumA.3 Work, Energy and PowerA.1 KinematicsHL

A single-leaf bascule bridge has a straight, uniform deck of length 24 m and mass 6.0 × 10⁴ kg. The deck can rotate in a vertical plane about a horizontal pivot 2.0 m from its rear end (Figure 1), so its tip is 22 m in front of the pivot and its centre of mass is 10 m in front of the pivot. A counterweight is fixed behind the pivot on an arm of negligible mass; the centre of mass of the counterweight is on the line of the deck, 6.0 m behind the pivot. The deck and counterweight rotate together as one rigid leaf.

When the bridge is closed the deck is horizontal. Friction at the pivot is negligible.

Figure 1: the closed bridge, side view (not to scale)

(a)(i)

Show that a counterweight of mass 1.0 × 10⁵ kg would exactly balance the horizontal deck, with no force needed at the tip.

HL onlyDERIVENotes

Show solution

rotational equilibrium about the pivot: m_cg × 6.0 = 6.0 × 10⁴ × g × 10	✓ 1	Clockwise torque = anticlockwise torque, stated or implied by the equation.
m_c = 6.0 × 10⁵/6.0 = 1.0 × 10⁵ kg	✓ 1

Answer: 1.0 × 10⁵ kg

(2)

(a)(ii)

Calculate the force exerted by the pivot on the leaf in this case.

HL onlyCALCNotes

Show solution

resultant force is zero, so the pivot force balances both weights: (6.0 × 10⁴ + 1.0 × 10⁵) × 9.8	✓ 1
= 1.6 × 10⁶ N, vertically upwards	✓ 1	Accept 1.57 × 10⁶ N. Direction needed.

Answer: 1.6 × 10⁶ N upwards

(2)

(b)

The counterweight actually fitted has a mass of 9.5 × 10⁴ kg, so that the tip of the closed deck rests on a support (Figure 1).

Determine the force exerted by the support on the deck.

HL onlyMULTINotes

Show solution

net torque of the weights about the pivot = 6.0 × 10⁴ × 9.8 × 10 − 9.5 × 10⁴ × 9.8 × 6.0 = 2.94 × 10⁵ N m	✓ 1
F = 2.94 × 10⁵/22 = 1.3 × 10⁴ N	✓ 1	Accept 1.34 × 10⁴ N. Using 10 m or 24 m as the support distance scores the first mark only.

Answer: 1.3 × 10⁴ N

(2)

(c)

The moment of inertia of the deck alone about the pivot axis is 8.9 × 10⁶ kg m². The counterweight may be treated as a point mass.

Show that the moment of inertia of the whole leaf about the pivot axis is about 1.2 × 10⁷ kg m².

HL onlyDERIVENotes

Show solution

counterweight: mr² = 9.5 × 10⁴ × 6.0² = 3.4 × 10⁶ kg m²	✓ 1
I = 8.9 × 10⁶ + 3.42 × 10⁶ = 1.23 × 10⁷ kg m²	✓ 1	Answer to 3 s.f. must be seen.

Answer: 1.23 × 10⁷ kg m²

(2)

(d)

To open the bridge, a motor applies a torque to the leaf. Determine the motor torque needed to give the leaf an initial angular acceleration of 5.0 × 10⁻³ rad s⁻² from rest, with the deck horizontal.

HL onlyMULTINotes

Show solution

resultant torque = Iα = 1.23 × 10⁷ × 5.0 × 10⁻³ = 6.2 × 10⁴ N m	✓ 1
motor torque = 2.94 × 10⁵ + 6.2 × 10⁴ = 3.6 × 10⁵ N m	✓ 1	Allow ECF from (b) and (c). Giving only Iα scores the first mark.

Answer: 3.6 × 10⁵ N m

(2)

(e)

As the bridge opens, the opening angle of the deck above the horizontal increases from 0° to 80°.

On the axes of Figure 2, sketch how the net gravitational torque on the leaf about the pivot varies with the opening angle from 0° to 80°.

Figure 2: axes for the sketch in (e)

HL onlySKETCHNotes

Show solution

curve starting at 2.9 × 10⁵ N m at 0° and decreasing ever more steeply as the angle increases (cosine shape, starting with zero gradient)	✓ 1	Both perpendicular distances are proportional to cos(angle), so the net torque is 2.94 × 10⁵ cos(angle).
ending at about 0.5 × 10⁵ N m at 80° (not zero)	✓ 1	Accept 0.4–0.6 × 10⁵ N m. Allow ECF from (b).

(2)

(f)

During an opening, the leaf accelerates uniformly from rest at 5.0 × 10⁻³ rad s⁻² until its angular velocity is 0.025 rad s⁻¹. It then rotates at this angular velocity before decelerating uniformly at 5.0 × 10⁻³ rad s⁻² to rest at an opening angle of 80°.

Show that the opening takes about 60 s.

HL onlyMULTINotes

Show solution

accelerating stage: t = 0.025/5.0 × 10⁻³ = 5.0 s and Δθ = ½ × 5.0 × 10⁻³ × 5.0² = 0.0625 rad; the decelerating stage is the same	✓ 1	Or Δθ = (0 + 0.025)/2 × 5.0.
80° = 1.396 rad, so the constant stage covers 1.396 − 0.125 = 1.271 rad and takes 1.271/0.025 = 50.9 s	✓ 1
total time = 5.0 + 50.9 + 5.0 = 60.9 s (about 60 s)	✓ 1	Answer to at least 2 s.f. (61 s) with working must be seen.

Answer: 60.9 s

(3)

(g)(i)

Determine the work done by the motor in opening the bridge from 0° to 80°.

HL onlyMULTINotes

Show solution

deck rises 10 sin 80° = 9.85 m: +5.79 × 10⁶ J; counterweight falls 6.0 sin 80° = 5.91 m: −5.50 × 10⁶ J	✓ 1	Or the area under the torque–angle graph: 2.94 × 10⁵ × sin 80°.
the leaf starts and ends at rest, so work done = net gain in E_p = 2.9 × 10⁵ J	✓ 1	Allow ECF from (b).

Answer: 2.9 × 10⁵ J

(2)

(g)(ii)

Calculate the average power output of the motor during the opening.

HL onlyCALCNotes

Show solution

P = 2.9 × 10⁵/60.9 = 4.8 × 10³ W

✓ 1

Allow ECF from (f) and (g)(i). Accept 60 s, giving 4.8 kW.

Answer: 4.8 kW

(1)

(g)(iii)

Explain why the counterweight greatly reduces the average power needed from the motor, even though it does not reduce the moment of inertia of the leaf.

HL onlyEXPLAINNotes

Show solution

as the deck rises the counterweight falls: its loss of E_p supplies almost all the energy to raise the deck, so the motor does much less work in the same time	✓ 1	Accept a comparison: without the counterweight the motor would need 5.8 × 10⁶ J, about 95 kW.
the extra moment of inertia only matters while the leaf speeds up: the rotational kinetic energy given to the leaf is very small (½Iω² = ½ × 1.23 × 10⁷ × 0.025² ≈ 4 × 10³ J) and is returned as it slows down, so the work done by the motor is essentially the net gain in E_p	✓ 1	Accept: the torque needed for the angular acceleration (6.2 × 10⁴ N m) is small compared with the unbalanced gravitational torque without a counterweight (5.9 × 10⁶ N m). Stating only that the counterweight increases I is not enough.

(2)

(h)

For an inspection, a 2.0 × 10³ kg lifting platform is fixed to the deck at its tip. The platform may be treated as a point mass.

Determine the motor torque now needed to give the same initial angular acceleration from rest, with the deck horizontal.

HL onlyMULTINotes

Show solution

extra gravitational torque = 2.0 × 10³ × 9.8 × 22 = 4.3 × 10⁵ N m	✓ 1
new I = 1.23 × 10⁷ + 2.0 × 10³ × 22² = 1.33 × 10⁷ kg m², so Iα = 6.6 × 10⁴ N m	✓ 1
motor torque = 2.94 × 10⁵ + 4.31 × 10⁵ + 6.6 × 10⁴ = 7.9 × 10⁵ N m	✓ 1	Allow ECF from (b), (c) and (d).

Answer: 7.9 × 10⁵ N m

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
rotational equilibrium about the pivot: m_cg × 6.0 = 6.0 × 10⁴ × g × 10	✓ 1	Clockwise torque = anticlockwise torque, stated or implied by the equation.
m_c = 6.0 × 10⁵/6.0 = 1.0 × 10⁵ kg	✓ 1
Part (a)(ii)
resultant force is zero, so the pivot force balances both weights: (6.0 × 10⁴ + 1.0 × 10⁵) × 9.8	✓ 1
= 1.6 × 10⁶ N, vertically upwards	✓ 1	Accept 1.57 × 10⁶ N. Direction needed.
Part (b)
net torque of the weights about the pivot = 6.0 × 10⁴ × 9.8 × 10 − 9.5 × 10⁴ × 9.8 × 6.0 = 2.94 × 10⁵ N m	✓ 1
F = 2.94 × 10⁵/22 = 1.3 × 10⁴ N	✓ 1	Accept 1.34 × 10⁴ N. Using 10 m or 24 m as the support distance scores the first mark only.
Part (c)
counterweight: mr² = 9.5 × 10⁴ × 6.0² = 3.4 × 10⁶ kg m²	✓ 1
I = 8.9 × 10⁶ + 3.42 × 10⁶ = 1.23 × 10⁷ kg m²	✓ 1	Answer to 3 s.f. must be seen.
Part (d)
resultant torque = Iα = 1.23 × 10⁷ × 5.0 × 10⁻³ = 6.2 × 10⁴ N m	✓ 1
motor torque = 2.94 × 10⁵ + 6.2 × 10⁴ = 3.6 × 10⁵ N m	✓ 1	Allow ECF from (b) and (c). Giving only Iα scores the first mark.
Part (e)
curve starting at 2.9 × 10⁵ N m at 0° and decreasing ever more steeply as the angle increases (cosine shape, starting with zero gradient)	✓ 1	Both perpendicular distances are proportional to cos(angle), so the net torque is 2.94 × 10⁵ cos(angle).
ending at about 0.5 × 10⁵ N m at 80° (not zero)	✓ 1	Accept 0.4–0.6 × 10⁵ N m. Allow ECF from (b).
Part (f)
accelerating stage: t = 0.025/5.0 × 10⁻³ = 5.0 s and Δθ = ½ × 5.0 × 10⁻³ × 5.0² = 0.0625 rad; the decelerating stage is the same	✓ 1	Or Δθ = (0 + 0.025)/2 × 5.0.
80° = 1.396 rad, so the constant stage covers 1.396 − 0.125 = 1.271 rad and takes 1.271/0.025 = 50.9 s	✓ 1
total time = 5.0 + 50.9 + 5.0 = 60.9 s (about 60 s)	✓ 1	Answer to at least 2 s.f. (61 s) with working must be seen.
Part (g)(i)
deck rises 10 sin 80° = 9.85 m: +5.79 × 10⁶ J; counterweight falls 6.0 sin 80° = 5.91 m: −5.50 × 10⁶ J	✓ 1	Or the area under the torque–angle graph: 2.94 × 10⁵ × sin 80°.
the leaf starts and ends at rest, so work done = net gain in E_p = 2.9 × 10⁵ J	✓ 1	Allow ECF from (b).
Part (g)(ii)
P = 2.9 × 10⁵/60.9 = 4.8 × 10³ W	✓ 1	Allow ECF from (f) and (g)(i). Accept 60 s, giving 4.8 kW.
Part (g)(iii)
as the deck rises the counterweight falls: its loss of E_p supplies almost all the energy to raise the deck, so the motor does much less work in the same time	✓ 1	Accept a comparison: without the counterweight the motor would need 5.8 × 10⁶ J, about 95 kW.
the extra moment of inertia only matters while the leaf speeds up: the rotational kinetic energy given to the leaf is very small (½Iω² = ½ × 1.23 × 10⁷ × 0.025² ≈ 4 × 10³ J) and is returned as it slows down, so the work done by the motor is essentially the net gain in E_p	✓ 1	Accept: the torque needed for the angular acceleration (6.2 × 10⁴ N m) is small compared with the unbalanced gravitational torque without a counterweight (5.9 × 10⁶ N m). Stating only that the counterweight increases I is not enough.
Part (h)
extra gravitational torque = 2.0 × 10³ × 9.8 × 22 = 4.3 × 10⁵ N m	✓ 1
new I = 1.23 × 10⁷ + 2.0 × 10³ × 22² = 1.33 × 10⁷ kg m², so Iα = 6.6 × 10⁴ N m	✓ 1
motor torque = 2.94 × 10⁵ + 4.31 × 10⁵ + 6.6 × 10⁴ = 7.9 × 10⁵ N m	✓ 1	Allow ECF from (b), (c) and (d).

Answers: (a)(i) 1.0 × 10⁵ kg · (a)(ii) 1.6 × 10⁶ N upwards · (b) 1.3 × 10⁴ N · (c) 1.23 × 10⁷ kg m² · (d) 3.6 × 10⁵ N m · (f) 60.9 s · (g)(i) 2.9 × 10⁵ J · (g)(ii) 4.8 kW · (h) 7.9 × 10⁵ N m

Common error (d) Using τ = Iα with the motor torque alone. Iα is the resultant torque: the motor must also overcome the unbalanced gravitational torque of the leaf.

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A.5 Galilean and Special Relativity

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A fast interstellar probe sending laser pulses home

A.5 Galilean and Special RelativityC.2 Wave ModelA.1 KinematicsHL

A space probe is launched from Earth and travels in a straight line at a constant speed of 0.60c relative to Earth towards a star. In the Earth frame the star is 4.2 ly from Earth. Earth and the star are at rest relative to each other.

Treat the launch as instantaneous. The launch is the event x = 0, t = 0 in the Earth frame and x′ = 0, t′ = 0 in the inertial frame of the probe, and the positive x-direction points from Earth towards the star. 1 ly is the distance light travels in 1 year, so c = 1 ly yr⁻¹.

During the journey the probe fires a short laser pulse back towards Earth every 30 days, as measured by the probe's clock.

Figure 1: the journey in the Earth frame (not to scale)

(a)

State the two postulates of special relativity.

HL onlyRECALLNotes

Show solution

the laws of physics are the same in all inertial frames of reference	✓ 1	Not enough: “the laws of motion” only (that is Galilean relativity).
the speed of light in a vacuum is the same for all inertial observers	✓ 1	Accept “independent of the motion of the source or of the observer”.

(2)

(b)(i)

Calculate the time taken for the journey in the Earth frame.

HL onlyCALCNotes

Show solution

4.2 ÷ 0.60 = 7.0 yr

✓ 1

Accept 2.2 × 10⁸ s.

Answer: 7.0 yr

(1)

(b)(ii)

Show that the journey takes 5.6 years according to the probe's clock.

HL onlyDERIVENotes

Show solution

launch and arrival both happen at the probe, so the probe's clock measures the proper time interval	✓ 1	Reason is required for this mark.
γ = 1/√(1 − 0.60²) = 1.25 and Δt₀ = 7.0 ÷ 1.25 = 5.6 yr	✓ 1	Allow ECF from (b)(i). Both steps must be seen.

Answer: 5.6 yr

(2)

(b)(iii)

Calculate the distance between Earth and the star in the frame of the probe.

HL onlyCALCNotes

Show solution

4.2 ÷ 1.25 = 3.4 ly

✓ 1

Accept 3.36 ly. Accept 0.60 × 5.6 = 3.4 ly.

Answer: 3.4 ly

(1)

(c)

Show that the space–time interval between the launch and the arrival at the star has the same value in the Earth frame and in the frame of the probe.

HL onlyDERIVENotes

Show solution

Earth frame: (Δs)² = 7.0² − 4.2² = 31.4 ly²	✓ 1	Accept Δs = 5.6 ly. Allow ECF from (b)(i).
probe frame: Δx′ = 0, so (Δs)² = 5.6² − 0 = 31.4 ly², the same value	✓ 1	Δx′ = 0 must be stated or used.

Answer: 31.4 ly² in both frames (Δs = 5.6 ly)

(2)

(d)(i)

One pulse is fired when the probe's clock reads 2.0 yr after the launch. This is event P.

Use the Lorentz transformations to determine the position and the time of event P in the Earth frame.

HL onlyCALCNotes

Show solution

x = γ(x′ + vt′) = 1.25 × (0 + 0.60 × 2.0) = 1.5 ly	✓ 1	Accept the Earth-frame time × 0.60 = 1.5 ly.
t = γ(t′ + vx′/c²) = 1.25 × 2.0 = 2.5 yr	✓ 1	Accept time dilation, 1.25 × 2.0.

Answer: x = 1.5 ly, t = 2.5 yr

(2)

(d)(ii)

Figure 2 is a space–time diagram for the Earth frame. The world line of the star is shown.

On Figure 2, draw the world line of the probe from the launch to its arrival at the star, mark event P, and draw the world line of the pulse fired at P. Hence determine the time, in the Earth frame, at which this pulse reaches Earth.

Figure 2: space–time diagram for the Earth frame (equal scales on both axes)

HL onlyGRAPHNotes

Show solution

straight line from the origin to (4.2 ly, 7.0 ly)	✓ 1	Line at about 31° to the ct axis (tan θ = 0.60).
P at (1.5 ly, 2.5 ly) on the probe's world line, with a line at 45° from P back to x = 0	✓ 1	Allow ECF from (d)(i).
pulse meets the ct axis at ct = 4.0 ly, so t = 4.0 yr	✓ 1	Accept 3.9–4.1 yr. Accept 2.5 + 1.5 = 4.0 yr.

Answer: 4.0 yr

(3)

(d)(iii)

Hence show that pulses fired every 30 days by the probe's clock arrive at Earth every 60 days.

HL onlyMULTINotes

Show solution

a pulse fired at probe time τ is fired at Earth time 1.25τ and position 0.75τ, so it arrives at 1.25τ + 0.75τ = 2.0τ	✓ 1	Accept: launch (τ = 0) arrives at 0 and P (τ = 2.0 yr) arrives at 4.0 yr, and both world lines are straight, so arrival time ∝ τ. A single ratio 4.0/2.0 with no argument scores 0 for this mark.
so the interval at Earth is 2.0 × 30 = 60 days	✓ 1

Answer: 60 days

(2)

(e)(i)

The laser emits light of wavelength 1064 nm, measured in the frame of the probe.

Calculate the frequency of this light in the frame of the probe, and state the region of the electromagnetic spectrum to which it belongs.

HL onlyCALCNotes

Show solution

f = 3.00 × 10⁸ ÷ 1064 × 10⁻⁹ = 2.82 × 10¹⁴ Hz	✓ 1	Accept 2.8 × 10¹⁴ Hz.
infrared	✓ 1	Reason not required.

Answer: 2.82 × 10¹⁴ Hz; infrared

(2)

(e)(ii)

Outline why the pulses can travel from the probe to Earth through the vacuum of interstellar space. Compare the speed of a pulse relative to Earth predicted by Galilean relativity with the speed measured on Earth.

HL onlyEXPLAINNotes

Show solution

light is an electromagnetic wave: oscillating electric and magnetic fields, so no medium is needed	✓ 1	Not enough: “light is a wave”.
Galilean: u = u′ + v = −c + 0.60c, so a speed of 0.40c (1.2 × 10⁸ m s⁻¹) towards Earth	✓ 1	Sign need not be shown.
measured: c = 3.00 × 10⁸ m s⁻¹, because the speed of light is the same for all inertial observers / independent of the motion of the source	✓ 1	Must refer to postulate 2 or its content.

(3)

(f)

When the probe reaches the star, a beacon at mission control on Earth flashes. In the Earth frame the flash and the arrival at the star are simultaneous, both at t = 7.0 yr.

Determine the time between these two events in the frame of the probe, and state which event happens first in that frame.

HL onlyREASONNotes

Show solution

Δt′ = γ(Δt − vΔx/c²) with Δt = 0 and Δx = 4.2 ly	✓ 1	Or Lorentz-transform each event separately: t′ = 8.75 yr (flash) and 5.6 yr (arrival).
\|Δt′\| = 1.25 × 0.60 × 4.2 = 3.2 yr	✓ 1	Accept 3.15 yr.
the arrival at the star happens first	✓ 1	Accept: the event further along the direction of the probe's motion happens first. Must be consistent with the sign used.

Answer: 3.2 yr; arrival at the star first

(3)

(g)

The mission is redesigned so that the probe travels at 0.80c relative to Earth. Everything else is unchanged.

Determine the journey time according to the probe's clock, and predict the interval between the pulses received at Earth.

HL onlyPREDICTNotes

Show solution

γ = 1/√(1 − 0.80²) = 1.67 or Earth-frame time = 4.2 ÷ 0.80 = 5.25 yr	✓ 1
proper time = 5.25 ÷ 1.67 = 3.2 yr	✓ 1	Accept 3.15 yr. Accept (4.2 ÷ 1.67) ÷ 0.80.
arrival time = γ(1 + 0.80)τ = 3.0τ, so pulses arrive every 90 days	✓ 1	Allow a method that repeats (d)(i)–(d)(iii) with 0.80c.

Answer: 3.2 yr; 90 days

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
the laws of physics are the same in all inertial frames of reference	✓ 1	Not enough: “the laws of motion” only (that is Galilean relativity).
the speed of light in a vacuum is the same for all inertial observers	✓ 1	Accept “independent of the motion of the source or of the observer”.
Part (b)(i)
4.2 ÷ 0.60 = 7.0 yr	✓ 1	Accept 2.2 × 10⁸ s.
Part (b)(ii)
launch and arrival both happen at the probe, so the probe's clock measures the proper time interval	✓ 1	Reason is required for this mark.
γ = 1/√(1 − 0.60²) = 1.25 and Δt₀ = 7.0 ÷ 1.25 = 5.6 yr	✓ 1	Allow ECF from (b)(i). Both steps must be seen.
Part (b)(iii)
4.2 ÷ 1.25 = 3.4 ly	✓ 1	Accept 3.36 ly. Accept 0.60 × 5.6 = 3.4 ly.
Part (c)
Earth frame: (Δs)² = 7.0² − 4.2² = 31.4 ly²	✓ 1	Accept Δs = 5.6 ly. Allow ECF from (b)(i).
probe frame: Δx′ = 0, so (Δs)² = 5.6² − 0 = 31.4 ly², the same value	✓ 1	Δx′ = 0 must be stated or used.
Part (d)(i)
x = γ(x′ + vt′) = 1.25 × (0 + 0.60 × 2.0) = 1.5 ly	✓ 1	Accept the Earth-frame time × 0.60 = 1.5 ly.
t = γ(t′ + vx′/c²) = 1.25 × 2.0 = 2.5 yr	✓ 1	Accept time dilation, 1.25 × 2.0.
Part (d)(ii)
straight line from the origin to (4.2 ly, 7.0 ly)	✓ 1	Line at about 31° to the ct axis (tan θ = 0.60).
P at (1.5 ly, 2.5 ly) on the probe's world line, with a line at 45° from P back to x = 0	✓ 1	Allow ECF from (d)(i).
pulse meets the ct axis at ct = 4.0 ly, so t = 4.0 yr	✓ 1	Accept 3.9–4.1 yr. Accept 2.5 + 1.5 = 4.0 yr.
Part (d)(iii)
a pulse fired at probe time τ is fired at Earth time 1.25τ and position 0.75τ, so it arrives at 1.25τ + 0.75τ = 2.0τ	✓ 1	Accept: launch (τ = 0) arrives at 0 and P (τ = 2.0 yr) arrives at 4.0 yr, and both world lines are straight, so arrival time ∝ τ. A single ratio 4.0/2.0 with no argument scores 0 for this mark.
so the interval at Earth is 2.0 × 30 = 60 days	✓ 1
Part (e)(i)
f = 3.00 × 10⁸ ÷ 1064 × 10⁻⁹ = 2.82 × 10¹⁴ Hz	✓ 1	Accept 2.8 × 10¹⁴ Hz.
infrared	✓ 1	Reason not required.
Part (e)(ii)
light is an electromagnetic wave: oscillating electric and magnetic fields, so no medium is needed	✓ 1	Not enough: “light is a wave”.
Galilean: u = u′ + v = −c + 0.60c, so a speed of 0.40c (1.2 × 10⁸ m s⁻¹) towards Earth	✓ 1	Sign need not be shown.
measured: c = 3.00 × 10⁸ m s⁻¹, because the speed of light is the same for all inertial observers / independent of the motion of the source	✓ 1	Must refer to postulate 2 or its content.
Part (f)
Δt′ = γ(Δt − vΔx/c²) with Δt = 0 and Δx = 4.2 ly	✓ 1	Or Lorentz-transform each event separately: t′ = 8.75 yr (flash) and 5.6 yr (arrival).
\|Δt′\| = 1.25 × 0.60 × 4.2 = 3.2 yr	✓ 1	Accept 3.15 yr.
the arrival at the star happens first	✓ 1	Accept: the event further along the direction of the probe's motion happens first. Must be consistent with the sign used.
Part (g)
γ = 1/√(1 − 0.80²) = 1.67 or Earth-frame time = 4.2 ÷ 0.80 = 5.25 yr	✓ 1
proper time = 5.25 ÷ 1.67 = 3.2 yr	✓ 1	Accept 3.15 yr. Accept (4.2 ÷ 1.67) ÷ 0.80.
arrival time = γ(1 + 0.80)τ = 3.0τ, so pulses arrive every 90 days	✓ 1	Allow a method that repeats (d)(i)–(d)(iii) with 0.80c.

Answers: (b)(i) 7.0 yr · (b)(ii) 5.6 yr · (b)(iii) 3.4 ly · (c) 31.4 ly² in both frames (Δs = 5.6 ly) · (d)(i) x = 1.5 ly, t = 2.5 yr · (d)(ii) 4.0 yr · (d)(iii) 60 days · (e)(i) 2.82 × 10¹⁴ Hz; infrared · (f) 3.2 yr; arrival at the star first · (g) 3.2 yr; 90 days

Exam tip (b)(ii) Before using γ, find the frame in which both events happen at the same place: that frame measures the proper time, and it is always the shortest time between the events.

Common error (f) Answering “they are simultaneous because they are at the same time on Earth”. Simultaneity of events at different places depends on the frame: here Δx ≠ 0, so Δt′ ≠ 0.

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A kaon beam in an accelerator: where to place the detector

A.5 Galilean and Special RelativityE.3 Radioactive DecayA.1 KinematicsHL

In a particle-physics laboratory, protons strike a target and produce positive kaons (K⁺), which are unstable particles. The kaons leave the target at a constant speed of 0.90c relative to the laboratory and travel along a straight evacuated beam pipe (Figure 1).

The half-life of kaons at rest is 8.58 ns. Each kaon decays at random.

Figure 1: the beam line in the laboratory frame (not to scale)

(a)

Identify, with a reason in each case, the frame in which

the proper length of the beam pipe is measured
the proper half-life of the kaons is measured.

HL onlyREASONNotes

Show solution

beam pipe: the laboratory frame, because the pipe is at rest in it	✓ 1
half-life: the rest frame of the kaons, because there the creation and the decay of a kaon happen at the same place / are timed by one clock present at both events	✓ 1	Not enough: “the observer at rest”.

(2)

(b)(i)

Show that the Lorentz factor γ for the kaons is about 2.3.

HL onlyDERIVENotes

Show solution

γ = 1/√(1 − 0.90²) = 1/√0.19 = 2.29

✓ 1

Full substitution or a value to at least 3 s.f. needed.

Answer: 2.29

(1)

(b)(ii)

Calculate the half-life of the kaons in the laboratory frame.

HL onlyCALCNotes

Show solution

2.29 × 8.58 = 19.7 ns

✓ 1

Allow ECF from (b)(i). Accept 19.6–19.7 ns.

Answer: 19.7 ns

(1)

(b)(iii)

A detector is placed at the distance L from the target, measured in the laboratory, at which 25.0 % of the kaons leaving the target remain in the beam.

Determine L.

HL onlyMULTINotes

Show solution

25.0 % remain after two half-lives, so the time of flight in the laboratory = 2 × 19.7 = 39.4 ns	✓ 1	Allow ECF from (b)(ii).
L = 0.90 × 3.00 × 10⁸ × 39.4 × 10⁻⁹ = 10.6 m	✓ 1	Accept 10.6–10.7 m.

Answer: 10.6 m

(2)

(b)(iv)

A student ignores relativistic effects and calculates the distance from the target at which 25.0 % of the kaons remain.

Calculate the student’s distance.

HL onlyCALCNotes

Show solution

time = 2 × 8.58 = 17.2 ns, using the half-life at rest	✓ 1
distance = 0.90 × 3.00 × 10⁸ × 17.2 × 10⁻⁹ = 4.63 m	✓ 1	Accept 4.6 m.

Answer: 4.63 m

(2)

(c)

In the rest frame of the kaons, the detector moves towards them.

Determine the distance between the target and the detector in this frame. Hence show that an observer in this frame also predicts that 25.0 % of the kaons reach the detector.

HL onlyDERIVENotes

Show solution

distance = 10.6 ÷ 2.29 = 4.63 m	✓ 1	Allow ECF from (b)(i) and (b)(iii). Accept 4.62–4.64 m.
time for the detector to reach the kaons = 4.63 ÷ (0.90 × 3.00 × 10⁸) = 17.2 ns	✓ 1	Accept 39.4 ns ÷ 2.29 only if it is linked to this frame.
17.2 ns = 2 × 8.58 ns, two proper half-lives, so (½)² = 25.0 % remain	✓ 1	The proper half-life must be used. Using the contracted distance with the dilated half-life scores 0 for this mark.

Answer: 4.63 m; two half-lives, so 25.0 %

(3)

(d)

The distance in (b)(iv) is equal to the distance in (c). Explain why the student’s calculation in (b)(iv) nevertheless gives the wrong position for the detector in the laboratory.

HL onlyEXPLAINNotes

Show solution

the student combines the proper half-life, which belongs to the kaon frame, with distances in the laboratory frame / mixes two frames	✓ 1	Accept: the student ignores time dilation.
4.63 m is the target–detector distance in the kaon frame, which is length-contracted; in the laboratory the kaon clocks run slow, and the distance is the proper length, γ × 4.63 = 10.6 m	✓ 1	Allow ECF from (b)(iii) and (c).

(2)

(e)(i)

The prediction is tested. Detectors placed at different distances d from the target measure the number N of kaons remaining in the beam. Figure 2 shows how N/N₀ varies with d, where N₀ is the number leaving the target.

Use Figure 2 to determine the distance over which the number of kaons in the beam halves.

Figure 2: fraction of kaons remaining in the beam against distance from the target (laboratory frame)

HL onlyGRAPHNotes

Show solution

reading from the graph, e.g. N/N₀ = 0.50 at 5.3 m, or 0.25 at 10.6 m, or 0.125 at 15.9 m	✓ 1	Reading at 0.25 or 0.125 and dividing by 2 or 3 gives a more precise value.
5.3 m	✓ 1	Accept 5.1–5.5 m.

Answer: 5.3 m

(2)

(e)(ii)

Use your answer to (e)(i) to determine the half-life of the kaons in the laboratory frame. Hence evaluate whether the data in Figure 2 support time dilation.

HL onlyMULTINotes

Show solution

half-life in the laboratory = 5.3 ÷ (0.90 × 3.00 × 10⁸) = 19.6 ns	✓ 1	Allow ECF from (e)(i). Accept 19–20 ns.
ratio to the half-life at rest = 19.6 ÷ 8.58 = 2.3	✓ 1	Or: compare 19.6 ns with the prediction of (b)(ii).
this equals γ (within the precision of the reading), so the data support time dilation; without it the beam would halve over only 2.3 m	✓ 1	A conclusion consistent with the numbers is required.

Answer: 19.6 ns; ratio 2.3 = γ, so yes

(3)

(f)

The energy of the beam is reduced, and the detector is moved to 9.00 m from the target. The time taken for the kaons to travel from the target to the detector is now measured in the laboratory as 45.0 ns.

Predict the fraction of the kaons leaving the target that now reach the detector.

HL onlyPREDICTNotes

Show solution

v = 9.00 ÷ 45.0 × 10⁻⁹ = 2.00 × 10⁸ m s⁻¹, so γ = 1/√(1 − 0.667²) = 1.34	✓ 1
dilated half-life = 1.34 × 8.58 = 11.5 ns, so 45.0 ÷ 11.5 = 3.9 half-lives	✓ 1	Accept the kaon-frame method: proper time 45.0 ÷ 1.34 = 33.5 ns = 3.9 proper half-lives.
fraction = (½)^3.91 = 0.067	✓ 1	Accept 0.066–0.068. Using e^−λt is equivalent.

Answer: 0.067 (about 7 %)

(3)

(g)

A kaon moving at 0.90c decays into a muon and a neutrino. In the rest frame of the kaon, the muon moves in the direction of the beam at 0.91c.

Calculate the speed of the muon in the laboratory frame predicted by Galilean relativity and by special relativity. Comment on your answers.

HL onlyREASONNotes

Show solution

Galilean: 0.91c + 0.90c = 1.81c	✓ 1
relativistic: u = (0.91c + 0.90c) ÷ (1 + 0.91 × 0.90) = 0.995c	✓ 1	Accept the data-booklet form with the laboratory moving at −0.90c relative to the kaon. Accept 0.99c.
the Galilean value exceeds c, which is impossible; the relativistic value is below c	✓ 1

Answer: Galilean 1.81c; relativistic 0.995c

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
beam pipe: the laboratory frame, because the pipe is at rest in it	✓ 1
half-life: the rest frame of the kaons, because there the creation and the decay of a kaon happen at the same place / are timed by one clock present at both events	✓ 1	Not enough: “the observer at rest”.
Part (b)(i)
γ = 1/√(1 − 0.90²) = 1/√0.19 = 2.29	✓ 1	Full substitution or a value to at least 3 s.f. needed.
Part (b)(ii)
2.29 × 8.58 = 19.7 ns	✓ 1	Allow ECF from (b)(i). Accept 19.6–19.7 ns.
Part (b)(iii)
25.0 % remain after two half-lives, so the time of flight in the laboratory = 2 × 19.7 = 39.4 ns	✓ 1	Allow ECF from (b)(ii).
L = 0.90 × 3.00 × 10⁸ × 39.4 × 10⁻⁹ = 10.6 m	✓ 1	Accept 10.6–10.7 m.
Part (b)(iv)
time = 2 × 8.58 = 17.2 ns, using the half-life at rest	✓ 1
distance = 0.90 × 3.00 × 10⁸ × 17.2 × 10⁻⁹ = 4.63 m	✓ 1	Accept 4.6 m.
Part (c)
distance = 10.6 ÷ 2.29 = 4.63 m	✓ 1	Allow ECF from (b)(i) and (b)(iii). Accept 4.62–4.64 m.
time for the detector to reach the kaons = 4.63 ÷ (0.90 × 3.00 × 10⁸) = 17.2 ns	✓ 1	Accept 39.4 ns ÷ 2.29 only if it is linked to this frame.
17.2 ns = 2 × 8.58 ns, two proper half-lives, so (½)² = 25.0 % remain	✓ 1	The proper half-life must be used. Using the contracted distance with the dilated half-life scores 0 for this mark.
Part (d)
the student combines the proper half-life, which belongs to the kaon frame, with distances in the laboratory frame / mixes two frames	✓ 1	Accept: the student ignores time dilation.
4.63 m is the target–detector distance in the kaon frame, which is length-contracted; in the laboratory the kaon clocks run slow, and the distance is the proper length, γ × 4.63 = 10.6 m	✓ 1	Allow ECF from (b)(iii) and (c).
Part (e)(i)
reading from the graph, e.g. N/N₀ = 0.50 at 5.3 m, or 0.25 at 10.6 m, or 0.125 at 15.9 m	✓ 1	Reading at 0.25 or 0.125 and dividing by 2 or 3 gives a more precise value.
5.3 m	✓ 1	Accept 5.1–5.5 m.
Part (e)(ii)
half-life in the laboratory = 5.3 ÷ (0.90 × 3.00 × 10⁸) = 19.6 ns	✓ 1	Allow ECF from (e)(i). Accept 19–20 ns.
ratio to the half-life at rest = 19.6 ÷ 8.58 = 2.3	✓ 1	Or: compare 19.6 ns with the prediction of (b)(ii).
this equals γ (within the precision of the reading), so the data support time dilation; without it the beam would halve over only 2.3 m	✓ 1	A conclusion consistent with the numbers is required.
Part (f)
v = 9.00 ÷ 45.0 × 10⁻⁹ = 2.00 × 10⁸ m s⁻¹, so γ = 1/√(1 − 0.667²) = 1.34	✓ 1
dilated half-life = 1.34 × 8.58 = 11.5 ns, so 45.0 ÷ 11.5 = 3.9 half-lives	✓ 1	Accept the kaon-frame method: proper time 45.0 ÷ 1.34 = 33.5 ns = 3.9 proper half-lives.
fraction = (½)^3.91 = 0.067	✓ 1	Accept 0.066–0.068. Using e^−λt is equivalent.
Part (g)
Galilean: 0.91c + 0.90c = 1.81c	✓ 1
relativistic: u = (0.91c + 0.90c) ÷ (1 + 0.91 × 0.90) = 0.995c	✓ 1	Accept the data-booklet form with the laboratory moving at −0.90c relative to the kaon. Accept 0.99c.
the Galilean value exceeds c, which is impossible; the relativistic value is below c	✓ 1

Answers: (b)(i) 2.29 · (b)(ii) 19.7 ns · (b)(iii) 10.6 m · (b)(iv) 4.63 m · (c) 4.63 m; two half-lives, so 25.0 % · (e)(i) 5.3 m · (e)(ii) 19.6 ns; ratio 2.3 = γ, so yes · (f) 0.067 (about 7 %) · (g) Galilean 1.81c; relativistic 0.995c

Common error (c) Using the contracted distance together with the dilated half-life. Each frame uses one effect only: the kaon frame has the proper half-life and the contracted distance.

More A.5 questions in the question bank

Theme B · The particulate nature of matter

● home subchapter · ○ linked subchapter · last column: links to other themes

Setup	B.1	B.2	B.3	B.4	B.5	Other
Steaming milk with an espresso machine	●		○		○	—
Heat loss through the wall of a low-energy house	●	○				A.3
A Mars base: energy balance and surface temperature	○	●				D.1
Melting Arctic sea ice and the enhanced greenhouse effect	○	●				C.2
A scuba cylinder and rising bubbles	○		●			A.2
Racing tyres warming up on track	○		●			A.3
A solar Stirling engine: a gas cycle between two isotherms	○		○	●		A.3
Ocean thermal energy conversion: power from a small temperature difference	○	○		●		A.3
A camping headlamp: LED, resistor and battery					●	A.3 E.1
An aquarium heater with a thermistor control circuit	○				●	A.3

B.1 Thermal Energy Transfers

+ Add to test✓ In your testCC-IBP-B1-123 marks~30 min

Steaming milk with an espresso machine

B.1 Thermal Energy TransfersB.3 Gas LawsB.5 Current and CircuitsSL

A barista heats milk with the steam wand of an espresso machine (Figure 1). Steam at 100 °C is blown into 0.25 kg of milk in a jug. The milk starts at 6 °C, straight from the fridge, and the barista stops when the milk reaches 65 °C. The steam condenses in the milk, and the water formed cools to the final temperature of the milk.

specific heat capacity of milk	3900 J kg⁻¹ K⁻¹
specific heat capacity of water	4180 J kg⁻¹ K⁻¹
specific latent heat of vaporization of water	2.26 × 10⁶ J kg⁻¹

Figure 1: the boiler and steam wand of the espresso machine (not to scale)

(a)

Describe, in terms of the molecular model, what happens to the water molecules when the steam condenses, and explain why energy is released although the temperature stays at 100 °C.

EXPLAINNotes

Show solution

the molecules move much closer together (from about ten diameters apart to about one) as intermolecular attractions bind them into a liquid	✓ 1	Accept: the separation of the molecules decreases.
the intermolecular potential energy decreases while the average (random) kinetic energy, and so the temperature, stays the same; the energy lost is transferred to the milk	✓ 1	Not enough: “bonds form” with no reference to potential energy.

(2)

(b)(i)

Assume that no energy is transferred to the jug or to the surroundings.

Calculate the energy gained by the milk.

CALCNotes

Show solution

0.25 × 3900 × (65 − 6) = 5.8 × 10⁴ J

✓ 1

Accept 57 500 J.

Answer: 5.8 × 10⁴ J

(1)

(b)(ii)

Show that about 24 g of steam condenses in the milk.

DERIVENotes

Show solution

energy released by mass m of steam = m(2.26 × 10⁶ + 4180 × (100 − 65))	✓ 1	Both the latent heat term and the cooling term are needed.
m × 2.41 × 10⁶ = 5.75 × 10⁴	✓ 1	Allow ECF from (b)(i).
m = 0.0239 kg ≈ 24 g	✓ 1	At least 3 s.f. or 23.9 g must be seen for a “show that”.

Answer: 0.0239 kg

(3)

(c)

Explain why steam at 100 °C that condenses on the skin causes a much more severe burn than the same mass of water at 100 °C.

EXPLAINNotes

Show solution

the steam first releases its latent heat, 2.26 × 10⁶ J for each kg, as it condenses at 100 °C	✓ 1	Accept a qualitative statement that latent heat is released in addition.
water cooling from 100 °C to skin temperature releases only about 4180 × 63 ≈ 2.6 × 10⁵ J kg⁻¹, so steam transfers far more energy (about ten times) to the skin	✓ 1	Any sensible skin temperature. A comparison is required.

(2)

(d)(i)

Figure 2 shows how the temperature of the milk varies with time t while the steam is supplied at a constant rate.

Use Figure 2 to determine the rate of increase of the temperature of the milk at t = 0.

Figure 2: temperature of the milk against time while it is steamed

GRAPHNotes

Show solution

tangent drawn to the curve at t = 0	✓ 1	A chord from the origin scores 0 for this mark.
gradient = 1.6 K s⁻¹	✓ 1	Accept 1.4–1.8 K s⁻¹. Accept °C s⁻¹.

Answer: 1.6 K s⁻¹

(2)

(d)(ii)

Hence calculate the rate at which the internal energy of the milk increases at t = 0. Ignore the mass of the condensed steam.

CALCNotes

Show solution

0.25 × 3900 × 1.6 = 1.6 × 10³ W

✓ 1

Allow ECF from (d)(i). Accept 1.4–1.8 kW.

Answer: 1.6 × 10³ W

(1)

(d)(iii)

The gradient of the graph decreases as the milk warms, although the steam is supplied at a constant rate. Explain why.

EXPLAINNotes

Show solution

as the milk becomes hotter than the surroundings, the rate of energy transfer from the milk and jug to the surroundings increases	✓ 1	Accept conduction through the jug, convection in the air, evaporation or radiation. Accept: at first the milk is colder than the room and gains energy from it. Also accept: as the milk gets hotter, the temperature difference between the steam and the milk decreases, so less of the steam condenses and some escapes, and the energy given to the milk per second falls.
so the net rate of energy gain by the milk falls, and the gradient (= net power ÷ mc) decreases	✓ 1	Accept as a minor extra: each kg of condensed water cools through a smaller temperature difference, and the mass of liquid increases.

(2)

(e)(i)

The boiler of the machine is a closed tank partly filled with water. The space above the water has a volume of 0.80 L and contains steam at a pressure of 2.0 × 10⁵ Pa and a temperature of 120 °C.

Treat the steam as an ideal gas. Calculate the amount of steam, in mol, in the space above the water.

CALCNotes

Show solution

T = 393 K and V = 8.0 × 10⁻⁴ m³	✓ 1
n = PV/RT = (2.0 × 10⁵ × 8.0 × 10⁻⁴) ÷ (8.31 × 393) = 0.049 mol	✓ 1	Using 120 K or 0.80 m³ scores 0 for this mark.

Answer: 0.049 mol

(2)

(e)(ii)

The molar mass of water is 18 g mol⁻¹. Deduce, using your answers to (b)(ii) and (e)(i), why the water in the boiler must keep boiling while the milk is being steamed.

REASONNotes

Show solution

mass of steam in the space = 0.049 × 18 = 0.88 g	✓ 1	Allow ECF from (e)(i).
this is far less than the 24 g needed (under 4 %), so new steam must be produced continuously by boiling water during steaming	✓ 1	A comparison with (b)(ii) is required.

(2)

(f)(i)

The heating element is connected to the 230 V mains supply. The current in the element is 5.0 A.

Calculate the resistance of the element and the power dissipated in it.

CALCNotes

Show solution

R = 230 ÷ 5.0 = 46 Ω	✓ 1
P = 230 × 5.0 = 1150 W	✓ 1	Accept 1.2 kW.

Answer: 46 Ω; 1150 W

(2)

(f)(ii)

Estimate the minimum time the element needs to boil enough water to replace the steam used for this jug of milk. Assume that the water in the boiler is already at its boiling point and that the specific latent heat of vaporization has the value given above.

MULTINotes

Show solution

energy = 0.024 × 2.26 × 10⁶ = 5.4 × 10⁴ J	✓ 1	Allow ECF from (b)(ii).
t = 5.4 × 10⁴ ÷ 1150 = 47 s	✓ 1	Allow ECF from (f)(i). Accept 46–48 s.

Answer: 47 s

(2)

(g)

The barista now steams 0.25 kg of milk that starts at 20 °C, and stops at 65 °C as before. Predict the mass of steam that condenses, assuming no energy losses.

PREDICTNotes

Show solution

energy gained by milk = 0.25 × 3900 × 45 = 4.4 × 10⁴ J	✓ 1
m = 4.39 × 10⁴ ÷ 2.41 × 10⁶ = 0.018 kg (18 g)	✓ 1	Allow ECF from the method of (b)(ii). Accept a ratio method: 24 × 45/59 = 18 g.

Answer: 0.018 kg

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the molecules move much closer together (from about ten diameters apart to about one) as intermolecular attractions bind them into a liquid	✓ 1	Accept: the separation of the molecules decreases.
the intermolecular potential energy decreases while the average (random) kinetic energy, and so the temperature, stays the same; the energy lost is transferred to the milk	✓ 1	Not enough: “bonds form” with no reference to potential energy.
Part (b)(i)
0.25 × 3900 × (65 − 6) = 5.8 × 10⁴ J	✓ 1	Accept 57 500 J.
Part (b)(ii)
energy released by mass m of steam = m(2.26 × 10⁶ + 4180 × (100 − 65))	✓ 1	Both the latent heat term and the cooling term are needed.
m × 2.41 × 10⁶ = 5.75 × 10⁴	✓ 1	Allow ECF from (b)(i).
m = 0.0239 kg ≈ 24 g	✓ 1	At least 3 s.f. or 23.9 g must be seen for a “show that”.
Part (c)
the steam first releases its latent heat, 2.26 × 10⁶ J for each kg, as it condenses at 100 °C	✓ 1	Accept a qualitative statement that latent heat is released in addition.
water cooling from 100 °C to skin temperature releases only about 4180 × 63 ≈ 2.6 × 10⁵ J kg⁻¹, so steam transfers far more energy (about ten times) to the skin	✓ 1	Any sensible skin temperature. A comparison is required.
Part (d)(i)
tangent drawn to the curve at t = 0	✓ 1	A chord from the origin scores 0 for this mark.
gradient = 1.6 K s⁻¹	✓ 1	Accept 1.4–1.8 K s⁻¹. Accept °C s⁻¹.
Part (d)(ii)
0.25 × 3900 × 1.6 = 1.6 × 10³ W	✓ 1	Allow ECF from (d)(i). Accept 1.4–1.8 kW.
Part (d)(iii)
as the milk becomes hotter than the surroundings, the rate of energy transfer from the milk and jug to the surroundings increases	✓ 1	Accept conduction through the jug, convection in the air, evaporation or radiation. Accept: at first the milk is colder than the room and gains energy from it. Also accept: as the milk gets hotter, the temperature difference between the steam and the milk decreases, so less of the steam condenses and some escapes, and the energy given to the milk per second falls.
so the net rate of energy gain by the milk falls, and the gradient (= net power ÷ mc) decreases	✓ 1	Accept as a minor extra: each kg of condensed water cools through a smaller temperature difference, and the mass of liquid increases.
Part (e)(i)
T = 393 K and V = 8.0 × 10⁻⁴ m³	✓ 1
n = PV/RT = (2.0 × 10⁵ × 8.0 × 10⁻⁴) ÷ (8.31 × 393) = 0.049 mol	✓ 1	Using 120 K or 0.80 m³ scores 0 for this mark.
Part (e)(ii)
mass of steam in the space = 0.049 × 18 = 0.88 g	✓ 1	Allow ECF from (e)(i).
this is far less than the 24 g needed (under 4 %), so new steam must be produced continuously by boiling water during steaming	✓ 1	A comparison with (b)(ii) is required.
Part (f)(i)
R = 230 ÷ 5.0 = 46 Ω	✓ 1
P = 230 × 5.0 = 1150 W	✓ 1	Accept 1.2 kW.
Part (f)(ii)
energy = 0.024 × 2.26 × 10⁶ = 5.4 × 10⁴ J	✓ 1	Allow ECF from (b)(ii).
t = 5.4 × 10⁴ ÷ 1150 = 47 s	✓ 1	Allow ECF from (f)(i). Accept 46–48 s.
Part (g)
energy gained by milk = 0.25 × 3900 × 45 = 4.4 × 10⁴ J	✓ 1
m = 4.39 × 10⁴ ÷ 2.41 × 10⁶ = 0.018 kg (18 g)	✓ 1	Allow ECF from the method of (b)(ii). Accept a ratio method: 24 × 45/59 = 18 g.

Answers: (b)(i) 5.8 × 10⁴ J · (b)(ii) 0.0239 kg · (d)(i) 1.6 K s⁻¹ · (d)(ii) 1.6 × 10³ W · (e)(i) 0.049 mol · (f)(i) 46 Ω; 1150 W · (f)(ii) 47 s · (g) 0.018 kg

Common error (b)(ii) Leaving out the cooling of the condensed water from 100 °C to 65 °C. Write one term for each stage: condensing (mL), then cooling (mcΔT).

More B.1 questions in the question bank

+ Add to test✓ In your testCC-IBP-B1-224 marks~30 min

Heat loss through the wall of a low-energy house

B.1 Thermal Energy TransfersB.2 Greenhouse EffectA.3 Work, Energy and PowerSL

An outside wall of a room in a house has an area of 12 m². It is made of a layer of brick on the inside and a layer of foam insulation on the outside (Figure 1).

layer	thickness / m	thermal conductivity / W m⁻¹ K⁻¹
brick	0.20	0.80
insulation	0.12	0.030

On a winter day the inner surface of the wall is at 21 °C and the outer surface is at −4 °C. These temperatures stay constant.

Figure 1: cross-section through the wall (layer thicknesses to scale)

(a)

Outline how thermal energy is conducted through the brick.

EXPLAINNotes

Show solution

particles near the warmer (inner) side vibrate with a greater average kinetic energy	✓ 1	Accept: larger amplitude of vibration.
they collide with / pass kinetic energy to neighbouring particles, so energy is passed through the brick towards the colder side	✓ 1	Not enough: “heat moves through the brick”. Free electrons are not needed for brick.

(2)

(b)(i)

Explain why the rate of thermal energy transfer through the brick is equal to the rate through the insulation.

REASONNotes

Show solution

the temperatures are constant, so no energy builds up at the boundary: all the energy conducted through the brick must pass through the insulation

✓ 1

Accept: steady state / conservation of energy at the boundary.

(1)

(b)(ii)

Show that the temperature at the boundary between the brick and the insulation is about 19.5 °C.

DERIVENotes

Show solution

0.80 × A × (21 − T) ÷ 0.20 = 0.030 × A × (T + 4) ÷ 0.12	✓ 1	Equal rates through the two layers with the correct ΔT for each.
4.0(21 − T) = 0.25(T + 4)	✓ 1	Or 84 − 4.0T = 0.25T + 1.0.
T = 83 ÷ 4.25 = 19.5 °C	✓ 1	19.53 °C or 83/4.25 must be seen.

Answer: 19.5 °C

(3)

(b)(iii)

Calculate the rate of thermal energy transfer through the wall.

CALCNotes

Show solution

0.80 × 12 × (21 − 19.53) ÷ 0.20 = 71 W

✓ 1

Accept 70–72 W (72 W from 19.5 °C). Accept the calculation through the insulation.

Answer: 71 W

(1)

(c)

On the axes of Figure 2, sketch a graph to show how the temperature varies with the distance x from the inner surface of the wall.

Figure 2: axes for the temperature profile through the wall

SKETCHNotes

Show solution

straight line from 21 °C at x = 0 to about 19.5 °C at x = 0.20 m	✓ 1	Allow ECF from (b)(ii).
steeper straight line from 19.5 °C at 0.20 m to −4 °C at 0.32 m	✓ 1	Curves in either layer score 0 for that layer.

(2)

(d)

The room is heated by a gas boiler with an efficiency of 0.90. The gas costs €0.12 per kWh of energy supplied to the boiler.

Determine the cost per day of replacing the energy lost through this wall.

MULTINotes

Show solution

energy lost per day = 71 × 24 × 3600 = 6.1 × 10⁶ J = 1.7 kWh	✓ 1	Allow ECF from (b)(iii).
energy supplied by gas = 1.69 ÷ 0.90 = 1.9 kWh	✓ 1	Multiplying by 0.90 scores 0 for this mark.
cost = 1.88 × 0.12 = €0.23	✓ 1	Accept €0.22–€0.23.

Answer: €0.23

(3)

(e)(i)

A thermal camera shows that the outer surface of the wall is at −4 °C.

Calculate the wavelength at which the outer surface emits the most power per unit wavelength, and state the region of the electromagnetic spectrum to which this wavelength belongs.

CALCNotes

Show solution

λ_max = 2.9 × 10⁻³ ÷ 269 = 1.1 × 10⁻⁵ m	✓ 1	Using −4 or 4 in place of 269 K scores 0.
infrared	✓ 1

Answer: 1.1 × 10⁻⁵ m; infrared

(2)

(e)(ii)

The emissivity of the outer surface is 0.92. Show that the outer surface emits radiation at a rate of about 3 kW.

DERIVENotes

Show solution

P = eσAT⁴ = 0.92 × 5.67 × 10⁻⁸ × 12 × 269⁴	✓ 1	Full substitution with T in kelvin.
= 3.3 × 10³ W	✓ 1	At least 2 s.f. must be seen.

Answer: 3.3 × 10³ W

(2)

(e)(iii)

Explain why the rate in (e)(ii) is much greater than the rate in (b)(iii), even though the temperatures of the wall are constant.

EXPLAINNotes

Show solution

the outer surface also absorbs radiation from its surroundings (ground, buildings, air, sky), which are at similar temperatures	✓ 1	Accept: the net radiated power is small.
in the steady state the net rate of energy loss from the outer surface (net radiation plus convection to the air) equals the 71 W conducted through the wall	✓ 1	Allow ECF from (b)(iii).

(2)

(f)

Some older houses have cavity walls: two layers of brick separated by a 50 mm gap filled with air. Air has a thermal conductivity of 0.025 W m⁻¹ K⁻¹, smaller than that of the foam insulation.

Explain why an air-filled cavity is much less effective at reducing energy loss than a 50 mm layer of foam insulation.

EXPLAINNotes

Show solution

air next to the warm inner brick is heated, expands and becomes less dense, so it rises	✓ 1	Not enough: “heat rises”.
air next to the cold outer brick cools, becomes denser and sinks, so a convection current carries energy across the cavity	✓ 1	Accept a description of the circulation.
in foam the air is trapped in small cells, so it cannot circulate and energy crosses mainly by slow conduction	✓ 1	Accept: radiation across the cavity also transfers energy.

(3)

(g)

The thickness of the insulation is doubled to 0.24 m. The surface temperatures stay at 21 °C and −4 °C.

Determine the new rate of thermal energy transfer through the wall, and explain why it is not half the value in (b)(iii).

PREDICTNotes

Show solution

new boundary temperature from 4.0(21 − T) = 0.125(T + 4): T = 20.2 °C	✓ 1	Or: (Δx/k) total = 0.25 + 8.0 = 8.25 m² K W⁻¹, so rate per m² = 25/8.25.
rate = 0.80 × 12 × (21 − 20.24) ÷ 0.20 = 36 W	✓ 1	Accept 35–37 W.
the brick layer is unchanged, so the overall opposition to the energy flow is less than doubled / the boundary temperature changes	✓ 1	Accept: the temperature difference across the insulation increases from 23.5 K to 24.2 K.

Answer: 36 W

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
particles near the warmer (inner) side vibrate with a greater average kinetic energy	✓ 1	Accept: larger amplitude of vibration.
they collide with / pass kinetic energy to neighbouring particles, so energy is passed through the brick towards the colder side	✓ 1	Not enough: “heat moves through the brick”. Free electrons are not needed for brick.
Part (b)(i)
the temperatures are constant, so no energy builds up at the boundary: all the energy conducted through the brick must pass through the insulation	✓ 1	Accept: steady state / conservation of energy at the boundary.
Part (b)(ii)
0.80 × A × (21 − T) ÷ 0.20 = 0.030 × A × (T + 4) ÷ 0.12	✓ 1	Equal rates through the two layers with the correct ΔT for each.
4.0(21 − T) = 0.25(T + 4)	✓ 1	Or 84 − 4.0T = 0.25T + 1.0.
T = 83 ÷ 4.25 = 19.5 °C	✓ 1	19.53 °C or 83/4.25 must be seen.
Part (b)(iii)
0.80 × 12 × (21 − 19.53) ÷ 0.20 = 71 W	✓ 1	Accept 70–72 W (72 W from 19.5 °C). Accept the calculation through the insulation.
Part (c)
straight line from 21 °C at x = 0 to about 19.5 °C at x = 0.20 m	✓ 1	Allow ECF from (b)(ii).
steeper straight line from 19.5 °C at 0.20 m to −4 °C at 0.32 m	✓ 1	Curves in either layer score 0 for that layer.
Part (d)
energy lost per day = 71 × 24 × 3600 = 6.1 × 10⁶ J = 1.7 kWh	✓ 1	Allow ECF from (b)(iii).
energy supplied by gas = 1.69 ÷ 0.90 = 1.9 kWh	✓ 1	Multiplying by 0.90 scores 0 for this mark.
cost = 1.88 × 0.12 = €0.23	✓ 1	Accept €0.22–€0.23.
Part (e)(i)
λ_max = 2.9 × 10⁻³ ÷ 269 = 1.1 × 10⁻⁵ m	✓ 1	Using −4 or 4 in place of 269 K scores 0.
infrared	✓ 1
Part (e)(ii)
P = eσAT⁴ = 0.92 × 5.67 × 10⁻⁸ × 12 × 269⁴	✓ 1	Full substitution with T in kelvin.
= 3.3 × 10³ W	✓ 1	At least 2 s.f. must be seen.
Part (e)(iii)
the outer surface also absorbs radiation from its surroundings (ground, buildings, air, sky), which are at similar temperatures	✓ 1	Accept: the net radiated power is small.
in the steady state the net rate of energy loss from the outer surface (net radiation plus convection to the air) equals the 71 W conducted through the wall	✓ 1	Allow ECF from (b)(iii).
Part (f)
air next to the warm inner brick is heated, expands and becomes less dense, so it rises	✓ 1	Not enough: “heat rises”.
air next to the cold outer brick cools, becomes denser and sinks, so a convection current carries energy across the cavity	✓ 1	Accept a description of the circulation.
in foam the air is trapped in small cells, so it cannot circulate and energy crosses mainly by slow conduction	✓ 1	Accept: radiation across the cavity also transfers energy.
Part (g)
new boundary temperature from 4.0(21 − T) = 0.125(T + 4): T = 20.2 °C	✓ 1	Or: (Δx/k) total = 0.25 + 8.0 = 8.25 m² K W⁻¹, so rate per m² = 25/8.25.
rate = 0.80 × 12 × (21 − 20.24) ÷ 0.20 = 36 W	✓ 1	Accept 35–37 W.
the brick layer is unchanged, so the overall opposition to the energy flow is less than doubled / the boundary temperature changes	✓ 1	Accept: the temperature difference across the insulation increases from 23.5 K to 24.2 K.

Answers: (b)(ii) 19.5 °C · (b)(iii) 71 W · (d) €0.23 · (e)(i) 1.1 × 10⁻⁵ m; infrared · (e)(ii) 3.3 × 10³ W · (g) 36 W

Exam tip (c) With the same rate through each layer, ΔT/Δx is inversely proportional to k: the poorer conductor has the steeper temperature gradient.

More B.1 questions in the question bank

B.2 Greenhouse Effect

+ Add to test✓ In your testCC-IBP-B2-120 marks~25 min

A Mars base: energy balance and surface temperature

B.2 Greenhouse EffectB.1 Thermal Energy TransfersD.1 Gravitational FieldsSL

A research team is planning a base on the surface of Mars. The Earth and Mars move around the Sun in orbits that are approximately circular (Figure 1).

Radius of the Earth's orbit = 1.50 × 10¹¹ m; orbital period of the Earth = 365 days
Orbital period of Mars = 687 days
Luminosity of the Sun, L = 3.83 × 10²⁶ W
Albedo of Mars = 0.20

Figure 1: the orbits of the Earth and Mars around the Sun (drawn to scale; the planets are not)

(a)

Show that the radius of the orbit of Mars is about 2.3 × 10¹¹ m.

CALCNotes

Show solution

T² ∝ r³, so r_Mars = 1.50 × 10¹¹ × (687/365)^2/3	✓ 1	Accept the same relation written as T²/r³ = constant, or a calculation of 4π²/GM from the Earth's data.
r_Mars = 2.29 × 10¹¹ m	✓ 1	A value to at least 3 significant figures is needed for a “show that”; 2.3 × 10¹¹ m alone does not score this mark. Using the exponent 3/2 instead of 2/3 (3.87 × 10¹¹ m) scores 0 for this mark.

Answer: 2.29 × 10¹¹ m

(2)

(b)

Calculate the intensity of the Sun's radiation at the orbit of Mars.

CALCNotes

Show solution

b = L/4πd² = 3.83 × 10²⁶ / (4π × (2.29 × 10¹¹)²)	✓ 1	Allow ECF from (a).
581 W m⁻²	✓ 1	Accept 575–590 W m⁻² (583 W m⁻² with the unrounded radius; 576 W m⁻² with 2.3 × 10¹¹ m).

Answer: 581 W m⁻²

(2)

(c)

Explain why the mean intensity of solar radiation over the whole surface of Mars is one quarter of the value in (b).

EXPLAINNotes

Show solution

Mars intercepts the Sun's radiation over a disc / its cross-sectional area πR²	✓ 1	“Projected area” is acceptable.
this power is spread (as the planet rotates) over the whole spherical surface 4πR², and πR²/4πR² = 1/4	✓ 1	“Half of Mars is in darkness” alone does not explain the factor 4.

(2)

(d)

Treat Mars as a black body with no atmosphere, at a uniform temperature.

Determine the equilibrium temperature of Mars.

MULTINotes

Show solution

mean absorbed intensity = (1 − 0.20) × 581/4 = 116 W m⁻²	✓ 1	Allow ECF from (b).
power absorbed = power emitted, so σT⁴ = 116 W m⁻²	✓ 1
T = (116 / 5.67 × 10⁻⁸)^1/4 = 213 K	✓ 1	Accept 212–213 K. Using α in place of (1 − α) gives 150 K: award the second mark only.

Answer: 213 K

(3)

(e)

Calculate the wavelength at which the radiation emitted by Mars has its greatest intensity, and state the region of the electromagnetic spectrum in which it lies.

CALCNotes

Show solution

λ_max = 2.9 × 10⁻³ / 213 = 1.4 × 10⁻⁵ m (14 μm)	✓ 1	Allow ECF from (d).
infrared	✓ 1

Answer: 1.4 × 10⁻⁵ m, infrared

(2)

(f)

Figure 2 shows the spectrum of the radiation emitted by a black body at 288 K, the mean surface temperature of the Earth.

On Figure 2, sketch the spectrum of the radiation emitted by Mars, treated as a black body at the temperature found in (d).

Figure 2: spectrum of a black body at 288 K

SKETCHNotes

Show solution

peak at a longer wavelength than 10 μm, near 14 μm	✓ 1	Allow ECF from (e).
curve lies below the 288 K curve at every wavelength, with a much lower peak (about one fifth of the height, near 6 W m⁻² μm⁻¹)	✓ 1	The peak height scales as T⁵: (213/288)⁵ ≈ 0.22. Accept any peak clearly lower, between about 3 and 8.
correct black-body shape: starts near zero at short wavelengths, rises steeply, then falls more gradually with a long tail	✓ 1	Curves that cross the 288 K curve lose this mark.

(3)

(g)

The atmosphere of Mars is about 95 % carbon dioxide, but its density at the surface is less than 2 % of the density of air at the Earth's surface, and it contains almost no water vapour. The observed mean surface temperature of Mars is close to the value found in (d).

Suggest why the greenhouse effect on Mars is very small.

EXPLAINNotes

Show solution

the thin atmosphere absorbs only a small fraction of the infrared radiation emitted by the surface; most escapes directly to space	✓ 1	Accept: almost no water vapour, the gas that contributes most to the natural greenhouse effect on the Earth.
so little infrared is re-emitted back down to the surface, which therefore stays close to its no-atmosphere equilibrium temperature	✓ 1	Answers that say greenhouse gases “trap” or “reflect” heat do not score this mark.

(2)

(h)(i)

A global dust storm raises the albedo of Mars to 0.30 for several weeks. Assume that Mars reaches a new equilibrium during the storm.

Determine the new equilibrium temperature of Mars.

PREDICTNotes

Show solution

absorbed intensity = 0.70 × 581/4 = 102 W m⁻² or T ∝ (1 − α)^1/4, so T = 213 × (0.70/0.80)^1/4	✓ 1	Allow ECF from (b) and (d).
206 K	✓ 1	Accept 205–207 K.

Answer: 206 K

(2)

(h)(ii)

The base is powered by an array of solar panels of total area 40 m², which converts 22 % of the radiation power incident on it into electrical power. At midday the array faces the Sun. In clear conditions 85 % of the intensity found in (b) reaches the array; during the dust storm only 30 % reaches it.

Calculate the decrease in the electrical power output of the array at midday caused by the storm.

MULTINotes

Show solution

power = intensity × area × 0.22: 0.85 × 581 × 40 × 0.22 = 4.35 kW and 0.30 × 581 × 40 × 0.22 = 1.53 kW	✓ 1	Or the fall in intensity (0.85 − 0.30) × 581 = 320 W m⁻². Allow ECF from (b).
decrease = 2.8 kW	✓ 1	Accept 2.8–2.9 kW.

Answer: 2.8 kW

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
T² ∝ r³, so r_Mars = 1.50 × 10¹¹ × (687/365)^2/3	✓ 1	Accept the same relation written as T²/r³ = constant, or a calculation of 4π²/GM from the Earth's data.
r_Mars = 2.29 × 10¹¹ m	✓ 1	A value to at least 3 significant figures is needed for a “show that”; 2.3 × 10¹¹ m alone does not score this mark. Using the exponent 3/2 instead of 2/3 (3.87 × 10¹¹ m) scores 0 for this mark.
Part (b)
b = L/4πd² = 3.83 × 10²⁶ / (4π × (2.29 × 10¹¹)²)	✓ 1	Allow ECF from (a).
581 W m⁻²	✓ 1	Accept 575–590 W m⁻² (583 W m⁻² with the unrounded radius; 576 W m⁻² with 2.3 × 10¹¹ m).
Part (c)
Mars intercepts the Sun's radiation over a disc / its cross-sectional area πR²	✓ 1	“Projected area” is acceptable.
this power is spread (as the planet rotates) over the whole spherical surface 4πR², and πR²/4πR² = 1/4	✓ 1	“Half of Mars is in darkness” alone does not explain the factor 4.
Part (d)
mean absorbed intensity = (1 − 0.20) × 581/4 = 116 W m⁻²	✓ 1	Allow ECF from (b).
power absorbed = power emitted, so σT⁴ = 116 W m⁻²	✓ 1
T = (116 / 5.67 × 10⁻⁸)^1/4 = 213 K	✓ 1	Accept 212–213 K. Using α in place of (1 − α) gives 150 K: award the second mark only.
Part (e)
λ_max = 2.9 × 10⁻³ / 213 = 1.4 × 10⁻⁵ m (14 μm)	✓ 1	Allow ECF from (d).
infrared	✓ 1
Part (f)
peak at a longer wavelength than 10 μm, near 14 μm	✓ 1	Allow ECF from (e).
curve lies below the 288 K curve at every wavelength, with a much lower peak (about one fifth of the height, near 6 W m⁻² μm⁻¹)	✓ 1	The peak height scales as T⁵: (213/288)⁵ ≈ 0.22. Accept any peak clearly lower, between about 3 and 8.
correct black-body shape: starts near zero at short wavelengths, rises steeply, then falls more gradually with a long tail	✓ 1	Curves that cross the 288 K curve lose this mark.
Part (g)
the thin atmosphere absorbs only a small fraction of the infrared radiation emitted by the surface; most escapes directly to space	✓ 1	Accept: almost no water vapour, the gas that contributes most to the natural greenhouse effect on the Earth.
so little infrared is re-emitted back down to the surface, which therefore stays close to its no-atmosphere equilibrium temperature	✓ 1	Answers that say greenhouse gases “trap” or “reflect” heat do not score this mark.
Part (h)(i)
absorbed intensity = 0.70 × 581/4 = 102 W m⁻² or T ∝ (1 − α)^1/4, so T = 213 × (0.70/0.80)^1/4	✓ 1	Allow ECF from (b) and (d).
206 K	✓ 1	Accept 205–207 K.
Part (h)(ii)
power = intensity × area × 0.22: 0.85 × 581 × 40 × 0.22 = 4.35 kW and 0.30 × 581 × 40 × 0.22 = 1.53 kW	✓ 1	Or the fall in intensity (0.85 − 0.30) × 581 = 320 W m⁻². Allow ECF from (b).
decrease = 2.8 kW	✓ 1	Accept 2.8–2.9 kW.

Answers: (a) 2.29 × 10¹¹ m · (b) 581 W m⁻² · (d) 213 K · (e) 1.4 × 10⁻⁵ m, infrared · (h)(i) 206 K · (h)(ii) 2.8 kW

Exam tip (a) The constant in Kepler's third law is the same for every body orbiting the Sun, so a ratio with the Earth's orbit avoids needing G or the mass of the Sun. Check the direction: the planet with the longer period must have the larger orbit.

Common error (d) Using α instead of (1 − α), or forgetting the factor 4. The planet absorbs (1 − α)S/4 per square metre of its surface.

Common error (f) Drawing the cooler body's curve crossing the hotter one. A black body at a lower temperature emits less at every wavelength, so its curve lies entirely underneath.

More B.2 questions in the question bank

+ Add to test✓ In your testCC-IBP-B2-224 marks~30 min

Melting Arctic sea ice and the enhanced greenhouse effect

B.2 Greenhouse EffectB.1 Thermal Energy TransfersC.2 Wave ModelSL

Figure 1 shows a one-layer energy-balance model of the Earth. All values are averages over the whole Earth and over a year.

The mean intensity of solar radiation arriving at the top of the atmosphere is 340 W m⁻².
The albedo of the Earth is 0.31.
The atmosphere is transparent to the incoming short-wave radiation. It absorbs a fraction ε = 0.80 of the infrared radiation emitted by the surface, and its emissivity is also 0.80.
The surface is a black body at temperature T_s. The atmosphere, at temperature T_a, radiates equally upwards and downwards.

Figure 1: the one-layer energy-balance model (not to scale)

(a)

Distinguish between the albedo of a surface and the emissivity of a surface.

RECALLNotes

Show solution

albedo: the fraction of the incident (radiation) power that is scattered / reflected	✓ 1	Accept total scattered power ÷ total incident power.
emissivity: power radiated per unit area ÷ power radiated per unit area by a black body at the same temperature	✓ 1	“At the same temperature” (or σT⁴ with the same T) is required.

(2)

(b)

Show that the model predicts a surface temperature T_s of about 288 K.

DERIVENotes

Show solution

intensity absorbed by the surface from the Sun = (1 − 0.31) × 340 = 235 W m⁻²	✓ 1
atmosphere in equilibrium: εσT_s⁴ = 2εσT_a⁴, so the downward intensity is ½εσT_s⁴ = 0.40σT_s⁴	✓ 1
surface: 235 + 0.40σT_s⁴ = σT_s⁴, so σT_s⁴ = 391 W m⁻² and T_s = 288 K	✓ 1	An answer of 288 K must be shown with at least one more significant figure (288.2 K) or full working.

Answer: 288 K

(3)

(c)(i)

The Sun's radiation has its greatest intensity at a wavelength of about 0.5 μm. The radiation emitted by the Earth's surface has its greatest intensity at about 10 μm.

State the region of the electromagnetic spectrum in which each of these wavelengths lies.

RECALLNotes

Show solution

0.5 μm: visible and 10 μm: infrared

✓ 1

Both needed.

Answer: visible; infrared

(1)

(c)(ii)

Figure 2 shows, for each wavelength, the fraction of the infrared radiation emitted by the surface that is absorbed by the atmosphere (solid line). The dashed line shows the spectrum of the radiation emitted by the surface, scaled so that its peak is 1.

Using Figure 2, identify the range of wavelengths between 7 μm and 14 μm over which the atmosphere absorbs less than 30 % of the surface's infrared radiation. Hence calculate the corresponding range of frequencies.

Figure 2: fraction of the surface's infrared radiation absorbed by the atmosphere (solid) and the surface's emission spectrum scaled to a peak of 1 (dashed). Simplified, schematic data.

GRAPHNotes

Show solution

range read as 8.0 μm to 13.0 μm	✓ 1	Accept 7.8–8.2 μm and 12.7–13.2 μm.
f = c/λ used with λ in metres, e.g. 3.00 × 10⁸ / 8.0 × 10⁻⁶	✓ 1
2.3 × 10¹³ Hz to 3.8 × 10¹³ Hz	✓ 1	Allow ECF from the readings. Accept 3.75 × 10¹³ Hz.

Answer: 8.0–13 μm; 2.3 × 10¹³ Hz to 3.8 × 10¹³ Hz

(3)

(d)

Explain why the range of wavelengths identified in (c)(ii) is important for the energy balance of the Earth's surface.

EXPLAINNotes

Show solution

the surface emits most strongly in (or near) this range / its emission peak at 10 μm lies inside it	✓ 1	Reference to the dashed curve or to the 10 μm peak is needed.
radiation in this range is hardly absorbed, so it escapes directly to space, removing energy from the surface (so T_s is lower than if all of it were absorbed)	✓ 1	Accept: this is why ε is less than 1.

(2)

(e)

Carbon dioxide absorbs strongly at wavelengths near 15 μm but hardly at all at visible wavelengths.

Explain this using the resonance model of a greenhouse-gas molecule.

EXPLAINNotes

Show solution

the bonds act like springs, so the molecule has a natural frequency of vibration equal to the frequency of 15 μm radiation (about 2 × 10¹³ Hz)	✓ 1
the oscillating electric field of the wave drives the (charged) atoms; when the driving frequency equals the natural frequency there is resonance: large-amplitude vibration, so energy is absorbed	✓ 1
visible light has a frequency (about 4–8 × 10¹⁴ Hz) far from the natural frequency, so the response is tiny and the light passes through	✓ 1

(3)

(f)(i)

In summer, the mean intensity of solar radiation reaching the surface in a region of the Arctic Ocean is 180 W m⁻². The region is covered with sea ice of albedo 0.60. As the climate warms, the sea ice is replaced by open ocean of albedo 0.10.

Calculate the increase in the intensity of solar radiation absorbed by the surface when the ice is replaced by open ocean.

CALCNotes

Show solution

absorbed by ice = 0.40 × 180 = 72 W m⁻² and by ocean = 0.90 × 180 = 162 W m⁻²	✓ 1	Or (0.60 − 0.10) × 180.
increase = 90 W m⁻²	✓ 1

Answer: 90 W m⁻²

(2)

(f)(ii)

The sea ice is 1.5 m thick and is at its melting point.

Density of ice = 920 kg m⁻³
Specific latent heat of fusion of ice = 3.3 × 10⁵ J kg⁻¹

Calculate the energy needed to melt the ice under 1.0 m² of the surface.

CALCNotes

Show solution

mass = 920 × 1.5 × 1.0 = 1380 kg	✓ 1
Q = mL = 1380 × 3.3 × 10⁵ = 4.6 × 10⁸ J	✓ 1

Answer: 4.6 × 10⁸ J

(2)

(f)(iii)

Assume that all the solar energy absorbed by the ice-covered surface is used to melt the ice.

Estimate the time, in days, needed to melt the ice.

MULTINotes

Show solution

t = 4.55 × 10⁸ / 72 = 6.3 × 10⁶ s	✓ 1	Allow ECF from (f)(i) and (f)(ii). Using 162 W m⁻² (open ocean) scores 0 for this mark.
73 days	✓ 1	Accept 72–74 days.

Answer: 73 days

(2)

(g)

The melting of sea ice is described as a positive feedback in the climate system.

Explain why.

EXPLAINNotes

Show solution

open ocean has a lower albedo, so it absorbs a larger fraction of the solar radiation (more than twice the power, from (f)(i)), and the region warms further	✓ 1	“More heat is absorbed” is acceptable if linked to the lower albedo.
the extra warming melts more ice, lowering the albedo further: the effect reinforces its own cause	✓ 1

(2)

(h)

Over many years, the loss of ice and snow is predicted to reduce the albedo of the whole Earth from 0.31 to 0.30. All other quantities in the model stay the same.

Determine the change in the surface temperature predicted by the model.

PREDICTNotes

Show solution

σT_s⁴ = 0.70 × 340 / 0.60 = 397 W m⁻² or T_s ∝ (1 − α)^1/4, so T_s = 288.2 × (0.70/0.69)^1/4	✓ 1	Allow ECF from (b).
T_s = 289.2 K, an increase of about 1.0 K	✓ 1	Accept 0.9–1.1 K.

Answer: an increase of about 1.0 K

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
albedo: the fraction of the incident (radiation) power that is scattered / reflected	✓ 1	Accept total scattered power ÷ total incident power.
emissivity: power radiated per unit area ÷ power radiated per unit area by a black body at the same temperature	✓ 1	“At the same temperature” (or σT⁴ with the same T) is required.
Part (b)
intensity absorbed by the surface from the Sun = (1 − 0.31) × 340 = 235 W m⁻²	✓ 1
atmosphere in equilibrium: εσT_s⁴ = 2εσT_a⁴, so the downward intensity is ½εσT_s⁴ = 0.40σT_s⁴	✓ 1
surface: 235 + 0.40σT_s⁴ = σT_s⁴, so σT_s⁴ = 391 W m⁻² and T_s = 288 K	✓ 1	An answer of 288 K must be shown with at least one more significant figure (288.2 K) or full working.
Part (c)(i)
0.5 μm: visible and 10 μm: infrared	✓ 1	Both needed.
Part (c)(ii)
range read as 8.0 μm to 13.0 μm	✓ 1	Accept 7.8–8.2 μm and 12.7–13.2 μm.
f = c/λ used with λ in metres, e.g. 3.00 × 10⁸ / 8.0 × 10⁻⁶	✓ 1
2.3 × 10¹³ Hz to 3.8 × 10¹³ Hz	✓ 1	Allow ECF from the readings. Accept 3.75 × 10¹³ Hz.
Part (d)
the surface emits most strongly in (or near) this range / its emission peak at 10 μm lies inside it	✓ 1	Reference to the dashed curve or to the 10 μm peak is needed.
radiation in this range is hardly absorbed, so it escapes directly to space, removing energy from the surface (so T_s is lower than if all of it were absorbed)	✓ 1	Accept: this is why ε is less than 1.
Part (e)
the bonds act like springs, so the molecule has a natural frequency of vibration equal to the frequency of 15 μm radiation (about 2 × 10¹³ Hz)	✓ 1
the oscillating electric field of the wave drives the (charged) atoms; when the driving frequency equals the natural frequency there is resonance: large-amplitude vibration, so energy is absorbed	✓ 1
visible light has a frequency (about 4–8 × 10¹⁴ Hz) far from the natural frequency, so the response is tiny and the light passes through	✓ 1
Part (f)(i)
absorbed by ice = 0.40 × 180 = 72 W m⁻² and by ocean = 0.90 × 180 = 162 W m⁻²	✓ 1	Or (0.60 − 0.10) × 180.
increase = 90 W m⁻²	✓ 1
Part (f)(ii)
mass = 920 × 1.5 × 1.0 = 1380 kg	✓ 1
Q = mL = 1380 × 3.3 × 10⁵ = 4.6 × 10⁸ J	✓ 1
Part (f)(iii)
t = 4.55 × 10⁸ / 72 = 6.3 × 10⁶ s	✓ 1	Allow ECF from (f)(i) and (f)(ii). Using 162 W m⁻² (open ocean) scores 0 for this mark.
73 days	✓ 1	Accept 72–74 days.
Part (g)
open ocean has a lower albedo, so it absorbs a larger fraction of the solar radiation (more than twice the power, from (f)(i)), and the region warms further	✓ 1	“More heat is absorbed” is acceptable if linked to the lower albedo.
the extra warming melts more ice, lowering the albedo further: the effect reinforces its own cause	✓ 1
Part (h)
σT_s⁴ = 0.70 × 340 / 0.60 = 397 W m⁻² or T_s ∝ (1 − α)^1/4, so T_s = 288.2 × (0.70/0.69)^1/4	✓ 1	Allow ECF from (b).
T_s = 289.2 K, an increase of about 1.0 K	✓ 1	Accept 0.9–1.1 K.

Answers: (b) 288 K · (c)(i) visible; infrared · (c)(ii) 8.0–13 μm; 2.3 × 10¹³ Hz to 3.8 × 10¹³ Hz · (f)(i) 90 W m⁻² · (f)(ii) 4.6 × 10⁸ J · (f)(iii) 73 days · (h) an increase of about 1.0 K

Exam tip (b) Apply power in = power out to the atmosphere and to the surface separately, and remember that the atmosphere radiates both upwards and downwards.

More B.2 questions in the question bank

B.3 Gas Laws

+ Add to test✓ In your testCC-IBP-B3-123 marks~30 min

A scuba cylinder and rising bubbles

B.3 Gas LawsB.1 Thermal Energy TransfersA.2 Forces and MomentumSL

A diver uses a steel scuba cylinder of internal volume 12 L (1.2 × 10⁻² m³). At a dive shop the cylinder is filled with air to a pressure of 2.0 × 10⁷ Pa, and the air inside is then at 40 °C. The cylinder is taken into the sea, where the water is at 14 °C, and the air in it cools to 14 °C before any of it is used.

Treat air as an ideal gas of molar mass 29 g mol⁻¹.

Figure 1: the cylinder and a rising bubble (not to scale)

(a)

Calculate the amount of air in the cylinder, in mol.

CALCNotes

Show solution

n = PV/RT = 2.0 × 10⁷ × 1.2 × 10⁻² / (8.31 × 313)	✓ 1	Temperature must be in kelvin (313 K).
92 mol	✓ 1	Accept 92.3 mol. Using 40 in place of 313 gives 722 mol: award 0 for this mark.

Answer: 92 mol

(2)

(b)

Determine the pressure of the air in the cylinder when it has cooled to 14 °C.

CALCNotes

Show solution

volume and amount constant, so P/T = constant with kelvin temperatures: P = 2.0 × 10⁷ × 287/313	✓ 1	Or nRT/V with the answer to (a).
1.8 × 10⁷ Pa	✓ 1	Accept 1.83 × 10⁷ Pa.

Answer: 1.8 × 10⁷ Pa

(2)

(c)

Calculate the density of the air in the cylinder.

CALCNotes

Show solution

mass = nM = 92.3 × 0.029 = 2.68 kg	✓ 1	Allow ECF from (a). The molar mass must be in kg mol⁻¹.
ρ = 2.68 / 1.2 × 10⁻² = 220 kg m⁻³	✓ 1	Accept 220–225 kg m⁻³. The density does not change as the air cools.

Answer: 220 kg m⁻³

(2)

(d)

Explain, in terms of the motion of the molecules, why the pressure of the air in the cylinder decreases as the air cools.

EXPLAINNotes

Show solution

lower temperature means a lower mean kinetic energy, so the molecules move more slowly	✓ 1	“Molecules move slower” alone is enough for this mark.
each collision with the wall gives a smaller change of momentum and collisions with each unit area of wall are less frequent	✓ 1	Both effects are needed for this mark.
the rate of change of momentum, which is the force on the walls, is smaller, so the force per unit area (pressure) is smaller	✓ 1	Collisions between molecules are not relevant and score nothing.

(3)

(e)

Figure 1 shows a bubble of air breathed out by the diver at a depth where the pressure is 3.0 × 10⁵ Pa. The volume of the bubble is 2.0 cm³ when it is released. It rises to the surface, where the pressure is 1.01 × 10⁵ Pa. Assume that the air in the bubble stays at 14 °C and that no air enters or leaves it.

Show that the volume of the bubble at the surface is about 6 cm³.

CALCNotes

Show solution

fixed amount at constant temperature: P₁V₁ = P₂V₂	✓ 1
V₂ = 3.0 × 10⁵ × 2.0 / 1.01 × 10⁵ = 5.9 cm³	✓ 1	A final value to at least 2 significant figures is needed.

Answer: 5.9 cm³

(2)

(f)(i)

To test the assumption in (e), a student traps a sample of air equal to the bubble in a syringe connected to a pressure sensor. The syringe is kept at 14 °C and the volume is changed slowly so that the air passes through the same pressures as the rising bubble. Figure 2 shows the student's graph of pressure P against 1/V, with a line of best fit.

Explain how Figure 2 shows that the air obeys Boyle's law.

Figure 2: pressure against 1/volume for the trapped air at 14 °C

GRAPHNotes

Show solution

the points lie on (or close to) a straight line that passes through the origin	✓ 1	Both “straight” and “through the origin” are needed.
so P ∝ 1/V (PV = constant) for a fixed amount of gas at constant temperature	✓ 1

(2)

(f)(ii)

Determine the gradient of the line in Figure 2. Hence determine the amount of air in the sample.

GRAPHNotes

Show solution

gradient = PV = e.g. 3.0 × 10⁵ / 5.0 × 10⁵ = 0.60 Pa m³ (J)	✓ 1	Accept 0.58–0.62. Powers of ten must be handled correctly.
n = gradient/RT = 0.60 / (8.31 × 287) = 2.5 × 10⁻⁴ mol	✓ 1	Allow ECF from the gradient.

Answer: 0.60 Pa m³; 2.5 × 10⁻⁴ mol

(2)

(g)(i)

The density of sea water is 1030 kg m⁻³.

Calculate the buoyancy force on the bubble just after it is released.

CALCNotes

Show solution

F_b = ρVg = 1030 × 2.0 × 10⁻⁶ × 9.8	✓ 1	Volume must be in m³.
0.020 N	✓ 1	Accept 2.0 × 10⁻² N.

Answer: 0.020 N

(2)

(g)(ii)

The bubble quickly reaches a constant speed.

Deduce the drag force on the bubble at this depth.

MULTINotes

Show solution

weight of the air = nMg = 2.5 × 10⁻⁴ × 0.029 × 9.8 = 7 × 10⁻⁵ N, which is negligible compared with F_b	✓ 1	Allow ECF from (f)(ii). Any valid argument that the weight is less than 1 % of F_b scores this mark.
constant speed so resultant force is zero: drag = F_b − weight ≈ 0.020 N, acting downwards	✓ 1	Direction needed. Allow ECF from (g)(i).

Answer: 0.020 N, downwards

(2)

(g)(iii)

Explain why the constant speed of the bubble increases as it rises.

EXPLAINNotes

Show solution

the pressure decreases as the bubble rises, so its volume increases	✓ 1	Reference to Boyle's law or to (e) is acceptable.
the buoyancy force (∝ V) increases, so a larger drag force is needed for a zero resultant force, which requires a larger speed	✓ 1

(2)

(h)

On another dive the water at the depth of release is at 8 °C, while the water near the surface is at 14 °C. A bubble of volume 2.0 cm³ is released at the same depth as before, and the air in it always has the same temperature as the water around it.

Determine, as a percentage, how much larger the volume of this bubble at the surface is than your answer to (e).

PREDICTNotes

Show solution

PV/T = constant with kelvin temperatures: V = 5.94 × 287/281 = 6.07 cm³ or ratio 287/281 = 1.021	✓ 1	Using Celsius (14/8) scores 0.
2.1 % larger	✓ 1	Accept 2 %.

Answer: about 2.1 % larger

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
n = PV/RT = 2.0 × 10⁷ × 1.2 × 10⁻² / (8.31 × 313)	✓ 1	Temperature must be in kelvin (313 K).
92 mol	✓ 1	Accept 92.3 mol. Using 40 in place of 313 gives 722 mol: award 0 for this mark.
Part (b)
volume and amount constant, so P/T = constant with kelvin temperatures: P = 2.0 × 10⁷ × 287/313	✓ 1	Or nRT/V with the answer to (a).
1.8 × 10⁷ Pa	✓ 1	Accept 1.83 × 10⁷ Pa.
Part (c)
mass = nM = 92.3 × 0.029 = 2.68 kg	✓ 1	Allow ECF from (a). The molar mass must be in kg mol⁻¹.
ρ = 2.68 / 1.2 × 10⁻² = 220 kg m⁻³	✓ 1	Accept 220–225 kg m⁻³. The density does not change as the air cools.
Part (d)
lower temperature means a lower mean kinetic energy, so the molecules move more slowly	✓ 1	“Molecules move slower” alone is enough for this mark.
each collision with the wall gives a smaller change of momentum and collisions with each unit area of wall are less frequent	✓ 1	Both effects are needed for this mark.
the rate of change of momentum, which is the force on the walls, is smaller, so the force per unit area (pressure) is smaller	✓ 1	Collisions between molecules are not relevant and score nothing.
Part (e)
fixed amount at constant temperature: P₁V₁ = P₂V₂	✓ 1
V₂ = 3.0 × 10⁵ × 2.0 / 1.01 × 10⁵ = 5.9 cm³	✓ 1	A final value to at least 2 significant figures is needed.
Part (f)(i)
the points lie on (or close to) a straight line that passes through the origin	✓ 1	Both “straight” and “through the origin” are needed.
so P ∝ 1/V (PV = constant) for a fixed amount of gas at constant temperature	✓ 1
Part (f)(ii)
gradient = PV = e.g. 3.0 × 10⁵ / 5.0 × 10⁵ = 0.60 Pa m³ (J)	✓ 1	Accept 0.58–0.62. Powers of ten must be handled correctly.
n = gradient/RT = 0.60 / (8.31 × 287) = 2.5 × 10⁻⁴ mol	✓ 1	Allow ECF from the gradient.
Part (g)(i)
F_b = ρVg = 1030 × 2.0 × 10⁻⁶ × 9.8	✓ 1	Volume must be in m³.
0.020 N	✓ 1	Accept 2.0 × 10⁻² N.
Part (g)(ii)
weight of the air = nMg = 2.5 × 10⁻⁴ × 0.029 × 9.8 = 7 × 10⁻⁵ N, which is negligible compared with F_b	✓ 1	Allow ECF from (f)(ii). Any valid argument that the weight is less than 1 % of F_b scores this mark.
constant speed so resultant force is zero: drag = F_b − weight ≈ 0.020 N, acting downwards	✓ 1	Direction needed. Allow ECF from (g)(i).
Part (g)(iii)
the pressure decreases as the bubble rises, so its volume increases	✓ 1	Reference to Boyle's law or to (e) is acceptable.
the buoyancy force (∝ V) increases, so a larger drag force is needed for a zero resultant force, which requires a larger speed	✓ 1
Part (h)
PV/T = constant with kelvin temperatures: V = 5.94 × 287/281 = 6.07 cm³ or ratio 287/281 = 1.021	✓ 1	Using Celsius (14/8) scores 0.
2.1 % larger	✓ 1	Accept 2 %.

Answers: (a) 92 mol · (b) 1.8 × 10⁷ Pa · (c) 220 kg m⁻³ · (e) 5.9 cm³ · (f)(ii) 0.60 Pa m³; 2.5 × 10⁻⁴ mol · (g)(i) 0.020 N · (g)(ii) 0.020 N, downwards · (h) about 2.1 % larger

Common error (a) Putting the Celsius temperature into PV = nRT. Every temperature in a gas equation must be in kelvin.

Exam tip (f)(ii) The gradient of a P against 1/V graph is PV, which equals nRT. Use a large triangle and read the powers of ten from the axis labels.

More B.3 questions in the question bank

+ Add to test✓ In your testCC-IBP-B3-224 marks~30 min

Racing tyres warming up on track

B.3 Gas LawsB.1 Thermal Energy TransfersA.3 Work, Energy and PowerSL

Before a race, a team sets the pressure in the tyres of a racing car. The air in one tyre has a volume of 0.030 m³, an absolute pressure of 2.4 × 10⁵ Pa and a temperature of 20 °C. During the first laps the air in the tyre warms up to 95 °C.

Treat the air as an ideal gas of molar mass 29 g mol⁻¹. Assume, until part (i), that the volume of the tyre does not change and that no air escapes.

(a)

Calculate the number of moles of air in the tyre.

CALCNotes

Show solution

n = PV/RT = 2.4 × 10⁵ × 0.030 / (8.31 × 293)	✓ 1	Kelvin temperature needed.
2.96 mol	✓ 1	Accept 3.0 mol.

Answer: 2.96 mol

(2)

(b)

Calculate the pressure of the air when it has warmed to 95 °C.

CALCNotes

Show solution

P/T = constant: P = 2.4 × 10⁵ × 368/293	✓ 1	Both temperatures in kelvin.
3.0 × 10⁵ Pa	✓ 1	Accept 3.01 × 10⁵ Pa.

Answer: 3.0 × 10⁵ Pa

(2)

(c)

Calculate the increase in the average translational kinetic energy of an air molecule as the air warms from 20 °C to 95 °C.

CALCNotes

Show solution

Ē_k = 3/2 k_BT, so ΔĒ_k = 3/2 × 1.38 × 10⁻²³ × 75	✓ 1	A temperature change of 75 K, the same as 75 °C.
1.6 × 10⁻²¹ J	✓ 1	Accept 1.55 × 10⁻²¹ J. Calculating both energies and subtracting is fine.

Answer: 1.6 × 10⁻²¹ J

(2)

(d)

Show that the root mean square speed of the air molecules increases by about 12 % as the air warms.

DERIVENotes

Show solution

½mv_rms² = 3/2 k_BT, so v_rms ∝ √T	✓ 1
√(368/293) = 1.12, an increase of 12 %	✓ 1	Ratio to at least 3 significant figures (1.121) or full working needed. √(95/20) scores 0.

Answer: factor 1.12

(2)

(e)(i)

Figure 1 shows the pressure of the air in the tyre at the two temperatures, plotted against Celsius temperature θ.

On Figure 1, sketch the graph of P against θ for the air in the tyre at constant volume, extended to P = 0.

Figure 1: pressure of the air in the tyre against Celsius temperature

SKETCHNotes

Show solution

straight line through both points	✓ 1	A curve scores 0.
line extended to meet the θ-axis at −273 °C	✓ 1	Accept −260 °C to −285 °C. The extended part may be dashed.

(2)

(e)(ii)

Explain why the intercept on the θ-axis would be the same for any fixed amount of ideal gas at any constant volume, and suggest why a real gas could not be cooled along this line to P = 0.

EXPLAINNotes

Show solution

for an ideal gas P ∝ T in kelvin, so P = 0 at T = 0 K (absolute zero, −273 °C) whatever the amount or volume	✓ 1	Accept: the intercept is absolute zero.
a real gas liquefies (its molecules attract each other and have a volume of their own) before reaching 0 K, so the line is only an extrapolation	✓ 1

(2)

(f)

Use P = ⅓ρv², where v² is the mean square speed of the molecules, to determine the root mean square speed of the air molecules in the cold tyre.

MULTINotes

Show solution

ρ = nM/V = 2.96 × 0.029 / 0.030 = 2.86 kg m⁻³	✓ 1	Allow ECF from (a). Molar mass in kg mol⁻¹.
mean square speed v² = 3P/ρ = 3 × 2.4 × 10⁵ / 2.86 = 2.5 × 10⁵ m² s⁻²	✓ 1
v_rms = 500 m s⁻¹	✓ 1	Accept 500–505 m s⁻¹. The same result follows from √(3RT/M).

Answer: 500 m s⁻¹

(3)

(g)(i)

The average force of rolling resistance on this tyre is 150 N. One lap of the circuit is 4.0 km long.

Calculate the work done against rolling resistance by this tyre in one lap.

CALCNotes

Show solution

W = Fs = 150 × 4.0 × 10³ = 6.0 × 10⁵ J

✓ 1

Answer: 6.0 × 10⁵ J

(1)

(g)(ii)

During the first lap, 60 % of this energy remains in the rubber of the tyre as internal energy. The rubber has a mass of 10 kg and a specific heat capacity of 1.8 × 10³ J kg⁻¹ K⁻¹.

Calculate the rise in temperature of the rubber during the first lap.

CALCNotes

Show solution

energy retained = 0.60 × 6.0 × 10⁵ = 3.6 × 10⁵ J	✓ 1	Allow ECF from (g)(i).
ΔT = 3.6 × 10⁵ / (10 × 1.8 × 10³) = 20 K	✓ 1	Accept 20 °C.

Answer: 20 K

(2)

(h)

Teams set the cold pressure of a tyre lower than the pressure they want during the race.

Justify this practice, using your answer to (b).

REASONNotes

Show solution

the pressure rises by about 25 % (from 2.4 × 10⁵ Pa to 3.0 × 10⁵ Pa) as the tyre warms up at (nearly) constant volume	✓ 1	Allow ECF from (b).
so the cold pressure is chosen lower so that the pressure reaches the required value only when the tyre is at its racing temperature (and does not exceed it)	✓ 1

(2)

(i)(i)

In fact, the tyre expands slightly as it warms, so that the volume of the hot air is 0.031 m³.

Determine the pressure of the hot air.

PREDICTNotes

Show solution

PV/T = constant: P₂ = P₁ × (T₂/T₁) × (V₁/V₂) = 2.4 × 10⁵ × (368/293) × (0.030/0.031)	✓ 1	Or P = nRT/V = 2.96 × 8.31 × 368 / 0.031, allowing ECF from (a). Both temperatures in kelvin.
2.92 × 10⁵ Pa	✓ 1	Accept 2.9 × 10⁵ Pa to 2.92 × 10⁵ Pa. 3.0 × 10⁵ Pa does not show the effect of the expansion.

Answer: 2.92 × 10⁵ Pa

(2)

(i)(ii)

Some teams fill their tyres with dry nitrogen (molar mass 28 g mol⁻¹) instead of air, using the same cold pressure, volume and temperature.

State and explain whether the hot pressure in (i)(i) would be different.

REASONNotes

Show solution

no, the hot pressure is the same	✓ 1	Mark only awarded with a valid explanation.
same P, V and T means the same number of moles (n = PV/RT); for an ideal gas P = nRT/V does not depend on the type of gas or its molar mass	✓ 1	Accept the statement that equal volumes of ideal gases at the same P and T contain equal numbers of molecules.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
n = PV/RT = 2.4 × 10⁵ × 0.030 / (8.31 × 293)	✓ 1	Kelvin temperature needed.
2.96 mol	✓ 1	Accept 3.0 mol.
Part (b)
P/T = constant: P = 2.4 × 10⁵ × 368/293	✓ 1	Both temperatures in kelvin.
3.0 × 10⁵ Pa	✓ 1	Accept 3.01 × 10⁵ Pa.
Part (c)
Ē_k = 3/2 k_BT, so ΔĒ_k = 3/2 × 1.38 × 10⁻²³ × 75	✓ 1	A temperature change of 75 K, the same as 75 °C.
1.6 × 10⁻²¹ J	✓ 1	Accept 1.55 × 10⁻²¹ J. Calculating both energies and subtracting is fine.
Part (d)
½mv_rms² = 3/2 k_BT, so v_rms ∝ √T	✓ 1
√(368/293) = 1.12, an increase of 12 %	✓ 1	Ratio to at least 3 significant figures (1.121) or full working needed. √(95/20) scores 0.
Part (e)(i)
straight line through both points	✓ 1	A curve scores 0.
line extended to meet the θ-axis at −273 °C	✓ 1	Accept −260 °C to −285 °C. The extended part may be dashed.
Part (e)(ii)
for an ideal gas P ∝ T in kelvin, so P = 0 at T = 0 K (absolute zero, −273 °C) whatever the amount or volume	✓ 1	Accept: the intercept is absolute zero.
a real gas liquefies (its molecules attract each other and have a volume of their own) before reaching 0 K, so the line is only an extrapolation	✓ 1
Part (f)
ρ = nM/V = 2.96 × 0.029 / 0.030 = 2.86 kg m⁻³	✓ 1	Allow ECF from (a). Molar mass in kg mol⁻¹.
mean square speed v² = 3P/ρ = 3 × 2.4 × 10⁵ / 2.86 = 2.5 × 10⁵ m² s⁻²	✓ 1
v_rms = 500 m s⁻¹	✓ 1	Accept 500–505 m s⁻¹. The same result follows from √(3RT/M).
Part (g)(i)
W = Fs = 150 × 4.0 × 10³ = 6.0 × 10⁵ J	✓ 1
Part (g)(ii)
energy retained = 0.60 × 6.0 × 10⁵ = 3.6 × 10⁵ J	✓ 1	Allow ECF from (g)(i).
ΔT = 3.6 × 10⁵ / (10 × 1.8 × 10³) = 20 K	✓ 1	Accept 20 °C.
Part (h)
the pressure rises by about 25 % (from 2.4 × 10⁵ Pa to 3.0 × 10⁵ Pa) as the tyre warms up at (nearly) constant volume	✓ 1	Allow ECF from (b).
so the cold pressure is chosen lower so that the pressure reaches the required value only when the tyre is at its racing temperature (and does not exceed it)	✓ 1
Part (i)(i)
PV/T = constant: P₂ = P₁ × (T₂/T₁) × (V₁/V₂) = 2.4 × 10⁵ × (368/293) × (0.030/0.031)	✓ 1	Or P = nRT/V = 2.96 × 8.31 × 368 / 0.031, allowing ECF from (a). Both temperatures in kelvin.
2.92 × 10⁵ Pa	✓ 1	Accept 2.9 × 10⁵ Pa to 2.92 × 10⁵ Pa. 3.0 × 10⁵ Pa does not show the effect of the expansion.
Part (i)(ii)
no, the hot pressure is the same	✓ 1	Mark only awarded with a valid explanation.
same P, V and T means the same number of moles (n = PV/RT); for an ideal gas P = nRT/V does not depend on the type of gas or its molar mass	✓ 1	Accept the statement that equal volumes of ideal gases at the same P and T contain equal numbers of molecules.

Answers: (a) 2.96 mol · (b) 3.0 × 10⁵ Pa · (c) 1.6 × 10⁻²¹ J · (d) factor 1.12 · (f) 500 m s⁻¹ · (g)(i) 6.0 × 10⁵ J · (g)(ii) 20 K · (i)(i) 2.92 × 10⁵ Pa

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B.4 Thermodynamics

+ Add to test✓ In your testCC-IBP-B4-124 marks~30 min

A solar Stirling engine: a gas cycle between two isotherms

B.4 ThermodynamicsB.3 Gas LawsA.3 Work, Energy and PowerB.1 Thermal Energy TransfersHL

A solar Stirling engine sits at the focus of a large parabolic mirror. Concentrated sunlight keeps the hot end of the engine at 627 °C, and cooling fins keep the cold end at 27 °C.

The working substance is 0.040 mol of helium, which may be treated as a monatomic ideal gas. In a simplified model the gas is taken round the cycle A → B → C → D → A shown in Figure 1:

A → B: isothermal expansion at 627 °C from 3.0 × 10⁻⁴ m³ to 5.0 × 10⁻⁴ m³
B → C: cooling at constant volume to 27 °C
C → D: isothermal compression at 27 °C back to 3.0 × 10⁻⁴ m³
D → A: heating at constant volume back to 627 °C.

Figure 1: P–V diagram of the cycle for 0.040 mol of helium (drawn to scale). The curves are isotherms at 627 °C and 27 °C.

(a)

Outline why the internal energy of the helium depends only on its temperature.

EXPLAINNotes

Show solution

in an ideal gas there are no intermolecular forces (except during collisions), so the atoms have no intermolecular potential energy: the internal energy is the total random kinetic energy of the atoms	✓ 1	Accept “internal energy is only kinetic energy”.
the mean kinetic energy of an atom is proportional to the absolute temperature, so U = ³/₂nRT changes only when T changes	✓ 1	Accept U = ³/₂Nk_BT quoted with a statement that N is fixed.

(2)

(b)(i)

Show that the pressure of the gas at A is about 1.0 × 10⁶ Pa.

DERIVENotes

Show solution

T = 627 + 273 = 900 K	✓ 1	Substituting a Celsius temperature scores 0 for the part.
P = nRT/V = 0.040 × 8.31 × 900 / 3.0 × 10⁻⁴ = 9.97 × 10⁵ Pa	✓ 1	The answer must be seen to more significant figures than the given value.

Answer: 9.97 × 10⁵ Pa

(2)

(b)(ii)

Calculate the pressure of the gas at C.

CALCNotes

Show solution

P = 0.040 × 8.31 × 300 / 5.0 × 10⁻⁴ = 2.0 × 10⁵ Pa

✓ 1

Accept 1.99 × 10⁵ Pa. Alternative: P_C = P_A × (300/900) × (3.0/5.0).

Answer: 2.0 × 10⁵ Pa

(1)

(c)

Estimate, using Figure 1, the work done by the gas during the expansion A → B.

GRAPHNotes

Show solution

work = area under the curve A → B between 3.0 × 10⁻⁴ m³ and 5.0 × 10⁻⁴ m³, found by counting squares or using trapezium strips	✓ 1	One grid square (0.5 × 10⁻⁴ m³ by 0.5 × 10⁵ Pa) represents 2.5 J; the area is about 61 squares.
≈ 150 J	✓ 1	Accept 145–160 J (a single trapezium gives 160 J). W = PΔV with one pressure (e.g. 200 J or 120 J) scores 0 for this mark.

Answer: ≈ 150 J (153 J)

(2)

(d)

For the stage B → C, determine the work done by the gas, the change in internal energy of the gas and the thermal energy transferred to the gas.

MULTINotes

Show solution

ΔU = ³/₂nRΔT = 1.5 × 0.040 × 8.31 × (300 − 900) = −299 J	✓ 1	Accept −3.0 × 10² J. The negative sign is required.
W = 0 (volume constant), so Q = ΔU + W = −299 J: 299 J of thermal energy leaves the gas	✓ 1	Allow ECF from ΔU. Both W = 0 and the value of Q are needed.

Answer: W = 0, ΔU = −299 J, Q = −299 J

(2)

(e)

By comparing the pressures on the two isothermal stages at the same volume, show that the net work done by the gas per cycle is about 100 J.

DERIVENotes

Show solution

no work is done in B → C or D → A (constant volume)	✓ 1
at the same volume P ∝ T, so at every volume the pressure on C → D is 300/900 = ¹/₃ of that on A → B; the work done on the gas in C → D is ¹/₃ × 153 ≈ 51 J	✓ 1	Allow ECF from (c). Accept an independent area estimate under C → D of 45–55 J.
net work = 153 − 51 ≈ 102 J (≈ 100 J)	✓ 1	Accept nRT ln(V₂/V₁) used for both stages: 152.8 − 50.9 = 102 J.

Answer: ≈ 102 J

(3)

(f)

Assume that all the thermal energy absorbed by the gas comes from the hot end of the engine. Determine the efficiency of the cycle.

MULTINotes

Show solution

A → B: ΔU = 0, so Q = W ≈ 153 J is absorbed	✓ 1	Allow ECF from (c).
D → A: Q = +299 J is absorbed, so the total thermal energy absorbed ≈ 153 + 299 = 452 J	✓ 1	Allow ECF from (d).
η = 102 / 452 = 0.23	✓ 1	Accept 0.22–0.23. Using the net thermal energy (102 J) or only 153 J as the input scores max 1.

Answer: 0.23 (23 %)

(3)

(g)

The engine completes 25 cycles per second and drives a generator of efficiency 0.85. Calculate the electrical power output.

CALCNotes

Show solution

mechanical power = 102 × 25 = 2.5 × 10³ W	✓ 1	Allow ECF from (e); 100 J per cycle gives 2.5 kW.
electrical power = 0.85 × 2.55 × 10³ = 2.2 × 10³ W	✓ 1	Accept 2.1–2.2 kW.

Answer: 2.2 kW

(2)

(h)(i)

Calculate the Carnot efficiency for reservoirs at the temperatures of the hot and cold ends.

CALCNotes

Show solution

η_Carnot = 1 − 300/900 = 0.67

✓ 1

Celsius values (1 − 27/627 = 0.96) score 0.

Answer: 0.67

(1)

(h)(ii)

Explain, with reference to the second law of thermodynamics, why no engine working between these two reservoirs can have an efficiency greater than your answer to (h)(i).

EXPLAINNotes

Show solution

a Carnot engine is reversible: the total entropy of the reservoirs is unchanged, so Q_h/T_h = Q_c/T_c	✓ 1
an engine with a greater efficiency would reject less thermal energy Q_c to the cold end for the same Q_h	✓ 1
the cold reservoir would then gain less entropy than the hot reservoir loses, so the total entropy of the isolated system would decrease, which the second law forbids	✓ 1	Accept −Q_h/T_h + Q_c/T_c < 0. A bare statement that “the Carnot engine is the most efficient” scores max 1.

(3)

(i)

The cooling fins become dusty and the cold end of the engine rises to 77 °C. The temperature of the hot end and the volumes at A and B do not change.

Determine the new net work done by the gas per cycle, and state the effect on the maximum possible efficiency of the engine.

PREDICTNotes

Show solution

T_c = 350 K; the work done in A → B is unchanged (≈ 153 J)	✓ 1
work done on the gas in C → D = 153 × 350/900 ≈ 59 J, so net work ≈ 153 − 59 = 93 J	✓ 1	Allow ECF from (c) and (e). Accept 92–95 J, or 102 × 550/600 = 93 J.
maximum efficiency falls: η_Carnot = 1 − 350/900 = 0.61	✓ 1

Answer: ≈ 93 J; maximum efficiency falls to 0.61

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
in an ideal gas there are no intermolecular forces (except during collisions), so the atoms have no intermolecular potential energy: the internal energy is the total random kinetic energy of the atoms	✓ 1	Accept “internal energy is only kinetic energy”.
the mean kinetic energy of an atom is proportional to the absolute temperature, so U = ³/₂nRT changes only when T changes	✓ 1	Accept U = ³/₂Nk_BT quoted with a statement that N is fixed.
Part (b)(i)
T = 627 + 273 = 900 K	✓ 1	Substituting a Celsius temperature scores 0 for the part.
P = nRT/V = 0.040 × 8.31 × 900 / 3.0 × 10⁻⁴ = 9.97 × 10⁵ Pa	✓ 1	The answer must be seen to more significant figures than the given value.
Part (b)(ii)
P = 0.040 × 8.31 × 300 / 5.0 × 10⁻⁴ = 2.0 × 10⁵ Pa	✓ 1	Accept 1.99 × 10⁵ Pa. Alternative: P_C = P_A × (300/900) × (3.0/5.0).
Part (c)
work = area under the curve A → B between 3.0 × 10⁻⁴ m³ and 5.0 × 10⁻⁴ m³, found by counting squares or using trapezium strips	✓ 1	One grid square (0.5 × 10⁻⁴ m³ by 0.5 × 10⁵ Pa) represents 2.5 J; the area is about 61 squares.
≈ 150 J	✓ 1	Accept 145–160 J (a single trapezium gives 160 J). W = PΔV with one pressure (e.g. 200 J or 120 J) scores 0 for this mark.
Part (d)
ΔU = ³/₂nRΔT = 1.5 × 0.040 × 8.31 × (300 − 900) = −299 J	✓ 1	Accept −3.0 × 10² J. The negative sign is required.
W = 0 (volume constant), so Q = ΔU + W = −299 J: 299 J of thermal energy leaves the gas	✓ 1	Allow ECF from ΔU. Both W = 0 and the value of Q are needed.
Part (e)
no work is done in B → C or D → A (constant volume)	✓ 1
at the same volume P ∝ T, so at every volume the pressure on C → D is 300/900 = ¹/₃ of that on A → B; the work done on the gas in C → D is ¹/₃ × 153 ≈ 51 J	✓ 1	Allow ECF from (c). Accept an independent area estimate under C → D of 45–55 J.
net work = 153 − 51 ≈ 102 J (≈ 100 J)	✓ 1	Accept nRT ln(V₂/V₁) used for both stages: 152.8 − 50.9 = 102 J.
Part (f)
A → B: ΔU = 0, so Q = W ≈ 153 J is absorbed	✓ 1	Allow ECF from (c).
D → A: Q = +299 J is absorbed, so the total thermal energy absorbed ≈ 153 + 299 = 452 J	✓ 1	Allow ECF from (d).
η = 102 / 452 = 0.23	✓ 1	Accept 0.22–0.23. Using the net thermal energy (102 J) or only 153 J as the input scores max 1.
Part (g)
mechanical power = 102 × 25 = 2.5 × 10³ W	✓ 1	Allow ECF from (e); 100 J per cycle gives 2.5 kW.
electrical power = 0.85 × 2.55 × 10³ = 2.2 × 10³ W	✓ 1	Accept 2.1–2.2 kW.
Part (h)(i)
η_Carnot = 1 − 300/900 = 0.67	✓ 1	Celsius values (1 − 27/627 = 0.96) score 0.
Part (h)(ii)
a Carnot engine is reversible: the total entropy of the reservoirs is unchanged, so Q_h/T_h = Q_c/T_c	✓ 1
an engine with a greater efficiency would reject less thermal energy Q_c to the cold end for the same Q_h	✓ 1
the cold reservoir would then gain less entropy than the hot reservoir loses, so the total entropy of the isolated system would decrease, which the second law forbids	✓ 1	Accept −Q_h/T_h + Q_c/T_c < 0. A bare statement that “the Carnot engine is the most efficient” scores max 1.
Part (i)
T_c = 350 K; the work done in A → B is unchanged (≈ 153 J)	✓ 1
work done on the gas in C → D = 153 × 350/900 ≈ 59 J, so net work ≈ 153 − 59 = 93 J	✓ 1	Allow ECF from (c) and (e). Accept 92–95 J, or 102 × 550/600 = 93 J.
maximum efficiency falls: η_Carnot = 1 − 350/900 = 0.61	✓ 1

Answers: (b)(i) 9.97 × 10⁵ Pa · (b)(ii) 2.0 × 10⁵ Pa · (c) ≈ 150 J (153 J) · (d) W = 0, ΔU = −299 J, Q = −299 J · (e) ≈ 102 J · (f) 0.23 (23 %) · (g) 2.2 kW · (h)(i) 0.67 · (i) ≈ 93 J; maximum efficiency falls to 0.61

Common error (c) Using W = PΔV for an isothermal change. The pressure changes along A → B, so the work is the area under the curve.

Exam tip (f) The input energy of an engine is the sum of all the positive values of Q in the cycle, not the net thermal energy (which equals the net work).

More B.4 questions in the question bank

+ Add to test✓ In your testCC-IBP-B4-223 marks~30 min

Ocean thermal energy conversion: power from a small temperature difference

B.4 ThermodynamicsB.1 Thermal Energy TransfersA.3 Work, Energy and PowerB.2 Greenhouse EffectHL

An ocean thermal energy conversion (OTEC) plant on a tropical island uses the temperature difference between warm surface sea water at 26 °C and cold sea water at 5 °C pumped up from a depth of about 1000 m (Figure 1). The warm water evaporates a working fluid that drives a turbine; the cold water condenses the fluid again.

The plant delivers an electrical power of 10 MW. Its overall efficiency (electrical power output ÷ rate at which thermal energy is taken from the warm water) is 0.030.

Treat the warm surface water and the cold deep water as thermal reservoirs at constant temperatures of 26 °C and 5 °C.

Specific heat capacity of sea water = 4.0 × 10³ J kg⁻¹ K⁻¹

Figure 1: the OTEC plant (not to scale)

(a)

Calculate the maximum possible efficiency of a heat engine operating between the two reservoirs.

CALCNotes

Show solution

T_h = 299 K and T_c = 278 K	✓ 1
η_Carnot = 1 − 278/299 = 0.070	✓ 1	Accept 0.0702 or 7.0 %. Celsius values (1 − 5/26 = 0.81) score 0 for the part.

Answer: 0.070 (7.0 %)

(2)

(b)(i)

Calculate the rate at which thermal energy is taken from the warm water.

CALCNotes

Show solution

P_in = 10 × 10⁶ / 0.030 = 3.3 × 10⁸ W

✓ 1

Answer: 3.3 × 10⁸ W

(1)

(b)(ii)

The warm water leaves the evaporator 3.0 K cooler than it entered. Determine the mass of warm water that must flow through the evaporator each second.

CALCNotes

Show solution

thermal energy per second = (mass per second) × cΔT	✓ 1
mass per second = 3.33 × 10⁸ / (4.0 × 10³ × 3.0) = 2.8 × 10⁴ kg s⁻¹	✓ 1	Allow ECF from (b)(i).

Answer: 2.8 × 10⁴ kg s⁻¹

(2)

(c)

Show that the total entropy of the two reservoirs increases by about 5 × 10⁴ J K⁻¹ every second.

DERIVENotes

Show solution

thermal energy given to the cold water per second = 3.33 × 10⁸ − 1.0 × 10⁷ = 3.23 × 10⁸ W	✓ 1	Taking the energy rejected as equal to the energy taken in scores max 1 for the part.
ΔS (warm) = −3.33 × 10⁸ / 299 = −1.11 × 10⁶ J K⁻¹ and ΔS (cold) = +3.23 × 10⁸ / 278 = +1.16 × 10⁶ J K⁻¹ (each second)	✓ 1	Both signs needed.
total ΔS = +4.8 × 10⁴ J K⁻¹ each second (≈ 5 × 10⁴)	✓ 1	Accept 4.7–4.9 × 10⁴ J K⁻¹; the value must be seen to 2 s.f.

Answer: +4.8 × 10⁴ J K⁻¹ per second

(3)

(d)

In a simplified model, the working fluid is replaced by a monatomic ideal gas taken round a Carnot cycle between 299 K and 278 K.

Sketch, on the axes of Figure 2, the P–V diagram for one cycle of this engine. Label the isothermal and adiabatic stages and show the direction in which the cycle is followed.

Figure 2: axes for the sketch of the Carnot cycle

SKETCHNotes

Show solution

closed loop of four curved stages, each with negative gradient: two isotherms joined by two adiabatics	✓ 1	Straight-line stages score max 1.
adiabatics drawn steeper than the isotherms, with the 299 K isotherm above and to the right of the 278 K isotherm; stages correctly labelled	✓ 1
arrows show a clockwise cycle (expansion along the 299 K isotherm)	✓ 1

(3)

(e)

One adiabatic stage of the cycle in (d) is a compression that raises the temperature of the gas from 278 K to 299 K. Show that the volume of the gas at the end of this stage is about 0.90 of its volume at the start.

DERIVENotes

Show solution

P₁V₁^5/3 = P₂V₂^5/3 and P₁V₁/T₁ = P₂V₂/T₂	✓ 1
eliminating P: T₁V₁^2/3 = T₂V₂^2/3	✓ 1	Accept TV^2/3 = constant quoted.
V₂/V₁ = (278/299)^3/2 = 0.897 ≈ 0.90	✓ 1	0.896–0.897 must be seen.

Answer: 0.897

(3)

(f)

The pumps that bring up the cold water move 2.5 × 10⁴ kg of water each second. The work they do on the water each second is equal to the work needed to raise this water through a height of 8.0 m. The pumps are driven by the plant's own electrical output and have an efficiency of 0.80.

Determine the net electrical power available from the plant and its net efficiency.

MULTINotes

Show solution

useful power of the pumps = 2.5 × 10⁴ × 9.8 × 8.0 = 1.96 × 10⁶ W	✓ 1
electrical input to the pumps = 1.96 × 10⁶ / 0.80 = 2.45 × 10⁶ W, so net power = 10 − 2.45 ≈ 7.6 MW	✓ 1	Accept 7.5–7.6 MW.
net efficiency = 7.55 × 10⁶ / 3.33 × 10⁸ = 0.023	✓ 1	Allow ECF from (b)(i).

Answer: 7.6 MW; 0.023

(3)

(g)

Discuss why OTEC plants are built only in tropical regions, and how they could help to reduce the enhanced greenhouse effect.

EXPLAINNotes

Show solution

tropical surface water stays warm all year, so the temperature difference with deep water (about 20 K) is as large as the oceans allow	✓ 1	Award one mark for each of three points, including at least one from each half of the question.
even then the Carnot efficiency is only about 0.07; with a smaller temperature difference the output would barely exceed the power needed for pumping	✓ 1	Also accept: the warm-water flow needed for a given output rises sharply as ΔT falls.
no fuel is burned, so no CO₂ (the main contributor to the enhanced greenhouse effect) is emitted while the plant operates; it can replace fossil-fuel (e.g. diesel) generation on islands	✓ 1	Also accept: unlike solar and wind it can run continuously, day and night, so less fossil-fuel back-up is needed.

(3)

(h)

In winter the surface water cools to 22 °C, while the deep water stays at 5 °C. Assume that the overall efficiency of the plant remains the same fraction of the Carnot efficiency and that the warm water still cools by 3.0 K.

Determine the percentage increase in the flow rate of warm water needed to keep the electrical output at 10 MW.

PREDICTNotes

Show solution

new η_Carnot = 1 − 278/295 = 0.058	✓ 1
for the same output, flow rate ∝ 1/η, so the ratio = 0.0702/0.0576 = 1.22	✓ 1	Alternative: new η = 0.030 × 0.0576/0.0702 = 0.0246, P_in = 4.1 × 10⁸ W, flow = 3.4 × 10⁴ kg s⁻¹.
increase ≈ 22 %	✓ 1	Accept 21–23 %. Allow ECF from (a).

Answer: ≈ 22 %

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
T_h = 299 K and T_c = 278 K	✓ 1
η_Carnot = 1 − 278/299 = 0.070	✓ 1	Accept 0.0702 or 7.0 %. Celsius values (1 − 5/26 = 0.81) score 0 for the part.
Part (b)(i)
P_in = 10 × 10⁶ / 0.030 = 3.3 × 10⁸ W	✓ 1
Part (b)(ii)
thermal energy per second = (mass per second) × cΔT	✓ 1
mass per second = 3.33 × 10⁸ / (4.0 × 10³ × 3.0) = 2.8 × 10⁴ kg s⁻¹	✓ 1	Allow ECF from (b)(i).
Part (c)
thermal energy given to the cold water per second = 3.33 × 10⁸ − 1.0 × 10⁷ = 3.23 × 10⁸ W	✓ 1	Taking the energy rejected as equal to the energy taken in scores max 1 for the part.
ΔS (warm) = −3.33 × 10⁸ / 299 = −1.11 × 10⁶ J K⁻¹ and ΔS (cold) = +3.23 × 10⁸ / 278 = +1.16 × 10⁶ J K⁻¹ (each second)	✓ 1	Both signs needed.
total ΔS = +4.8 × 10⁴ J K⁻¹ each second (≈ 5 × 10⁴)	✓ 1	Accept 4.7–4.9 × 10⁴ J K⁻¹; the value must be seen to 2 s.f.
Part (d)
closed loop of four curved stages, each with negative gradient: two isotherms joined by two adiabatics	✓ 1	Straight-line stages score max 1.
adiabatics drawn steeper than the isotherms, with the 299 K isotherm above and to the right of the 278 K isotherm; stages correctly labelled	✓ 1
arrows show a clockwise cycle (expansion along the 299 K isotherm)	✓ 1
Part (e)
P₁V₁^5/3 = P₂V₂^5/3 and P₁V₁/T₁ = P₂V₂/T₂	✓ 1
eliminating P: T₁V₁^2/3 = T₂V₂^2/3	✓ 1	Accept TV^2/3 = constant quoted.
V₂/V₁ = (278/299)^3/2 = 0.897 ≈ 0.90	✓ 1	0.896–0.897 must be seen.
Part (f)
useful power of the pumps = 2.5 × 10⁴ × 9.8 × 8.0 = 1.96 × 10⁶ W	✓ 1
electrical input to the pumps = 1.96 × 10⁶ / 0.80 = 2.45 × 10⁶ W, so net power = 10 − 2.45 ≈ 7.6 MW	✓ 1	Accept 7.5–7.6 MW.
net efficiency = 7.55 × 10⁶ / 3.33 × 10⁸ = 0.023	✓ 1	Allow ECF from (b)(i).
Part (g)
tropical surface water stays warm all year, so the temperature difference with deep water (about 20 K) is as large as the oceans allow	✓ 1	Award one mark for each of three points, including at least one from each half of the question.
even then the Carnot efficiency is only about 0.07; with a smaller temperature difference the output would barely exceed the power needed for pumping	✓ 1	Also accept: the warm-water flow needed for a given output rises sharply as ΔT falls.
no fuel is burned, so no CO₂ (the main contributor to the enhanced greenhouse effect) is emitted while the plant operates; it can replace fossil-fuel (e.g. diesel) generation on islands	✓ 1	Also accept: unlike solar and wind it can run continuously, day and night, so less fossil-fuel back-up is needed.
Part (h)
new η_Carnot = 1 − 278/295 = 0.058	✓ 1
for the same output, flow rate ∝ 1/η, so the ratio = 0.0702/0.0576 = 1.22	✓ 1	Alternative: new η = 0.030 × 0.0576/0.0702 = 0.0246, P_in = 4.1 × 10⁸ W, flow = 3.4 × 10⁴ kg s⁻¹.
increase ≈ 22 %	✓ 1	Accept 21–23 %. Allow ECF from (a).

Answers: (a) 0.070 (7.0 %) · (b)(i) 3.3 × 10⁸ W · (b)(ii) 2.8 × 10⁴ kg s⁻¹ · (c) +4.8 × 10⁴ J K⁻¹ per second · (e) 0.897 · (f) 7.6 MW; 0.023 · (h) ≈ 22 %

Common error (c) Giving the warm water a positive entropy change. It loses thermal energy, so ΔS is negative; the cold water gains more entropy because its energy arrives at a lower temperature.

More B.4 questions in the question bank

B.5 Current and Circuits

+ Add to test✓ In your testCC-IBP-B5-122 marks~25 min

A camping headlamp: LED, resistor and battery

B.5 Current and CircuitsA.3 Work, Energy and PowerE.1 Structure of the AtomSL

A camping headlamp contains a battery of three identical cells connected in series. Each cell has an emf of 1.5 V and an internal resistance of 0.20 Ω. The battery is connected in series with a switch S, a fixed resistor of resistance R and a light-emitting diode (LED), as shown in Figure 1. The LED emits blue light of wavelength 450 nm.

Figure 2 shows the current–potential difference (I–V) characteristic of the LED.

Figure 1: the headlamp circuit

Figure 2: I–V characteristic of the LED (the current is zero for all pds below 2.0 V)

(a)(i)

Use Figure 2 to state the potential difference across the LED when the current in it is 30 mA.

GRAPHNotes

Show solution

3.2 V

✓ 1

Accept 3.18–3.22 V.

Answer: 3.2 V

(1)

(a)(ii)

State and explain whether the LED is an ohmic conductor.

EXPLAINNotes

Show solution

not ohmic	✓ 1	The judgement mark needs a correct reason.
the I–V graph is not a straight line through the origin: the current is not proportional to the pd (no current below about 2.6 V, then a steep rise), so V/I is not constant	✓ 1

(2)

(b)

Switch S is closed and the current in the circuit is 30 mA. Show that R is about 43 Ω.

DERIVENotes

Show solution

emf of the battery = 4.5 V and its internal resistance = 0.60 Ω	✓ 1
terminal pd = 4.5 − 0.030 × 0.60 = 4.48 V	✓ 1	Or ε = I(R + r) + V_LED written correctly.
R = (4.48 − 3.2) / 0.030 = 42.7 Ω (≈ 43 Ω)	✓ 1	Allow ECF from (a)(i). Ignoring the internal resistance (43.3 Ω) scores max 2.

Answer: 42.7 Ω

(3)

(c)

Calculate the power dissipated in the resistor and the power dissipated in the internal resistance of the battery.

CALCNotes

Show solution

resistor: P = I²R = 0.030² × 42.7 = 0.038 W	✓ 1	Accept 0.039 W from 43 Ω, or IV = 0.030 × 1.28.
internal resistance: P = 0.030² × 0.60 = 5.4 × 10⁻⁴ W	✓ 1

Answer: 38 mW; 0.54 mW

(2)

(d)(i)

Calculate, in eV, the energy of one photon emitted by the LED.

CALCNotes

Show solution

E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 450 × 10⁻⁹ = 4.42 × 10⁻¹⁹ J	✓ 1
E = 4.42 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.8 eV	✓ 1	Accept 2.76 eV. Award [2] for direct use of hc = 1.24 × 10⁻⁶ eV m.

Answer: 2.8 eV

(2)

(d)(ii)

The LED emits light energy at a rate of 27 mW. Determine the number of photons emitted per second.

CALCNotes

Show solution

N = 0.027 / 4.42 × 10⁻¹⁹ = 6.1 × 10¹⁶ s⁻¹

✓ 1

Allow ECF from (d)(i).

Answer: 6.1 × 10¹⁶ s⁻¹

(1)

(e)

The capacity of each cell is 1200 mA h. Determine the efficiency with which the energy transferred by the battery is converted into light, and estimate how long the headlamp can operate before the battery is exhausted.

MULTINotes

Show solution

power converted by the battery = εI = 4.5 × 0.030 = 0.135 W	✓ 1
η = 0.027 / 0.135 = 0.20	✓ 1	Accept 0.20 from the terminal pd (0.134 W).
time = 1200 mA h / 30 mA = 40 h	✓ 1	Using 3 × 1200 mA h (120 h) scores 0 for this mark: the same charge passes through every cell in series.

Answer: 0.20; 40 h

(3)

(f)

As the cells age, their internal resistance increases. Explain why the headlamp becomes dimmer.

EXPLAINNotes

Show solution

the total resistance of the circuit increases, so the current decreases (ε = I(R + r) + V_LED)	✓ 1
more of the emf is lost across the internal resistance (larger Ir), so the terminal pd falls and less pd is available for the resistor and LED	✓ 1
the LED current falls, so fewer photons are emitted per second / the light power (IV in the LED) falls	✓ 1

(3)

(g)(i)

A second, identical LED is connected in parallel with the first. The resistor and the battery are unchanged. A student predicts that each LED will now carry 30 mA.

Explain, without calculation, why each LED must carry less than 30 mA.

REASONNotes

Show solution

the total current through the resistor and battery increases (two paths for the current), so the pd across R and r increases	✓ 1
the pd across the LEDs is therefore less than 3.2 V, so from Figure 2 the current in each LED is less than 30 mA	✓ 1	Accept: if each carried 30 mA, 60 mA in R would leave far less than 3.2 V for the LEDs.

(2)

(g)(ii)

The current in each LED is in fact about 17 mA. Use Figure 2 to show that this is consistent with the circuit.

GRAPHNotes

Show solution

from Figure 2, pd across each LED at 17 mA ≈ 3.05 V	✓ 1	Accept 3.0–3.1 V.
total current = 34 mA, so pd across resistor and internal resistance = 0.034 × (43 + 0.60) = 1.48 V	✓ 1
pd left for the LEDs = 4.5 − 1.48 = 3.02 V, which agrees with the graph value (within the precision of the reading)	✓ 1

Answer: ≈ 3.0 V from both

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
3.2 V	✓ 1	Accept 3.18–3.22 V.
Part (a)(ii)
not ohmic	✓ 1	The judgement mark needs a correct reason.
the I–V graph is not a straight line through the origin: the current is not proportional to the pd (no current below about 2.6 V, then a steep rise), so V/I is not constant	✓ 1
Part (b)
emf of the battery = 4.5 V and its internal resistance = 0.60 Ω	✓ 1
terminal pd = 4.5 − 0.030 × 0.60 = 4.48 V	✓ 1	Or ε = I(R + r) + V_LED written correctly.
R = (4.48 − 3.2) / 0.030 = 42.7 Ω (≈ 43 Ω)	✓ 1	Allow ECF from (a)(i). Ignoring the internal resistance (43.3 Ω) scores max 2.
Part (c)
resistor: P = I²R = 0.030² × 42.7 = 0.038 W	✓ 1	Accept 0.039 W from 43 Ω, or IV = 0.030 × 1.28.
internal resistance: P = 0.030² × 0.60 = 5.4 × 10⁻⁴ W	✓ 1
Part (d)(i)
E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 450 × 10⁻⁹ = 4.42 × 10⁻¹⁹ J	✓ 1
E = 4.42 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.8 eV	✓ 1	Accept 2.76 eV. Award [2] for direct use of hc = 1.24 × 10⁻⁶ eV m.
Part (d)(ii)
N = 0.027 / 4.42 × 10⁻¹⁹ = 6.1 × 10¹⁶ s⁻¹	✓ 1	Allow ECF from (d)(i).
Part (e)
power converted by the battery = εI = 4.5 × 0.030 = 0.135 W	✓ 1
η = 0.027 / 0.135 = 0.20	✓ 1	Accept 0.20 from the terminal pd (0.134 W).
time = 1200 mA h / 30 mA = 40 h	✓ 1	Using 3 × 1200 mA h (120 h) scores 0 for this mark: the same charge passes through every cell in series.
Part (f)
the total resistance of the circuit increases, so the current decreases (ε = I(R + r) + V_LED)	✓ 1
more of the emf is lost across the internal resistance (larger Ir), so the terminal pd falls and less pd is available for the resistor and LED	✓ 1
the LED current falls, so fewer photons are emitted per second / the light power (IV in the LED) falls	✓ 1
Part (g)(i)
the total current through the resistor and battery increases (two paths for the current), so the pd across R and r increases	✓ 1
the pd across the LEDs is therefore less than 3.2 V, so from Figure 2 the current in each LED is less than 30 mA	✓ 1	Accept: if each carried 30 mA, 60 mA in R would leave far less than 3.2 V for the LEDs.
Part (g)(ii)
from Figure 2, pd across each LED at 17 mA ≈ 3.05 V	✓ 1	Accept 3.0–3.1 V.
total current = 34 mA, so pd across resistor and internal resistance = 0.034 × (43 + 0.60) = 1.48 V	✓ 1
pd left for the LEDs = 4.5 − 1.48 = 3.02 V, which agrees with the graph value (within the precision of the reading)	✓ 1

Answers: (a)(i) 3.2 V · (b) 42.7 Ω · (c) 38 mW; 0.54 mW · (d)(i) 2.8 eV · (d)(ii) 6.1 × 10¹⁶ s⁻¹ · (e) 0.20; 40 h · (g)(ii) ≈ 3.0 V from both

Common error (e) Adding the capacities of cells in series. The capacity (in mA h) is a charge, and the same charge flows through each cell in a series battery, so three cells in series last as long as one.

More B.5 questions in the question bank

+ Add to test✓ In your testCC-IBP-B5-220 marks~25 min

An aquarium heater with a thermistor control circuit

B.5 Current and CircuitsB.1 Thermal Energy TransfersA.3 Work, Energy and PowerSL

A tropical aquarium holds 60 kg of water. The water is warmed by a 150 W heater that runs from the 230 V mains supply. The heater is switched on and off by a control unit connected to the potential divider shown in Figure 1.

The potential divider consists of a variable resistor and an NTC thermistor in series across a 9.0 V supply of negligible internal resistance. The thermistor is immersed in the water. The control unit draws negligible current. It switches the heater on when the pd V_out across the thermistor is greater than 4.5 V, and off when V_out is less than 4.5 V.

Figure 2 shows how the resistance of the thermistor varies with temperature.

Figure 1: the potential divider and control unit

Figure 2: resistance of the thermistor against temperature

(a)

Calculate the current in the heater and its resistance when it is operating.

CALCNotes

Show solution

I = P/V = 150/230 = 0.65 A	✓ 1
R = V/I = 230/0.652 = 350 Ω	✓ 1	Accept 353 Ω, or V²/P.

Answer: 0.65 A; 350 Ω

(2)

(b)(i)

Use Figure 2 to determine the resistance of the thermistor at 25 °C.

GRAPHNotes

Show solution

4.5 kΩ

✓ 1

Accept 4.4–4.6 kΩ.

Answer: 4.5 kΩ

(1)

(b)(ii)

The variable resistor is set to 3.0 kΩ. Calculate V_out when the water is at 25 °C.

CALCNotes

Show solution

current = 9.0 / (4.5 + 3.0) × 10³ = 1.2 × 10⁻³ A, or V_out = 9.0 × 4.5/(4.5 + 3.0)	✓ 1
V_out = 5.4 V	✓ 1	Allow ECF from (b)(i).

Answer: 5.4 V

(2)

(b)(iii)

Hence, use Figure 2 to determine the temperature at which the heater switches off with this setting.

GRAPHNotes

Show solution

V_out = 4.5 V is half of 9.0 V, so the thermistor resistance must equal that of the variable resistor, 3.0 kΩ	✓ 1
from the graph, 35 °C	✓ 1	Accept 34–36 °C.

Answer: 35 °C

(2)

(c)

The variable resistor is adjusted so that the heater switches off when the water reaches 25 °C. Show that the current in the potential divider is then about 1 mA, and calculate the power dissipated in the potential divider.

DERIVENotes

Show solution

the variable resistor must also be 4.5 kΩ (equal resistances share 9.0 V equally)	✓ 1
I = 9.0 / 9.0 × 10³ = 1.0 × 10⁻³ A	✓ 1
P = VI = 9.0 × 1.0 × 10⁻³ = 9.0 × 10⁻³ W	✓ 1	Allow ECF for the current.

Answer: 1.0 mA; 9.0 mW

(3)

(d)

Explain, in terms of charge carriers, why the resistance of the thermistor decreases as the temperature of the water increases.

EXPLAINNotes

Show solution

the thermistor is a semiconductor: at a higher temperature more charge carriers (electrons) are freed and become mobile	✓ 1
this increase in the number of charge carriers outweighs the extra collisions with the more strongly vibrating lattice ions, so the current for a given pd is larger and R = V/I is smaller	✓ 1	Do not accept “the electrons move faster” on its own.

(2)

(e)

After a partial water change the water in the tank is at 22 °C. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹.

Calculate the minimum time for the heater to raise the temperature of the water to 25 °C.

CALCNotes

Show solution

Q = mcΔT = 60 × 4180 × 3.0 = 7.5 × 10⁵ J	✓ 1
t = Q/P = 7.52 × 10⁵ / 150 = 5.0 × 10³ s	✓ 1	Accept 84 min.

Answer: 5.0 × 10³ s (84 min)

(2)

(f)

In practice, the water takes 2.0 hours to warm from 22 °C to 25 °C. Determine the efficiency of the heating process and suggest one way in which thermal energy is lost from the aquarium.

MULTINotes

Show solution

energy supplied = 150 × 2.0 × 3600 = 1.08 × 10⁶ J	✓ 1
η = 7.52 × 10⁵ / 1.08 × 10⁶ = 0.70	✓ 1	Allow ECF from (e).
any one: evaporation from the open water surface / conduction through the glass walls and convection in the air around them / radiation from the tank	✓ 1

Answer: 0.70

(3)

(g)

The thermistor and the variable resistor are accidentally connected the other way round, so that V_out is now the pd across the variable resistor, still set as in (c). The water is at 24 °C when this happens, and the room is cooler than 25 °C.

Explain how the heater now responds to the water temperature, and predict what happens to the temperature of the water from now on.

PREDICTNotes

Show solution

below 25 °C the thermistor resistance is greater than 4.5 kΩ, so it takes more than half of the 9.0 V: V_out < 4.5 V and the heater is off when the water is cool	✓ 1
above 25 °C the thermistor resistance is less than 4.5 kΩ, so V_out > 4.5 V and the heater is on only when the water is warmer than 25 °C	✓ 1
at 24 °C V_out < 4.5 V, so the heater stays off: the water is never heated and cools towards room temperature (below 25 °C), well below the 25 °C wanted	✓ 1	Do not accept “the water overheats”: it is already below 25 °C, so the heater never switches on.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
I = P/V = 150/230 = 0.65 A	✓ 1
R = V/I = 230/0.652 = 350 Ω	✓ 1	Accept 353 Ω, or V²/P.
Part (b)(i)
4.5 kΩ	✓ 1	Accept 4.4–4.6 kΩ.
Part (b)(ii)
current = 9.0 / (4.5 + 3.0) × 10³ = 1.2 × 10⁻³ A, or V_out = 9.0 × 4.5/(4.5 + 3.0)	✓ 1
V_out = 5.4 V	✓ 1	Allow ECF from (b)(i).
Part (b)(iii)
V_out = 4.5 V is half of 9.0 V, so the thermistor resistance must equal that of the variable resistor, 3.0 kΩ	✓ 1
from the graph, 35 °C	✓ 1	Accept 34–36 °C.
Part (c)
the variable resistor must also be 4.5 kΩ (equal resistances share 9.0 V equally)	✓ 1
I = 9.0 / 9.0 × 10³ = 1.0 × 10⁻³ A	✓ 1
P = VI = 9.0 × 1.0 × 10⁻³ = 9.0 × 10⁻³ W	✓ 1	Allow ECF for the current.
Part (d)
the thermistor is a semiconductor: at a higher temperature more charge carriers (electrons) are freed and become mobile	✓ 1
this increase in the number of charge carriers outweighs the extra collisions with the more strongly vibrating lattice ions, so the current for a given pd is larger and R = V/I is smaller	✓ 1	Do not accept “the electrons move faster” on its own.
Part (e)
Q = mcΔT = 60 × 4180 × 3.0 = 7.5 × 10⁵ J	✓ 1
t = Q/P = 7.52 × 10⁵ / 150 = 5.0 × 10³ s	✓ 1	Accept 84 min.
Part (f)
energy supplied = 150 × 2.0 × 3600 = 1.08 × 10⁶ J	✓ 1
η = 7.52 × 10⁵ / 1.08 × 10⁶ = 0.70	✓ 1	Allow ECF from (e).
any one: evaporation from the open water surface / conduction through the glass walls and convection in the air around them / radiation from the tank	✓ 1
Part (g)
below 25 °C the thermistor resistance is greater than 4.5 kΩ, so it takes more than half of the 9.0 V: V_out < 4.5 V and the heater is off when the water is cool	✓ 1
above 25 °C the thermistor resistance is less than 4.5 kΩ, so V_out > 4.5 V and the heater is on only when the water is warmer than 25 °C	✓ 1
at 24 °C V_out < 4.5 V, so the heater stays off: the water is never heated and cools towards room temperature (below 25 °C), well below the 25 °C wanted	✓ 1	Do not accept “the water overheats”: it is already below 25 °C, so the heater never switches on.

Answers: (a) 0.65 A; 350 Ω · (b)(i) 4.5 kΩ · (b)(ii) 5.4 V · (b)(iii) 35 °C · (c) 1.0 mA; 9.0 mW · (e) 5.0 × 10³ s (84 min) · (f) 0.70

Exam tip (g) For a potential divider, ask which resistor takes the larger share of the supply pd: the larger resistance always does. Then decide what happens to that resistance as the condition changes.

More B.5 questions in the question bank

Theme C · Wave behaviour

● home subchapter · ○ linked subchapter · last column: links to other themes

Setup	C.1	C.2	C.3	C.4	C.5	Other
A pendulum clock taken to the Moon: period, field strength and energy	●					D.1 A.3
Car suspension and evenly spaced speed bumps: SHM, resonance and damping	●			○		A.2 A.1
A sonar fish-finder on a small boat: wavelength, echoes and refraction		●	○			A.1
Locating an earthquake with P and S waves		●	○			A.1
Total internal reflection in a fibre-optic broadband link		○	●			A.1
Water waves passing through gaps in a harbour wall		○	●	○		—
A violin string and the resonance of its body	○	○		●		—
A wobbling footbridge: pedestrians, resonance and dampers	○			●		A.2 A.3
Measuring a tennis serve with a radar gun		○			●	A.1 A.2
Finding an exoplanet from a star’s wobble					●	D.1 E.1 E.5 A.2

C.1 Simple Harmonic Motion

+ Add to test✓ In your testCC-IBP-C1-124 marks~30 min

A pendulum clock taken to the Moon: period, field strength and energy

C.1 Simple Harmonic MotionD.1 Gravitational FieldsA.3 Work, Energy and PowerSL

A clock is controlled by a simple pendulum of length 0.994 m with a bob of mass 0.40 kg (Figure 1). The clock mechanism moves the hands forward by 2.00 s for every complete oscillation of the pendulum, so the clock keeps correct time on Earth, where g = 9.8 m s⁻². The bob swings with an amplitude of 5.0 cm, measured horizontally from the equilibrium position. Energy losses are replaced by the clock mechanism, so the amplitude stays constant.

Data for the Moon: mass = 7.35 × 10²² kg, radius = 1.74 × 10⁶ m.

Figure 1: the pendulum of the clock (not to scale; the angle of swing is exaggerated)

(a)

State the two conditions on the acceleration of a body for its motion to be simple harmonic.

RECALLNotes

Show solution

the acceleration is proportional to the displacement from the equilibrium position	✓ 1	Accept restoring force in place of acceleration. Not “proportional to distance”.
the acceleration is directed towards the equilibrium position / opposite to the displacement	✓ 1	Award both marks for a = −ω²x with x identified as displacement from equilibrium and the minus sign explained.

(2)

(b)

Show that the period of the pendulum on Earth is 2.00 s.

CALCNotes

Show solution

T = 2π√(l/g) = 2π√(0.994/9.8)	✓ 1	Substitution must be seen.
T = 2.001 s	✓ 1	At least 4 s.f. (2.001 s) needed, since the answer is given. 2.00 s alone with no working: 0.

Answer: 2.001 s

(2)

(c)(i)

Figure 2 shows how the displacement x of the bob varies with time t on Earth.

Write down the amplitude and the period shown by Figure 2.

Figure 2: displacement x of the bob against time t on Earth

GRAPHNotes

Show solution

amplitude = 5.0 cm	✓ 1	Accept 0.050 m. Not 10 cm.
period = 2.0 s	✓ 1	Read between two successive maxima, e.g. 0 s and 2.0 s.

Answer: 5.0 cm; 2.0 s

(2)

(c)(ii)

On the axes of Figure 3, sketch a graph to show how the velocity v of the bob varies with time from t = 0 to t = 4.0 s. No values are required on the velocity axis.

Figure 3: axes for the velocity–time sketch (same time scale as Figure 2)

SKETCHNotes

Show solution

sinusoidal curve starting at v = 0 at t = 0 and becoming negative first	✓ 1	The bob is released from +x₀ and moves towards negative x: v = −v_max sin(πt).
same period as Figure 2: v = 0 at 0, 1.0, 2.0, 3.0 and 4.0 s, with two complete cycles	✓ 1	Extremes of v at 0.5, 1.5, 2.5 and 3.5 s. Allow a positive-first curve for this mark only.

(2)

(d)(i)

By considering the height through which the bob rises between the equilibrium position and an extreme of the swing, show that the maximum speed of the bob is about 0.16 m s⁻¹.

DERIVENotes

Show solution

rise in height h = l − √(l² − x₀²) = 0.994 − √(0.994² − 0.050²) = 1.26 × 10⁻³ m	✓ 1	Accept h ≈ x₀²/2l = 1.26 × 10⁻³ m, or l(1 − cos θ) with θ = 0.0503 rad.
mgh = ½mv² so v = √(2gh)	✓ 1	Energy conservation stated or used: all the gravitational potential energy at the extreme becomes kinetic energy at equilibrium.
v = √(2 × 9.8 × 1.26 × 10⁻³) = 0.157 m s⁻¹	✓ 1	At least 3 s.f. needed. An answer using v_max = ωx₀ = (2π/2.00) × 0.050 = 0.157 m s⁻¹ scores this mark only.

Answer: 0.157 m s⁻¹

(3)

(d)(ii)

Calculate the maximum kinetic energy of the bob.

CALCNotes

Show solution

E_k = ½ × 0.40 × 0.157² = 4.9 × 10⁻³ J

✓ 1

Accept mgh = 0.40 × 9.8 × 1.26 × 10⁻³ = 4.9 × 10⁻³ J. Using 0.16 m s⁻¹ gives 5.1 × 10⁻³ J: accept.

Answer: 4.9 × 10⁻³ J

(1)

(e)

The maximum angle between the string and the vertical is about 3°. Explain why it is important for the timekeeping of the clock that this angle is small.

EXPLAINNotes

Show solution

for small angles sin θ ≈ θ, so the restoring force (mg sin θ) is proportional to the displacement and the motion is simple harmonic	✓ 1	Must link small angle to proportionality / SHM.
so the period is independent of the amplitude / T = 2π√(l/g) is valid; the clock keeps time even if the amplitude changes slightly	✓ 1	For a large swing the motion is not SHM and the period depends on the amplitude.

(2)

(f)(i)

The clock is taken to the Moon. Show that the gravitational field strength at the surface of the Moon is about 1.6 N kg⁻¹.

CALCNotes

Show solution

g = GM/r² = 6.67 × 10⁻¹¹ × 7.35 × 10²² / (1.74 × 10⁶)²	✓ 1	Substitution must be seen.
g = 1.62 N kg⁻¹	✓ 1	At least 3 s.f. Accept 1.619 N kg⁻¹.

Answer: 1.62 N kg⁻¹

(2)

(f)(ii)

Determine the period of the pendulum on the Moon.

CALCNotes

Show solution

T = 2π√(0.994/1.62) = 4.92 s

✓ 1

Allow ECF from (f)(i). Accept 4.9 s. Accept 2.00 × √(9.8/1.62).

Answer: 4.92 s

(1)

(f)(iii)

The clock is set to the correct time and then left running on the Moon for 24.0 hours. Determine the time interval, in hours, that the clock shows.

MULTINotes

Show solution

number of oscillations = 24.0 × 3600 / 4.92 = 1.76 × 10⁴	✓ 1	Or: time shown is in the ratio T_Earth/T_Moon = 2.00/4.92 = 0.407 of the real time.
time shown = 1.76 × 10⁴ × 2.00 s = 3.51 × 10⁴ s = 9.8 h	✓ 1	Allow ECF from (f)(ii). Accept 9.75–9.76 h. Award 1 mark max for 59 h (ratio inverted).

Answer: 9.8 h

(2)

(g)

The length of the pendulum is now adjusted so that the clock keeps correct time on the Moon. Calculate the new length.

CALCNotes

Show solution

the period must again be 2.00 s: l = gT²/4π²	✓ 1	Or l ∝ g: l = 0.994 × 1.62/9.8.
l = 1.62 × 2.00²/4π² = 0.164 m	✓ 1	Allow ECF from (f)(i). Accept 0.16 m.

Answer: 0.164 m

(2)

(h)

The pendulum is returned to its original length and to Earth. Its displacement is given by x = x₀ sin(ωt), where t = 0 is an instant at which the bob passes through the equilibrium position.

Determine the shortest time taken for the bob to move from the equilibrium position to a displacement of 2.5 cm.

HL onlyMULTINotes

Show solution

ω = 2π/T = 2π/2.00 = 3.14 rad s⁻¹	✓ 1
0.025 = 0.050 sin(ωt) so ωt = π/6 (0.524 rad)	✓ 1	Calculator in radians. sin⁻¹(0.5) = 30° converted correctly is fine.
t = 0.524/3.14 = 0.17 s	✓ 1	Accept 0.167 s, or T/12. Not T/8 = 0.25 s (half the amplitude is not reached in half the time to the extreme).

Answer: 0.17 s

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
the acceleration is proportional to the displacement from the equilibrium position	✓ 1	Accept restoring force in place of acceleration. Not “proportional to distance”.
the acceleration is directed towards the equilibrium position / opposite to the displacement	✓ 1	Award both marks for a = −ω²x with x identified as displacement from equilibrium and the minus sign explained.
Part (b)
T = 2π√(l/g) = 2π√(0.994/9.8)	✓ 1	Substitution must be seen.
T = 2.001 s	✓ 1	At least 4 s.f. (2.001 s) needed, since the answer is given. 2.00 s alone with no working: 0.
Part (c)(i)
amplitude = 5.0 cm	✓ 1	Accept 0.050 m. Not 10 cm.
period = 2.0 s	✓ 1	Read between two successive maxima, e.g. 0 s and 2.0 s.
Part (c)(ii)
sinusoidal curve starting at v = 0 at t = 0 and becoming negative first	✓ 1	The bob is released from +x₀ and moves towards negative x: v = −v_max sin(πt).
same period as Figure 2: v = 0 at 0, 1.0, 2.0, 3.0 and 4.0 s, with two complete cycles	✓ 1	Extremes of v at 0.5, 1.5, 2.5 and 3.5 s. Allow a positive-first curve for this mark only.
Part (d)(i)
rise in height h = l − √(l² − x₀²) = 0.994 − √(0.994² − 0.050²) = 1.26 × 10⁻³ m	✓ 1	Accept h ≈ x₀²/2l = 1.26 × 10⁻³ m, or l(1 − cos θ) with θ = 0.0503 rad.
mgh = ½mv² so v = √(2gh)	✓ 1	Energy conservation stated or used: all the gravitational potential energy at the extreme becomes kinetic energy at equilibrium.
v = √(2 × 9.8 × 1.26 × 10⁻³) = 0.157 m s⁻¹	✓ 1	At least 3 s.f. needed. An answer using v_max = ωx₀ = (2π/2.00) × 0.050 = 0.157 m s⁻¹ scores this mark only.
Part (d)(ii)
E_k = ½ × 0.40 × 0.157² = 4.9 × 10⁻³ J	✓ 1	Accept mgh = 0.40 × 9.8 × 1.26 × 10⁻³ = 4.9 × 10⁻³ J. Using 0.16 m s⁻¹ gives 5.1 × 10⁻³ J: accept.
Part (e)
for small angles sin θ ≈ θ, so the restoring force (mg sin θ) is proportional to the displacement and the motion is simple harmonic	✓ 1	Must link small angle to proportionality / SHM.
so the period is independent of the amplitude / T = 2π√(l/g) is valid; the clock keeps time even if the amplitude changes slightly	✓ 1	For a large swing the motion is not SHM and the period depends on the amplitude.
Part (f)(i)
g = GM/r² = 6.67 × 10⁻¹¹ × 7.35 × 10²² / (1.74 × 10⁶)²	✓ 1	Substitution must be seen.
g = 1.62 N kg⁻¹	✓ 1	At least 3 s.f. Accept 1.619 N kg⁻¹.
Part (f)(ii)
T = 2π√(0.994/1.62) = 4.92 s	✓ 1	Allow ECF from (f)(i). Accept 4.9 s. Accept 2.00 × √(9.8/1.62).
Part (f)(iii)
number of oscillations = 24.0 × 3600 / 4.92 = 1.76 × 10⁴	✓ 1	Or: time shown is in the ratio T_Earth/T_Moon = 2.00/4.92 = 0.407 of the real time.
time shown = 1.76 × 10⁴ × 2.00 s = 3.51 × 10⁴ s = 9.8 h	✓ 1	Allow ECF from (f)(ii). Accept 9.75–9.76 h. Award 1 mark max for 59 h (ratio inverted).
Part (g)
the period must again be 2.00 s: l = gT²/4π²	✓ 1	Or l ∝ g: l = 0.994 × 1.62/9.8.
l = 1.62 × 2.00²/4π² = 0.164 m	✓ 1	Allow ECF from (f)(i). Accept 0.16 m.
Part (h)
ω = 2π/T = 2π/2.00 = 3.14 rad s⁻¹	✓ 1
0.025 = 0.050 sin(ωt) so ωt = π/6 (0.524 rad)	✓ 1	Calculator in radians. sin⁻¹(0.5) = 30° converted correctly is fine.
t = 0.524/3.14 = 0.17 s	✓ 1	Accept 0.167 s, or T/12. Not T/8 = 0.25 s (half the amplitude is not reached in half the time to the extreme).

Answers: (b) 2.001 s · (c)(i) 5.0 cm; 2.0 s · (d)(i) 0.157 m s⁻¹ · (d)(ii) 4.9 × 10⁻³ J · (f)(i) 1.62 N kg⁻¹ · (f)(ii) 4.92 s · (f)(iii) 9.8 h · (g) 0.164 m · (h) 0.17 s

Exam tip (b) In a “show that”, write the substitution and give one more significant figure than the value quoted.

Common error (c)(ii) Starting the velocity graph at its maximum. At t = 0 the bob is at an extreme, where the gradient of the x–t graph, and so the velocity, is zero.

More C.1 questions in the question bank

+ Add to test✓ In your testCC-IBP-C1-222 marks~25 min

Car suspension and evenly spaced speed bumps: SHM, resonance and damping

C.1 Simple Harmonic MotionC.4 Standing Waves and ResonanceA.2 Forces and MomentumA.1 KinematicsSL

The body of a car has a mass of 1200 kg and is supported by four identical springs, one above each wheel (Figure 1). When a load of 80 kg is placed in the car, the body moves down by 1.0 cm. Treat the four springs as a single spring that obeys Hooke's law, and treat the car body as the only oscillating mass. Take g = 9.8 m s⁻².

The 80 kg load is then removed.

Figure 1: the car body on its springs, approaching a road with speed bumps (not to scale)

(a)

Show that the combined spring constant of the four springs is about 8 × 10⁴ N m⁻¹.

CALCNotes

Show solution

extra force = weight of load = 80 × 9.8 = 784 N	✓ 1	Hooke's law F = kΔx applied to the extra compression.
k = 784/0.010 = 7.84 × 10⁴ N m⁻¹	✓ 1	Answer must be seen as 7.8 × 10⁴ or more precisely. Using 1.0 (cm) unconverted: 0 for this mark.

Answer: 7.84 × 10⁴ N m⁻¹

(2)

(b)

The car body is displaced vertically and released. Show that the natural frequency of its vertical oscillations is about 1.3 Hz.

CALCNotes

Show solution

T = 2π√(m/k) = 2π√(1200/7.84 × 10⁴) = 0.777 s	✓ 1	Allow ECF from (a). Using m = 1280 kg: max 1 mark (the load has been removed).
f = 1/T = 1.29 Hz	✓ 1	At least 3 s.f.

Answer: 1.29 Hz

(2)

(c)

Explain why the weight of the car body does not affect the period of its vertical oscillations.

EXPLAINNotes

Show solution

at the equilibrium position the spring force balances the weight (kx_e = mg), so the weight only sets / shifts the equilibrium position	✓ 1	Accept: the springs are already compressed by the weight at equilibrium.
for a further displacement x the resultant (restoring) force is kx, which does not involve g, so a = −(k/m)x and T = 2π√(m/k)	✓ 1	Not enough: “g is not in the equation” with no physics.

(2)

(d)(i)

With worn shock absorbers, the car drives over a pothole. Figure 2 shows how the vertical displacement x of the car body varies with time t afterwards.

Use Figure 2 to determine the period of the oscillations, and comment on whether your value is consistent with your answer to (b).

Figure 2: vertical displacement x of the car body against time t after the pothole (worn shock absorbers)

GRAPHNotes

Show solution

time for several oscillations read from successive maxima, e.g. 4 oscillations in 3.12 s, giving T = 0.78 s	✓ 1	Or from zero crossings, e.g. 0.20 s to 2.93 s is 3.5 periods. Accept 0.76–0.80 s. Using one cycle only is acceptable if the value is in range.
consistent: f = 1/0.78 = 1.28 Hz, about the same as 1.29 Hz / light damping hardly changes the period	✓ 1	Allow ECF from (b).

Answer: 0.78 s

(2)

(d)(ii)

At an extreme of the oscillation all the energy of the oscillation is stored as elastic potential energy, ½kx², measured from the equilibrium position.

Use Figure 2 to determine the fraction of the energy of the oscillation that is dissipated during the first complete oscillation.

GRAPHNotes

Show solution

amplitudes read as 4.0 cm at t = 0 and 2.0 cm after one period; energy ∝ x₀² so the energy falls to (2.0/4.0)² = 0.25 of its initial value	✓ 1
fraction dissipated = 0.75	✓ 1	Answer 0.50 (amplitude ratio used as energy ratio): 1 mark max.

Answer: 0.75

(2)

(d)(iii)

Describe the energy changes of the oscillating car body during one oscillation in Figure 2.

EXPLAINNotes

Show solution

kinetic energy is maximum at the equilibrium position and zero at the extremes, where the (elastic) potential energy is maximum; the two interchange twice per oscillation	✓ 1
the total energy decreases each cycle: the shock absorbers / resistive forces transfer energy to internal (thermal) energy	✓ 1	Must mention dissipation as well as the interchange.

(2)

(e)(i)

The car is driven at constant speed along a road with speed bumps spaced 12 m apart (Figure 1). At one particular speed the car body bounces with a very large amplitude.

Explain why this happens.

EXPLAINNotes

Show solution

the bumps provide a periodic driving force on the car body	✓ 1	Accept: the bumps force the body to oscillate at the rate at which they are hit.
resonance: when the driving frequency (bumps hit per second) equals the natural frequency of the body, energy is transferred most effectively and the amplitude is a maximum	✓ 1

(2)

(e)(ii)

Calculate the speed at which this happens.

CALCNotes

Show solution

time between bumps = natural period = 0.777 s	✓ 1	Or driving frequency = 1.29 Hz. Allow ECF from (b).
v = 12/0.777 = 15 m s⁻¹	✓ 1	Accept 15.4 or 15.5 m s⁻¹ (about 56 km h⁻¹).

Answer: 15 m s⁻¹

(2)

(f)(i)

Explain the purpose of the shock absorbers and state the type of damping that is most desirable for a car suspension.

EXPLAINNotes

Show solution

shock absorbers dissipate the energy of the oscillation, reducing the amplitude, including the amplitude at resonance	✓ 1
critical damping: the body returns to equilibrium in the shortest time without oscillating / bouncing	✓ 1	Accept “close to critical”. Not heavy damping.

(2)

(f)(ii)

On Figure 2, sketch the variation of x with t if the car body, starting from the same initial displacement, were critically damped.

SKETCHNotes

Show solution

curve starting at x = 4.0 cm at t = 0 and falling to zero without crossing the time axis	✓ 1	No oscillation / overshoot.
reaches (very close to) zero within about one period of the original oscillation, i.e. by about 0.8 s	✓ 1	Not a curve that takes several periods to settle (that would be heavy damping).

(2)

(g)

Four passengers with a total mass of 300 kg get into the car. Determine the new speed at which the car body bounces with a very large amplitude on the same road.

MULTINotes

Show solution

T = 2π√(1500/7.84 × 10⁴) = 0.869 s	✓ 1	Or f = 1.15 Hz. Allow ECF from (a).
v = 12/0.869 = 14 m s⁻¹	✓ 1	Accept 13.8 m s⁻¹. Allow ECF from (e)(ii) by scaling: v ∝ 1/√m, 15.4 × √(1200/1500) = 13.8 m s⁻¹.

Answer: 14 m s⁻¹

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
extra force = weight of load = 80 × 9.8 = 784 N	✓ 1	Hooke's law F = kΔx applied to the extra compression.
k = 784/0.010 = 7.84 × 10⁴ N m⁻¹	✓ 1	Answer must be seen as 7.8 × 10⁴ or more precisely. Using 1.0 (cm) unconverted: 0 for this mark.
Part (b)
T = 2π√(m/k) = 2π√(1200/7.84 × 10⁴) = 0.777 s	✓ 1	Allow ECF from (a). Using m = 1280 kg: max 1 mark (the load has been removed).
f = 1/T = 1.29 Hz	✓ 1	At least 3 s.f.
Part (c)
at the equilibrium position the spring force balances the weight (kx_e = mg), so the weight only sets / shifts the equilibrium position	✓ 1	Accept: the springs are already compressed by the weight at equilibrium.
for a further displacement x the resultant (restoring) force is kx, which does not involve g, so a = −(k/m)x and T = 2π√(m/k)	✓ 1	Not enough: “g is not in the equation” with no physics.
Part (d)(i)
time for several oscillations read from successive maxima, e.g. 4 oscillations in 3.12 s, giving T = 0.78 s	✓ 1	Or from zero crossings, e.g. 0.20 s to 2.93 s is 3.5 periods. Accept 0.76–0.80 s. Using one cycle only is acceptable if the value is in range.
consistent: f = 1/0.78 = 1.28 Hz, about the same as 1.29 Hz / light damping hardly changes the period	✓ 1	Allow ECF from (b).
Part (d)(ii)
amplitudes read as 4.0 cm at t = 0 and 2.0 cm after one period; energy ∝ x₀² so the energy falls to (2.0/4.0)² = 0.25 of its initial value	✓ 1
fraction dissipated = 0.75	✓ 1	Answer 0.50 (amplitude ratio used as energy ratio): 1 mark max.
Part (d)(iii)
kinetic energy is maximum at the equilibrium position and zero at the extremes, where the (elastic) potential energy is maximum; the two interchange twice per oscillation	✓ 1
the total energy decreases each cycle: the shock absorbers / resistive forces transfer energy to internal (thermal) energy	✓ 1	Must mention dissipation as well as the interchange.
Part (e)(i)
the bumps provide a periodic driving force on the car body	✓ 1	Accept: the bumps force the body to oscillate at the rate at which they are hit.
resonance: when the driving frequency (bumps hit per second) equals the natural frequency of the body, energy is transferred most effectively and the amplitude is a maximum	✓ 1
Part (e)(ii)
time between bumps = natural period = 0.777 s	✓ 1	Or driving frequency = 1.29 Hz. Allow ECF from (b).
v = 12/0.777 = 15 m s⁻¹	✓ 1	Accept 15.4 or 15.5 m s⁻¹ (about 56 km h⁻¹).
Part (f)(i)
shock absorbers dissipate the energy of the oscillation, reducing the amplitude, including the amplitude at resonance	✓ 1
critical damping: the body returns to equilibrium in the shortest time without oscillating / bouncing	✓ 1	Accept “close to critical”. Not heavy damping.
Part (f)(ii)
curve starting at x = 4.0 cm at t = 0 and falling to zero without crossing the time axis	✓ 1	No oscillation / overshoot.
reaches (very close to) zero within about one period of the original oscillation, i.e. by about 0.8 s	✓ 1	Not a curve that takes several periods to settle (that would be heavy damping).
Part (g)
T = 2π√(1500/7.84 × 10⁴) = 0.869 s	✓ 1	Or f = 1.15 Hz. Allow ECF from (a).
v = 12/0.869 = 14 m s⁻¹	✓ 1	Accept 13.8 m s⁻¹. Allow ECF from (e)(ii) by scaling: v ∝ 1/√m, 15.4 × √(1200/1500) = 13.8 m s⁻¹.

Answers: (a) 7.84 × 10⁴ N m⁻¹ · (b) 1.29 Hz · (d)(i) 0.78 s · (d)(ii) 0.75 · (e)(ii) 15 m s⁻¹ · (g) 14 m s⁻¹

Common error (d)(ii) Halving the amplitude does not halve the energy. The energy depends on the amplitude squared, so it falls to a quarter.

More C.1 questions in the question bank

C.2 Wave Model

+ Add to test✓ In your testCC-IBP-C2-120 marks~25 min

A sonar fish-finder on a small boat: wavelength, echoes and refraction

C.2 Wave ModelC.3 Wave PhenomenaA.1 KinematicsSL

A fish-finder on a small boat has a transducer under the hull that sends short pulses of ultrasound vertically downwards (Figure 1). The frequency of the ultrasound is 200 kHz. The fish-finder detects the echoes and calculates depths assuming that the speed of sound in sea water is 1500 m s⁻¹.

Figure 1: the fish-finder sends pulses down to a shoal and the seabed (not to scale)

(a)

Ultrasound is a longitudinal wave. Describe how the water particles move as a pulse passes through the water.

RECALLNotes

Show solution

the particles oscillate about fixed equilibrium positions / there is no net movement of the water	✓ 1	Not “the particles move with the wave”.
the oscillations are parallel to the direction of energy transfer / vertical here	✓ 1	Accept: forming compressions and rarefactions.

(2)

(b)(i)

Calculate the wavelength of the ultrasound in sea water.

CALCNotes

Show solution

λ = v/f = 1500/(200 × 10³) = 7.5 × 10⁻³ m

✓ 1

Accept 7.5 mm. Power-of-ten error: 0.

Answer: 7.5 × 10⁻³ m

(1)

(b)(ii)

Before the trip, the transducer is switched on in air, where the speed of sound is 340 m s⁻¹. Calculate the wavelength of the ultrasound in air.

CALCNotes

Show solution

same frequency (set by the source): λ = 340/(200 × 10³) = 1.7 × 10⁻³ m

✓ 1

Accept 1.7 mm.

Answer: 1.7 × 10⁻³ m

(1)

(c)

The echo from a shoal of fish arrives 24 ms after a pulse is emitted. Calculate the depth of the shoal below the transducer.

CALCNotes

Show solution

total distance travelled = 1500 × 24 × 10⁻³ = 36 m	✓ 1	There and back.
depth = 36/2 = 18 m	✓ 1	Answer 36 m (no factor of 2): 1 mark max.

Answer: 18 m

(2)

(d)(i)

Figure 2 shows the displacement of the water particles against distance below the transducer at one instant. A positive displacement is downwards, the direction in which the wave travels.

Show that Figure 2 is consistent with your answer to (b)(i).

Figure 2: displacement of the water particles against distance x below the transducer at one instant (positive = displaced downwards)

GRAPHNotes

Show solution

two complete wavelengths span 15.0 mm, so λ = 7.5 mm, as calculated

✓ 1

Must refer to a reading from the graph, e.g. P to S = 7.5 mm.

Answer: 7.5 mm

(1)

(d)(ii)

Identify which of the labelled points P, Q, R or S is at the centre of a compression. Explain your answer.

REASONNotes

Show solution

Q	✓ 1	No mark for Q without some reasoning.
particles just above Q are displaced downwards (positive) and particles just below Q are displaced upwards (negative), so both move towards Q / Q is where the graph crosses zero going down	✓ 1	Not: “Q has zero displacement” alone (R also has zero displacement and is a rarefaction).

Answer: Q

(2)

(e)

Two fish swim one above the other, 2.0 m apart in depth. Show that their echoes from the same pulse arrive about 2.7 ms apart.

CALCNotes

Show solution

extra distance travelled by the deeper echo = 2 × 2.0 = 4.0 m	✓ 1
Δt = 4.0/1500 = 2.67 × 10⁻³ s	✓ 1	At least 3 s.f. Answer 1.3 ms (no factor 2): 1 mark max.

Answer: 2.67 ms

(2)

(f)(i)

Part of a pulse reaches the flat seabed with an angle of incidence of 20°. The speed of sound in the seabed sediment is 1800 m s⁻¹.

Calculate the angle of refraction of the sound that enters the sediment.

CALCNotes

Show solution

sin θ₂/sin 20° = v₂/v₁ = 1800/1500 so sin θ₂ = 0.410	✓ 1
θ₂ = 24°	✓ 1	Accept 24.2°. 16.5° (ratio inverted): 1 mark max.

Answer: 24°

(2)

(f)(ii)

Sound that reaches the seabed at an angle of incidence of 60° does not enter the sediment. Explain why, supporting your answer with a calculation.

EXPLAINNotes

Show solution

critical angle: sin θ_c = 1500/1800 so θ_c = 56°	✓ 1	Accept 56.4°.
the sound travels towards a medium in which it is faster and 60° is greater than the critical angle, so it is totally reflected	✓ 1	Both conditions needed for the mark. Allow ECF for the critical angle.

(2)

(g)

The fish-finder can also operate at 50 kHz. Sound of higher frequency is absorbed more strongly by sea water. Discuss the advantage of each setting, 200 kHz and 50 kHz.

EXPLAINNotes

Show solution

at 50 kHz the wavelength is 30 mm (four times larger)	✓ 1	Accept a qualitative statement: lower frequency, longer wavelength (same speed).
200 kHz: shorter wavelength, so waves diffract less around small objects / small fish or details comparable to a few mm are detected; finer detail	✓ 1	Must link wavelength to the size of the object.
50 kHz: less absorbed, so the pulse and its echo remain detectable from greater depths / longer range	✓ 1

(3)

(h)

The boat moves into warm, brackish water in which the speed of sound is 1480 m s⁻¹. The fish-finder still assumes a speed of 1500 m s⁻¹. An echo from the seabed arrives 24 ms after the pulse is emitted.

Determine the percentage error in the depth displayed, and state whether the displayed depth is too large or too small.

MULTINotes

Show solution

true depth = 1480 × 0.012 = 17.8 m; displayed depth = 18.0 m	✓ 1	Allow ECF from (c).
percentage error = (18.0 − 17.76)/17.76 × 100 = 1.4 %, displayed depth too large	✓ 1	Accept 1.3–1.4 %. Both the value and “too large” needed.

Answer: 1.4 %, too large

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the particles oscillate about fixed equilibrium positions / there is no net movement of the water	✓ 1	Not “the particles move with the wave”.
the oscillations are parallel to the direction of energy transfer / vertical here	✓ 1	Accept: forming compressions and rarefactions.
Part (b)(i)
λ = v/f = 1500/(200 × 10³) = 7.5 × 10⁻³ m	✓ 1	Accept 7.5 mm. Power-of-ten error: 0.
Part (b)(ii)
same frequency (set by the source): λ = 340/(200 × 10³) = 1.7 × 10⁻³ m	✓ 1	Accept 1.7 mm.
Part (c)
total distance travelled = 1500 × 24 × 10⁻³ = 36 m	✓ 1	There and back.
depth = 36/2 = 18 m	✓ 1	Answer 36 m (no factor of 2): 1 mark max.
Part (d)(i)
two complete wavelengths span 15.0 mm, so λ = 7.5 mm, as calculated	✓ 1	Must refer to a reading from the graph, e.g. P to S = 7.5 mm.
Part (d)(ii)
Q	✓ 1	No mark for Q without some reasoning.
particles just above Q are displaced downwards (positive) and particles just below Q are displaced upwards (negative), so both move towards Q / Q is where the graph crosses zero going down	✓ 1	Not: “Q has zero displacement” alone (R also has zero displacement and is a rarefaction).
Part (e)
extra distance travelled by the deeper echo = 2 × 2.0 = 4.0 m	✓ 1
Δt = 4.0/1500 = 2.67 × 10⁻³ s	✓ 1	At least 3 s.f. Answer 1.3 ms (no factor 2): 1 mark max.
Part (f)(i)
sin θ₂/sin 20° = v₂/v₁ = 1800/1500 so sin θ₂ = 0.410	✓ 1
θ₂ = 24°	✓ 1	Accept 24.2°. 16.5° (ratio inverted): 1 mark max.
Part (f)(ii)
critical angle: sin θ_c = 1500/1800 so θ_c = 56°	✓ 1	Accept 56.4°.
the sound travels towards a medium in which it is faster and 60° is greater than the critical angle, so it is totally reflected	✓ 1	Both conditions needed for the mark. Allow ECF for the critical angle.
Part (g)
at 50 kHz the wavelength is 30 mm (four times larger)	✓ 1	Accept a qualitative statement: lower frequency, longer wavelength (same speed).
200 kHz: shorter wavelength, so waves diffract less around small objects / small fish or details comparable to a few mm are detected; finer detail	✓ 1	Must link wavelength to the size of the object.
50 kHz: less absorbed, so the pulse and its echo remain detectable from greater depths / longer range	✓ 1
Part (h)
true depth = 1480 × 0.012 = 17.8 m; displayed depth = 18.0 m	✓ 1	Allow ECF from (c).
percentage error = (18.0 − 17.76)/17.76 × 100 = 1.4 %, displayed depth too large	✓ 1	Accept 1.3–1.4 %. Both the value and “too large” needed.

Answers: (b)(i) 7.5 × 10⁻³ m · (b)(ii) 1.7 × 10⁻³ m · (c) 18 m · (d)(i) 7.5 mm · (d)(ii) Q · (e) 2.67 ms · (f)(i) 24° · (h) 1.4 %, too large

Common error (c) Forgetting that the echo time covers the journey down and back up. Halve the distance.

Exam tip (d)(ii) Compressions and rarefactions are where the displacement is zero, not at the peaks. Look at which way the neighbouring particles are displaced.

More C.2 questions in the question bank

+ Add to test✓ In your testCC-IBP-C2-221 marks~25 min

Locating an earthquake with P and S waves

C.2 Wave ModelC.3 Wave PhenomenaA.1 KinematicsSL

An earthquake produces two types of seismic wave that travel through the Earth's crust: P waves, which are longitudinal, and S waves, which are transverse. In the crust the P waves travel at 6.0 km s⁻¹ and the S waves at 3.5 km s⁻¹. Both types leave the earthquake at the same instant.

Figure 1 shows a simplified seismogram, the record of the ground motion at a seismic station X.

Figure 1: simplified seismogram recorded at station X; t is measured from the start of the recording

(a)

Distinguish between a longitudinal wave and a transverse wave.

RECALLNotes

Show solution

longitudinal: the particles of the medium oscillate parallel to the direction of energy transfer	✓ 1
transverse: the particles oscillate perpendicular to the direction of energy transfer	✓ 1	Not “perpendicular to the wave”. Award 1 mark max if “direction of energy transfer / travel” is missing throughout.

(2)

(b)

Use Figure 1 to determine the time interval between the arrival of the P waves and the arrival of the S waves at station X.

GRAPHNotes

Show solution

P arrives at 18 s, S at 43 s, interval = 25 s

✓ 1

Accept 24–26 s.

Answer: 25 s

(1)

(c)

Show that the distance from station X to the earthquake is about 210 km.

DERIVENotes

Show solution

t_S − t_P = d/3.5 − d/6.0	✓ 1	Any correct equation linking the lag to the distance.
25 = d(1/3.5 − 1/6.0) = 0.119d	✓ 1	Allow ECF from (b).
d = 210 km	✓ 1	Must show 210 or a value such as 210.0 from working. A check that 210/3.5 − 210/6.0 = 25 s, fully set out, scores 2 marks max.

Answer: 210 km

(3)

(d)

Determine the time at which the earthquake occurred, relative to t = 0 on Figure 1.

MULTINotes

Show solution

P-wave travel time = 210/6.0 = 35 s	✓ 1	Or S-wave travel time = 210/3.5 = 60 s.
earthquake at t = 18 − 35 = −17 s, i.e. 17 s before the recording started	✓ 1	Allow ECF from (b) and (c).

Answer: 17 s before t = 0

(2)

(e)(i)

The S waves contain a range of frequencies. Calculate the wavelength in the crust of an S-wave component of frequency 2.0 Hz.

CALCNotes

Show solution

λ = 3.5 × 10³/2.0 = 1.8 × 10³ m

✓ 1

Accept 1750 m or 1.75 km.

Answer: 1.8 × 10³ m

(1)

(e)(ii)

This S wave passes into a basin of soft sediment, where S waves travel at 1.4 km s⁻¹. State its frequency in the sediment and determine its wavelength there.

CALCNotes

Show solution

frequency unchanged, 2.0 Hz (set by the source)	✓ 1
λ = 1.4 × 10³/2.0 = 700 m	✓ 1	Accept 0.70 km.

Answer: 2.0 Hz; 700 m

(2)

(f)

Stations on the opposite side of the Earth detect P waves but no S waves that have passed through the Earth's liquid outer core. Explain why S waves cannot travel through the liquid outer core.

EXPLAINNotes

Show solution

S waves are transverse: each layer of the medium must pull its neighbour sideways, perpendicular to the direction of travel	✓ 1	Accept: a transverse mechanical wave needs a restoring force on particles displaced sideways.
a liquid cannot provide a sideways (shear) restoring force, so the transverse oscillation is not passed on; P waves pass because a liquid resists compression	✓ 1

(2)

(g)(i)

Figure 2 shows a P wave in the crust meeting the boundary with the mantle at an angle of incidence of 30°. P waves travel at 8.0 km s⁻¹ in the mantle.

Calculate the angle of refraction of the P wave in the mantle.

Figure 2: a P wave meeting the crust–mantle boundary (not to scale)

CALCNotes

Show solution

sin θ₂ = sin 30° × 8.0/6.0 = 0.667	✓ 1
θ₂ = 42°	✓ 1	Accept 41.8°. 22° (ratio inverted): 1 mark max.

Answer: 42°

(2)

(g)(ii)

Within the mantle, the wave speed increases gradually with depth. Explain why the paths of seismic waves through the mantle are curved.

EXPLAINNotes

Show solution

the wave is refracted continuously as it moves into deeper rock where it travels faster	✓ 1
it bends away from the normal / towards the horizontal at each depth, so the path curves round and back up towards the surface	✓ 1	Accept a reference to the wavefronts swinging round because their deeper part travels faster.

(2)

(h)

Justify why the time intervals measured at three separate stations are needed to locate the earthquake.

REASONNotes

Show solution

one station gives only the distance, so the earthquake lies somewhere on a circle around that station	✓ 1	Accept “in any direction”.
two circles generally cross at two points; the circle from a third station identifies which point is the earthquake	✓ 1

(2)

(i)

At a second station, Y, the time interval between the arrival of the P waves and the S waves is 40 s. Determine the distance from station Y to the earthquake.

CALCNotes

Show solution

d ∝ time interval, or 40 = d(1/3.5 − 1/6.0)	✓ 1
d = 210 × 40/25 = 336 km	✓ 1	Accept 340 km. Allow ECF from (c).

Answer: 336 km

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
longitudinal: the particles of the medium oscillate parallel to the direction of energy transfer	✓ 1
transverse: the particles oscillate perpendicular to the direction of energy transfer	✓ 1	Not “perpendicular to the wave”. Award 1 mark max if “direction of energy transfer / travel” is missing throughout.
Part (b)
P arrives at 18 s, S at 43 s, interval = 25 s	✓ 1	Accept 24–26 s.
Part (c)
t_S − t_P = d/3.5 − d/6.0	✓ 1	Any correct equation linking the lag to the distance.
25 = d(1/3.5 − 1/6.0) = 0.119d	✓ 1	Allow ECF from (b).
d = 210 km	✓ 1	Must show 210 or a value such as 210.0 from working. A check that 210/3.5 − 210/6.0 = 25 s, fully set out, scores 2 marks max.
Part (d)
P-wave travel time = 210/6.0 = 35 s	✓ 1	Or S-wave travel time = 210/3.5 = 60 s.
earthquake at t = 18 − 35 = −17 s, i.e. 17 s before the recording started	✓ 1	Allow ECF from (b) and (c).
Part (e)(i)
λ = 3.5 × 10³/2.0 = 1.8 × 10³ m	✓ 1	Accept 1750 m or 1.75 km.
Part (e)(ii)
frequency unchanged, 2.0 Hz (set by the source)	✓ 1
λ = 1.4 × 10³/2.0 = 700 m	✓ 1	Accept 0.70 km.
Part (f)
S waves are transverse: each layer of the medium must pull its neighbour sideways, perpendicular to the direction of travel	✓ 1	Accept: a transverse mechanical wave needs a restoring force on particles displaced sideways.
a liquid cannot provide a sideways (shear) restoring force, so the transverse oscillation is not passed on; P waves pass because a liquid resists compression	✓ 1
Part (g)(i)
sin θ₂ = sin 30° × 8.0/6.0 = 0.667	✓ 1
θ₂ = 42°	✓ 1	Accept 41.8°. 22° (ratio inverted): 1 mark max.
Part (g)(ii)
the wave is refracted continuously as it moves into deeper rock where it travels faster	✓ 1
it bends away from the normal / towards the horizontal at each depth, so the path curves round and back up towards the surface	✓ 1	Accept a reference to the wavefronts swinging round because their deeper part travels faster.
Part (h)
one station gives only the distance, so the earthquake lies somewhere on a circle around that station	✓ 1	Accept “in any direction”.
two circles generally cross at two points; the circle from a third station identifies which point is the earthquake	✓ 1
Part (i)
d ∝ time interval, or 40 = d(1/3.5 − 1/6.0)	✓ 1
d = 210 × 40/25 = 336 km	✓ 1	Accept 340 km. Allow ECF from (c).

Answers: (b) 25 s · (c) 210 km · (d) 17 s before t = 0 · (e)(i) 1.8 × 10³ m · (e)(ii) 2.0 Hz; 700 m · (g)(i) 42° · (i) 336 km

Exam tip (c) Both waves cover the same distance. Write each travel time as distance ÷ speed and subtract, rather than subtracting the speeds.

More C.2 questions in the question bank

C.3 Wave Phenomena

+ Add to test✓ In your testCC-IBP-C3-122 marks~25 min

Total internal reflection in a fibre-optic broadband link

C.3 Wave PhenomenaC.2 Wave ModelA.1 KinematicsSL

A broadband link carries pulses of light along a glass optical fibre 5.0 km long. The fibre has a glass core surrounded by a glass cladding of refractive index 1.46. Light enters the flat end face of the core from air (Figure 1).

Take the speed of light in a vacuum and in air to be 3.00 × 10⁸ m s⁻¹.

Figure 1: a ray entering the fibre and guided along the core (not to scale)

(a)

State two conditions needed for total internal reflection to occur at the core–cladding boundary.

RECALLNotes

Show solution

light travels from a medium of higher refractive index towards one of lower refractive index / from the core towards the cladding	✓ 1	Accept “from the denser to the less dense medium” only if refractive index or optical density is meant.
the angle of incidence is greater than the critical angle	✓ 1	Do not accept “equal to the critical angle”.

(2)

(b)(i)

To find the refractive index of the core glass, a student shines a ray from air into a block of the same glass and measures the angle of incidence i and the angle of refraction r for several rays. Figure 2 shows sin i plotted against sin r, with the line of best fit.

Show, using Figure 2, that the refractive index of the core glass is about 1.48.

Figure 2: sin i against sin r for rays entering the core glass from air

GRAPHNotes

Show solution

Snell's law with air: sin i = n sin r, so n = gradient	✓ 1	Must be stated or clearly used.
gradient from a large triangle, e.g. 0.888 ÷ 0.600 = 1.48	✓ 1	Accept 1.46 to 1.50. Reading points must lie on the line, not data points off it.

Answer: n = 1.48

(2)

(b)(ii)

Outline how Figure 2 shows that the data obey Snell's law.

DATANotes

Show solution

the points lie on a straight line through the origin, so sin i is proportional to sin r

✓ 1

Both “straight line” and “through the origin” needed.

(1)

(c)

Calculate the speed of light in the core.

CALCNotes

Show solution

v = c/n = 3.00 × 10⁸ ÷ 1.48 = 2.03 × 10⁸ m s⁻¹

✓ 1

Answer: 2.03 × 10⁸ m s⁻¹

(1)

(d)(i)

Calculate the critical angle at the core–cladding boundary.

CALCNotes

Show solution

sin θ_c = 1.46 ÷ 1.48 (= 0.986)	✓ 1	Award for correct substitution in n₁ sin θ_c = n₂ sin 90°.
θ_c = 80.6°	✓ 1	Accept 81°.

Answer: 80.6°

(2)

(d)(ii)

Determine the largest angle θ_a at which a ray can enter the end face from air and still be guided along the core by total internal reflection.

MULTINotes

Show solution

angle between the ray in the core and the axis = 90° − 80.6° = 9.4°	✓ 1	Allow ECF from (d)(i).
at the end face: 1.00 × sin θ_a = 1.48 × sin 9.4°	✓ 1	Angle of refraction at the end face is measured from the axis, which is the normal to the face.
θ_a = 14°	✓ 1	Accept 13.9° to 14.1°. Unrounded working (80.57°) gives 14.1°; 80.6° gives 14.0°.

Answer: 14° (14.1°)

(3)

(e)

A short pulse enters the fibre. Light travelling along the axis and light following the steepest guided path, which meets the core–cladding boundary at exactly the critical angle, both travel at the speed found in (c).

Show that, over the 5.0 km link, the steepest guided path arrives about 0.34 μs after the axial path.

DERIVENotes

Show solution

axial time = 5.0 × 10³ ÷ 2.03 × 10⁸ = 24.67 μs	✓ 1	Allow ECF from (c).
steepest path length = 5.0 km ÷ sin θ_c = 5.0 × 1.48 ÷ 1.46 = 5.068 km or its time = 24.67 × 1.48 ÷ 1.46 = 25.005 μs	✓ 1	Path length = L/cos 9.4° is equivalent.
difference = 25.005 − 24.667 = 0.338 μs ≈ 0.34 μs or Δt = 24.67 × (1.48 ÷ 1.46 − 1) = 0.338 μs	✓ 1	Intermediate times must be kept to at least 5 s.f. (rounding both to 25.00 and 24.67 gives 0.33 μs), or use the ratio method. 0.338 μs must be seen, not just the given 0.34 μs.

Answer: 0.34 μs (0.338 μs)

(3)

(f)

Explain, with reference to (e), why this fibre limits the rate at which separate pulses can be sent along the link.

EXPLAINNotes

Show solution

light in one pulse travels along paths of different length, so the pulse is spread out in time / longer when it arrives	✓ 1	Idea of different arrival times needed.
pulses closer together than about 0.34 μs would overlap and could not be told apart	✓ 1	Accept “maximum rate of about 3 × 10⁶ pulses per second”.

(2)

(g)

The fibre is replaced by one with the same core but a cladding of refractive index 1.40.

Determine the new largest entry angle θ_a, and deduce how the spreading of a pulse over 5.0 km changes.

PREDICTNotes

Show solution

new θ_c: sin θ_c = 1.40 ÷ 1.48 → θ_c = 71.1°; ray at 18.9° to the axis	✓ 1	Or sin θ_a = √(1.48² − 1.40²) = 0.48.
θ_a = 29° (28.7°)	✓ 1
the steepest path is now longer, so the spreading increases (to 24.67 × (1.48 ÷ 1.40 − 1) = 1.4 μs)	✓ 1	A qualitative “increases, because steeper rays travel further” is enough; the value is not required.

Answer: 29°; spreading increases (to about 1.4 μs)

(3)

(h)

Several signals with different wavelengths share the fibre. At the receiving end, light containing wavelengths of 1530 nm and 1550 nm is incident normally on a diffraction grating with 600 lines per millimetre.

Determine the angular separation of the two first-order maxima.

HL onlyCALCNotes

Show solution

d = 1 ÷ 600 mm = 1.67 × 10⁻⁶ m	✓ 1
sin θ = 1530 × 10⁻⁹ ÷ 1.667 × 10⁻⁶ → θ = 66.6° and sin θ = 1550 × 10⁻⁹ ÷ 1.667 × 10⁻⁶ → θ = 68.4°	✓ 1	Both angles needed.
separation = 1.8°	✓ 1	Accept 1.7° to 1.9°. Using the small-angle approximation scores at most the first mark.

Answer: 1.8°

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
light travels from a medium of higher refractive index towards one of lower refractive index / from the core towards the cladding	✓ 1	Accept “from the denser to the less dense medium” only if refractive index or optical density is meant.
the angle of incidence is greater than the critical angle	✓ 1	Do not accept “equal to the critical angle”.
Part (b)(i)
Snell's law with air: sin i = n sin r, so n = gradient	✓ 1	Must be stated or clearly used.
gradient from a large triangle, e.g. 0.888 ÷ 0.600 = 1.48	✓ 1	Accept 1.46 to 1.50. Reading points must lie on the line, not data points off it.
Part (b)(ii)
the points lie on a straight line through the origin, so sin i is proportional to sin r	✓ 1	Both “straight line” and “through the origin” needed.
Part (c)
v = c/n = 3.00 × 10⁸ ÷ 1.48 = 2.03 × 10⁸ m s⁻¹	✓ 1
Part (d)(i)
sin θ_c = 1.46 ÷ 1.48 (= 0.986)	✓ 1	Award for correct substitution in n₁ sin θ_c = n₂ sin 90°.
θ_c = 80.6°	✓ 1	Accept 81°.
Part (d)(ii)
angle between the ray in the core and the axis = 90° − 80.6° = 9.4°	✓ 1	Allow ECF from (d)(i).
at the end face: 1.00 × sin θ_a = 1.48 × sin 9.4°	✓ 1	Angle of refraction at the end face is measured from the axis, which is the normal to the face.
θ_a = 14°	✓ 1	Accept 13.9° to 14.1°. Unrounded working (80.57°) gives 14.1°; 80.6° gives 14.0°.
Part (e)
axial time = 5.0 × 10³ ÷ 2.03 × 10⁸ = 24.67 μs	✓ 1	Allow ECF from (c).
steepest path length = 5.0 km ÷ sin θ_c = 5.0 × 1.48 ÷ 1.46 = 5.068 km or its time = 24.67 × 1.48 ÷ 1.46 = 25.005 μs	✓ 1	Path length = L/cos 9.4° is equivalent.
difference = 25.005 − 24.667 = 0.338 μs ≈ 0.34 μs or Δt = 24.67 × (1.48 ÷ 1.46 − 1) = 0.338 μs	✓ 1	Intermediate times must be kept to at least 5 s.f. (rounding both to 25.00 and 24.67 gives 0.33 μs), or use the ratio method. 0.338 μs must be seen, not just the given 0.34 μs.
Part (f)
light in one pulse travels along paths of different length, so the pulse is spread out in time / longer when it arrives	✓ 1	Idea of different arrival times needed.
pulses closer together than about 0.34 μs would overlap and could not be told apart	✓ 1	Accept “maximum rate of about 3 × 10⁶ pulses per second”.
Part (g)
new θ_c: sin θ_c = 1.40 ÷ 1.48 → θ_c = 71.1°; ray at 18.9° to the axis	✓ 1	Or sin θ_a = √(1.48² − 1.40²) = 0.48.
θ_a = 29° (28.7°)	✓ 1
the steepest path is now longer, so the spreading increases (to 24.67 × (1.48 ÷ 1.40 − 1) = 1.4 μs)	✓ 1	A qualitative “increases, because steeper rays travel further” is enough; the value is not required.
Part (h)
d = 1 ÷ 600 mm = 1.67 × 10⁻⁶ m	✓ 1
sin θ = 1530 × 10⁻⁹ ÷ 1.667 × 10⁻⁶ → θ = 66.6° and sin θ = 1550 × 10⁻⁹ ÷ 1.667 × 10⁻⁶ → θ = 68.4°	✓ 1	Both angles needed.
separation = 1.8°	✓ 1	Accept 1.7° to 1.9°. Using the small-angle approximation scores at most the first mark.

Answers: (b)(i) n = 1.48 · (c) 2.03 × 10⁸ m s⁻¹ · (d)(i) 80.6° · (d)(ii) 14° (14.1°) · (e) 0.34 μs (0.338 μs) · (g) 29°; spreading increases (to about 1.4 μs) · (h) 1.8°

Common error (a) Saying only “the angle must be large enough” earns nothing: name the critical angle and say the angle of incidence must be greater than it.

Exam tip (d)(ii) At the end face the normal is the fibre axis, but at the core–cladding boundary the normal is perpendicular to the axis. Draw both normals before writing Snell's law.

More C.3 questions in the question bank

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Water waves passing through gaps in a harbour wall

C.3 Wave PhenomenaC.2 Wave ModelC.4 Standing Waves and ResonanceSL

Figure 1 is a plan view of a harbour. A straight breakwater has two gaps, A and B, each 20 m wide, with their centres 120 m apart. The shore inside the harbour is straight, parallel to the breakwater and 400 m from it. O is the point on the shore directly opposite the midpoint of AB.

Water waves of period 8.0 s and wavelength 60 m approach from the sea with their wavefronts parallel to the breakwater. Assume that the water depth, and so the wave speed, is the same everywhere in and near the harbour unless a part states otherwise.

Figure 1: plan view of the harbour (drawn to scale)

(a)

Calculate the speed of the waves.

CALCNotes

Show solution

v = λ/T = 60 ÷ 8.0 = 7.5 m s⁻¹

✓ 1

Answer: 7.5 m s⁻¹

(1)

(b)

Describe what happens to the waves as they pass through one of the gaps, and explain why the effect is significant here.

EXPLAINNotes

Show solution

the waves diffract / spread out (into almost semicircular wavefronts) beyond the gap, with the wavelength unchanged	✓ 1	“Spread out” or “diffract” needed.
because the gap width (20 m) is comparable to / smaller than the wavelength (60 m)	✓ 1	Comparison of the two sizes needed.

(2)

(c)(i)

Point P on the shore is 104 m from O, 432 m from A and 402 m from B.

Show that the water at P is calm.

MULTINotes

Show solution

path difference = 432 − 402 = 30 m	✓ 1
30 m = 0.5λ / (n + ½)λ with n = 0, so the waves arrive in antiphase and interfere destructively	✓ 1	Conclusion with reason needed; “waves cancel” alone is not enough without the half-wavelength link.

Answer: path difference = ½λ: destructive interference

(2)

(c)(ii)

Explain why the waves leaving A and B can produce a stable interference pattern.

REASONNotes

Show solution

both come from the same incoming wavefront / arrive at A and B in phase, so they are coherent (constant phase difference, same frequency)

✓ 1

“Same frequency” alone is not enough; the constant phase difference is needed.

(1)

(d)

The next point beyond P where the amplitude is large is 233 m from O.

On the axes in Figure 2, sketch how the amplitude of the water waves varies with distance along the shore from O, between −250 m and +250 m.

Figure 2: axes for the sketch

SKETCHNotes

Show solution

maximum amplitude at O (0 m)	✓ 1
minima (amplitude near zero) at about ±104 m, the pattern symmetrical about O	✓ 1	Accept minima between 90 m and 120 m from O on both sides.
further maxima at about ±233 m, rising again beyond the minima	✓ 1	Allow the outer maxima slightly lower than the central one.

(3)

(e)(i)

During a storm, waves of period 12 s arrive. Their speed near the harbour is unchanged.

Calculate the wavelength of the storm waves.

CALCNotes

Show solution

λ = vT = 7.5 × 12 = 90 m

✓ 1

Allow ECF from (a).

Answer: 90 m

(1)

(e)(ii)

Predict whether the water at P is calm during the storm. Give a reason.

PREDICTNotes

Show solution

path difference at P is still 30 m, which is ⅓λ, not a half-wavelength (or a whole number of wavelengths)	✓ 1	Allow ECF from (e)(i).
so P is not calm (the interference is only partly destructive); the calm points move further from O	✓ 1	Conclusion must follow from the reasoning.

Answer: not calm

(2)

(f)(i)

Out at sea the original waves (period 8.0 s) cross a sandbank, where the shallower water reduces their speed to 4.4 m s⁻¹.

Show that the wavelength over the sandbank is about 35 m.

DERIVENotes

Show solution

the frequency / period is unchanged, set by the source	✓ 1	Must be stated.
λ = 4.4 × 8.0 = 35.2 m (≈ 35 m)	✓ 1

Answer: 35.2 m

(2)

(f)(ii)

The wavefronts meet the edge of the sandbank with an angle of incidence of 40°.

Calculate the angle of refraction.

CALCNotes

Show solution

sin θ₂ = (4.4 ÷ 7.5) × sin 40° (= 0.377)	✓ 1	Use of sin θ₂/sin θ₁ = v₂/v₁.
θ₂ = 22°	✓ 1	22.2°; bent towards the normal.

Answer: 22°

(2)

(g)(i)

Waves arriving at the solid seaward face of the breakwater are reflected straight back.

Outline how a standing wave forms in front of the breakwater.

EXPLAINNotes

Show solution

the reflected wave has the same frequency, wavelength and speed (and similar amplitude) as the incoming wave and travels in the opposite direction	✓ 1	Opposite direction needed.
the two waves superpose, producing nodes and antinodes at fixed positions	✓ 1	Accept “interfere” for superpose.

(2)

(g)(ii)

State the distance between adjacent nodes of this standing wave.

RECALLNotes

Show solution

λ/2 = 30 m

✓ 1

Answer: 30 m

(1)

(h)

Real sea waves are not all of one period. Explain why the calm regions along the shore are much less clear than the pattern described in (c) and (d).

EXPLAINNotes

Show solution

different periods have different wavelengths, so the positions of the minima (where path difference = ½λ) are different for each	✓ 1	Link between wavelength and position of the minima needed.
the patterns overlap, so a point calm for one wavelength has large waves of another; the minima are filled in	✓ 1	Accept “the pattern is smeared / averaged out”.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
v = λ/T = 60 ÷ 8.0 = 7.5 m s⁻¹	✓ 1
Part (b)
the waves diffract / spread out (into almost semicircular wavefronts) beyond the gap, with the wavelength unchanged	✓ 1	“Spread out” or “diffract” needed.
because the gap width (20 m) is comparable to / smaller than the wavelength (60 m)	✓ 1	Comparison of the two sizes needed.
Part (c)(i)
path difference = 432 − 402 = 30 m	✓ 1
30 m = 0.5λ / (n + ½)λ with n = 0, so the waves arrive in antiphase and interfere destructively	✓ 1	Conclusion with reason needed; “waves cancel” alone is not enough without the half-wavelength link.
Part (c)(ii)
both come from the same incoming wavefront / arrive at A and B in phase, so they are coherent (constant phase difference, same frequency)	✓ 1	“Same frequency” alone is not enough; the constant phase difference is needed.
Part (d)
maximum amplitude at O (0 m)	✓ 1
minima (amplitude near zero) at about ±104 m, the pattern symmetrical about O	✓ 1	Accept minima between 90 m and 120 m from O on both sides.
further maxima at about ±233 m, rising again beyond the minima	✓ 1	Allow the outer maxima slightly lower than the central one.
Part (e)(i)
λ = vT = 7.5 × 12 = 90 m	✓ 1	Allow ECF from (a).
Part (e)(ii)
path difference at P is still 30 m, which is ⅓λ, not a half-wavelength (or a whole number of wavelengths)	✓ 1	Allow ECF from (e)(i).
so P is not calm (the interference is only partly destructive); the calm points move further from O	✓ 1	Conclusion must follow from the reasoning.
Part (f)(i)
the frequency / period is unchanged, set by the source	✓ 1	Must be stated.
λ = 4.4 × 8.0 = 35.2 m (≈ 35 m)	✓ 1
Part (f)(ii)
sin θ₂ = (4.4 ÷ 7.5) × sin 40° (= 0.377)	✓ 1	Use of sin θ₂/sin θ₁ = v₂/v₁.
θ₂ = 22°	✓ 1	22.2°; bent towards the normal.
Part (g)(i)
the reflected wave has the same frequency, wavelength and speed (and similar amplitude) as the incoming wave and travels in the opposite direction	✓ 1	Opposite direction needed.
the two waves superpose, producing nodes and antinodes at fixed positions	✓ 1	Accept “interfere” for superpose.
Part (g)(ii)
λ/2 = 30 m	✓ 1
Part (h)
different periods have different wavelengths, so the positions of the minima (where path difference = ½λ) are different for each	✓ 1	Link between wavelength and position of the minima needed.
the patterns overlap, so a point calm for one wavelength has large waves of another; the minima are filled in	✓ 1	Accept “the pattern is smeared / averaged out”.

Answers: (a) 7.5 m s⁻¹ · (c)(i) path difference = ½λ: destructive interference · (e)(i) 90 m · (e)(ii) not calm · (f)(i) 35.2 m · (f)(ii) 22° · (g)(ii) 30 m

More C.3 questions in the question bank

C.4 Standing Waves and Resonance

+ Add to test✓ In your testCC-IBP-C4-122 marks~25 min

A violin string and the resonance of its body

C.4 Standing Waves and ResonanceC.2 Wave ModelC.1 Simple Harmonic MotionSL

A violin string is fixed at the nut and at the bridge, which are 0.330 m apart. When the string is bowed it vibrates in its first harmonic with a frequency of 440 Hz (Figure 1).

The speed of sound in air is 340 m s⁻¹.

Figure 1: the string vibrating in its first harmonic (vertical scale exaggerated)

(a)

Explain how a standing wave forms on the string.

EXPLAINNotes

Show solution

waves travel along the string and are reflected at the fixed ends (nut and bridge)	✓ 1
the incident and reflected waves, with the same frequency, wavelength and speed travelling in opposite directions, superpose to give nodes and antinodes in fixed positions	✓ 1	Must mention superposition / interference of waves travelling in opposite directions.

(2)

(b)

On Figure 2, sketch the standing wave on the string for the second harmonic and for the third harmonic.

Figure 2: strings for the sketches

SKETCHNotes

Show solution

second harmonic: two loops, with nodes at both ends and one node at the centre	✓ 1	Envelope (two curves) or one snapshot accepted.
third harmonic: three equal loops, with nodes at both ends and at one-third and two-thirds of the length	✓ 1

(2)

(c)(i)

Calculate the speed of the transverse waves on the string.

CALCNotes

Show solution

first harmonic: λ = 2L = 2 × 0.330 = 0.660 m	✓ 1
v = fλ = 440 × 0.660 = 290 m s⁻¹	✓ 1	290.4 m s⁻¹. Using λ = 0.330 m (145 m s⁻¹) scores 0.

Answer: 290 m s⁻¹

(2)

(c)(ii)

Calculate the wavelength of the sound produced in the air.

CALCNotes

Show solution

the sound has the same frequency as the string, 440 Hz	✓ 1	May be implied by the calculation.
λ = 340 ÷ 440 = 0.773 m	✓ 1	Accept 0.77 m. 0.660 m (the string wavelength) scores 0.

Answer: 0.773 m

(2)

(d)

The violinist presses the string firmly against the fingerboard, so that the vibrating length becomes 0.250 m. The tension in the string does not change.

Show that the frequency of the first harmonic is now about 580 Hz.

DERIVENotes

Show solution

the wave speed is unchanged (same string, same tension) and the new wavelength is 2 × 0.250 = 0.500 m	✓ 1
f = 290.4 ÷ 0.500 = 581 Hz (≈ 580 Hz)	✓ 1	Or f = 440 × 0.330 ÷ 0.250 = 581 Hz. Allow ECF from (c)(i). Must see a value to at least 3 s.f.

Answer: 581 Hz

(2)

(e)

The full-length string (0.330 m) is now made to vibrate in its second harmonic. Points X, Y and Z on the string are 0.050 m, 0.120 m and 0.250 m from the nut.

Deduce the phase difference between X and Y, and between X and Z.

REASONNotes

Show solution

X and Y: zero / in phase, because both are between the same pair of adjacent nodes (0 and 0.165 m)	✓ 1	Node position need not be quoted for the mark if the reasoning is clear.
X and Z: π rad / 180° / antiphase, because there is one node (at 0.165 m) between them	✓ 1

Answer: X–Y: 0; X–Z: π rad

(2)

(f)(i)

In the first harmonic the midpoint of the string moves with simple harmonic motion of amplitude 0.80 mm.

Calculate the maximum acceleration of the midpoint.

CALCNotes

Show solution

ω = 2π × 440 = 2.76 × 10³ rad s⁻¹	✓ 1
a_max = ω²x₀ = (2.76 × 10³)² × 0.80 × 10⁻³ = 6.1 × 10³ m s⁻²	✓ 1	Award [1 max] for 6.1 × 10⁶ (amplitude left in mm) or for using ω = 440.

Answer: 6.1 × 10³ m s⁻²

(2)

(f)(ii)

Calculate the maximum speed of the midpoint.

HL onlyCALCNotes

Show solution

v_max = ωx₀ = 2.76 × 10³ × 0.80 × 10⁻³ = 2.2 m s⁻¹

✓ 1

Allow ECF for ω from (f)(i).

Answer: 2.2 m s⁻¹

(1)

(g)(i)

The string drives the wooden body of the violin through the bridge. Figure 3 shows how the amplitude of vibration of the body varies with the driving frequency.

Write down the two natural frequencies of the body shown in Figure 3.

Figure 3: amplitude of the violin body against driving frequency

GRAPHNotes

Show solution

280 Hz and 470 Hz

✓ 1

Accept 275 to 285 Hz and 460 to 480 Hz; both needed.

Answer: 280 Hz and 470 Hz

(1)

(g)(ii)

Identify which of the two resonances is more heavily damped. Give two pieces of evidence from Figure 3.

GRAPHNotes

Show solution

the resonance at 470 Hz and its peak amplitude is lower	✓ 1	Identification alone scores 0.
its peak is broader / the body responds strongly over a wider range of frequencies around it	✓ 1

Answer: the 470 Hz resonance

(2)

(g)(iii)

Use Figure 3 to explain why the body vibrates with a much larger amplitude when the string plays 440 Hz than when it plays 350 Hz.

REASONNotes

Show solution

440 Hz is close to the natural frequency of 470 Hz, so the body is driven near resonance (amplitude about 0.47 compared with about 0.24 at 350 Hz)	✓ 1	Accept readings within ±0.05; readings not required for the mark if the reasoning is clear.
350 Hz is far from both natural frequencies, so less energy is transferred to the body per cycle	✓ 1

(2)

(h)

The violinist tightens the string so that the wave speed on it increases by 2.0%. The length is 0.330 m again.

Calculate the new frequency of the first harmonic and the new wavelength of the sound in air.

PREDICTNotes

Show solution

f ∝ v for a fixed length: f = 1.020 × 440 = 449 Hz	✓ 1
λ = 340 ÷ 449 = 0.758 m	✓ 1	Allow ECF from their frequency. The speed of sound in air is unchanged.

Answer: 449 Hz; 0.758 m

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
waves travel along the string and are reflected at the fixed ends (nut and bridge)	✓ 1
the incident and reflected waves, with the same frequency, wavelength and speed travelling in opposite directions, superpose to give nodes and antinodes in fixed positions	✓ 1	Must mention superposition / interference of waves travelling in opposite directions.
Part (b)
second harmonic: two loops, with nodes at both ends and one node at the centre	✓ 1	Envelope (two curves) or one snapshot accepted.
third harmonic: three equal loops, with nodes at both ends and at one-third and two-thirds of the length	✓ 1
Part (c)(i)
first harmonic: λ = 2L = 2 × 0.330 = 0.660 m	✓ 1
v = fλ = 440 × 0.660 = 290 m s⁻¹	✓ 1	290.4 m s⁻¹. Using λ = 0.330 m (145 m s⁻¹) scores 0.
Part (c)(ii)
the sound has the same frequency as the string, 440 Hz	✓ 1	May be implied by the calculation.
λ = 340 ÷ 440 = 0.773 m	✓ 1	Accept 0.77 m. 0.660 m (the string wavelength) scores 0.
Part (d)
the wave speed is unchanged (same string, same tension) and the new wavelength is 2 × 0.250 = 0.500 m	✓ 1
f = 290.4 ÷ 0.500 = 581 Hz (≈ 580 Hz)	✓ 1	Or f = 440 × 0.330 ÷ 0.250 = 581 Hz. Allow ECF from (c)(i). Must see a value to at least 3 s.f.
Part (e)
X and Y: zero / in phase, because both are between the same pair of adjacent nodes (0 and 0.165 m)	✓ 1	Node position need not be quoted for the mark if the reasoning is clear.
X and Z: π rad / 180° / antiphase, because there is one node (at 0.165 m) between them	✓ 1
Part (f)(i)
ω = 2π × 440 = 2.76 × 10³ rad s⁻¹	✓ 1
a_max = ω²x₀ = (2.76 × 10³)² × 0.80 × 10⁻³ = 6.1 × 10³ m s⁻²	✓ 1	Award [1 max] for 6.1 × 10⁶ (amplitude left in mm) or for using ω = 440.
Part (f)(ii)
v_max = ωx₀ = 2.76 × 10³ × 0.80 × 10⁻³ = 2.2 m s⁻¹	✓ 1	Allow ECF for ω from (f)(i).
Part (g)(i)
280 Hz and 470 Hz	✓ 1	Accept 275 to 285 Hz and 460 to 480 Hz; both needed.
Part (g)(ii)
the resonance at 470 Hz and its peak amplitude is lower	✓ 1	Identification alone scores 0.
its peak is broader / the body responds strongly over a wider range of frequencies around it	✓ 1
Part (g)(iii)
440 Hz is close to the natural frequency of 470 Hz, so the body is driven near resonance (amplitude about 0.47 compared with about 0.24 at 350 Hz)	✓ 1	Accept readings within ±0.05; readings not required for the mark if the reasoning is clear.
350 Hz is far from both natural frequencies, so less energy is transferred to the body per cycle	✓ 1
Part (h)
f ∝ v for a fixed length: f = 1.020 × 440 = 449 Hz	✓ 1
λ = 340 ÷ 449 = 0.758 m	✓ 1	Allow ECF from their frequency. The speed of sound in air is unchanged.

Answers: (c)(i) 290 m s⁻¹ · (c)(ii) 0.773 m · (d) 581 Hz · (e) X–Y: 0; X–Z: π rad · (f)(i) 6.1 × 10³ m s⁻² · (f)(ii) 2.2 m s⁻¹ · (g)(i) 280 Hz and 470 Hz · (g)(ii) the 470 Hz resonance · (h) 449 Hz; 0.758 m

Exam tip (b) Start every standing-wave sketch at the ends: a fixed end is always a node. Then fit in a whole number of loops of equal length.

Common error (c)(i) The length of the string in the first harmonic is half a wavelength, not one wavelength: one loop is λ/2.

More C.4 questions in the question bank

+ Add to test✓ In your testCC-IBP-C4-221 marks~25 min

A wobbling footbridge: pedestrians, resonance and dampers

C.4 Standing Waves and ResonanceC.1 Simple Harmonic MotionA.2 Forces and MomentumA.3 Work, Energy and PowerSL

A steel footbridge sways sideways when crowds cross it. For sideways oscillations the deck behaves like a mass m = 1.2 × 10⁵ kg on a spring of spring constant k (Figure 1). The natural frequency of this sideways oscillation is 1.0 Hz.

People walk at about 2.0 steps per second. With each step the walker pushes sideways on the deck, alternately to the left (left foot) and to the right (right foot).

Figure 1: the swaying deck and its mass–spring model (not to scale)

(a)

Explain why people walking at 2.0 steps per second can make the deck sway with a large amplitude.

EXPLAINNotes

Show solution

the sideways force repeats every two steps, so the deck is driven at 1.0 Hz	✓ 1	Must deduce the driving frequency, not just quote 2.0 Hz.
this driving frequency equals the natural frequency of the deck, so resonance occurs	✓ 1
energy is transferred to the deck in every cycle (maximum transfer per cycle), so the amplitude builds up until the energy dissipated per cycle equals the energy supplied	✓ 1	Accept “the driving force is in step with the motion of the deck”.

(3)

(b)

Show that k is about 4.7 × 10⁶ N m⁻¹.

DERIVENotes

Show solution

T = 1/f = 1.0 s and T = 2π√(m/k) rearranged: k = 4π²m/T²	✓ 1
k = 4π² × 1.2 × 10⁵ ÷ 1.0² = 4.74 × 10⁶ N m⁻¹	✓ 1	Must see a value to at least 3 s.f. or the unrounded expression.

Answer: 4.74 × 10⁶ N m⁻¹

(2)

(c)

To move their body sideways with each step, every walker needs a sideways force of up to 25 N from the deck. On a crowded day 400 people walk in step with the sway.

Explain why the deck experiences a sideways force, and show that the largest total sideways force on the deck is 10 kN.

REASONNotes

Show solution

by Newton's third law each walker exerts an equal and opposite force of 25 N on the deck	✓ 1	Reference to Newton's third law (or “equal and opposite force on the deck”) needed.
the forces are in step so they add: 400 × 25 N = 1.0 × 10⁴ N = 10 kN	✓ 1

Answer: 10 kN

(2)

(d)

Dampers are fitted to the deck. Figure 2 shows how the amplitude of the sway varies with the driving frequency of the crowd's sideways force, without dampers (A) and with dampers (B).

Compare the two curves.

Figure 2: amplitude of the sway against driving frequency

GRAPHNotes

Show solution

B has a much lower maximum amplitude (about 8 mm compared with 70 mm)	✓ 1	Values not required but the comparison must be clear.
B has a broader / flatter peak	✓ 1
the peak of B is at a slightly lower frequency than the peak of A (just below 1.0 Hz)	✓ 1	Do not accept “B peaks at a higher frequency”. Accept “the curves become similar far from 1.0 Hz” as an alternative to any one of the three marks.

(3)

(e)

Without dampers the amplitude at 1.0 Hz is 70 mm.

Calculate the maximum sideways acceleration of the deck.

CALCNotes

Show solution

ω = 2π × 1.0 = 6.28 rad s⁻¹	✓ 1
a_max = ω²x₀ = 6.28² × 0.070 = 2.8 m s⁻²	✓ 1	2.76 m s⁻². Award [1 max] for 2.8 × 10³ (amplitude left in mm).

Answer: 2.8 m s⁻²

(2)

(f)(i)

With dampers fitted, the crowd still drives the deck at 1.0 Hz and the amplitude stays at 8.0 mm.

State how the energy dissipated by the dampers in one cycle compares with the work done by the crowd on the deck in one cycle. Give a reason.

REASONNotes

Show solution

they are equal, because the amplitude (and so the energy of the deck) is constant

✓ 1

Both the comparison and the reason needed.

Answer: equal

(1)

(f)(ii)

Assume that the crowd's total force of 10 kN always acts in the direction in which the deck is moving.

Estimate the energy dissipated by the dampers in one cycle, and the mean power dissipated.

MULTINotes

Show solution

distance moved by the deck in one cycle = 4 × 8.0 mm = 0.032 m	✓ 1	Using 2 × 8.0 mm or 8.0 mm scores 0 for this mark but allow ECF.
W = Fs = 1.0 × 10⁴ × 0.032 = 320 J (per cycle)	✓ 1	Allow ECF from (c).
P = 320 J ÷ 1.0 s = 320 W	✓ 1	Allow ECF from their energy.

Answer: 320 J; 320 W

(3)

(g)

At a festival a large crowd marches across the bridge at 1.6 steps per second.

Determine the amplitude of the sway with the dampers fitted, and comment on how much difference the dampers make at this frequency.

PREDICTNotes

Show solution

sideways driving frequency = 1.6 ÷ 2 = 0.80 Hz	✓ 1
from curve B: amplitude ≈ 6 mm	✓ 1	Accept 5 to 7 mm. Allow ECF for a wrong driving frequency read correctly from B.
curves A and B are close together at 0.80 Hz (A about 7.5 mm), so the dampers make only a small difference far from resonance	✓ 1	The idea that damping matters mainly near the natural frequency is needed.

Answer: about 6 mm

(3)

(h)

An engineer considers making the deck's sideways motion critically damped instead.

Outline how the deck would move after a single sideways push if it were lightly damped, and if it were critically damped.

EXPLAINNotes

Show solution

lightly damped: it oscillates about the equilibrium position with an amplitude that decreases gradually	✓ 1	Must mention oscillation and decreasing amplitude.
critically damped: it returns to the equilibrium position in the shortest possible time without oscillating	✓ 1	Both “no oscillation / overshoot” and “shortest time” needed.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the sideways force repeats every two steps, so the deck is driven at 1.0 Hz	✓ 1	Must deduce the driving frequency, not just quote 2.0 Hz.
this driving frequency equals the natural frequency of the deck, so resonance occurs	✓ 1
energy is transferred to the deck in every cycle (maximum transfer per cycle), so the amplitude builds up until the energy dissipated per cycle equals the energy supplied	✓ 1	Accept “the driving force is in step with the motion of the deck”.
Part (b)
T = 1/f = 1.0 s and T = 2π√(m/k) rearranged: k = 4π²m/T²	✓ 1
k = 4π² × 1.2 × 10⁵ ÷ 1.0² = 4.74 × 10⁶ N m⁻¹	✓ 1	Must see a value to at least 3 s.f. or the unrounded expression.
Part (c)
by Newton's third law each walker exerts an equal and opposite force of 25 N on the deck	✓ 1	Reference to Newton's third law (or “equal and opposite force on the deck”) needed.
the forces are in step so they add: 400 × 25 N = 1.0 × 10⁴ N = 10 kN	✓ 1
Part (d)
B has a much lower maximum amplitude (about 8 mm compared with 70 mm)	✓ 1	Values not required but the comparison must be clear.
B has a broader / flatter peak	✓ 1
the peak of B is at a slightly lower frequency than the peak of A (just below 1.0 Hz)	✓ 1	Do not accept “B peaks at a higher frequency”. Accept “the curves become similar far from 1.0 Hz” as an alternative to any one of the three marks.
Part (e)
ω = 2π × 1.0 = 6.28 rad s⁻¹	✓ 1
a_max = ω²x₀ = 6.28² × 0.070 = 2.8 m s⁻²	✓ 1	2.76 m s⁻². Award [1 max] for 2.8 × 10³ (amplitude left in mm).
Part (f)(i)
they are equal, because the amplitude (and so the energy of the deck) is constant	✓ 1	Both the comparison and the reason needed.
Part (f)(ii)
distance moved by the deck in one cycle = 4 × 8.0 mm = 0.032 m	✓ 1	Using 2 × 8.0 mm or 8.0 mm scores 0 for this mark but allow ECF.
W = Fs = 1.0 × 10⁴ × 0.032 = 320 J (per cycle)	✓ 1	Allow ECF from (c).
P = 320 J ÷ 1.0 s = 320 W	✓ 1	Allow ECF from their energy.
Part (g)
sideways driving frequency = 1.6 ÷ 2 = 0.80 Hz	✓ 1
from curve B: amplitude ≈ 6 mm	✓ 1	Accept 5 to 7 mm. Allow ECF for a wrong driving frequency read correctly from B.
curves A and B are close together at 0.80 Hz (A about 7.5 mm), so the dampers make only a small difference far from resonance	✓ 1	The idea that damping matters mainly near the natural frequency is needed.
Part (h)
lightly damped: it oscillates about the equilibrium position with an amplitude that decreases gradually	✓ 1	Must mention oscillation and decreasing amplitude.
critically damped: it returns to the equilibrium position in the shortest possible time without oscillating	✓ 1	Both “no oscillation / overshoot” and “shortest time” needed.

Answers: (b) 4.74 × 10⁶ N m⁻¹ · (c) 10 kN · (e) 2.8 m s⁻² · (f)(i) equal · (f)(ii) 320 J; 320 W · (g) about 6 mm

Exam tip (d) When comparing resonance curves, use three features: height of the peak, width of the peak and the frequency at which the peak occurs.

Common error (f)(ii) In one full oscillation the deck travels four amplitudes (out, back, out the other way, back), not one.

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C.5 Doppler Effect

+ Add to test✓ In your testCC-IBP-C5-123 marks~30 min

Measuring a tennis serve with a radar gun

C.5 Doppler EffectC.2 Wave ModelA.1 KinematicsA.2 Forces and MomentumSL

A radar speed gun is used to measure the speed of tennis serves. The gun stands behind the far baseline, in line with the path of the ball (Figure 1). It emits microwaves of frequency 24.1 GHz and detects the microwaves reflected by the ball. The gun measures the difference Δf between the reflected frequency and the emitted frequency, and uses it to calculate the speed of the ball.

Assume that, until it bounces, the ball moves directly towards the gun. The speed of electromagnetic waves is c = 3.00 × 10⁸ m s⁻¹.

Figure 1: side view of the court and the radar gun (not to scale)

(a)

Calculate the wavelength of the microwaves emitted by the gun.

CALCNotes

Show solution

λ = c/f = 3.00 × 10⁸ / 24.1 × 10⁹ = 1.24 × 10⁻² m

✓ 1

Accept 1.2 × 10⁻² m or 1.24 cm.

Answer: 1.24 × 10⁻² m

(1)

(b)

Explain, by considering the wavefronts that reach the ball and the wavefronts that the ball sends back, why the reflected microwaves arrive at the gun with a frequency greater than 24.1 GHz.

EXPLAINNotes

Show solution

the ball moves towards the gun, so it meets the wavefronts more often than they are emitted «it receives a frequency higher than 24.1 GHz»	✓ 1	Accept “the ball acts as a moving observer”.
the ball reflects the waves as a moving source: it moves towards the gun between reflecting successive wavefronts, so the reflected wavefronts are closer together «shorter wavelength»	✓ 1
the waves still travel at c, so by c = fλ more wavefronts reach the gun each second «the detected frequency is higher»	✓ 1	Do not accept “the ball emits a higher frequency” with no reference to the spacing or rate of arrival of wavefronts.

(3)

(c)

Show that, for a ball moving directly towards the gun at 50 m s⁻¹, Δf is about 8 kHz.

CALCNotes

Show solution

each of the two steps changes the frequency by a fraction ≈ u/c, so Δf ≈ 2uf/c	✓ 1	May be seen as a substitution with the factor 2.
Δf = 2 × 50 × 24.1 × 10⁹ / 3.00 × 10⁸ = 8.03 × 10³ Hz	✓ 1	A value more precise than the given 8 kHz must be seen. 4.0 kHz (factor 2 omitted) scores 0.

Answer: 8.03 kHz

(2)

(d)

For one serve the gun measures Δf = 8.68 kHz just after the ball leaves the racket.

Determine the speed of the ball, in km h⁻¹, that the gun displays.

CALCNotes

Show solution

u = cΔf/2f = 3.00 × 10⁸ × 8.68 × 10³ / (2 × 24.1 × 10⁹) = 54.0 m s⁻¹	✓ 1
54.0 × 3.6 = 194 km h⁻¹	✓ 1	Allow ECF for the conversion. 389 km h⁻¹ (factor 2 omitted) scores [1].

Answer: 194 km h⁻¹ (54.0 m s⁻¹)

(2)

(e)

The mass of the ball is 58 g. When the racket strikes it, the ball is momentarily at rest at the top of its toss. The racket is in contact with the ball for 4.0 ms.

Calculate the average force exerted by the racket on the ball.

CALCNotes

Show solution

Δp = 0.058 × 54.0 = 3.13 N s	✓ 1	Allow ECF from (d).
F = Δp/Δt = 3.13 / 4.0 × 10⁻³ = 780 N	✓ 1	Accept 783 N. Ignore the weight of the ball.

Answer: 780 N

(2)

(f)(i)

Figure 2 shows how Δf varies with time t from the moment the ball leaves the racket (t = 0) until just before it bounces, at t = 0.40 s.

Use Figure 2 to determine the average acceleration of the ball between t = 0 and t = 0.40 s.

Figure 2: frequency difference Δf measured by the gun against time t

GRAPHNotes

Show solution

Δf at t = 0.40 s read as 6.0 kHz	✓ 1	Accept 5.9 to 6.1 kHz.
speed at 0.40 s = 3.00 × 10⁸ × 6.0 × 10³ / (2 × 24.1 × 10⁹) = 37.3 m s⁻¹	✓ 1	Or by proportion: 54.0 × 6.0/8.68. Allow ECF from (d).
a = (37.3 − 54.0)/0.40 = −42 m s⁻²	✓ 1	Accept −40 to −44 m s⁻², or a deceleration of 40 to 44 m s⁻². A positive answer with no indication that the ball slows down scores [2] max.

Answer: about −42 m s⁻² (a deceleration)

(3)

(f)(ii)

Explain, with reference to the shape of the graph, how the drag force on the ball changes during its flight.

EXPLAINNotes

Show solution

Δf is proportional to the speed, and the gradient of the graph decreases in magnitude, so the deceleration of the ball decreases	✓ 1
the drag force decreases «F = ma», because drag depends on speed and the speed is falling	✓ 1	Do not accept “drag decreases” with no reference to the gradient or the deceleration.

(2)

(f)(iii)

Estimate the distance travelled by the ball between t = 0 and t = 0.40 s.

GRAPHNotes

Show solution

distance = area under the speed–time graph, found from the area under Figure 2 multiplied by c/2f «6.22 × 10⁻³ m s⁻¹ per Hz», or from the mean speed ≈ (54.0 + 37.3)/2 m s⁻¹	✓ 1
distance ≈ 18 m	✓ 1	Accept 17 to 19 m. The mean of the end speeds gives 18.3 m, slightly too large because the curve lies below the straight line joining its ends. Allow ECF from (f)(i).

Answer: about 18 m

(2)

(g)

The gun is moved to the side of the court. Just after the ball leaves the racket, the line from the gun to the ball makes an angle of 20° with the path of the ball. The serve in (d) is repeated.

Calculate the speed that the gun now displays, and explain why it differs from the true speed of the ball.

REASONNotes

Show solution

only the component of the ball’s velocity along the line joining the ball and the gun produces a Doppler shift	✓ 1
component = 54.0 × cos 20° = 50.7 m s⁻¹ «183 km h⁻¹»	✓ 1	Allow ECF from (d).
the measured Δf is smaller, and the gun treats it as motion directly towards it, so it displays a speed lower than the true speed	✓ 1

Answer: 50.7 m s⁻¹ (183 km h⁻¹)

(3)

(h)

Between points, a loudspeaker on the umpire’s chair emits a steady tone of frequency 880 Hz. A ball kid runs directly towards the chair at 6.0 m s⁻¹. The speed of sound in air is 340 m s⁻¹.

Calculate the frequency heard by the ball kid, and state the wavelength of the sound that reaches the ball kid.

HL onlyCALCNotes

Show solution

moving observer approaching: f′ = f(v + u_o)/v	✓ 1	The moving-source equation scores 0 for this mark, even though it happens to give a similar value here.
f′ = 880 × 346/340 = 896 Hz	✓ 1
wavelength unchanged, because the source is at rest: λ = 340/880 = 0.386 m	✓ 1	Accept 0.39 m. 0.393 m «346/880» scores 0.

Answer: 896 Hz; 0.386 m

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
λ = c/f = 3.00 × 10⁸ / 24.1 × 10⁹ = 1.24 × 10⁻² m	✓ 1	Accept 1.2 × 10⁻² m or 1.24 cm.
Part (b)
the ball moves towards the gun, so it meets the wavefronts more often than they are emitted «it receives a frequency higher than 24.1 GHz»	✓ 1	Accept “the ball acts as a moving observer”.
the ball reflects the waves as a moving source: it moves towards the gun between reflecting successive wavefronts, so the reflected wavefronts are closer together «shorter wavelength»	✓ 1
the waves still travel at c, so by c = fλ more wavefronts reach the gun each second «the detected frequency is higher»	✓ 1	Do not accept “the ball emits a higher frequency” with no reference to the spacing or rate of arrival of wavefronts.
Part (c)
each of the two steps changes the frequency by a fraction ≈ u/c, so Δf ≈ 2uf/c	✓ 1	May be seen as a substitution with the factor 2.
Δf = 2 × 50 × 24.1 × 10⁹ / 3.00 × 10⁸ = 8.03 × 10³ Hz	✓ 1	A value more precise than the given 8 kHz must be seen. 4.0 kHz (factor 2 omitted) scores 0.
Part (d)
u = cΔf/2f = 3.00 × 10⁸ × 8.68 × 10³ / (2 × 24.1 × 10⁹) = 54.0 m s⁻¹	✓ 1
54.0 × 3.6 = 194 km h⁻¹	✓ 1	Allow ECF for the conversion. 389 km h⁻¹ (factor 2 omitted) scores [1].
Part (e)
Δp = 0.058 × 54.0 = 3.13 N s	✓ 1	Allow ECF from (d).
F = Δp/Δt = 3.13 / 4.0 × 10⁻³ = 780 N	✓ 1	Accept 783 N. Ignore the weight of the ball.
Part (f)(i)
Δf at t = 0.40 s read as 6.0 kHz	✓ 1	Accept 5.9 to 6.1 kHz.
speed at 0.40 s = 3.00 × 10⁸ × 6.0 × 10³ / (2 × 24.1 × 10⁹) = 37.3 m s⁻¹	✓ 1	Or by proportion: 54.0 × 6.0/8.68. Allow ECF from (d).
a = (37.3 − 54.0)/0.40 = −42 m s⁻²	✓ 1	Accept −40 to −44 m s⁻², or a deceleration of 40 to 44 m s⁻². A positive answer with no indication that the ball slows down scores [2] max.
Part (f)(ii)
Δf is proportional to the speed, and the gradient of the graph decreases in magnitude, so the deceleration of the ball decreases	✓ 1
the drag force decreases «F = ma», because drag depends on speed and the speed is falling	✓ 1	Do not accept “drag decreases” with no reference to the gradient or the deceleration.
Part (f)(iii)
distance = area under the speed–time graph, found from the area under Figure 2 multiplied by c/2f «6.22 × 10⁻³ m s⁻¹ per Hz», or from the mean speed ≈ (54.0 + 37.3)/2 m s⁻¹	✓ 1
distance ≈ 18 m	✓ 1	Accept 17 to 19 m. The mean of the end speeds gives 18.3 m, slightly too large because the curve lies below the straight line joining its ends. Allow ECF from (f)(i).
Part (g)
only the component of the ball’s velocity along the line joining the ball and the gun produces a Doppler shift	✓ 1
component = 54.0 × cos 20° = 50.7 m s⁻¹ «183 km h⁻¹»	✓ 1	Allow ECF from (d).
the measured Δf is smaller, and the gun treats it as motion directly towards it, so it displays a speed lower than the true speed	✓ 1
Part (h)
moving observer approaching: f′ = f(v + u_o)/v	✓ 1	The moving-source equation scores 0 for this mark, even though it happens to give a similar value here.
f′ = 880 × 346/340 = 896 Hz	✓ 1
wavelength unchanged, because the source is at rest: λ = 340/880 = 0.386 m	✓ 1	Accept 0.39 m. 0.393 m «346/880» scores 0.

Answers: (a) 1.24 × 10⁻² m · (c) 8.03 kHz · (d) 194 km h⁻¹ (54.0 m s⁻¹) · (e) 780 N · (f)(i) about −42 m s⁻² (a deceleration) · (f)(iii) about 18 m · (g) 50.7 m s⁻¹ (183 km h⁻¹) · (h) 896 Hz; 0.386 m

Common error (c) Leaving out the factor 2. The ball first receives the microwaves as a moving observer and then reflects them as a moving source, so the frequency is shifted twice.

Exam tip (f)(iii) When the plotted quantity is proportional to speed, the area under the graph is proportional to distance. Convert either the area or the scale before you use it.

More C.5 questions in the question bank

+ Add to test✓ In your testCC-IBP-C5-225 marks~30 min

Finding an exoplanet from a star’s wobble

C.5 Doppler EffectD.1 Gravitational FieldsE.1 Structure of the AtomE.5 Fusion and StarsA.2 Forces and MomentumSL

A main-sequence star has the same luminosity and the same surface temperature as the Sun. Its spectrum, recorded with a high-resolution spectrograph, contains a dark absorption line of hydrogen whose wavelength measured in a laboratory is 656.28 nm.

Over many nights the observed wavelength of this line varies periodically about a mean value of 656.306 nm. The variation is caused by an unseen planet: the star and the planet orbit their common centre of mass (Figure 1), so the star moves alternately towards and away from the Earth.

Assume that the orbits are circular and, unless told otherwise, that the Earth lies in the plane of the orbits. Mass of the Sun = 2.0 × 10³⁰ kg.

Figure 1: the star and the planet seen from above the plane of their orbits (not to scale)

(a)

Outline how hydrogen in the outer layers of the star produces a dark line at 656.28 nm in the star’s spectrum.

EXPLAINNotes

Show solution

light with a continuous spectrum from the hotter layers below passes through cooler hydrogen gas in the outer layers	✓ 1
hydrogen atoms absorb photons whose energy equals the difference between two energy levels «ΔE = hc/λ», so fewer photons of this wavelength travel on towards the Earth	✓ 1	Accept “the absorbed energy is re-emitted in all directions”.

(2)

(b)

Calculate, in eV, the energy of a photon of wavelength 656.28 nm.

CALCNotes

Show solution

E = hc/λ = 1.99 × 10⁻²⁵ / 656.28 × 10⁻⁹ = 3.03 × 10⁻¹⁹ J	✓ 1
3.03 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.89 eV	✓ 1	Accept 1.9 eV. Allow ECF for the conversion.

Answer: 1.89 eV

(2)

(c)

Using the mean wavelength of the line, determine the velocity of the star–planet system relative to the Earth.

CALCNotes

Show solution

Δλ = 656.306 − 656.28 = 0.026 nm	✓ 1
v = cΔλ/λ = 3.00 × 10⁸ × 0.026/656.28 = 1.2 × 10⁴ m s⁻¹, away from the Earth	✓ 1	The direction (moving away, red-shift) is needed for this mark.

Answer: 1.2 × 10⁴ m s⁻¹ away from the Earth

(2)

(d)(i)

Figure 2 shows how the shift of the line from its mean wavelength of 656.306 nm varies with time t. The points are measurements and the curve is the best fit.

Use Figure 2 to determine the orbital period of the planet, in days.

Figure 2: shift of the line from its mean wavelength against time t

GRAPHNotes

Show solution

T = 4.2 days

✓ 1

Accept 4.1 to 4.3 days. Best found from several cycles, e.g. three cycles in 12.6 days.

Answer: 4.2 days

(1)

(d)(ii)

Calculate the maximum speed of the star along the line of sight.

CALCNotes

Show solution

maximum shift read from Figure 2 as 1.10 × 10⁻³ nm	✓ 1	Accept 1.05 to 1.15 × 10⁻³ nm.
v = 3.00 × 10⁸ × 1.10 × 10⁻³ / 656.28 = 503 m s⁻¹	✓ 1	Accept 480 to 530 m s⁻¹. Using 656.306 nm makes no difference.

Answer: 503 m s⁻¹

(2)

(d)(iii)

Explain why the motion of the system found in (c) does not affect the values found in (d)(i) and (d)(ii).

REASONNotes

Show solution

the system recedes at a constant velocity, so it adds the same shift «0.026 nm» to every measurement	✓ 1
the period and the size of the oscillation about the mean wavelength depend only on the star’s orbital motion	✓ 1

(2)

(e)

Explain why the mass of the star can be taken to be equal to the mass of the Sun.

EXPLAINNotes

Show solution

the star has the same luminosity and surface temperature as the Sun, so it is at the same position on the main sequence of the HR diagram	✓ 1
the position of a star on the main sequence is determined by its mass	✓ 1

(2)

(f)

Assume that the planet orbits the centre of the star and that the mass of the planet is much smaller than the mass of the star.

Show that the radius of the planet’s orbit is about 0.05 AU. (1 AU = 1.50 × 10¹¹ m)

DERIVENotes

Show solution

gravitational force provides the centripetal force: GMm/r² = 4π²mr/T², so r³ = GMT²/4π²	✓ 1
T = 4.2 × 86 400 = 3.63 × 10⁵ s substituted with M = 2.0 × 10³⁰ kg	✓ 1	Allow ECF from (d)(i).
r = 7.63 × 10⁹ m = 0.0509 AU	✓ 1	A value more precise than 0.05 AU must be seen.

Answer: 7.6 × 10⁹ m = 0.051 AU

(3)

(g)

The star and the planet have momenta that are equal in magnitude and opposite in direction.

Estimate the mass of the planet.

MULTINotes

Show solution

orbital speed of the planet v_p = 2πr/T = 2π × 7.63 × 10⁹ / 3.63 × 10⁵ = 1.32 × 10⁵ m s⁻¹	✓ 1	Or from √(GM/r).
m_pv_p = Mv_star	✓ 1
m_p = 2.0 × 10³⁰ × 503 / 1.32 × 10⁵ = 7.6 × 10²⁷ kg	✓ 1	Allow ECF from (d)(ii) and (f).

Answer: 7.6 × 10²⁷ kg

(3)

(h)

Later observations show that the Earth does not lie in the plane of the orbits: the line of sight makes an angle of 30° with that plane.

Determine the mass of the planet, and explain why the estimate in (g) is a minimum value.

REASONNotes

Show solution

only the component of the star’s velocity along the line of sight is measured, so 503 m s⁻¹ = v_star cos 30°	✓ 1
v_star = 581 m s⁻¹, with r and v_p unchanged, so m_p = 8.8 × 10²⁷ kg	✓ 1	Allow ECF from (g): mass in (g) ÷ cos 30°.
for any tilt the true speed is at least the measured component, so the mass in (g) is a lower limit	✓ 1

Answer: 8.8 × 10²⁷ kg

(3)

(i)

The parallax angle of the star is 0.0200 arc-seconds.

Determine the largest angle between the planet and the star as seen from the Earth, in arc-seconds, and suggest why the planet has not been seen directly in images.

REASONNotes

Show solution

d = 1/0.0200 = 50.0 pc	✓ 1
1 AU seen from 1 pc subtends 1 arc-second, so 0.0509 AU at 50.0 pc subtends 0.0509/50.0 = 1.0 × 10⁻³ arc-seconds	✓ 1	Or in metres: 7.63 × 10⁹ / (50.0 × 3.26 × 9.46 × 10¹⁵) = 4.9 × 10⁻⁹ rad. Allow ECF from (f).
the angle is far too small to resolve the planet from the star «and the faint planet is lost in the light of the much brighter star»	✓ 1

Answer: 1.0 × 10⁻³ arc-seconds

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
light with a continuous spectrum from the hotter layers below passes through cooler hydrogen gas in the outer layers	✓ 1
hydrogen atoms absorb photons whose energy equals the difference between two energy levels «ΔE = hc/λ», so fewer photons of this wavelength travel on towards the Earth	✓ 1	Accept “the absorbed energy is re-emitted in all directions”.
Part (b)
E = hc/λ = 1.99 × 10⁻²⁵ / 656.28 × 10⁻⁹ = 3.03 × 10⁻¹⁹ J	✓ 1
3.03 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.89 eV	✓ 1	Accept 1.9 eV. Allow ECF for the conversion.
Part (c)
Δλ = 656.306 − 656.28 = 0.026 nm	✓ 1
v = cΔλ/λ = 3.00 × 10⁸ × 0.026/656.28 = 1.2 × 10⁴ m s⁻¹, away from the Earth	✓ 1	The direction (moving away, red-shift) is needed for this mark.
Part (d)(i)
T = 4.2 days	✓ 1	Accept 4.1 to 4.3 days. Best found from several cycles, e.g. three cycles in 12.6 days.
Part (d)(ii)
maximum shift read from Figure 2 as 1.10 × 10⁻³ nm	✓ 1	Accept 1.05 to 1.15 × 10⁻³ nm.
v = 3.00 × 10⁸ × 1.10 × 10⁻³ / 656.28 = 503 m s⁻¹	✓ 1	Accept 480 to 530 m s⁻¹. Using 656.306 nm makes no difference.
Part (d)(iii)
the system recedes at a constant velocity, so it adds the same shift «0.026 nm» to every measurement	✓ 1
the period and the size of the oscillation about the mean wavelength depend only on the star’s orbital motion	✓ 1
Part (e)
the star has the same luminosity and surface temperature as the Sun, so it is at the same position on the main sequence of the HR diagram	✓ 1
the position of a star on the main sequence is determined by its mass	✓ 1
Part (f)
gravitational force provides the centripetal force: GMm/r² = 4π²mr/T², so r³ = GMT²/4π²	✓ 1
T = 4.2 × 86 400 = 3.63 × 10⁵ s substituted with M = 2.0 × 10³⁰ kg	✓ 1	Allow ECF from (d)(i).
r = 7.63 × 10⁹ m = 0.0509 AU	✓ 1	A value more precise than 0.05 AU must be seen.
Part (g)
orbital speed of the planet v_p = 2πr/T = 2π × 7.63 × 10⁹ / 3.63 × 10⁵ = 1.32 × 10⁵ m s⁻¹	✓ 1	Or from √(GM/r).
m_pv_p = Mv_star	✓ 1
m_p = 2.0 × 10³⁰ × 503 / 1.32 × 10⁵ = 7.6 × 10²⁷ kg	✓ 1	Allow ECF from (d)(ii) and (f).
Part (h)
only the component of the star’s velocity along the line of sight is measured, so 503 m s⁻¹ = v_star cos 30°	✓ 1
v_star = 581 m s⁻¹, with r and v_p unchanged, so m_p = 8.8 × 10²⁷ kg	✓ 1	Allow ECF from (g): mass in (g) ÷ cos 30°.
for any tilt the true speed is at least the measured component, so the mass in (g) is a lower limit	✓ 1
Part (i)
d = 1/0.0200 = 50.0 pc	✓ 1
1 AU seen from 1 pc subtends 1 arc-second, so 0.0509 AU at 50.0 pc subtends 0.0509/50.0 = 1.0 × 10⁻³ arc-seconds	✓ 1	Or in metres: 7.63 × 10⁹ / (50.0 × 3.26 × 9.46 × 10¹⁵) = 4.9 × 10⁻⁹ rad. Allow ECF from (f).
the angle is far too small to resolve the planet from the star «and the faint planet is lost in the light of the much brighter star»	✓ 1

Answers: (b) 1.89 eV · (c) 1.2 × 10⁴ m s⁻¹ away from the Earth · (d)(i) 4.2 days · (d)(ii) 503 m s⁻¹ · (f) 7.6 × 10⁹ m = 0.051 AU · (g) 7.6 × 10²⁷ kg · (h) 8.8 × 10²⁷ kg · (i) 1.0 × 10⁻³ arc-seconds

More C.5 questions in the question bank

Theme D · Fields

● home subchapter · ○ linked subchapter · last column: links to other themes

Setup	D.1	D.2	D.3	D.4	Other
Low-orbit internet satellites	●				A.2 C.2 A.3
Weighing Jupiter from its moons	●				A.2 E.5
Electrostatic paint spraying in a car factory		●	○		A.2 B.5
A thundercloud and a lightning strike		●			B.5 A.3
Separating carbon isotopes in a mass spectrometer		○	●		E.3 E.1
A laboratory railgun: forces on a current			●		A.1 B.5 A.3
A bicycle hub dynamo powering a lamp				●	A.4 B.5 C.1
Magnetic brakes on a roller coaster			○	●	A.3 B.5

D.1 Gravitational Fields

+ Add to test✓ In your testCC-IBP-D1-124 marks~30 min

Low-orbit internet satellites

D.1 Gravitational FieldsA.2 Forces and MomentumC.2 Wave ModelA.3 Work, Energy and PowerSL

A company operates a large group of broadband internet satellites. Each satellite moves in a circular orbit at a height of 550 km above the Earth’s surface (Figure 1).

Mass of the Earth = 5.97 × 10²⁴ kg. Radius of the Earth = 6.37 × 10⁶ m. Treat the Earth as a uniform sphere and ignore its rotation.

Figure 1: a satellite in its orbit (not to scale)

(a)

State Newton’s universal law of gravitation.

RECALLNotes

Show solution

every point mass attracts every other point mass with a force proportional to the product of their masses	✓ 1
and inversely proportional to the square of their separation	✓ 1	Accept F = Gm₁m₂/r² with every symbol defined for [2]. “Separation of their centres” is acceptable for spheres.

(2)

(b)

Calculate the gravitational field strength due to the Earth at the height of the orbit.

CALCNotes

Show solution

r = 6.37 × 10⁶ + 5.50 × 10⁵ = 6.92 × 10⁶ m	✓ 1
g = GM/r² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (6.92 × 10⁶)² = 8.32 N kg⁻¹	✓ 1	Using r = 550 km gives 1.3 × 10³ N kg⁻¹ and scores 0.

Answer: 8.32 N kg⁻¹

(2)

(c)

Objects inside an orbiting spacecraft are often described as “weightless”. Explain why a loose object floats inside a spacecraft at this height, even though the gravitational field strength there is about 85 % of its value at the Earth’s surface.

EXPLAINNotes

Show solution

the spacecraft and the object are both in free fall, with the same acceleration «8.3 m s⁻²» towards the centre of the Earth	✓ 1
so no contact «normal» force acts between the object and the spacecraft; the object does not move relative to the spacecraft	✓ 1	Do not accept “there is no gravity in orbit”.

(2)

(d)

Show that the orbital period of a satellite is about 96 minutes.

DERIVENotes

Show solution

the gravitational force provides the centripetal force: GMm/r² = 4π²mr/T²	✓ 1
T = 2π√(r³/GM) = 2π√((6.92 × 10⁶)³ / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴))	✓ 1	Allow ECF for r from (b).
T = 5.73 × 10³ s = 95.5 min	✓ 1	A value more precise than 96 min must be seen.

Answer: 5.73 × 10³ s (95.5 min)

(3)

(e)

Calculate the orbital speed of a satellite.

CALCNotes

Show solution

v = 2πr/T = 2π × 6.92 × 10⁶ / 5.73 × 10³ = 7.59 × 10³ m s⁻¹

✓ 1

Or v = √(GM/r). Accept 7.5 to 7.6 × 10³ m s⁻¹ (96 min gives 7.55 × 10³ m s⁻¹).

Answer: 7.59 × 10³ m s⁻¹

(1)

(f)(i)

Television satellites move in geostationary orbits, with an orbital period of 24.0 hours.

Use Kepler’s third law and your answer to (d) to determine the radius of a geostationary orbit.

CALCNotes

Show solution

r_geo = r(T_geo/T)^2/3 = 6.92 × 10⁶ × (86 400/5730)^2/3	✓ 1	Or r³ = GMT²/4π² with T = 86 400 s.
r_geo = 4.22 × 10⁷ m	✓ 1	Using 96 min gives 4.21 × 10⁷ m; accept.

Answer: 4.22 × 10⁷ m

(2)

(f)(ii)

Internet signals travel as radio waves between a ground station and a satellite directly above it.

Compare the minimum time for a signal to travel up to the satellite and back down for a satellite at 550 km and for a geostationary satellite.

CALCNotes

Show solution

550 km: t = 2 × 5.50 × 10⁵ / 3.00 × 10⁸ = 3.7 × 10⁻³ s	✓ 1
geostationary: height = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.59 × 10⁷ m, t = 0.24 s, about 65 times longer	✓ 1	Allow ECF from (f)(i). Using the orbital radius instead of the height (0.28 s) scores [1] max for this question part.

Answer: 3.7 ms and 0.24 s (about 65 times longer)

(2)

(g)

On the axes in Figure 2, sketch a graph to show how the gravitational field strength g due to the Earth varies with the distance r from the centre of the Earth, from the Earth’s surface to the geostationary orbit.

Figure 2: axes for (g)

SKETCHNotes

Show solution

curve starts at r = 0.64 × 10⁷ m with g ≈ 9.8 N kg⁻¹, with nothing drawn at smaller r	✓ 1
decreasing curve with a gradient that decreases in magnitude, passing close to g ≈ 2.5 N kg⁻¹ at r ≈ 1.3 × 10⁷ m «inverse-square: a quarter of the surface value at twice the radius»	✓ 1	A straight line scores 0 for this mark.
curve ends at r ≈ 4.2 × 10⁷ m with g ≈ 0.2 N kg⁻¹, above the axis	✓ 1	Allow ECF from (f)(i) for the end point.

(3)

(h)

A satellite is moved to a new circular orbit at a height of 1100 km.

Determine the percentage change in its orbital speed.

PREDICTNotes

Show solution

v = √(GM/r) with r = 7.47 × 10⁶ m gives 7.30 × 10³ m s⁻¹	✓ 1	Or by ratio: v ∝ r^−1/2, √(6.92/7.47) = 0.962.
change = (7.30 − 7.59)/7.59 × 100 = −3.8 % «a decrease»	✓ 1	Accept −3.7 % to −3.8 %. A decrease must be clear.

Answer: −3.8 % (a decrease)

(2)

(i)(i)

The mass of the satellite is 300 kg.

Show that the minimum energy needed to move the satellite from the 550 km orbit to the 1100 km orbit is about 6.4 × 10⁸ J.

HL onlyDERIVENotes

Show solution

total energy in orbit E_T = −GMm/2r	✓ 1
E_T = −8.631 × 10⁹ J at 550 km and −7.996 × 10⁹ J at 1100 km	✓ 1	Or ΔE = (GMm/2)(1/r₁ − 1/r₂) with values substituted.
ΔE = 6.36 × 10⁸ J	✓ 1	A value more precise than 6.4 × 10⁸ J must be seen.

Answer: 6.36 × 10⁸ J

(3)

(i)(ii)

The satellite’s electric thrusters are powered by solar panels that supply 2.5 kW. The thrusters transfer energy to the satellite’s orbital energy with an overall efficiency of 25 %.

Calculate the minimum time, in days, needed to raise the orbit.

CALCNotes

Show solution

energy input = 6.4 × 10⁸ / 0.25 = 2.6 × 10⁹ J	✓ 1
t = 2.6 × 10⁹ / 2500 = 1.0 × 10⁶ s ≈ 12 days	✓ 1	Accept 11.8 days. Multiplying by the efficiency instead of dividing (0.74 days) scores [1] max.

Answer: about 12 days

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
every point mass attracts every other point mass with a force proportional to the product of their masses	✓ 1
and inversely proportional to the square of their separation	✓ 1	Accept F = Gm₁m₂/r² with every symbol defined for [2]. “Separation of their centres” is acceptable for spheres.
Part (b)
r = 6.37 × 10⁶ + 5.50 × 10⁵ = 6.92 × 10⁶ m	✓ 1
g = GM/r² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (6.92 × 10⁶)² = 8.32 N kg⁻¹	✓ 1	Using r = 550 km gives 1.3 × 10³ N kg⁻¹ and scores 0.
Part (c)
the spacecraft and the object are both in free fall, with the same acceleration «8.3 m s⁻²» towards the centre of the Earth	✓ 1
so no contact «normal» force acts between the object and the spacecraft; the object does not move relative to the spacecraft	✓ 1	Do not accept “there is no gravity in orbit”.
Part (d)
the gravitational force provides the centripetal force: GMm/r² = 4π²mr/T²	✓ 1
T = 2π√(r³/GM) = 2π√((6.92 × 10⁶)³ / (6.67 × 10⁻¹¹ × 5.97 × 10²⁴))	✓ 1	Allow ECF for r from (b).
T = 5.73 × 10³ s = 95.5 min	✓ 1	A value more precise than 96 min must be seen.
Part (e)
v = 2πr/T = 2π × 6.92 × 10⁶ / 5.73 × 10³ = 7.59 × 10³ m s⁻¹	✓ 1	Or v = √(GM/r). Accept 7.5 to 7.6 × 10³ m s⁻¹ (96 min gives 7.55 × 10³ m s⁻¹).
Part (f)(i)
r_geo = r(T_geo/T)^2/3 = 6.92 × 10⁶ × (86 400/5730)^2/3	✓ 1	Or r³ = GMT²/4π² with T = 86 400 s.
r_geo = 4.22 × 10⁷ m	✓ 1	Using 96 min gives 4.21 × 10⁷ m; accept.
Part (f)(ii)
550 km: t = 2 × 5.50 × 10⁵ / 3.00 × 10⁸ = 3.7 × 10⁻³ s	✓ 1
geostationary: height = 4.22 × 10⁷ − 6.37 × 10⁶ = 3.59 × 10⁷ m, t = 0.24 s, about 65 times longer	✓ 1	Allow ECF from (f)(i). Using the orbital radius instead of the height (0.28 s) scores [1] max for this question part.
Part (g)
curve starts at r = 0.64 × 10⁷ m with g ≈ 9.8 N kg⁻¹, with nothing drawn at smaller r	✓ 1
decreasing curve with a gradient that decreases in magnitude, passing close to g ≈ 2.5 N kg⁻¹ at r ≈ 1.3 × 10⁷ m «inverse-square: a quarter of the surface value at twice the radius»	✓ 1	A straight line scores 0 for this mark.
curve ends at r ≈ 4.2 × 10⁷ m with g ≈ 0.2 N kg⁻¹, above the axis	✓ 1	Allow ECF from (f)(i) for the end point.
Part (h)
v = √(GM/r) with r = 7.47 × 10⁶ m gives 7.30 × 10³ m s⁻¹	✓ 1	Or by ratio: v ∝ r^−1/2, √(6.92/7.47) = 0.962.
change = (7.30 − 7.59)/7.59 × 100 = −3.8 % «a decrease»	✓ 1	Accept −3.7 % to −3.8 %. A decrease must be clear.
Part (i)(i)
total energy in orbit E_T = −GMm/2r	✓ 1
E_T = −8.631 × 10⁹ J at 550 km and −7.996 × 10⁹ J at 1100 km	✓ 1	Or ΔE = (GMm/2)(1/r₁ − 1/r₂) with values substituted.
ΔE = 6.36 × 10⁸ J	✓ 1	A value more precise than 6.4 × 10⁸ J must be seen.
Part (i)(ii)
energy input = 6.4 × 10⁸ / 0.25 = 2.6 × 10⁹ J	✓ 1
t = 2.6 × 10⁹ / 2500 = 1.0 × 10⁶ s ≈ 12 days	✓ 1	Accept 11.8 days. Multiplying by the efficiency instead of dividing (0.74 days) scores [1] max.

Answers: (b) 8.32 N kg⁻¹ · (d) 5.73 × 10³ s (95.5 min) · (e) 7.59 × 10³ m s⁻¹ · (f)(i) 4.22 × 10⁷ m · (f)(ii) 3.7 ms and 0.24 s (about 65 times longer) · (h) −3.8 % (a decrease) · (i)(i) 6.36 × 10⁸ J · (i)(ii) about 12 days

Common error (b) Using the height above the surface as r. In every gravitation equation r is measured from the centre of the Earth.

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+ Add to test✓ In your testCC-IBP-D1-224 marks~30 min

Weighing Jupiter from its moons

D.1 Gravitational FieldsA.2 Forces and MomentumE.5 Fusion and StarsSL

The table gives orbital data for four of Jupiter’s moons. Each moon moves in a circular orbit of radius r about the centre of Jupiter, with orbital period T.

Moon	r / 10⁸ m	T / days	r³ / 10²⁴ m³	T² / 10¹⁰ s²
Amalthea	1.81	0.498	5.93	0.185
Io	4.22	1.77	75.2	2.34
Europa	6.71	3.55	302	9.41
Ganymede	10.7	7.15

1 day = 86 400 s. Figure 1, in part (c), shows the first three moons plotted on a graph of T² against r³.

(a)

State Kepler’s third law.

RECALLNotes

Show solution

the square of the orbital period is proportional to the cube of the «mean» orbital radius «for bodies orbiting the same central mass»

✓ 1

Accept T² ∝ r³ with symbols defined.

(1)

(b)(i)

Show that, for a moon in a circular orbit, T² = (4π²/GM)r³, where M is the mass of Jupiter.

DERIVENotes

Show solution

the gravitational force on the moon provides the centripetal force	✓ 1
GMm/r² = m·4π²r/T²	✓ 1	Or GMm/r² = mv²/r with v = 2πr/T.
rearrangement to T² = 4π²r³/GM shown	✓ 1	The answer is given: every step must be seen.

(3)

(b)(ii)

Explain why the masses of the moons do not affect the graph of T² against r³.

REASONNotes

Show solution

the moon’s mass appears in both the gravitational force and the centripetal force, so it cancels: T depends only on M and r

✓ 1

(1)

(c)(i)

Calculate the values of r³ and T² for Ganymede, in the units used in the table.

CALCNotes

Show solution

r³ = (10.7 × 10⁸)³ = 1.23 × 10²⁷ m³ → 1225 «× 10²⁴ m³»	✓ 1	Accept 1230.
T = 7.15 × 86 400 = 6.18 × 10⁵ s, T² = 3.82 × 10¹¹ s² → 38.2 «× 10¹⁰ s²»	✓ 1

Answer: 1225 (× 10²⁴ m³) and 38.2 (× 10¹⁰ s²)

(2)

(c)(ii)

Plot the point for Ganymede on Figure 1 and draw the line of best fit.

Figure 1: T² against r³ for the moons of Jupiter (Amalthea, Io and Europa plotted)

GRAPHNotes

Show solution

point plotted at (1225, 38.2) to within half a small square	✓ 1	Allow ECF from (c)(i).
single straight line through the origin passing through or very close to all four points	✓ 1

(2)

(c)(iii)

Use your graph to determine the mass of Jupiter.

MULTINotes

Show solution

gradient from a large triangle, e.g. 38.2 × 10¹⁰ / 1225 × 10²⁴ = 3.12 × 10⁻¹⁶ s² m⁻³	✓ 1	Accept 3.05 to 3.20 × 10⁻¹⁶ s² m⁻³. The powers of ten from the axis labels must be included.
M = 4π²/(G × gradient)	✓ 1
M = 1.9 × 10²⁷ kg	✓ 1	Accept 1.85 to 1.94 × 10²⁷ kg. Allow ECF from the gradient.

Answer: 1.9 × 10²⁷ kg

(3)

(d)(i)

Use a mass of 1.90 × 10²⁷ kg for Jupiter in the rest of this question. The radius of Jupiter is 6.99 × 10⁷ m.

Calculate the gravitational field strength at the surface of Jupiter.

CALCNotes

Show solution

g = GM/R² = 6.67 × 10⁻¹¹ × 1.90 × 10²⁷ / (6.99 × 10⁷)²	✓ 1
g = 25.9 N kg⁻¹	✓ 1	Accept 26 N kg⁻¹.

Answer: 25.9 N kg⁻¹

(2)

(d)(ii)

Calculate the gravitational field strength due to Jupiter at the orbit of Io, and state its direction.

CALCNotes

Show solution

g = 6.67 × 10⁻¹¹ × 1.90 × 10²⁷ / (4.22 × 10⁸)² = 0.712 N kg⁻¹	✓ 1	Or by ratio from (d)(i): 25.9 × (6.99/42.2)².
directed towards the centre of Jupiter	✓ 1

Answer: 0.712 N kg⁻¹, towards the centre of Jupiter

(2)

(e)

A space probe is placed in a circular orbit around Jupiter at twice the orbital radius of Io.

Predict the orbital period of the probe, in days.

PREDICTNotes

Show solution

T ∝ r^3/2, so T = 1.77 × 2^3/2	✓ 1
T = 5.0 days	✓ 1	Accept 5.01 days. The probe’s mass is irrelevant.

Answer: 5.0 days

(2)

(f)

A body needs a mass of at least about 0.08 times the mass of the Sun to sustain hydrogen fusion in its core. The mass of the Sun is 2.0 × 10³⁰ kg.

Compare the mass of Jupiter with this minimum mass and explain, with reference to the conditions needed for fusion, why Jupiter is not a star.

EXPLAINNotes

Show solution

minimum mass = 1.6 × 10²⁹ kg, so Jupiter has about 1/84 «0.012» of this mass	✓ 1
fusion needs a very high core temperature «and density» so that nuclei have enough kinetic energy to overcome their electrostatic repulsion	✓ 1
Jupiter’s gravity is too weak to compress and heat its core enough «too little gravitational potential energy is transferred as it contracts»	✓ 1

(3)

(g)

Ignore the gravitational field of Io itself.

Determine the escape speed from Jupiter’s gravitational field at the orbital radius of Io, and hence the minimum increase in speed needed by a probe moving with Io in its orbit to escape from Jupiter.

HL onlyMULTINotes

Show solution

v_esc = √(2GM/r) = √(2 × 6.67 × 10⁻¹¹ × 1.90 × 10²⁷ / 4.22 × 10⁸)	✓ 1
v_esc = 2.45 × 10⁴ m s⁻¹	✓ 1
orbital speed = √(GM/r) = 1.73 × 10⁴ m s⁻¹, so increase = 7.2 × 10³ m s⁻¹	✓ 1	Or orbital speed = 2πr/T from the table data. Allow ECF.

Answer: 2.45 × 10⁴ m s⁻¹; increase 7.2 × 10³ m s⁻¹

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
the square of the orbital period is proportional to the cube of the «mean» orbital radius «for bodies orbiting the same central mass»	✓ 1	Accept T² ∝ r³ with symbols defined.
Part (b)(i)
the gravitational force on the moon provides the centripetal force	✓ 1
GMm/r² = m·4π²r/T²	✓ 1	Or GMm/r² = mv²/r with v = 2πr/T.
rearrangement to T² = 4π²r³/GM shown	✓ 1	The answer is given: every step must be seen.
Part (b)(ii)
the moon’s mass appears in both the gravitational force and the centripetal force, so it cancels: T depends only on M and r	✓ 1
Part (c)(i)
r³ = (10.7 × 10⁸)³ = 1.23 × 10²⁷ m³ → 1225 «× 10²⁴ m³»	✓ 1	Accept 1230.
T = 7.15 × 86 400 = 6.18 × 10⁵ s, T² = 3.82 × 10¹¹ s² → 38.2 «× 10¹⁰ s²»	✓ 1
Part (c)(ii)
point plotted at (1225, 38.2) to within half a small square	✓ 1	Allow ECF from (c)(i).
single straight line through the origin passing through or very close to all four points	✓ 1
Part (c)(iii)
gradient from a large triangle, e.g. 38.2 × 10¹⁰ / 1225 × 10²⁴ = 3.12 × 10⁻¹⁶ s² m⁻³	✓ 1	Accept 3.05 to 3.20 × 10⁻¹⁶ s² m⁻³. The powers of ten from the axis labels must be included.
M = 4π²/(G × gradient)	✓ 1
M = 1.9 × 10²⁷ kg	✓ 1	Accept 1.85 to 1.94 × 10²⁷ kg. Allow ECF from the gradient.
Part (d)(i)
g = GM/R² = 6.67 × 10⁻¹¹ × 1.90 × 10²⁷ / (6.99 × 10⁷)²	✓ 1
g = 25.9 N kg⁻¹	✓ 1	Accept 26 N kg⁻¹.
Part (d)(ii)
g = 6.67 × 10⁻¹¹ × 1.90 × 10²⁷ / (4.22 × 10⁸)² = 0.712 N kg⁻¹	✓ 1	Or by ratio from (d)(i): 25.9 × (6.99/42.2)².
directed towards the centre of Jupiter	✓ 1
Part (e)
T ∝ r^3/2, so T = 1.77 × 2^3/2	✓ 1
T = 5.0 days	✓ 1	Accept 5.01 days. The probe’s mass is irrelevant.
Part (f)
minimum mass = 1.6 × 10²⁹ kg, so Jupiter has about 1/84 «0.012» of this mass	✓ 1
fusion needs a very high core temperature «and density» so that nuclei have enough kinetic energy to overcome their electrostatic repulsion	✓ 1
Jupiter’s gravity is too weak to compress and heat its core enough «too little gravitational potential energy is transferred as it contracts»	✓ 1
Part (g)
v_esc = √(2GM/r) = √(2 × 6.67 × 10⁻¹¹ × 1.90 × 10²⁷ / 4.22 × 10⁸)	✓ 1
v_esc = 2.45 × 10⁴ m s⁻¹	✓ 1
orbital speed = √(GM/r) = 1.73 × 10⁴ m s⁻¹, so increase = 7.2 × 10³ m s⁻¹	✓ 1	Or orbital speed = 2πr/T from the table data. Allow ECF.

Answers: (c)(i) 1225 (× 10²⁴ m³) and 38.2 (× 10¹⁰ s²) · (c)(iii) 1.9 × 10²⁷ kg · (d)(i) 25.9 N kg⁻¹ · (d)(ii) 0.712 N kg⁻¹, towards the centre of Jupiter · (e) 5.0 days · (g) 2.45 × 10⁴ m s⁻¹; increase 7.2 × 10³ m s⁻¹

Exam tip (c)(iii) Use a gradient triangle that covers most of the line, and include the powers of ten from the axis labels before you substitute into 4π²/GM.

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D.2 Electric and Magnetic Fields

+ Add to test✓ In your testCC-IBP-D2-128 marks~35 min

Electrostatic paint spraying in a car factory

D.2 Electric and Magnetic FieldsD.3 Motion in Electromagnetic FieldsA.2 Forces and MomentumB.5 Current and CircuitsSL

In a car factory, body panels are painted with an electrostatic spray gun (Figure 1). The paint leaves the nozzle as a fine spray of droplets. Each droplet becomes negatively charged as it is forced through the nozzle. The metal nozzle tip is connected to a high-voltage supply that keeps it negatively charged. The car panel is earthed.

Data for one droplet:

mass = 2.0 × 10⁻⁹ kg
charge = −4.0 × 10⁻¹² C

Close to the gun, the nozzle tip can be treated as a small charged sphere. Air resistance is ignored unless a part says otherwise.

Figure 1: the spray gun and the earthed car panel (not to scale)

(a)

The droplets become negatively charged by friction with the inside of the nozzle. Outline, in terms of electrons, how the droplets become negatively charged, and state what happens to the total charge.

RECALLNotes

Show solution

electrons are transferred from the nozzle (material) to the paint droplets	✓ 1	Must refer to electrons moving. Do not accept positive charge moving from the droplets.
the total charge is conserved: the nozzle material loses as many electrons as the droplets gain	✓ 1	Accept: charge is not created, only transferred / the nozzle material gains an equal positive charge (which the supply removes).

(2)

(b)

On Figure 1, sketch the electric field pattern between the nozzle tip and the car panel. Draw at least five field lines.

SKETCHNotes

Show solution

lines start on the panel and end on the nozzle tip, with arrows pointing towards the nozzle tip	✓ 1	Arrows on the lines are required for this mark; direction must be towards the negative tip.
lines meet the panel (and the tip) at right angles	✓ 1	Judge by eye on at least most lines.
lines closest together near the tip and spreading out towards the panel, with no lines crossing	✓ 1	Radial pattern near the tip.

(3)

(c)(i)

Figure 2 shows how the electric field strength E varies with distance r from the centre of the nozzle tip, along the line from the tip towards the panel.

Use Figure 2 to show that E is inversely proportional to r².

Figure 2: electric field strength E against distance r from the centre of the nozzle tip

GRAPHNotes

Show solution

two correct readings, e.g. E = 1.5 × 10⁵ V m⁻¹ at r = 0.10 m and E = 0.375 × 10⁵ V m⁻¹ at r = 0.20 m	✓ 1	Any pair of points; readings within half a small square. Other pairs: 6.0 at 0.05 m, 0.24 at 0.25 m.
doubling r reduces E to a quarter / Er² is constant (= 1.5 × 10³ V m) for both points	✓ 1	Must show the comparison numerically. Accept a check of Er² at three points.

Answer: Er² = 1.5 × 10³ V m at every point

(2)

(c)(ii)

Determine the magnitude of the charge on the nozzle tip.

CALCNotes

Show solution

E = kQ/r² so Q = Er²/k = 1.5 × 10⁵ × 0.10² / 8.99 × 10⁹	✓ 1	Any point from the graph may be used.
Q = 1.7 × 10⁻⁷ C	✓ 1	Accept 1.6–1.7 × 10⁻⁷ C. Allow ECF from (c)(i) readings. Sign not required.

Answer: 1.7 × 10⁻⁷ C (negative)

(2)

(d)

A droplet is at a distance of 0.10 m from the centre of the nozzle tip. Show that the electric force on the droplet is about 30 times its weight.

CALCNotes

Show solution

electric force F = qE = 4.0 × 10⁻¹² × 1.5 × 10⁵ = 6.0 × 10⁻⁷ N	✓ 1	E read from Figure 2 or calculated from the charge in (c)(ii).
weight = mg = 2.0 × 10⁻⁹ × 9.8 = 1.96 × 10⁻⁸ N, so F/mg = 31 ≈ 30	✓ 1	The ratio must be seen, to at least 2 s.f.; answer given.

Answer: 6.0 × 10⁻⁷ N ÷ 1.96 × 10⁻⁸ N ≈ 31

(2)

(e)(i)

Calculate the acceleration of the droplet at this position due to the electric force alone.

CALCNotes

Show solution

a = qE/m = 6.0 × 10⁻⁷ / 2.0 × 10⁻⁹	✓ 1	Allow ECF from (d).
a = 3.0 × 10² m s⁻²	✓ 1	Direction not required.

Answer: 3.0 × 10² m s⁻²

(2)

(e)(ii)

In practice, air resistance acts on the moving droplet. As the droplet moves towards the panel, its acceleration decreases. Explain why, giving two reasons.

EXPLAINNotes

Show solution

the field strength (and so the electric force qE) decreases as the distance from the tip increases / inverse-square decrease	✓ 1	Accept reference to Figure 2.
air resistance (drag) increases as the droplet's speed increases, so the resultant force decreases	✓ 1	Must link drag to increasing speed.

(2)

(f)

Two neighbouring droplets in the spray are 2.0 mm apart. Calculate the electric force between them and state its effect on the spray.

CALCNotes

Show solution

F = kq₁q₂/r² = 8.99 × 10⁹ × (4.0 × 10⁻¹²)² / (2.0 × 10⁻³)²	✓ 1	Correct substitution with r in metres.
F = 3.6 × 10⁻⁸ N	✓ 1
the force is repulsive (like charges), so the droplets spread out into an even cloud / do not clump together	✓ 1	Accept: gives a uniform coat.

Answer: 3.6 × 10⁻⁸ N, repulsive

(3)

(g)

The car panel is uncharged before spraying starts. Explain why droplets are attracted to the earthed panel, and why paint also reaches the surface of the panel that faces away from the gun.

EXPLAINNotes

Show solution

the negative charges (nozzle and droplets) repel free electrons in the panel to earth	✓ 1	Electrons must move to earth, not positive charge from earth.
leaving the panel positively charged (by induction), so it attracts the negative droplets	✓ 1
field lines curve round the edges and end on every surface of the panel at right angles, and the droplets are pushed along the field lines, so they reach the back surface	✓ 1	Accept: the force on a droplet is along the field, which curves round to the rear of the panel.

(3)

(h)

The gun sprays paint at a rate of 3.0 g s⁻¹. Every droplet has the mass and charge given above. Calculate the electric current carried by the droplets.

MULTINotes

Show solution

number of droplets per second = 3.0 × 10⁻³ / 2.0 × 10⁻⁹ = 1.5 × 10⁶ s⁻¹	✓ 1	Or charge per unit mass = 2.0 × 10⁻³ C kg⁻¹.
I = Δq/Δt = 1.5 × 10⁶ × 4.0 × 10⁻¹² = 6.0 × 10⁻⁶ A	✓ 1	Accept 6.0 μA. Do not accept a negative sign without comment.

Answer: 6.0 × 10⁻⁶ A

(2)

(i)

On a humid day, charge leaks away from the droplets and each droplet now carries a charge of −1.0 × 10⁻¹² C. The paint flow rate is unchanged.

Predict the effect on the current carried by the droplets and on how well the hidden surfaces of the panel are coated.

PREDICTNotes

Show solution

current falls to a quarter: 1.5 × 10⁻⁶ A	✓ 1	Allow ECF from (h).
the electric force falls to a quarter (now only about 8 times the weight at 0.10 m), so weight and air resistance have a larger effect and fewer droplets follow the field round to the hidden surfaces / more paint is wasted	✓ 1	Answer must refer to the smaller electric force.

Answer: 1.5 × 10⁻⁶ A; poorer coverage of hidden surfaces

(2)

(j)

The droplet in (d) moves from r = 0.10 m to r = 0.25 m. Ignoring air resistance and gravity, determine the kinetic energy gained by the droplet.

HL onlyMULTINotes

Show solution

V_e = kQ/r: magnitudes 1.5 × 10⁴ V at 0.10 m and 6.0 × 10³ V at 0.25 m	✓ 1	Signs: both potentials are negative (−1.5 × 10⁴ V and −6.0 × 10³ V). Allow ECF from (c)(ii).
potential difference ΔV_e = 9.0 × 10³ V	✓ 1	The droplet moves to higher (less negative) potential.
E_k = \|q\|ΔV_e = 4.0 × 10⁻¹² × 9.0 × 10³ = 3.6 × 10⁻⁸ J	✓ 1	A method using E × distance scores 0 for this row, as the field is not uniform. Accept 3.5–3.6 × 10⁻⁸ J.

Answer: 3.6 × 10⁻⁸ J

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
electrons are transferred from the nozzle (material) to the paint droplets	✓ 1	Must refer to electrons moving. Do not accept positive charge moving from the droplets.
the total charge is conserved: the nozzle material loses as many electrons as the droplets gain	✓ 1	Accept: charge is not created, only transferred / the nozzle material gains an equal positive charge (which the supply removes).
Part (b)
lines start on the panel and end on the nozzle tip, with arrows pointing towards the nozzle tip	✓ 1	Arrows on the lines are required for this mark; direction must be towards the negative tip.
lines meet the panel (and the tip) at right angles	✓ 1	Judge by eye on at least most lines.
lines closest together near the tip and spreading out towards the panel, with no lines crossing	✓ 1	Radial pattern near the tip.
Part (c)(i)
two correct readings, e.g. E = 1.5 × 10⁵ V m⁻¹ at r = 0.10 m and E = 0.375 × 10⁵ V m⁻¹ at r = 0.20 m	✓ 1	Any pair of points; readings within half a small square. Other pairs: 6.0 at 0.05 m, 0.24 at 0.25 m.
doubling r reduces E to a quarter / Er² is constant (= 1.5 × 10³ V m) for both points	✓ 1	Must show the comparison numerically. Accept a check of Er² at three points.
Part (c)(ii)
E = kQ/r² so Q = Er²/k = 1.5 × 10⁵ × 0.10² / 8.99 × 10⁹	✓ 1	Any point from the graph may be used.
Q = 1.7 × 10⁻⁷ C	✓ 1	Accept 1.6–1.7 × 10⁻⁷ C. Allow ECF from (c)(i) readings. Sign not required.
Part (d)
electric force F = qE = 4.0 × 10⁻¹² × 1.5 × 10⁵ = 6.0 × 10⁻⁷ N	✓ 1	E read from Figure 2 or calculated from the charge in (c)(ii).
weight = mg = 2.0 × 10⁻⁹ × 9.8 = 1.96 × 10⁻⁸ N, so F/mg = 31 ≈ 30	✓ 1	The ratio must be seen, to at least 2 s.f.; answer given.
Part (e)(i)
a = qE/m = 6.0 × 10⁻⁷ / 2.0 × 10⁻⁹	✓ 1	Allow ECF from (d).
a = 3.0 × 10² m s⁻²	✓ 1	Direction not required.
Part (e)(ii)
the field strength (and so the electric force qE) decreases as the distance from the tip increases / inverse-square decrease	✓ 1	Accept reference to Figure 2.
air resistance (drag) increases as the droplet's speed increases, so the resultant force decreases	✓ 1	Must link drag to increasing speed.
Part (f)
F = kq₁q₂/r² = 8.99 × 10⁹ × (4.0 × 10⁻¹²)² / (2.0 × 10⁻³)²	✓ 1	Correct substitution with r in metres.
F = 3.6 × 10⁻⁸ N	✓ 1
the force is repulsive (like charges), so the droplets spread out into an even cloud / do not clump together	✓ 1	Accept: gives a uniform coat.
Part (g)
the negative charges (nozzle and droplets) repel free electrons in the panel to earth	✓ 1	Electrons must move to earth, not positive charge from earth.
leaving the panel positively charged (by induction), so it attracts the negative droplets	✓ 1
field lines curve round the edges and end on every surface of the panel at right angles, and the droplets are pushed along the field lines, so they reach the back surface	✓ 1	Accept: the force on a droplet is along the field, which curves round to the rear of the panel.
Part (h)
number of droplets per second = 3.0 × 10⁻³ / 2.0 × 10⁻⁹ = 1.5 × 10⁶ s⁻¹	✓ 1	Or charge per unit mass = 2.0 × 10⁻³ C kg⁻¹.
I = Δq/Δt = 1.5 × 10⁶ × 4.0 × 10⁻¹² = 6.0 × 10⁻⁶ A	✓ 1	Accept 6.0 μA. Do not accept a negative sign without comment.
Part (i)
current falls to a quarter: 1.5 × 10⁻⁶ A	✓ 1	Allow ECF from (h).
the electric force falls to a quarter (now only about 8 times the weight at 0.10 m), so weight and air resistance have a larger effect and fewer droplets follow the field round to the hidden surfaces / more paint is wasted	✓ 1	Answer must refer to the smaller electric force.
Part (j)
V_e = kQ/r: magnitudes 1.5 × 10⁴ V at 0.10 m and 6.0 × 10³ V at 0.25 m	✓ 1	Signs: both potentials are negative (−1.5 × 10⁴ V and −6.0 × 10³ V). Allow ECF from (c)(ii).
potential difference ΔV_e = 9.0 × 10³ V	✓ 1	The droplet moves to higher (less negative) potential.
E_k = \|q\|ΔV_e = 4.0 × 10⁻¹² × 9.0 × 10³ = 3.6 × 10⁻⁸ J	✓ 1	A method using E × distance scores 0 for this row, as the field is not uniform. Accept 3.5–3.6 × 10⁻⁸ J.

Answers: (c)(i) Er² = 1.5 × 10³ V m at every point · (c)(ii) 1.7 × 10⁻⁷ C (negative) · (d) 6.0 × 10⁻⁷ N ÷ 1.96 × 10⁻⁸ N ≈ 31 · (e)(i) 3.0 × 10² m s⁻² · (f) 3.6 × 10⁻⁸ N, repulsive · (h) 6.0 × 10⁻⁶ A · (i) 1.5 × 10⁻⁶ A; poorer coverage of hidden surfaces · (j) 3.6 × 10⁻⁸ J

Common error (a) A common error is to describe positive charges (protons) moving. In charging by friction only electrons move; a body becomes positive by losing electrons.

Exam tip (d) In a “show that” part, give the final value to one more significant figure than the value quoted in the question (31, not just 30), so it is clear that you calculated it.

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+ Add to test✓ In your testCC-IBP-D2-225 marks~30 min

A thundercloud and a lightning strike

D.2 Electric and Magnetic FieldsB.5 Current and CircuitsA.3 Work, Energy and PowerSL

The base of a thundercloud is flat and horizontal, 1.2 km above flat ground (Figure 1). Collisions between ice particles inside the cloud leave the cloud base negatively charged, and the ground below becomes positively charged. The cloud base and the ground can be modelled as a pair of large parallel plates.

The potential difference between the cloud base and the ground is 1.8 × 10⁸ V. Air breaks down (becomes conducting) when the electric field strength reaches 3.0 × 10⁶ V m⁻¹.

Figure 1: the cloud base and the ground (not to scale)

(a)

Explain why the ground below the cloud becomes positively charged.

EXPLAINNotes

Show solution

free electrons in the ground are repelled by the negative charge on the cloud base	✓ 1	Must refer to electrons.
they move away (deeper / sideways into the Earth), leaving the surface below the cloud with a positive charge / charging by induction	✓ 1	Do not accept positive charge moving up.

(2)

(b)

On Figure 1, draw the electric field pattern between the cloud base and the ground, including the regions near the edges of the cloud.

SKETCHNotes

Show solution

straight, vertical, parallel and equally spaced lines in the central region	✓ 1	At least four lines.
arrows pointing upwards, from the ground to the cloud base	✓ 1	Arrows from positive to negative.
lines curving outwards (bulging) near the edges of the cloud	✓ 1	Lines must still start and end on the two surfaces.

(3)

(c)(i)

Calculate the average electric field strength between the cloud base and the ground.

CALCNotes

Show solution

E = V/d = 1.8 × 10⁸ / 1200 = 1.5 × 10⁵ V m⁻¹

✓ 1

Accept N C⁻¹.

Answer: 1.5 × 10⁵ V m⁻¹

(1)

(c)(ii)

Comment on whether this field is strong enough to make the air conduct.

REASONNotes

Show solution

the average field is 20 times smaller than the breakdown field (1.5 × 10⁵ < 3.0 × 10⁶ V m⁻¹)	✓ 1	Numerical comparison required. Allow ECF from (c)(i).
so breakdown can only start in small regions where the field is much stronger than average (e.g. near sharp points or charge concentrations), not across the whole gap at once	✓ 1	Accept: the field is not uniform in reality / it is locally stronger.

(2)

(d)

Calculate the magnitude of the force on a free electron between the cloud and the ground, and state its direction.

CALCNotes

Show solution

F = eE = 1.60 × 10⁻¹⁹ × 1.5 × 10⁵ = 2.4 × 10⁻¹⁴ N	✓ 1	Allow ECF from (c)(i).
vertically downwards / towards the ground	✓ 1	Opposite to the field because the electron is negative.

Answer: 2.4 × 10⁻¹⁴ N, downwards

(2)

(e)(i)

A lightning stroke between the cloud and the ground lasts 0.30 ms. Figure 2 is a simplified model of how the current in the stroke varies with time.

Use Figure 2 to show that the charge transferred by the stroke is about 15 C.

Figure 2: current I in the lightning stroke against time t (simplified model)

GRAPHNotes

Show solution

charge = area under the current–time graph	✓ 1	Stated or implied by the calculation.
½ × 0.30 × 10⁻³ × 1.0 × 10⁵ = 15 C	✓ 1	Unit conversion of ms and kA must be correct. Answer given, so working must be complete.

Answer: 15 C

(2)

(e)(ii)

Calculate the mean current in the stroke.

CALCNotes

Show solution

I = Δq/Δt = 15 / 0.30 × 10⁻³	✓ 1
I = 5.0 × 10⁴ A	✓ 1	Accept 50 kA. Do not accept the peak value of 1.0 × 10⁵ A.

Answer: 5.0 × 10⁴ A

(2)

(f)(i)

Show that the energy transferred by the stroke is about 3 × 10⁹ J. Assume that the potential difference stays at 1.8 × 10⁸ V during the stroke.

CALCNotes

Show solution

W = qV (from V = W/q)	✓ 1
W = 15 × 1.8 × 10⁸ = 2.7 × 10⁹ J	✓ 1	Value to at least 2 s.f.

Answer: 2.7 × 10⁹ J

(2)

(f)(ii)

A household uses 3000 kWh of electrical energy in one year. Compare the energy transferred by the stroke with this.

MULTINotes

Show solution

3000 kWh = 3000 × 1000 × 3600 = 1.08 × 10¹⁰ J	✓ 1
the stroke transfers about a quarter of the household's annual use / about three months' worth	✓ 1	Allow ECF from (f)(i). Ratio 0.25 or equivalent.

Answer: about ¼ of a year's use (≈ 3 months)

(2)

(g)

A tall building near the storm has a metal lightning rod with a sharp point at the top, connected to the ground by a thick copper strip. Explain why the rod is pointed and why it is connected to the ground.

EXPLAINNotes

Show solution

field lines crowd together at the sharp point, so the field strength there is very large	✓ 1	Accept: induced charge concentrates at the point.
so the field near the point can reach the breakdown value and the air there becomes conducting / the strike is attracted to the rod	✓ 1	Link to the breakdown field of (c)(ii).
the earth connection carries the charge (a large current) safely to the ground instead of through the building	✓ 1	Accept: low-resistance path to earth.

(3)

(h)

Later, the cloud base lowers to 0.80 km above the ground, with the same potential difference. Determine the new average field strength and suggest the effect on the likelihood of a strike.

PREDICTNotes

Show solution

E = 1.8 × 10⁸ / 800 = 2.3 × 10⁵ V m⁻¹ (1.5 times larger)	✓ 1	Accept 2.25 × 10⁵ V m⁻¹.
the field is closer to the breakdown value, so a strike becomes more likely	✓ 1	Must refer to the larger field.

Answer: 2.3 × 10⁵ V m⁻¹; strike more likely

(2)

(i)

Return to the original situation, with the cloud base 1.2 km above the ground. The ground is at zero potential. Determine the electric potential at a height of 400 m, and the work done by the electric field on an electron that moves from this height to the ground. Give the work in eV.

HL onlyMULTINotes

Show solution

the potential falls linearly from 0 at the ground to −1.8 × 10⁸ V at the cloud base, so V_e = −1.5 × 10⁵ × 400 = −6.0 × 10⁷ V	✓ 1	Sign required for this mark.
work done by the field = 6.0 × 10⁷ eV (positive: the electron moves to higher potential)	✓ 1	Accept 9.6 × 10⁻¹² J if given as well as eV. Allow ECF for the value of V_e.

Answer: −6.0 × 10⁷ V; 6.0 × 10⁷ eV

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
free electrons in the ground are repelled by the negative charge on the cloud base	✓ 1	Must refer to electrons.
they move away (deeper / sideways into the Earth), leaving the surface below the cloud with a positive charge / charging by induction	✓ 1	Do not accept positive charge moving up.
Part (b)
straight, vertical, parallel and equally spaced lines in the central region	✓ 1	At least four lines.
arrows pointing upwards, from the ground to the cloud base	✓ 1	Arrows from positive to negative.
lines curving outwards (bulging) near the edges of the cloud	✓ 1	Lines must still start and end on the two surfaces.
Part (c)(i)
E = V/d = 1.8 × 10⁸ / 1200 = 1.5 × 10⁵ V m⁻¹	✓ 1	Accept N C⁻¹.
Part (c)(ii)
the average field is 20 times smaller than the breakdown field (1.5 × 10⁵ < 3.0 × 10⁶ V m⁻¹)	✓ 1	Numerical comparison required. Allow ECF from (c)(i).
so breakdown can only start in small regions where the field is much stronger than average (e.g. near sharp points or charge concentrations), not across the whole gap at once	✓ 1	Accept: the field is not uniform in reality / it is locally stronger.
Part (d)
F = eE = 1.60 × 10⁻¹⁹ × 1.5 × 10⁵ = 2.4 × 10⁻¹⁴ N	✓ 1	Allow ECF from (c)(i).
vertically downwards / towards the ground	✓ 1	Opposite to the field because the electron is negative.
Part (e)(i)
charge = area under the current–time graph	✓ 1	Stated or implied by the calculation.
½ × 0.30 × 10⁻³ × 1.0 × 10⁵ = 15 C	✓ 1	Unit conversion of ms and kA must be correct. Answer given, so working must be complete.
Part (e)(ii)
I = Δq/Δt = 15 / 0.30 × 10⁻³	✓ 1
I = 5.0 × 10⁴ A	✓ 1	Accept 50 kA. Do not accept the peak value of 1.0 × 10⁵ A.
Part (f)(i)
W = qV (from V = W/q)	✓ 1
W = 15 × 1.8 × 10⁸ = 2.7 × 10⁹ J	✓ 1	Value to at least 2 s.f.
Part (f)(ii)
3000 kWh = 3000 × 1000 × 3600 = 1.08 × 10¹⁰ J	✓ 1
the stroke transfers about a quarter of the household's annual use / about three months' worth	✓ 1	Allow ECF from (f)(i). Ratio 0.25 or equivalent.
Part (g)
field lines crowd together at the sharp point, so the field strength there is very large	✓ 1	Accept: induced charge concentrates at the point.
so the field near the point can reach the breakdown value and the air there becomes conducting / the strike is attracted to the rod	✓ 1	Link to the breakdown field of (c)(ii).
the earth connection carries the charge (a large current) safely to the ground instead of through the building	✓ 1	Accept: low-resistance path to earth.
Part (h)
E = 1.8 × 10⁸ / 800 = 2.3 × 10⁵ V m⁻¹ (1.5 times larger)	✓ 1	Accept 2.25 × 10⁵ V m⁻¹.
the field is closer to the breakdown value, so a strike becomes more likely	✓ 1	Must refer to the larger field.
Part (i)
the potential falls linearly from 0 at the ground to −1.8 × 10⁸ V at the cloud base, so V_e = −1.5 × 10⁵ × 400 = −6.0 × 10⁷ V	✓ 1	Sign required for this mark.
work done by the field = 6.0 × 10⁷ eV (positive: the electron moves to higher potential)	✓ 1	Accept 9.6 × 10⁻¹² J if given as well as eV. Allow ECF for the value of V_e.

Answers: (c)(i) 1.5 × 10⁵ V m⁻¹ · (d) 2.4 × 10⁻¹⁴ N, downwards · (e)(i) 15 C · (e)(ii) 5.0 × 10⁴ A · (f)(i) 2.7 × 10⁹ J · (f)(ii) about ¼ of a year's use (≈ 3 months) · (h) 2.3 × 10⁵ V m⁻¹; strike more likely · (i) −6.0 × 10⁷ V; 6.0 × 10⁷ eV

Exam tip (b) Field-line sketches score on details: arrows on every line, lines meeting conducting surfaces at right angles, equal spacing in a uniform region and curved lines at the edges.

More D.2 questions in the question bank

D.3 Motion in Electromagnetic Fields

+ Add to test✓ In your testCC-IBP-D3-124 marks~30 min

Separating carbon isotopes in a mass spectrometer

D.3 Motion in Electromagnetic FieldsD.2 Electric and Magnetic FieldsE.3 Radioactive DecayE.1 Structure of the AtomSL

A mass spectrometer is used to find the proportion of carbon-14 in a sample of ancient wood. Carbon atoms from the sample are made into singly charged positive ions, C⁺. The ions leave the ion source with a range of speeds and pass through a velocity selector, then through slit S into a region of uniform magnetic field (Figure 1). Each ion moves through a semicircle and lands on the detector plate.

velocity selector: plates 3.0 cm apart, magnetic field 0.050 T into the page
speed of the ions that pass through S: 4.0 × 10⁵ m s⁻¹
magnetic field beyond S: 0.25 T into the page
mass of a carbon-12 ion = 12.0 u; mass of a carbon-14 ion = 14.0 u; 1 u = 1.661 × 10⁻²⁷ kg

The motion takes place in a vacuum, and the weight of the ions is negligible.

Figure 1: the mass spectrometer, seen from above (not to scale)

(a)

Carbon-12 and carbon-14 are isotopes of carbon, which has proton number 6. State what is meant by isotopes, and give the number of neutrons in a carbon-14 nucleus.

RECALLNotes

Show solution

nuclei (atoms) with the same proton number but different nucleon numbers / different numbers of neutrons	✓ 1	Accept: same element, different number of neutrons.
14 − 6 = 8 neutrons	✓ 1

Answer: 8 neutrons

(2)

(b)(i)

Identify which plate of the velocity selector, P or Q, must be positive. Explain your answer.

REASONNotes

Show solution

the magnetic force on the positive ions moving to the right in a field into the page is upwards (left-hand rule)	✓ 1	Direction of the magnetic force must be stated.
so the electric force must be downwards / the electric field must point from P to Q: plate P is positive	✓ 1	No mark for “P” without a reason.

Answer: plate P

(2)

(b)(ii)

Determine the potential difference between the plates of the velocity selector.

CALCNotes

Show solution

qE = qvB so E = vB = 4.0 × 10⁵ × 0.050 = 2.0 × 10⁴ V m⁻¹	✓ 1
V = Ed = 2.0 × 10⁴ × 0.030 = 600 V	✓ 1	Accept 6.0 × 10² V.

Answer: 600 V

(2)

(c)

Explain why an ion that enters the velocity selector with a speed greater than 4.0 × 10⁵ m s⁻¹ does not pass through slit S.

EXPLAINNotes

Show solution

the magnetic force qvB increases with speed but the electric force qE does not, so the magnetic force is larger than the electric force	✓ 1	Must compare the two forces.
there is a resultant (upward) force, so the ion is deflected (upwards) and misses the slit / hits the upper plate	✓ 1	Allow ECF from (b)(i) for the direction.

(2)

(d)

Show that the radius of the path of a carbon-12 ion beyond S is about 0.20 m.

DERIVENotes

Show solution

magnetic force provides the centripetal force: qvB = mv²/r, so r = mv/qB	✓ 1	Starting equation must be seen.
m = 12.0 × 1.661 × 10⁻²⁷ = 1.99 × 10⁻²⁶ kg and q = 1.60 × 10⁻¹⁹ C	✓ 1
r = 1.99 × 10⁻²⁶ × 4.0 × 10⁵ / (1.60 × 10⁻¹⁹ × 0.25) = 0.199 m	✓ 1	Final value to at least 3 s.f., or with full substitution seen.

Answer: 0.199 m

(3)

(e)

Explain why the speed of an ion stays constant while it moves in the magnetic field beyond S.

EXPLAINNotes

Show solution

the magnetic force is always perpendicular to the velocity (direction of motion)	✓ 1
so it does no work on the ion (W = Fs cos 90° = 0), and the kinetic energy and speed do not change	✓ 1	“The force is centripetal” alone is not enough.

(2)

(f)

Determine the distance between the points where the carbon-12 and carbon-14 ions land on the detector plate.

MULTINotes

Show solution

r ∝ m at the same speed, so r₁₄ = 0.199 × 14.0/12.0 = 0.233 m	✓ 1	Allow ECF from (d).
separation = 2(r₁₄ − r₁₂) = 2 × (0.233 − 0.199) = 0.066 m	✓ 1	Accept 0.066–0.067 m. Using radius instead of diameter (0.033 m) scores the first mark only.

Answer: 0.066 m (6.6 cm)

(2)

(g)(i)

Figure 2 shows the radius r of the path against the mass number A for the singly charged ions carbon-12, carbon-13 and carbon-14.

Explain why the graph is a straight line through the origin.

Figure 2: radius r of the path against mass number A for singly charged carbon ions

GRAPHNotes

Show solution

r = mv/qB, and v, q and B are the same for all the ions, so r ∝ m	✓ 1	Must state which quantities are constant, including the speed (from the selector).
the mass of an ion is (very nearly) proportional to its mass number, m ≈ A u, so r ∝ A	✓ 1

(2)

(g)(ii)

Some carbon atoms become doubly charged ions, C²⁺. These ions also pass through the velocity selector and enter the magnetic field. On Figure 2, draw the line that shows r against A for C²⁺ ions.

SKETCHNotes

Show solution

straight line through the origin	✓ 1	Ruled, below the original line.
with half the gradient of the original line, e.g. passing through r ≈ 0.10 m at A = 12	✓ 1	The selected speed is unchanged because q cancels in qE = qvB.

Answer: line through the origin with half the gradient

(2)

(h)

The velocity selector is now removed. Instead, the C⁺ ions are accelerated from rest through a potential difference of 10.0 kV before entering the same 0.25 T field.

Determine the new distance between the points where the carbon-12 and carbon-14 ions land, and compare it with your answer to (f).

MULTINotes

Show solution

½mv² = qV: v₁₂ = 4.0 × 10⁵ m s⁻¹ and v₁₄ = 3.7 × 10⁵ m s⁻¹ / heavier ions are slower, v ∝ 1/√m	✓ 1	Accept the result r = √(2mV/q)/B, so r ∝ √m.
r₁₂ = 0.200 m and r₁₄ = 0.216 m (ratio √(14/12) = 1.08)	✓ 1
separation = 2 × (0.216 − 0.200) = 0.032 m, about half of the separation in (f), so the isotopes are harder to separate	✓ 1	Allow ECF from (f) for the comparison.

Answer: 0.032 m, about half the separation with the selector

(3)

(i)

In living wood the ratio (number of carbon-14 atoms)/(number of carbon-12 atoms) is 1.2 × 10⁻¹². For the ancient wood the spectrometer gives a ratio of 3.0 × 10⁻¹³. The half-life of carbon-14 is 5730 years. Determine the age of the wood.

CALCNotes

Show solution

3.0 × 10⁻¹³ / 1.2 × 10⁻¹² = ¼, which is 2 half-lives	✓ 1	Assume the number of carbon-12 atoms is unchanged.
age = 2 × 5730 = 1.1 × 10⁴ years	✓ 1	Accept 11 460 years.

Answer: 1.1 × 10⁴ years

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
nuclei (atoms) with the same proton number but different nucleon numbers / different numbers of neutrons	✓ 1	Accept: same element, different number of neutrons.
14 − 6 = 8 neutrons	✓ 1
Part (b)(i)
the magnetic force on the positive ions moving to the right in a field into the page is upwards (left-hand rule)	✓ 1	Direction of the magnetic force must be stated.
so the electric force must be downwards / the electric field must point from P to Q: plate P is positive	✓ 1	No mark for “P” without a reason.
Part (b)(ii)
qE = qvB so E = vB = 4.0 × 10⁵ × 0.050 = 2.0 × 10⁴ V m⁻¹	✓ 1
V = Ed = 2.0 × 10⁴ × 0.030 = 600 V	✓ 1	Accept 6.0 × 10² V.
Part (c)
the magnetic force qvB increases with speed but the electric force qE does not, so the magnetic force is larger than the electric force	✓ 1	Must compare the two forces.
there is a resultant (upward) force, so the ion is deflected (upwards) and misses the slit / hits the upper plate	✓ 1	Allow ECF from (b)(i) for the direction.
Part (d)
magnetic force provides the centripetal force: qvB = mv²/r, so r = mv/qB	✓ 1	Starting equation must be seen.
m = 12.0 × 1.661 × 10⁻²⁷ = 1.99 × 10⁻²⁶ kg and q = 1.60 × 10⁻¹⁹ C	✓ 1
r = 1.99 × 10⁻²⁶ × 4.0 × 10⁵ / (1.60 × 10⁻¹⁹ × 0.25) = 0.199 m	✓ 1	Final value to at least 3 s.f., or with full substitution seen.
Part (e)
the magnetic force is always perpendicular to the velocity (direction of motion)	✓ 1
so it does no work on the ion (W = Fs cos 90° = 0), and the kinetic energy and speed do not change	✓ 1	“The force is centripetal” alone is not enough.
Part (f)
r ∝ m at the same speed, so r₁₄ = 0.199 × 14.0/12.0 = 0.233 m	✓ 1	Allow ECF from (d).
separation = 2(r₁₄ − r₁₂) = 2 × (0.233 − 0.199) = 0.066 m	✓ 1	Accept 0.066–0.067 m. Using radius instead of diameter (0.033 m) scores the first mark only.
Part (g)(i)
r = mv/qB, and v, q and B are the same for all the ions, so r ∝ m	✓ 1	Must state which quantities are constant, including the speed (from the selector).
the mass of an ion is (very nearly) proportional to its mass number, m ≈ A u, so r ∝ A	✓ 1
Part (g)(ii)
straight line through the origin	✓ 1	Ruled, below the original line.
with half the gradient of the original line, e.g. passing through r ≈ 0.10 m at A = 12	✓ 1	The selected speed is unchanged because q cancels in qE = qvB.
Part (h)
½mv² = qV: v₁₂ = 4.0 × 10⁵ m s⁻¹ and v₁₄ = 3.7 × 10⁵ m s⁻¹ / heavier ions are slower, v ∝ 1/√m	✓ 1	Accept the result r = √(2mV/q)/B, so r ∝ √m.
r₁₂ = 0.200 m and r₁₄ = 0.216 m (ratio √(14/12) = 1.08)	✓ 1
separation = 2 × (0.216 − 0.200) = 0.032 m, about half of the separation in (f), so the isotopes are harder to separate	✓ 1	Allow ECF from (f) for the comparison.
Part (i)
3.0 × 10⁻¹³ / 1.2 × 10⁻¹² = ¼, which is 2 half-lives	✓ 1	Assume the number of carbon-12 atoms is unchanged.
age = 2 × 5730 = 1.1 × 10⁴ years	✓ 1	Accept 11 460 years.

Answers: (a) 8 neutrons · (b)(i) plate P · (b)(ii) 600 V · (d) 0.199 m · (f) 0.066 m (6.6 cm) · (g)(ii) line through the origin with half the gradient · (h) 0.032 m, about half the separation with the selector · (i) 1.1 × 10⁴ years

Exam tip (d) The equation r = mv/qB is not in the data booklet. In a “show that” part, start from qvB = mv²/r; that line usually earns a mark of its own.

More D.3 questions in the question bank

+ Add to test✓ In your testCC-IBP-D3-222 marks~25 min

A laboratory railgun: forces on a current

D.3 Motion in Electromagnetic FieldsA.1 KinematicsB.5 Current and CircuitsA.3 Work, Energy and PowerSL

Figure 1 is a plan view of a small laboratory railgun. Two straight, parallel, horizontal metal rails are 0.10 m apart and are connected to a supply. A metal bar of mass 25 g rests across the rails at right angles to them and completes the circuit. The current in the circuit is 120 A. The rails lie in a uniform vertical magnetic field of 0.80 T, directed into the page in Figure 1.

The bar starts from rest and leaves the end of the rails after moving 1.5 m. Friction and air resistance are negligible, and the bar stays at right angles to the rails.

Figure 1: the railgun seen from above (not to scale)

(a)

Determine the direction of the magnetic force on the bar. Explain how you decided.

REASONNotes

Show solution

Fleming's left-hand rule: first finger along the field (into the page), second finger along the current in the bar (down the page)	✓ 1	Must name or describe the rule used.
force is to the right, along the rails, away from the supply	✓ 1

Answer: to the right, along the rails

(2)

(b)

Calculate the magnitude of the force on the bar and its acceleration.

CALCNotes

Show solution

F = BIL = 0.80 × 120 × 0.10 = 9.6 N	✓ 1
a = F/m = 9.6/0.025 = 3.8 × 10² m s⁻²	✓ 1	Accept 384 m s⁻². Allow ECF for F.

Answer: 9.6 N; 3.8 × 10² m s⁻²

(2)

(c)(i)

Calculate the speed of the bar as it leaves the rails.

CALCNotes

Show solution

v² = u² + 2as = 0 + 2 × 384 × 1.5	✓ 1	Allow ECF from (b).
v = 34 m s⁻¹	✓ 1	Accept 33.9 m s⁻¹.

Answer: 34 m s⁻¹

(2)

(c)(ii)

Show that the bar takes about 90 ms to travel along the rails.

CALCNotes

Show solution

t = v/a = 33.9/384 or t = √(2s/a) = √(2 × 1.5/384)	✓ 1	Any correct suvat route.
t = 0.088 s = 88 ms	✓ 1	At least 2 s.f. required.

Answer: 88 ms

(2)

(d)

On Figure 2, sketch the velocity–time graph of the bar while it is on the rails.

The supply is then replaced by one whose current starts at 120 A and decreases steadily while the bar moves. On the same axes, sketch the graph for this case, also ending when the bar leaves the rails.

Figure 2: axes for the velocity–time graph of the bar

SKETCHNotes

Show solution

first graph: straight line from the origin ending at about 88 ms and 34 m s⁻¹	✓ 1	Allow ECF from (c).
second graph: starts at the origin with the same initial gradient as the first	✓ 1	Tangent at t = 0 coincides with the first line.
second graph: gradient decreasing, ending later than 88 ms and at a lower speed than 34 m s⁻¹	✓ 1	The bar still travels 1.5 m (equal areas), so it must take longer and leave more slowly.

(3)

(e)(i)

Calculate the kinetic energy of the bar as it leaves the rails.

CALCNotes

Show solution

E_k = ½ × 0.025 × 33.9² = 14 J

✓ 1

Accept 14.4 J. Allow ECF from (c)(i); or E_k = Fs = 9.6 × 1.5.

Answer: 14 J

(1)

(e)(ii)

The average potential difference across the supply while the bar moves is 12 V. Determine the efficiency of the railgun.

MULTINotes

Show solution

power input P = IV = 120 × 12 = 1440 W	✓ 1
energy input = Pt = 1440 × 0.088 = 127 J	✓ 1	Allow ECF from (c)(ii); using 0.090 s gives 130 J.
efficiency = 14.4/127 = 0.11 (11 %)	✓ 1	Accept 0.11–0.12. Allow ECF.

Answer: 0.11 (11 %)

(3)

(f)

The two rails carry currents in opposite directions. Explain why the rails repel each other.

EXPLAINNotes

Show solution

each rail is in the magnetic field produced by the current in the other rail (right-hand grip rule)	✓ 1
applying the left-hand rule to opposite currents gives a force on each rail directed away from the other rail	✓ 1	Accept: currents in opposite directions repel, if linked to the field of the other rail.

(2)

(g)

Calculate the force per unit length between the rails, and comment on its size compared with the force on the bar.

CALCNotes

Show solution

F/L = μ₀I₁I₂/2πr = 4π × 10⁻⁷ × 120²/(2π × 0.10) = 0.029 N m⁻¹	✓ 1	Accept 2.9 × 10⁻² N m⁻¹.
over 1.5 m this is about 0.04 N, roughly 200 times smaller than the 9.6 N force on the bar, so it is negligible here	✓ 1	Any valid numerical comparison. Allow ECF from (b).

Answer: 0.029 N m⁻¹; negligible

(2)

(h)

The railgun is rebuilt so that the magnetic field is doubled and the current is halved. Predict the effect on the exit speed of the bar and on the efficiency. Assume that the main energy loss is heating in the resistance of the circuit, which does not change.

PREDICTNotes

Show solution

F = BIL is unchanged (2 × ½ = 1)	✓ 1
so the acceleration, exit speed (34 m s⁻¹) and time on the rails are unchanged	✓ 1	Allow ECF from (c).
heating power I²R falls to a quarter, so less energy is wasted and the efficiency increases	✓ 1	Must refer to I²R or the smaller current.

Answer: same exit speed; higher efficiency

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
Fleming's left-hand rule: first finger along the field (into the page), second finger along the current in the bar (down the page)	✓ 1	Must name or describe the rule used.
force is to the right, along the rails, away from the supply	✓ 1
Part (b)
F = BIL = 0.80 × 120 × 0.10 = 9.6 N	✓ 1
a = F/m = 9.6/0.025 = 3.8 × 10² m s⁻²	✓ 1	Accept 384 m s⁻². Allow ECF for F.
Part (c)(i)
v² = u² + 2as = 0 + 2 × 384 × 1.5	✓ 1	Allow ECF from (b).
v = 34 m s⁻¹	✓ 1	Accept 33.9 m s⁻¹.
Part (c)(ii)
t = v/a = 33.9/384 or t = √(2s/a) = √(2 × 1.5/384)	✓ 1	Any correct suvat route.
t = 0.088 s = 88 ms	✓ 1	At least 2 s.f. required.
Part (d)
first graph: straight line from the origin ending at about 88 ms and 34 m s⁻¹	✓ 1	Allow ECF from (c).
second graph: starts at the origin with the same initial gradient as the first	✓ 1	Tangent at t = 0 coincides with the first line.
second graph: gradient decreasing, ending later than 88 ms and at a lower speed than 34 m s⁻¹	✓ 1	The bar still travels 1.5 m (equal areas), so it must take longer and leave more slowly.
Part (e)(i)
E_k = ½ × 0.025 × 33.9² = 14 J	✓ 1	Accept 14.4 J. Allow ECF from (c)(i); or E_k = Fs = 9.6 × 1.5.
Part (e)(ii)
power input P = IV = 120 × 12 = 1440 W	✓ 1
energy input = Pt = 1440 × 0.088 = 127 J	✓ 1	Allow ECF from (c)(ii); using 0.090 s gives 130 J.
efficiency = 14.4/127 = 0.11 (11 %)	✓ 1	Accept 0.11–0.12. Allow ECF.
Part (f)
each rail is in the magnetic field produced by the current in the other rail (right-hand grip rule)	✓ 1
applying the left-hand rule to opposite currents gives a force on each rail directed away from the other rail	✓ 1	Accept: currents in opposite directions repel, if linked to the field of the other rail.
Part (g)
F/L = μ₀I₁I₂/2πr = 4π × 10⁻⁷ × 120²/(2π × 0.10) = 0.029 N m⁻¹	✓ 1	Accept 2.9 × 10⁻² N m⁻¹.
over 1.5 m this is about 0.04 N, roughly 200 times smaller than the 9.6 N force on the bar, so it is negligible here	✓ 1	Any valid numerical comparison. Allow ECF from (b).
Part (h)
F = BIL is unchanged (2 × ½ = 1)	✓ 1
so the acceleration, exit speed (34 m s⁻¹) and time on the rails are unchanged	✓ 1	Allow ECF from (c).
heating power I²R falls to a quarter, so less energy is wasted and the efficiency increases	✓ 1	Must refer to I²R or the smaller current.

Answers: (a) to the right, along the rails · (b) 9.6 N; 3.8 × 10² m s⁻² · (c)(i) 34 m s⁻¹ · (c)(ii) 88 ms · (e)(i) 14 J · (e)(ii) 0.11 (11 %) · (g) 0.029 N m⁻¹; negligible · (h) same exit speed; higher efficiency

Common error (d) A common error is to end the second graph at the same time as the first. With a smaller force the bar takes longer to cover the same 1.5 m, so the line must extend further to the right.

More D.3 questions in the question bank

D.4 Induction

+ Add to test✓ In your testCC-IBP-D4-125 marks~30 min

A bicycle hub dynamo powering a lamp

D.4 InductionA.4 Rigid Body MechanicsB.5 Current and CircuitsC.1 Simple Harmonic MotionHL

A bicycle lamp is powered by a dynamo built into the hub of the front wheel. The wheel has a radius of 0.35 m and rolls along the road without slipping.

In a simplified model of the dynamo, a flat coil of 200 turns, each of area 4.0 cm², is geared so that it makes 35 revolutions for every revolution of the wheel. It rotates about an axis perpendicular to a uniform magnetic field of strength 0.12 T (Figure 1). The coil is connected to a lamp of constant resistance 12 Ω. The resistance of the coil is negligible.

The bicycle moves at a constant speed of 6.0 m s⁻¹.

Figure 1: the front wheel and (right) the simplified model of the dynamo coil (not to scale)

Figure 2: axes for the emf of the coil (for parts (d) and (h)(i))

(a)

Calculate the maximum flux linkage of the coil.

CALCNotes

Show solution

NΦ = NBA = 200 × 0.12 × 4.0 × 10⁻⁴ = 9.6 × 10⁻³ Wb

✓ 1

Accept Wb turns. Award 0 if the area is used in cm² (gives 96 Wb).

Answer: 9.6 × 10⁻³ Wb

(1)

(b)(i)

Show that the angular speed of the coil is 600 rad s⁻¹.

CALCNotes

Show solution

angular speed of the wheel ω = v/r = 6.0/0.35 = 17.1 rad s⁻¹	✓ 1	Rolling without slipping. Allow the frequency of the wheel 2.73 Hz.
coil: 35 × 17.14 = 600 rad s⁻¹	✓ 1	Must see the factor 35 applied to an unrounded value or a value given to at least 3 s.f. (35 × 17.1 = 599 is acceptable).

Answer: 600 rad s⁻¹

(2)

(b)(ii)

Show that the peak emf induced in the coil is about 5.8 V.

CALCNotes

Show solution

ε₀ = NBAω	✓ 1	Accept 2πfNBA with f = 95.5 Hz.
= 9.6 × 10⁻³ × 600 = 5.76 V	✓ 1	Answer must be given to at least 3 s.f. or with full substitution seen. Allow ECF from (a).

Answer: 5.76 V

(2)

(c)

State the orientation of the plane of the coil relative to the magnetic field when the emf is zero. Explain your answer.

EXPLAINNotes

Show solution

plane of the coil perpendicular to the field / normal to the coil parallel to the field	✓ 1	Accept “coil facing the poles”.
flux linkage is at a maximum there, so its rate of change is zero, so by Faraday's law the emf is zero	✓ 1	Must refer to the rate of change of flux (linkage), not just to “maximum flux”.

(2)

(d)

At t = 0 the plane of the coil is perpendicular to the magnetic field. On the axes of Figure 2, sketch a graph showing how the emf ε varies with time t from t = 0 to t = 25 ms while the bicycle moves at 6.0 m s⁻¹.

SKETCHNotes

Show solution

sinusoidal curve starting at ε = 0 at t = 0	✓ 1	A sine shape, not a cosine. Either initial sign is acceptable.
peak value of about 5.8 V (between 5.5 V and 6.0 V) in both directions	✓ 1	Allow ECF from (b)(ii).
period of about 10.5 ms (between 10 ms and 11 ms), so about 2.4 cycles in 25 ms	✓ 1	T = 2π/ω = 2π/600 = 10.5 ms. Look for zeros near 5.2, 10.5, 15.7 and 20.9 ms.

Answer: sine curve, amplitude 5.8 V, period 10.5 ms

(3)

(e)(i)

The flux linkage of the coil varies with time in the same way as the displacement of an object that performs simple harmonic motion. State the phase difference between the flux linkage and the emf, and identify the quantity in simple harmonic motion whose variation with time has the same form as the emf.

REASONNotes

Show solution

phase difference of π/2 rad / a quarter of a cycle	✓ 1	Accept 90°. Do not accept π rad.
velocity (of the oscillating object)	✓ 1	Accept “the rate of change of displacement”. The emf is (minus) the rate of change of flux linkage, as velocity is the rate of change of displacement.

Answer: π/2 rad; velocity

(2)

(e)(ii)

Determine the magnitude of the emf at t = 1.0 ms.

CALCNotes

Show solution

ε = ε₀ sin ωt with ωt = 600 × 1.0 × 10⁻³ = 0.60 rad	✓ 1	Phase angle must be in radians.
ε = 5.76 × sin 0.60 = 3.3 V	✓ 1	Accept 3.2–3.3 V. Using degrees gives 0.060 V: award the first mark only if 0.60 rad is seen. Allow ECF from (b)(ii).

Answer: 3.3 V

(2)

(f)(i)

Calculate the peak current in the lamp and the peak power dissipated in the lamp.

CALCNotes

Show solution

I₀ = ε₀/R = 5.76/12 = 0.48 A	✓ 1	Allow ECF from (b)(ii). Using 5.8 V gives 0.48 A.
P₀ = I₀²R = 0.48² × 12 = 2.8 W	✓ 1	Accept ε₀²/R or ε₀I₀; 2.76–2.80 W.

Answer: 0.48 A; 2.8 W

(2)

(f)(ii)

The mean power dissipated in the lamp is half of the peak power. All the energy delivered to the lamp is supplied by work done against the opposing torque that the dynamo exerts on the wheel. By considering the work done by a torque, determine the mean opposing torque on the wheel.

MULTINotes

Show solution

mean power = 2.76/2 = 1.38 W, and work done per second = τ × angle turned per second, so P = τω	✓ 1	Allow ECF from (f)(i). Accept a method based on one revolution: energy per revolution = τ × 2π.
τ = 1.38/17.1 = 0.081 N m	✓ 1	Using the coil's angular speed (600 rad s⁻¹) gives the torque on the coil, 2.3 × 10⁻³ N m: award the first mark only.

Answer: 0.081 N m

(2)

(g)

The cyclist notices that it is harder to pedal at the same speed when the lamp is switched on than when it is switched off. Explain this observation.

EXPLAINNotes

Show solution

with the lamp switched on the circuit is complete, so the induced emf drives a current in the coil	✓ 1	With the lamp off there is an emf but no current.
by Lenz's law the current opposes the change producing it / the force on the current-carrying coil in the field produces a torque opposing the rotation	✓ 1
the cyclist must do (extra) work against this opposing torque, and this work becomes the electrical energy transferred to the lamp	✓ 1	A statement about direction alone does not earn this mark; an energy argument is needed.

(3)

(h)(i)

The cyclist slows down to a constant speed of 3.0 m s⁻¹.

On the axes of Figure 2, sketch, with a dashed line, the variation of the emf with time at this speed, again with the plane of the coil perpendicular to the field at t = 0.

SKETCHNotes

Show solution

sinusoidal curve starting at zero with peak value about 2.9 V (half the original)	✓ 1	Allow ECF from (d).
period about 21 ms (double the original), so about 1.2 cycles in 25 ms	✓ 1	Both changes are needed for full marks; each change earns one mark.

Answer: amplitude 2.9 V, period 21 ms

(2)

(h)(ii)

Deduce the mean power dissipated in the lamp at 3.0 m s⁻¹.

PREDICTNotes

Show solution

the peak emf halves, and power ∝ ε² for a constant resistance, so the power falls to one quarter	✓ 1	Accept a full recalculation: ε₀ = 2.88 V, P₀ = 0.69 W.
mean power = 1.38/4 = 0.35 W	✓ 1	Allow ECF from (f)(ii). Accept 0.34–0.35 W.

Answer: 0.35 W

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
NΦ = NBA = 200 × 0.12 × 4.0 × 10⁻⁴ = 9.6 × 10⁻³ Wb	✓ 1	Accept Wb turns. Award 0 if the area is used in cm² (gives 96 Wb).
Part (b)(i)
angular speed of the wheel ω = v/r = 6.0/0.35 = 17.1 rad s⁻¹	✓ 1	Rolling without slipping. Allow the frequency of the wheel 2.73 Hz.
coil: 35 × 17.14 = 600 rad s⁻¹	✓ 1	Must see the factor 35 applied to an unrounded value or a value given to at least 3 s.f. (35 × 17.1 = 599 is acceptable).
Part (b)(ii)
ε₀ = NBAω	✓ 1	Accept 2πfNBA with f = 95.5 Hz.
= 9.6 × 10⁻³ × 600 = 5.76 V	✓ 1	Answer must be given to at least 3 s.f. or with full substitution seen. Allow ECF from (a).
Part (c)
plane of the coil perpendicular to the field / normal to the coil parallel to the field	✓ 1	Accept “coil facing the poles”.
flux linkage is at a maximum there, so its rate of change is zero, so by Faraday's law the emf is zero	✓ 1	Must refer to the rate of change of flux (linkage), not just to “maximum flux”.
Part (d)
sinusoidal curve starting at ε = 0 at t = 0	✓ 1	A sine shape, not a cosine. Either initial sign is acceptable.
peak value of about 5.8 V (between 5.5 V and 6.0 V) in both directions	✓ 1	Allow ECF from (b)(ii).
period of about 10.5 ms (between 10 ms and 11 ms), so about 2.4 cycles in 25 ms	✓ 1	T = 2π/ω = 2π/600 = 10.5 ms. Look for zeros near 5.2, 10.5, 15.7 and 20.9 ms.
Part (e)(i)
phase difference of π/2 rad / a quarter of a cycle	✓ 1	Accept 90°. Do not accept π rad.
velocity (of the oscillating object)	✓ 1	Accept “the rate of change of displacement”. The emf is (minus) the rate of change of flux linkage, as velocity is the rate of change of displacement.
Part (e)(ii)
ε = ε₀ sin ωt with ωt = 600 × 1.0 × 10⁻³ = 0.60 rad	✓ 1	Phase angle must be in radians.
ε = 5.76 × sin 0.60 = 3.3 V	✓ 1	Accept 3.2–3.3 V. Using degrees gives 0.060 V: award the first mark only if 0.60 rad is seen. Allow ECF from (b)(ii).
Part (f)(i)
I₀ = ε₀/R = 5.76/12 = 0.48 A	✓ 1	Allow ECF from (b)(ii). Using 5.8 V gives 0.48 A.
P₀ = I₀²R = 0.48² × 12 = 2.8 W	✓ 1	Accept ε₀²/R or ε₀I₀; 2.76–2.80 W.
Part (f)(ii)
mean power = 2.76/2 = 1.38 W, and work done per second = τ × angle turned per second, so P = τω	✓ 1	Allow ECF from (f)(i). Accept a method based on one revolution: energy per revolution = τ × 2π.
τ = 1.38/17.1 = 0.081 N m	✓ 1	Using the coil's angular speed (600 rad s⁻¹) gives the torque on the coil, 2.3 × 10⁻³ N m: award the first mark only.
Part (g)
with the lamp switched on the circuit is complete, so the induced emf drives a current in the coil	✓ 1	With the lamp off there is an emf but no current.
by Lenz's law the current opposes the change producing it / the force on the current-carrying coil in the field produces a torque opposing the rotation	✓ 1
the cyclist must do (extra) work against this opposing torque, and this work becomes the electrical energy transferred to the lamp	✓ 1	A statement about direction alone does not earn this mark; an energy argument is needed.
Part (h)(i)
sinusoidal curve starting at zero with peak value about 2.9 V (half the original)	✓ 1	Allow ECF from (d).
period about 21 ms (double the original), so about 1.2 cycles in 25 ms	✓ 1	Both changes are needed for full marks; each change earns one mark.
Part (h)(ii)
the peak emf halves, and power ∝ ε² for a constant resistance, so the power falls to one quarter	✓ 1	Accept a full recalculation: ε₀ = 2.88 V, P₀ = 0.69 W.
mean power = 1.38/4 = 0.35 W	✓ 1	Allow ECF from (f)(ii). Accept 0.34–0.35 W.

Answers: (a) 9.6 × 10⁻³ Wb · (b)(i) 600 rad s⁻¹ · (b)(ii) 5.76 V · (d) sine curve, amplitude 5.8 V, period 10.5 ms · (e)(i) π/2 rad; velocity · (e)(ii) 3.3 V · (f)(i) 0.48 A; 2.8 W · (f)(ii) 0.081 N m · (h)(i) amplitude 2.9 V, period 21 ms · (h)(ii) 0.35 W

Common error (c) A common error is to say that the emf is zero when the flux is zero. The emf depends on how fast the flux linkage changes, and that rate is zero where the flux linkage is greatest.

Exam tip (e)(ii) In ε = ε₀ sin ωt the phase angle ωt is in radians, exactly as in the SHM equations. Set your calculator to radians before evaluating the sine.

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Magnetic brakes on a roller coaster

D.4 InductionA.3 Work, Energy and PowerD.3 Motion in Electromagnetic FieldsB.5 Current and CircuitsHL

A roller-coaster car of mass 4000 kg is slowed by magnetic brakes. A thin vertical copper fin is fixed underneath the car. At the end of the ride the car enters a horizontal braking section, 50 m long, at a speed of 25 m s⁻¹. Pairs of permanent magnets are fixed close together along the whole of this section, so that at every instant the fin is between the poles of 16 magnet pairs.

In a simple model, at each magnet pair a vertical strip of the fin of height 0.25 m moves at right angles to a uniform magnetic field of strength 0.80 T, and the induced current flows round a path in the fin of total resistance 0.80 mΩ (Figure 1). The 16 regions act independently and identically. Ignore friction and air resistance in the braking section.

Figure 1: the fin passing one pair of magnets, seen from the side (not to scale)

Figure 2: speed v of the car against time t while it is in the braking section (t = 0 when it enters)

(a)(i)

Show that the emf induced in the strip of fin at one magnet pair is 5.0 V when the car enters the braking section.

CALCNotes

Show solution

ε = BvL = 0.80 × 25 × 0.25 = 5.0 V

✓ 1

The substitution must be seen.

Answer: 5.0 V

(1)

(a)(ii)

Calculate the induced current in the fin at one magnet pair at this instant.

CALCNotes

Show solution

I = ε/R = 5.0/(0.80 × 10⁻³) = 6.3 × 10³ A

✓ 1

Accept 6250 A. Award 0 for 6.25 A (mΩ not converted).

Answer: 6.3 × 10³ A

(1)

(b)

Determine the total braking force on the car as it enters the braking section.

CALCNotes

Show solution

force at one pair F = BIL = 0.80 × 6250 × 0.25 = 1250 N	✓ 1	Allow ECF from (a)(ii).
total = 16 × 1250 = 2.0 × 10⁴ N	✓ 1

Answer: 2.0 × 10⁴ N

(2)

(c)

Show that the total braking force F on the car is proportional to its speed v, and that F = kv with k = 800 N s m⁻¹.

DERIVENotes

Show solution

ε = BvL and I = BvL/R, so at one pair F = BIL = B²L²v/R, which is proportional to v since B, L and R are constant	✓ 1	Algebraic chain needed; a numerical check at one speed alone scores 0 for this mark.
k = 16B²L²/R = 16 × 0.80² × 0.25²/(0.80 × 10⁻³) = 800 N s m⁻¹	✓ 1	Accept 2.0 × 10⁴/25 = 800 only if proportionality has been shown.

Answer: k = 800 N s m⁻¹

(2)

(d)

Explain, with reference to Lenz's law, why the force on the fin acts opposite to the direction of motion of the car.

EXPLAINNotes

Show solution

the induced current opposes the change producing it, which is the motion of the fin through the field (the change of flux through the current path)	✓ 1	Do not accept “opposes the field” or “opposes the flux”.
so the force BIL on the current in the field acts against the velocity; a force in the direction of motion would increase the kinetic energy and the electrical energy together, breaking energy conservation	✓ 1	Accept a correct argument in terms of the direction of the force on the current from F = BIL.

(2)

(e)(i)

Figure 2 shows how the speed of the car varies with time while it is in the braking section.

Use Figure 2 to determine the deceleration of the car at t = 1.0 s, and show that your value is consistent with the result of (c).

GRAPHNotes

Show solution

tangent drawn at t = 1.0 s and its gradient found from a large triangle	✓ 1	Gradient magnitude 3.7–4.5 m s⁻².
deceleration ≈ 4.1 m s⁻²	✓ 1	Accept 3.7–4.5 m s⁻².
at t = 1.0 s, v ≈ 20.5 m s⁻¹, so a = kv/m = 800 × 20.5/4000 = 4.1 m s⁻², which agrees	✓ 1	Allow ECF from (c). Accept a speed reading of 20–21 m s⁻¹.

Answer: about 4.1 m s⁻²

(3)

(e)(ii)

Explain the shape of the graph in Figure 2.

REASONNotes

Show solution

the gradient (deceleration) decreases as time goes on	✓ 1
because the braking force is proportional to the speed, so as the car slows the force and hence the deceleration become smaller	✓ 1	Must link the decreasing force to the decreasing speed.

(2)

(f)(i)

The car leaves the braking section at a speed of 15 m s⁻¹.

Calculate the kinetic energy transferred to internal energy in the braking section.

CALCNotes

Show solution

ΔE_k = ½ × 4000 × (25² − 15²)	✓ 1	Award 0 for ½ × 4000 × 10².
= 8.0 × 10⁵ J	✓ 1

Answer: 8.0 × 10⁵ J

(2)

(f)(ii)

Use Figure 2 to determine the mean rate at which energy is dissipated in the fin while the car is in the braking section.

MULTINotes

Show solution

time in section read from the end of the graph: 2.5–2.6 s	✓ 1
P = 8.0 × 10⁵/2.55 = 3.1 × 10⁵ W	✓ 1	Accept 3.0–3.2 × 10⁵ W. Allow ECF from (f)(i).

Answer: 3.1 × 10⁵ W

(2)

(g)

Show that, as the car enters the braking section, the rate at which electrical energy is dissipated in the fin is equal to the rate at which the car loses kinetic energy.

MULTINotes

Show solution

rate of loss of kinetic energy = Fv = 2.0 × 10⁴ × 25 = 5.0 × 10⁵ W	✓ 1	Allow ECF from (b).
electrical power = 16I²R = 16 × 6250² × 0.80 × 10⁻³ = 5.0 × 10⁵ W, so the two are equal	✓ 1	Accept 16εI = 16 × 5.0 × 6250. Both values must be calculated.

Answer: both 5.0 × 10⁵ W

(2)

(h)

Explain why the magnetic brakes alone cannot bring the car to rest and hold it there, so that a friction brake is also needed at the end of the section.

EXPLAINNotes

Show solution

the braking force is proportional to speed, so it becomes very small as the car slows and the speed only approaches zero gradually	✓ 1
when the car is at rest there is no change of flux, no induced emf or current, and so no force to hold it (for example on a slope)	✓ 1

(2)

(i)

The magnets are replaced with magnets that produce a field of twice the strength. Deduce the deceleration of the car as it enters the braking section, and suggest why this design may be unsuitable.

PREDICTNotes

Show solution

emf and current both double, so the force BIL increases by a factor of 4 (F ∝ B²): a = 4 × 5.0 = 20 m s⁻²	✓ 1	Accept 8.0 × 10⁴/4000 = 20 m s⁻². Allow ECF from (b).
about 2g, a deceleration that is uncomfortable or unsafe for the riders	✓ 1	Accept any sensible consequence of the large deceleration, e.g. strain on the car or the riders' necks.

Answer: 20 m s⁻²

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
ε = BvL = 0.80 × 25 × 0.25 = 5.0 V	✓ 1	The substitution must be seen.
Part (a)(ii)
I = ε/R = 5.0/(0.80 × 10⁻³) = 6.3 × 10³ A	✓ 1	Accept 6250 A. Award 0 for 6.25 A (mΩ not converted).
Part (b)
force at one pair F = BIL = 0.80 × 6250 × 0.25 = 1250 N	✓ 1	Allow ECF from (a)(ii).
total = 16 × 1250 = 2.0 × 10⁴ N	✓ 1
Part (c)
ε = BvL and I = BvL/R, so at one pair F = BIL = B²L²v/R, which is proportional to v since B, L and R are constant	✓ 1	Algebraic chain needed; a numerical check at one speed alone scores 0 for this mark.
k = 16B²L²/R = 16 × 0.80² × 0.25²/(0.80 × 10⁻³) = 800 N s m⁻¹	✓ 1	Accept 2.0 × 10⁴/25 = 800 only if proportionality has been shown.
Part (d)
the induced current opposes the change producing it, which is the motion of the fin through the field (the change of flux through the current path)	✓ 1	Do not accept “opposes the field” or “opposes the flux”.
so the force BIL on the current in the field acts against the velocity; a force in the direction of motion would increase the kinetic energy and the electrical energy together, breaking energy conservation	✓ 1	Accept a correct argument in terms of the direction of the force on the current from F = BIL.
Part (e)(i)
tangent drawn at t = 1.0 s and its gradient found from a large triangle	✓ 1	Gradient magnitude 3.7–4.5 m s⁻².
deceleration ≈ 4.1 m s⁻²	✓ 1	Accept 3.7–4.5 m s⁻².
at t = 1.0 s, v ≈ 20.5 m s⁻¹, so a = kv/m = 800 × 20.5/4000 = 4.1 m s⁻², which agrees	✓ 1	Allow ECF from (c). Accept a speed reading of 20–21 m s⁻¹.
Part (e)(ii)
the gradient (deceleration) decreases as time goes on	✓ 1
because the braking force is proportional to the speed, so as the car slows the force and hence the deceleration become smaller	✓ 1	Must link the decreasing force to the decreasing speed.
Part (f)(i)
ΔE_k = ½ × 4000 × (25² − 15²)	✓ 1	Award 0 for ½ × 4000 × 10².
= 8.0 × 10⁵ J	✓ 1
Part (f)(ii)
time in section read from the end of the graph: 2.5–2.6 s	✓ 1
P = 8.0 × 10⁵/2.55 = 3.1 × 10⁵ W	✓ 1	Accept 3.0–3.2 × 10⁵ W. Allow ECF from (f)(i).
Part (g)
rate of loss of kinetic energy = Fv = 2.0 × 10⁴ × 25 = 5.0 × 10⁵ W	✓ 1	Allow ECF from (b).
electrical power = 16I²R = 16 × 6250² × 0.80 × 10⁻³ = 5.0 × 10⁵ W, so the two are equal	✓ 1	Accept 16εI = 16 × 5.0 × 6250. Both values must be calculated.
Part (h)
the braking force is proportional to speed, so it becomes very small as the car slows and the speed only approaches zero gradually	✓ 1
when the car is at rest there is no change of flux, no induced emf or current, and so no force to hold it (for example on a slope)	✓ 1
Part (i)
emf and current both double, so the force BIL increases by a factor of 4 (F ∝ B²): a = 4 × 5.0 = 20 m s⁻²	✓ 1	Accept 8.0 × 10⁴/4000 = 20 m s⁻². Allow ECF from (b).
about 2g, a deceleration that is uncomfortable or unsafe for the riders	✓ 1	Accept any sensible consequence of the large deceleration, e.g. strain on the car or the riders' necks.

Answers: (a)(i) 5.0 V · (a)(ii) 6.3 × 10³ A · (b) 2.0 × 10⁴ N · (c) k = 800 N s m⁻¹ · (e)(i) about 4.1 m s⁻² · (f)(i) 8.0 × 10⁵ J · (f)(ii) 3.1 × 10⁵ W · (g) both 5.0 × 10⁵ W · (i) 20 m s⁻²

Common error (f)(i) A common error is to use the change in speed, ½m(25 − 15)². Kinetic energy depends on v², so find the difference of the two squares.

More D.4 questions in the question bank

Theme E · Nuclear and quantum physics

● home subchapter · ○ linked subchapter · last column: links to other themes

Setup	E.1	E.2	E.3	E.4	E.5	Other
Firing alpha particles at a thin foil in a teaching lab	●		○			D.2 A.3
Inside a fluorescent lamp: mercury atoms and a phosphor	●					C.2 A.3 B.5
A potassium photocell: stopping voltage, the Planck constant and the work function	○	●				B.5 C.2
Electron waves in a transmission electron microscope		●				D.3 C.3 A.2
A PET scan tracer: making, transporting and detecting fluorine-18	○		●	○		—
Americium-241 in a smoke detector: decay, binding energy and the ionisation current	○		●			B.5 D.2
The reactor of a nuclear submarine			○	●		A.3 A.2
Oklo: a natural nuclear reactor two billion years ago			○	●		B.1 A.3
Fusion in the Sun's core			○		●	B.1
Betelgeuse: measuring a red supergiant					●	B.1 C.5

E.1 Structure of the Atom

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Firing alpha particles at a thin foil in a teaching lab

E.1 Structure of the AtomD.2 Electric and Magnetic FieldsA.3 Work, Energy and PowerE.3 Radioactive DecaySL

In a school laboratory, α particles are fired at a thin gold foil, ¹⁹⁷₇₉Au, inside an evacuated chamber (Figure 1). The α particles come from a sealed source of americium-241 and reach the foil with a kinetic energy of 5.0 MeV. A detector counts the α particles scattered through an angle θ from the direction of the beam.

Mass of an α particle = 6.64 × 10⁻²⁷ kg

Figure 1: the scattering apparatus seen from above (not to scale)

Figure 2: count rate against scattering angle θ (logarithmic vertical scale)

(a)

Americium-241, ²⁴¹₉₅Am, decays by α emission to an isotope of neptunium, Np. Write down the nuclear equation for this decay.

RECALLNotes

Show solution

²⁴¹₉₅Am → ²³⁷₉₃Np + …	✓ 1	Np with nucleon number 237 and proton number 93.
… + ⁴₂α	✓ 1	Accept ⁴₂He. Both numbers must be correct.

Answer: ²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α

(2)

(b)

State the number of protons and the number of neutrons in a nucleus of gold-197.

RECALLNotes

Show solution

79 protons and 118 neutrons

✓ 1

Both needed.

Answer: 79 protons, 118 neutrons

(1)

(c)

Show that the speed of the α particles as they reach the foil is about 1.6 × 10⁷ m s⁻¹.

CALCNotes

Show solution

E_k = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J	✓ 1
v = √(2E_k/m) = √(2 × 8.0 × 10⁻¹³/6.64 × 10⁻²⁷) = 1.55 × 10⁷ m s⁻¹	✓ 1	Answer to at least 3 s.f. or full substitution seen.

Answer: 1.55 × 10⁷ m s⁻¹

(2)

(d)

Explain why the experiment must be carried out in a vacuum and why the gold foil must be very thin.

EXPLAINNotes

Show solution

vacuum: α particles have a range of only a few cm in air; collisions with air molecules would stop them / change their energy and direction	✓ 1
thin foil: so that each α particle is scattered by at most one nucleus (single scattering) and loses little energy in the foil	✓ 1	Accept “so that the α particles are not absorbed in the foil” for this mark only if single scattering is not mentioned elsewhere.

(2)

(e)(i)

Figure 2 shows the count rate measured at different angles θ.

Describe what Figure 2 shows, and state what each feature reveals about the structure of the gold atom.

DATANotes

Show solution

the count rate is largest at small angles, so most α particles are deflected very little: the atom is mostly empty space	✓ 1
the count rate falls steeply as θ increases, but some α particles are scattered through large angles: the positive charge is concentrated in a very small region	✓ 1
some are scattered back (θ > 90°): this small region contains most of the mass of the atom	✓ 1	Award marks for features paired with conclusions only.

(3)

(e)(ii)

Rutherford's model predicts that the count rate is proportional to 1/sin⁴(θ/2). Use Figure 2 to show that the count rates at θ = 60° and θ = 120° are consistent with this prediction.

GRAPHNotes

Show solution

predicted ratio = sin⁴60°/sin⁴30° = 0.5625/0.0625 = 9.0	✓ 1
readings about 160 h⁻¹ and 18 h⁻¹, giving a ratio of about 9 — consistent	✓ 1	Accept readings 150–170 h⁻¹ and 16–20 h⁻¹ (ratio 8–10).

Answer: ratio ≈ 9 in both cases

(2)

(f)

Calculate the magnitude of the electric force between an α particle and a gold nucleus when their centres are 1.0 × 10⁻¹³ m apart.

CALCNotes

Show solution

F = kq₁q₂/r² with q₁ = 2e and q₂ = 79e	✓ 1	Award 0 for this mark if 4 and 197 are used for the charges.
F = 8.99 × 10⁹ × 2 × 79 × (1.60 × 10⁻¹⁹)²/(1.0 × 10⁻¹³)² = 3.6 N	✓ 1

Answer: 3.6 N

(2)

(g)

Show that the distance of closest approach of an α particle to a gold nucleus in a head-on collision is about 4.5 × 10⁻¹⁴ m. Assume that the gold nucleus does not move.

HL onlyDERIVENotes

Show solution

at closest approach the α particle is momentarily at rest, so its initial kinetic energy equals the electric potential energy: E_k = kq₁q₂/d	✓ 1	Energy conservation must be stated or clearly used.
d = 8.99 × 10⁹ × 2 × 79 × (1.60 × 10⁻¹⁹)²/(8.0 × 10⁻¹³)	✓ 1	Allow ECF for E_k from (c).
d = 4.55 × 10⁻¹⁴ m	✓ 1	Answer to at least 3 s.f. (4.54 or 4.55 × 10⁻¹⁴ m); 4.5 × 10⁻¹⁴ m alone is not enough.

Answer: 4.55 × 10⁻¹⁴ m

(3)

(h)(i)

The gold foil is replaced by an aluminium foil, ²⁷₁₃Al. The α particles have the same energy.

Calculate the distance of closest approach in a head-on collision with an aluminium nucleus.

HL onlyCALCNotes

Show solution

d = 4.55 × 10⁻¹⁴ × 13/79 = 7.5 × 10⁻¹⁵ m

✓ 1

Accept a full recalculation. Allow ECF from (g).

Answer: 7.5 × 10⁻¹⁵ m

(1)

(h)(ii)

The aluminium foil contains the same number of nuclei per unit area as the gold foil. Suggest two reasons why fewer α particles are scattered through large angles.

PREDICTNotes

Show solution

the aluminium nucleus has a smaller charge (13e rather than 79e), so the repulsive electric force at any separation is smaller and deflections are smaller	✓ 1
the aluminium nucleus has a much smaller mass (27 u, only about 7 times the α particle), so it recoils and is less able to turn the α particle back	✓ 1	Accept: the fixed-nucleus assumption is less valid.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
²⁴¹₉₅Am → ²³⁷₉₃Np + …	✓ 1	Np with nucleon number 237 and proton number 93.
… + ⁴₂α	✓ 1	Accept ⁴₂He. Both numbers must be correct.
Part (b)
79 protons and 118 neutrons	✓ 1	Both needed.
Part (c)
E_k = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J	✓ 1
v = √(2E_k/m) = √(2 × 8.0 × 10⁻¹³/6.64 × 10⁻²⁷) = 1.55 × 10⁷ m s⁻¹	✓ 1	Answer to at least 3 s.f. or full substitution seen.
Part (d)
vacuum: α particles have a range of only a few cm in air; collisions with air molecules would stop them / change their energy and direction	✓ 1
thin foil: so that each α particle is scattered by at most one nucleus (single scattering) and loses little energy in the foil	✓ 1	Accept “so that the α particles are not absorbed in the foil” for this mark only if single scattering is not mentioned elsewhere.
Part (e)(i)
the count rate is largest at small angles, so most α particles are deflected very little: the atom is mostly empty space	✓ 1
the count rate falls steeply as θ increases, but some α particles are scattered through large angles: the positive charge is concentrated in a very small region	✓ 1
some are scattered back (θ > 90°): this small region contains most of the mass of the atom	✓ 1	Award marks for features paired with conclusions only.
Part (e)(ii)
predicted ratio = sin⁴60°/sin⁴30° = 0.5625/0.0625 = 9.0	✓ 1
readings about 160 h⁻¹ and 18 h⁻¹, giving a ratio of about 9 — consistent	✓ 1	Accept readings 150–170 h⁻¹ and 16–20 h⁻¹ (ratio 8–10).
Part (f)
F = kq₁q₂/r² with q₁ = 2e and q₂ = 79e	✓ 1	Award 0 for this mark if 4 and 197 are used for the charges.
F = 8.99 × 10⁹ × 2 × 79 × (1.60 × 10⁻¹⁹)²/(1.0 × 10⁻¹³)² = 3.6 N	✓ 1
Part (g)
at closest approach the α particle is momentarily at rest, so its initial kinetic energy equals the electric potential energy: E_k = kq₁q₂/d	✓ 1	Energy conservation must be stated or clearly used.
d = 8.99 × 10⁹ × 2 × 79 × (1.60 × 10⁻¹⁹)²/(8.0 × 10⁻¹³)	✓ 1	Allow ECF for E_k from (c).
d = 4.55 × 10⁻¹⁴ m	✓ 1	Answer to at least 3 s.f. (4.54 or 4.55 × 10⁻¹⁴ m); 4.5 × 10⁻¹⁴ m alone is not enough.
Part (h)(i)
d = 4.55 × 10⁻¹⁴ × 13/79 = 7.5 × 10⁻¹⁵ m	✓ 1	Accept a full recalculation. Allow ECF from (g).
Part (h)(ii)
the aluminium nucleus has a smaller charge (13e rather than 79e), so the repulsive electric force at any separation is smaller and deflections are smaller	✓ 1
the aluminium nucleus has a much smaller mass (27 u, only about 7 times the α particle), so it recoils and is less able to turn the α particle back	✓ 1	Accept: the fixed-nucleus assumption is less valid.

Answers: (a) ²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α · (b) 79 protons, 118 neutrons · (c) 1.55 × 10⁷ m s⁻¹ · (e)(ii) ratio ≈ 9 in both cases · (f) 3.6 N · (g) 4.55 × 10⁻¹⁴ m · (h)(i) 7.5 × 10⁻¹⁵ m

Exam tip (e)(i) Pair each observation with its conclusion: few deflected → mostly empty space; large deflections → positive charge concentrated; back-scattering → most of the mass in that region.

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Inside a fluorescent lamp: mercury atoms and a phosphor

E.1 Structure of the AtomC.2 Wave ModelA.3 Work, Energy and PowerB.5 Current and CircuitsSL

A fluorescent lamp is a glass tube containing mercury vapour at low pressure. Electrons accelerated along the tube collide with mercury atoms and excite them. The excited atoms emit mostly ultraviolet photons of wavelength 254 nm, together with some visible light. A phosphor coating on the inside of the glass absorbs the ultraviolet photons and emits visible light.

Figure 1 shows four of the energy levels of a mercury atom.

Figure 1: four energy levels of a mercury atom (drawn to scale)

Figure 2: visible spectrum of the light emitted by the lamp

(a)

Show that the energy of a photon of wavelength 254 nm is about 4.9 eV.

CALCNotes

Show solution

E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/(254 × 10⁻⁹) = 7.83 × 10⁻¹⁹ J	✓ 1
7.83 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 4.89 eV	✓ 1	Answer to at least 3 s.f. Accept 1.24 × 10⁻⁶/254 × 10⁻⁹ = 4.88 eV.

Answer: 4.89 eV

(2)

(b)

Identify, using Figure 1, the transition that produces the 254 nm photons, and state the region of the electromagnetic spectrum to which they belong.

DATANotes

Show solution

from −5.55 eV to −10.44 eV (the ground state)	✓ 1	The gap is 4.89 eV. Accept an arrow drawn downwards between these levels.
ultraviolet	✓ 1

Answer: −5.55 eV → −10.44 eV; ultraviolet

(2)

(c)

Calculate the wavelength of the photon emitted when an atom makes a transition from the −2.71 eV level to the −4.98 eV level.

CALCNotes

Show solution

ΔE = 2.27 eV = 2.27 × 1.60 × 10⁻¹⁹ = 3.63 × 10⁻¹⁹ J	✓ 1
λ = hc/ΔE = 5.48 × 10⁻⁷ m	✓ 1	Accept 546–548 nm. Award 0 if a single level value is used instead of the difference.

Answer: 548 nm

(2)

(d)

State the number of different photon energies that could be emitted by atoms making downward transitions between the four levels in Figure 1.

RECALLNotes

Show solution

✓ 1

4 × 3/2 = 6.

Answer: 6

(1)

(e)(i)

Figure 2 shows the visible spectrum of the light emitted by the lamp. It contains two narrow lines, P and Q, and a broad continuous band.

Identify which line, P or Q, is produced by the transition in (c).

GRAPHNotes

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✓ 1

Q is at about 546 nm. Allow ECF from (c).

Answer: Q

(1)

(e)(ii)

Explain why the light from the mercury vapour forms narrow lines, whereas the light from the phosphor forms a broad continuous band.

EXPLAINNotes

Show solution

mercury atoms in a low-pressure gas have discrete energy levels, so they emit only photons of energies equal to the gaps, giving particular wavelengths	✓ 1
the phosphor is a solid in which the atoms interact strongly, so there are very many closely spaced energy levels and photons of a continuous range of energies are emitted	✓ 1	Accept “the phosphor is dense matter, which gives a continuous spectrum” for this mark.

(2)

(f)

White light from a filament lamp passes through a tube of cool mercury vapour and is then viewed with a spectrometer. Explain why the spectrum observed contains dark lines, including one at 254 nm.

EXPLAINNotes

Show solution

mercury atoms in the ground state absorb only photons whose energy equals a gap between levels, e.g. 4.89 eV photons with wavelength 254 nm	✓ 1	Must link the absorbed energy to a specific wavelength.
the excited atoms re-emit photons in all directions, so the intensity in the original direction is much lower at those wavelengths: dark lines	✓ 1

(2)

(g)

Determine the minimum speed an electron must have in order to raise a mercury atom from its ground state to the −5.55 eV level in a collision.

CALCNotes

Show solution

minimum kinetic energy = 4.89 eV = 4.89 × 1.60 × 10⁻¹⁹ = 7.82 × 10⁻¹⁹ J	✓ 1
v = √(2 × 7.82 × 10⁻¹⁹/9.11 × 10⁻³¹) = 1.3 × 10⁶ m s⁻¹	✓ 1

Answer: 1.3 × 10⁶ m s⁻¹

(2)

(h)(i)

The lamp takes a power of 36 W from the 230 V mains supply. Treat the supply as a constant potential difference.

Calculate the current in the lamp circuit.

CALCNotes

Show solution

I = P/V = 36/230 = 0.16 A

✓ 1

Accept 0.157 A.

Answer: 0.16 A

(1)

(h)(ii)

The lamp emits 3.0 × 10¹⁹ visible photons per second. Take the mean wavelength of these photons to be 550 nm. Show that the efficiency of the lamp is about 30 %.

MULTINotes

Show solution

energy of one photon = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/(550 × 10⁻⁹) = 3.62 × 10⁻¹⁹ J	✓ 1
light power = 3.0 × 10¹⁹ × 3.62 × 10⁻¹⁹ = 10.8 W	✓ 1
efficiency = 10.8/36 = 0.301 = 30.1 %	✓ 1	Answer to at least 3 s.f., or full working seen.

Answer: 30.1 %

(3)

(h)(iii)

In the phosphor, each 254 nm photon absorbed produces at most one visible photon of mean wavelength 550 nm. Deduce the maximum fraction of the ultraviolet energy that the phosphor can convert into visible light, and hence suggest one reason why the efficiency of the lamp is low.

REASONNotes

Show solution

fraction = (energy of visible photon)/(energy of UV photon) = 254/550 = 0.46 (or 2.26 eV/4.89 eV)	✓ 1	Accept 46 %.
more than half of the energy of each UV photon is transferred to internal energy of the phosphor (and further energy is lost heating the gas and electrodes), so the overall efficiency is low	✓ 1

Answer: 0.46

(2)

(h)(iv)

The fluorescent lamp is replaced by an LED lamp of efficiency 45 % that gives the same visible light output. Determine the power taken by the LED lamp and the current it takes from the 230 V supply.

PREDICTNotes

Show solution

power input = 10.8/0.45 = 24 W	✓ 1	Allow ECF from (h)(ii). Accept 36 × 30.1/45 = 24 W.
current = 24/230 = 0.10 A	✓ 1	Accept 0.105 A.

Answer: 24 W; 0.10 A

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/(254 × 10⁻⁹) = 7.83 × 10⁻¹⁹ J	✓ 1
7.83 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 4.89 eV	✓ 1	Answer to at least 3 s.f. Accept 1.24 × 10⁻⁶/254 × 10⁻⁹ = 4.88 eV.
Part (b)
from −5.55 eV to −10.44 eV (the ground state)	✓ 1	The gap is 4.89 eV. Accept an arrow drawn downwards between these levels.
ultraviolet	✓ 1
Part (c)
ΔE = 2.27 eV = 2.27 × 1.60 × 10⁻¹⁹ = 3.63 × 10⁻¹⁹ J	✓ 1
λ = hc/ΔE = 5.48 × 10⁻⁷ m	✓ 1	Accept 546–548 nm. Award 0 if a single level value is used instead of the difference.
Part (d)
6	✓ 1	4 × 3/2 = 6.
Part (e)(i)
Q	✓ 1	Q is at about 546 nm. Allow ECF from (c).
Part (e)(ii)
mercury atoms in a low-pressure gas have discrete energy levels, so they emit only photons of energies equal to the gaps, giving particular wavelengths	✓ 1
the phosphor is a solid in which the atoms interact strongly, so there are very many closely spaced energy levels and photons of a continuous range of energies are emitted	✓ 1	Accept “the phosphor is dense matter, which gives a continuous spectrum” for this mark.
Part (f)
mercury atoms in the ground state absorb only photons whose energy equals a gap between levels, e.g. 4.89 eV photons with wavelength 254 nm	✓ 1	Must link the absorbed energy to a specific wavelength.
the excited atoms re-emit photons in all directions, so the intensity in the original direction is much lower at those wavelengths: dark lines	✓ 1
Part (g)
minimum kinetic energy = 4.89 eV = 4.89 × 1.60 × 10⁻¹⁹ = 7.82 × 10⁻¹⁹ J	✓ 1
v = √(2 × 7.82 × 10⁻¹⁹/9.11 × 10⁻³¹) = 1.3 × 10⁶ m s⁻¹	✓ 1
Part (h)(i)
I = P/V = 36/230 = 0.16 A	✓ 1	Accept 0.157 A.
Part (h)(ii)
energy of one photon = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/(550 × 10⁻⁹) = 3.62 × 10⁻¹⁹ J	✓ 1
light power = 3.0 × 10¹⁹ × 3.62 × 10⁻¹⁹ = 10.8 W	✓ 1
efficiency = 10.8/36 = 0.301 = 30.1 %	✓ 1	Answer to at least 3 s.f., or full working seen.
Part (h)(iii)
fraction = (energy of visible photon)/(energy of UV photon) = 254/550 = 0.46 (or 2.26 eV/4.89 eV)	✓ 1	Accept 46 %.
more than half of the energy of each UV photon is transferred to internal energy of the phosphor (and further energy is lost heating the gas and electrodes), so the overall efficiency is low	✓ 1
Part (h)(iv)
power input = 10.8/0.45 = 24 W	✓ 1	Allow ECF from (h)(ii). Accept 36 × 30.1/45 = 24 W.
current = 24/230 = 0.10 A	✓ 1	Accept 0.105 A.

Answers: (a) 4.89 eV · (b) −5.55 eV → −10.44 eV; ultraviolet · (c) 548 nm · (d) 6 · (e)(i) Q · (g) 1.3 × 10⁶ m s⁻¹ · (h)(i) 0.16 A · (h)(ii) 30.1 % · (h)(iii) 0.46 · (h)(iv) 24 W; 0.10 A

Common error (c) A common error is to use the energy of one level, such as 2.71 eV, instead of the difference between two levels. The photon carries the energy gap.

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E.2 Quantum Physics

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A potassium photocell: stopping voltage, the Planck constant and the work function

E.2 Quantum PhysicsB.5 Current and CircuitsE.1 Structure of the AtomC.2 Wave ModelHL

A student investigates the photoelectric effect with the photocell circuit in Figure 1. The cathode C is coated with potassium. Light from a mercury lamp passes through a filter that transmits only one wavelength λ. The variable supply can make the anode A negative relative to the cathode. For each filter, the student increases this reverse potential difference from zero until the reading on the microammeter just falls to zero. This potential difference is the stopping voltage V_s.

λ / nm	f / 10¹⁴ Hz	V_s / V
436	6.88	0.59
405	7.41	0.82
365	?	1.14
334	8.98	1.47
313	9.58	1.71

(e = 1.60 × 10⁻¹⁹ C, c = 3.00 × 10⁸ m s⁻¹, m_e = 9.110 × 10⁻³¹ kg)

Figure 1: the photocell circuit (not to scale)

(a)

Define the work function of a metal.

RECALLNotes

Show solution

the minimum energy needed to remove an electron from the surface of the metal

✓ 1

Must include “minimum” or “least”. Do not accept “the energy of the photon needed”.

(1)

(b)

Show that the frequency of the light of wavelength 365 nm is about 8.2 × 10¹⁴ Hz.

DERIVENotes

Show solution

f = c/λ = 3.00 × 10⁸ / 365 × 10⁻⁹ = 8.22 × 10¹⁴ Hz

✓ 1

Must see the substitution OR an answer to at least 3 s.f.

Answer: 8.22 × 10¹⁴ Hz

(1)

(c)

Explain why no photoelectrons reach the anode when the reverse potential difference is equal to V_s. Hence state the relation between V_s and the maximum kinetic energy E_max of the photoelectrons.

EXPLAINNotes

Show solution

the electric field between A and C opposes the motion of the electrons, so an electron loses kinetic energy eV in moving from C towards A; at V_s even the most energetic electrons are brought to rest before they reach A	✓ 1	OWTTE. The idea that the fastest electrons are just stopped is needed.
E_max = eV_s	✓ 1	Accept “E_max in eV is numerically equal to V_s in V”.

Answer: E_max = eV_s

(2)

(d)

Calculate the maximum speed of the photoelectrons released by light of wavelength 365 nm.

MULTINotes

Show solution

E_max = eV_s = 1.60 × 10⁻¹⁹ × 1.14 = 1.82 × 10⁻¹⁹ J	✓ 1	Allow ECF from (c).
v = √(2E_max/m_e) = √(2 × 1.82 × 10⁻¹⁹ / 9.110 × 10⁻³¹) = 6.3 × 10⁵ m s⁻¹	✓ 1	Award [1 max] for 4.5 × 10⁵ m s⁻¹ (factor of 2 omitted).

Answer: 6.3 × 10⁵ m s⁻¹

(2)

(e)(i)

Figure 2 shows the data plotted as a graph of V_s against f.

Draw the line of best fit and use it to determine a value for the Planck constant.

Figure 2: stopping voltage against frequency (drawn to scale)

GRAPHNotes

Show solution

single straight line of best fit passing through or very close to all five points	✓ 1	Do not accept a line joining the first and last points only if it clearly misses the others.
gradient found from two widely separated points on the line ≈ 4.1 × 10⁻¹⁵ V s	✓ 1	Accept 4.0–4.3 × 10⁻¹⁵ V s (or V Hz⁻¹).
h = e × gradient = 1.60 × 10⁻¹⁹ × 4.14 × 10⁻¹⁵ ≈ 6.6 × 10⁻³⁴ J s	✓ 1	Accept 6.4–6.9 × 10⁻³⁴ J s. Allow ECF from their gradient. Unit needed.

Answer: about 6.6 × 10⁻³⁴ J s

(3)

(e)(ii)

Show that the threshold frequency for potassium is about 5.5 × 10¹⁴ Hz.

GRAPHNotes

Show solution

line extrapolated to V_s = 0, meeting the f axis at ≈ 5.4–5.5 × 10¹⁴ Hz

✓ 1

Accept a calculation from the gradient and a point on the line, e.g. f₀ = 9.58 × 10¹⁴ − 1.71 / 4.14 × 10⁻¹⁵ = 5.45 × 10¹⁴ Hz.

Answer: ≈ 5.45 × 10¹⁴ Hz

(1)

(e)(iii)

Determine the work function of potassium, in eV.

CALCNotes

Show solution

Φ = hf₀ = 6.63 × 10⁻³⁴ × 5.45 × 10¹⁴ = 3.61 × 10⁻¹⁹ J	✓ 1	Allow ECF from (e)(i) and (e)(ii). Accept the data-booklet value of h.
Φ = 3.61 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.26 eV	✓ 1	Accept 2.2–2.3 eV. Alternative: Φ = hf/e − V_s for any point on the line, e.g. 4.14 × 10⁻¹⁵ × 9.58 × 10¹⁴ − 1.71 = 2.26 eV.

Answer: 2.26 eV (2.2–2.3 eV)

(2)

(f)

With a filter that transmits light of wavelength 578 nm, there is no current at all, even when the anode is made positive relative to the cathode and the lamp is moved much closer to the photocell.

Explain this observation, and explain why the wave model of light cannot account for it.

EXPLAINNotes

Show solution

photon energy = hc/λ = 3.44 × 10⁻¹⁹ J = 2.15 eV, which is less than the work function «2.26 eV» / f = 5.19 × 10¹⁴ Hz is below the threshold frequency	✓ 1	A comparison with their value from (e) is needed. Allow ECF.
one photon gives all its energy to one electron, so no electron can gain enough energy to escape; moving the lamp closer increases only the number of photons per second, not the energy of each photon	✓ 1	Do not accept “the light is not strong enough”.
in the wave model the energy delivered to the surface increases with intensity «and with time», so emission would be expected at any frequency if the light were intense enough or shone for long enough	✓ 1	OWTTE.

(3)

(g)

With the 405 nm filter and the anode positive, the light delivers a power of 20 μW to the cathode and the photocurrent is 0.16 μA.

Determine the fraction of the incident photons that release a photoelectron. (h = 6.63 × 10⁻³⁴ J s)

MULTINotes

Show solution

energy of one photon = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 405 × 10⁻⁹ = 4.91 × 10⁻¹⁹ J	✓ 1
photons per second = 20 × 10⁻⁶ / 4.91 × 10⁻¹⁹ = 4.1 × 10¹³ s⁻¹ and electrons per second = I/e = 0.16 × 10⁻⁶ / 1.60 × 10⁻¹⁹ = 1.0 × 10¹² s⁻¹	✓ 1	Both values needed for this mark.
fraction = 1.0 × 10¹² / 4.07 × 10¹³ = 0.025 «2.5 %»	✓ 1	Accept 0.024–0.025. Allow ECF.

Answer: 0.025 (about 2.5 %)

(3)

(h)

The intensity of the 405 nm light is doubled. State and explain the effect on the photocurrent with the anode positive and on the stopping voltage.

PREDICTNotes

Show solution

photocurrent doubles «to 0.32 μA» because twice as many photons arrive per second, so twice as many photoelectrons are released per second	✓ 1
stopping voltage unchanged «0.82 V» because the energy of each photon, hf, and so E_max, is unchanged	✓ 1	Statement with a reason needed for each mark.

Answer: current doubles; V_s unchanged

(2)

(i)(i)

The potassium cathode is replaced by a cathode made of a metal with a work function of 2.95 eV.

Explain how the graph of V_s against f for the new cathode compares with the line in Figure 2.

REASONNotes

Show solution

same gradient / parallel line, because the gradient is h/e, which does not depend on the metal	✓ 1
line shifted to higher frequency: every V_s is lower by 2.95 − 2.26 ≈ 0.69 V / threshold frequency increases to ≈ 7.1 × 10¹⁴ Hz	✓ 1	Allow ECF from (e)(iii). Accept 0.65–0.75 V.

Answer: parallel line, about 0.69 V lower (threshold ≈ 7.1 × 10¹⁴ Hz)

(2)

(i)(ii)

Deduce which of the five wavelengths in the table release photoelectrons from the new cathode, and determine the stopping voltage for light of wavelength 313 nm.

PREDICTNotes

Show solution

405, 365, 334 and 313 nm release photoelectrons but 436 nm does not: 0.59 − 0.69 < 0 / photon energy at 436 nm is 2.85 eV, less than 2.95 eV / 6.88 × 10¹⁴ Hz is below the new threshold	✓ 1	A reason is needed.
V_s = 1.71 − 0.69 = 1.02 V	✓ 1	Accept 0.95–1.05 V, e.g. hf/e − 2.95 = 3.97 − 2.95 = 1.02 V. Allow ECF from (e)(iii).

Answer: all except 436 nm; V_s ≈ 1.0 V

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the minimum energy needed to remove an electron from the surface of the metal	✓ 1	Must include “minimum” or “least”. Do not accept “the energy of the photon needed”.
Part (b)
f = c/λ = 3.00 × 10⁸ / 365 × 10⁻⁹ = 8.22 × 10¹⁴ Hz	✓ 1	Must see the substitution OR an answer to at least 3 s.f.
Part (c)
the electric field between A and C opposes the motion of the electrons, so an electron loses kinetic energy eV in moving from C towards A; at V_s even the most energetic electrons are brought to rest before they reach A	✓ 1	OWTTE. The idea that the fastest electrons are just stopped is needed.
E_max = eV_s	✓ 1	Accept “E_max in eV is numerically equal to V_s in V”.
Part (d)
E_max = eV_s = 1.60 × 10⁻¹⁹ × 1.14 = 1.82 × 10⁻¹⁹ J	✓ 1	Allow ECF from (c).
v = √(2E_max/m_e) = √(2 × 1.82 × 10⁻¹⁹ / 9.110 × 10⁻³¹) = 6.3 × 10⁵ m s⁻¹	✓ 1	Award [1 max] for 4.5 × 10⁵ m s⁻¹ (factor of 2 omitted).
Part (e)(i)
single straight line of best fit passing through or very close to all five points	✓ 1	Do not accept a line joining the first and last points only if it clearly misses the others.
gradient found from two widely separated points on the line ≈ 4.1 × 10⁻¹⁵ V s	✓ 1	Accept 4.0–4.3 × 10⁻¹⁵ V s (or V Hz⁻¹).
h = e × gradient = 1.60 × 10⁻¹⁹ × 4.14 × 10⁻¹⁵ ≈ 6.6 × 10⁻³⁴ J s	✓ 1	Accept 6.4–6.9 × 10⁻³⁴ J s. Allow ECF from their gradient. Unit needed.
Part (e)(ii)
line extrapolated to V_s = 0, meeting the f axis at ≈ 5.4–5.5 × 10¹⁴ Hz	✓ 1	Accept a calculation from the gradient and a point on the line, e.g. f₀ = 9.58 × 10¹⁴ − 1.71 / 4.14 × 10⁻¹⁵ = 5.45 × 10¹⁴ Hz.
Part (e)(iii)
Φ = hf₀ = 6.63 × 10⁻³⁴ × 5.45 × 10¹⁴ = 3.61 × 10⁻¹⁹ J	✓ 1	Allow ECF from (e)(i) and (e)(ii). Accept the data-booklet value of h.
Φ = 3.61 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.26 eV	✓ 1	Accept 2.2–2.3 eV. Alternative: Φ = hf/e − V_s for any point on the line, e.g. 4.14 × 10⁻¹⁵ × 9.58 × 10¹⁴ − 1.71 = 2.26 eV.
Part (f)
photon energy = hc/λ = 3.44 × 10⁻¹⁹ J = 2.15 eV, which is less than the work function «2.26 eV» / f = 5.19 × 10¹⁴ Hz is below the threshold frequency	✓ 1	A comparison with their value from (e) is needed. Allow ECF.
one photon gives all its energy to one electron, so no electron can gain enough energy to escape; moving the lamp closer increases only the number of photons per second, not the energy of each photon	✓ 1	Do not accept “the light is not strong enough”.
in the wave model the energy delivered to the surface increases with intensity «and with time», so emission would be expected at any frequency if the light were intense enough or shone for long enough	✓ 1	OWTTE.
Part (g)
energy of one photon = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 405 × 10⁻⁹ = 4.91 × 10⁻¹⁹ J	✓ 1
photons per second = 20 × 10⁻⁶ / 4.91 × 10⁻¹⁹ = 4.1 × 10¹³ s⁻¹ and electrons per second = I/e = 0.16 × 10⁻⁶ / 1.60 × 10⁻¹⁹ = 1.0 × 10¹² s⁻¹	✓ 1	Both values needed for this mark.
fraction = 1.0 × 10¹² / 4.07 × 10¹³ = 0.025 «2.5 %»	✓ 1	Accept 0.024–0.025. Allow ECF.
Part (h)
photocurrent doubles «to 0.32 μA» because twice as many photons arrive per second, so twice as many photoelectrons are released per second	✓ 1
stopping voltage unchanged «0.82 V» because the energy of each photon, hf, and so E_max, is unchanged	✓ 1	Statement with a reason needed for each mark.
Part (i)(i)
same gradient / parallel line, because the gradient is h/e, which does not depend on the metal	✓ 1
line shifted to higher frequency: every V_s is lower by 2.95 − 2.26 ≈ 0.69 V / threshold frequency increases to ≈ 7.1 × 10¹⁴ Hz	✓ 1	Allow ECF from (e)(iii). Accept 0.65–0.75 V.
Part (i)(ii)
405, 365, 334 and 313 nm release photoelectrons but 436 nm does not: 0.59 − 0.69 < 0 / photon energy at 436 nm is 2.85 eV, less than 2.95 eV / 6.88 × 10¹⁴ Hz is below the new threshold	✓ 1	A reason is needed.
V_s = 1.71 − 0.69 = 1.02 V	✓ 1	Accept 0.95–1.05 V, e.g. hf/e − 2.95 = 3.97 − 2.95 = 1.02 V. Allow ECF from (e)(iii).

Answers: (b) 8.22 × 10¹⁴ Hz · (c) E_max = eV_s · (d) 6.3 × 10⁵ m s⁻¹ · (e)(i) about 6.6 × 10⁻³⁴ J s · (e)(ii) ≈ 5.45 × 10¹⁴ Hz · (e)(iii) 2.26 eV (2.2–2.3 eV) · (g) 0.025 (about 2.5 %) · (h) current doubles; V_s unchanged · (i)(i) parallel line, about 0.69 V lower (threshold ≈ 7.1 × 10¹⁴ Hz) · (i)(ii) all except 436 nm; V_s ≈ 1.0 V

Exam tip (e)(i) The gradient of a V_s–f graph is h/e, not h. Multiply by e before quoting the Planck constant, and read the gradient from points on your line, not from two data points.

Common error (f) Saying that brighter light gives each electron more energy. Intensity sets the number of photons arriving per second; the energy of each photon depends only on its frequency.

More E.2 questions in the question bank

+ Add to test✓ In your testCC-IBP-E2-221 marks~25 min

Electron waves in a transmission electron microscope

E.2 Quantum PhysicsD.3 Motion in Electromagnetic FieldsC.3 Wave PhenomenaA.2 Forces and MomentumHL

In a transmission electron microscope (Figure 1), electrons leave a hot filament with negligible speed and are accelerated through a potential difference of 50.0 kV. The beam then passes through a very thin specimen and strikes a fluorescent screen. Use non-relativistic mechanics throughout.

(e = 1.60 × 10⁻¹⁹ C, m_e = 9.110 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹)

Figure 1: the electron microscope, with a front view of the screen (not to scale)

(a)

Show that the kinetic energy of an electron after acceleration is 8.0 × 10⁻¹⁵ J.

DERIVENotes

Show solution

E_k = eV = 1.60 × 10⁻¹⁹ × 50.0 × 10³ = 8.0 × 10⁻¹⁵ J

✓ 1

Must see the substitution.

Answer: 8.0 × 10⁻¹⁵ J

(1)

(b)

Calculate the momentum of an electron after acceleration.

CALCNotes

Show solution

p = √(2m_eE_k) = √(2 × 9.110 × 10⁻³¹ × 8.0 × 10⁻¹⁵) = 1.21 × 10⁻²² kg m s⁻¹

✓ 1

Accept via v = √(2E_k/m_e) and p = m_ev.

Answer: 1.21 × 10⁻²² kg m s⁻¹

(1)

(c)

Show that the de Broglie wavelength of the electrons is about 5.5 pm.

DERIVENotes

Show solution

λ = h/p = 6.63 × 10⁻³⁴ / 1.21 × 10⁻²² = 5.49 × 10⁻¹² m

✓ 1

Must see the substitution OR an answer to at least 3 s.f. Allow ECF from (b).

Answer: 5.49 × 10⁻¹² m

(1)

(d)

Calculate the speed of the electrons. Comment on the use of non-relativistic mechanics in this question.

REASONNotes

Show solution

v = p/m_e = 1.21 × 10⁻²² / 9.110 × 10⁻³¹ = 1.3 × 10⁸ m s⁻¹	✓ 1	Accept 1.33 × 10⁸ m s⁻¹.
this is about 0.44c, a large fraction of the speed of light, so non-relativistic mechanics is not accurate here and the values in (b) and (c) are only approximate	✓ 1	A comparison with c is needed. Do not accept “the speed is less than c, so it is valid”. Credit a comment that a relativistic calculation gives a momentum about 2 % larger and a wavelength about 2 % shorter «5.36 pm».

Answer: 1.3 × 10⁸ m s⁻¹ (0.44c)

(2)

(e)

The specimen contains detail on a scale of about 0.20 nm, the spacing of its atoms.

Using θ ≈ λ/b with b = 0.20 nm, estimate the angle through which the electron waves are diffracted by detail of this size. Hence explain why this detail can be seen with the electron microscope but not with a light microscope using light of wavelength 500 nm.

EXPLAINNotes

Show solution

θ ≈ 5.49 × 10⁻¹² / 0.20 × 10⁻⁹ ≈ 0.027 rad «≈ 1.6°»	✓ 1	Allow ECF from (c).
for light, λ is about 2500 times larger than the detail «λ/b ≫ 1», so the light diffracts strongly in all directions and carries no information about detail this small	✓ 1	OWTTE. Accept “detail much smaller than the wavelength cannot be resolved”.
the electron wavelength is much smaller than the detail, so the diffraction spreading is small and the detail is resolved	✓ 1

Answer: θ ≈ 0.027 rad

(3)

(f)(i)

To test the microscope, a thin polycrystalline metal film is placed in the beam and a pattern of concentric rings forms on the screen. The radius r of one particular ring is measured for several accelerating potential differences V.

V / kV	20.0	25.0	30.0	40.0	50.0	60.0
r / mm	19.0	16.9	15.6	13.4	12.0	11.0

Figure 2 shows r plotted against 1/√V. Explain how the graph supports the hypothesis that the electrons behave as waves with λ = h/p.

Figure 2: ring radius r against 1/√V (drawn to scale)

GRAPHNotes

Show solution

p = √(2m_eeV), so λ = h/√(2m_eeV) ∝ 1/√V	✓ 1
the diffraction angle, and so «for small angles» the ring radius, is proportional to λ, so the hypothesis predicts r ∝ 1/√V	✓ 1
the points lie on a straight line through the origin «within the scatter», so r ∝ 1/√V as predicted	✓ 1	Must refer to both straight line AND origin. Accept a check that r√V is constant «≈ 85 mm kV^½ for every point».

(3)

(f)(ii)

The accelerating potential difference is reduced from 50.0 kV to one quarter of this value. Predict the new de Broglie wavelength of the electrons and the new radius of the ring.

PREDICTNotes

Show solution

λ ∝ 1/√V, so λ doubles: λ = 2 × 5.49 × 10⁻¹² = 1.1 × 10⁻¹¹ m	✓ 1	Allow ECF from (c).
r doubles: 2 × 12.0 = 24 mm	✓ 1	Accept 23–25 mm.

Answer: 1.1 × 10⁻¹¹ m; 24 mm

(2)

(g)(i)

When the beam strikes the specimen, X-rays are also produced. The most energetic X-ray photon is produced when an electron transfers all of its kinetic energy to a single photon.

Show that the shortest wavelength of the X-rays is about 25 pm.

DERIVENotes

Show solution

photon energy = electron kinetic energy = 8.0 × 10⁻¹⁵ J	✓ 1
λ = hc/E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 8.0 × 10⁻¹⁵ = 2.49 × 10⁻¹¹ m	✓ 1	Must see the substitution OR an answer to at least 3 s.f.

Answer: 2.49 × 10⁻¹¹ m

(2)

(g)(ii)

One of these 25 pm photons is scattered through 60° by a free electron that is initially at rest. Calculate the wavelength of the scattered photon.

CALCNotes

Show solution

λ_f = λ_i + (h/m_ec)(1 − cos 60°) = 2.49 × 10⁻¹¹ + 2.43 × 10⁻¹² × 0.500 = 2.61 × 10⁻¹¹ m

✓ 1

Accept 2.62 × 10⁻¹¹ m from λ_i = 25 pm. Award [0] for 2.73 × 10⁻¹¹ m (cos 60° term omitted).

Answer: 2.61 × 10⁻¹¹ m

(1)

(g)(iii)

Determine the magnitude of the momentum of the recoiling electron.

MULTINotes

Show solution

p_i = h/λ_i = 2.66 × 10⁻²³ kg m s⁻¹ and p_f = h/λ_f = 2.54 × 10⁻²³ kg m s⁻¹	✓ 1	Allow ECF from (g)(i) and (g)(ii).
momentum is conserved as a vector: p_e = p_i − p_f, with 60° between p_i and p_f, so p_e² = p_i² + p_f² − 2p_ip_f cos 60°	✓ 1	May be shown as a vector triangle, or with components: along the beam p_i − p_f cos 60° = 1.39 × 10⁻²³, at right angles p_f sin 60° = 2.20 × 10⁻²³ kg m s⁻¹.
p_e = 2.6 × 10⁻²³ kg m s⁻¹	✓ 1	Award [1 max] for p_i − p_f = 1.2 × 10⁻²⁴ kg m s⁻¹ (vector nature ignored).

Answer: 2.6 × 10⁻²³ kg m s⁻¹

(3)

(h)

Discuss how the behaviour of the electrons and of the X-ray photons in this question illustrates wave–particle duality.

EXPLAINNotes

Show solution

electrons show wave behaviour when they are diffracted by the film to form rings «whose size matches λ = h/p», yet show particle behaviour when accelerated by the electric field as charged particles of definite mass and charge	✓ 1	Both aspects of the electron needed.
X-rays are electromagnetic waves, yet in Compton scattering a photon behaves as a particle in a collision, carrying momentum h/λ that is conserved; neither model alone describes either entity	✓ 1	OWTTE.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
E_k = eV = 1.60 × 10⁻¹⁹ × 50.0 × 10³ = 8.0 × 10⁻¹⁵ J	✓ 1	Must see the substitution.
Part (b)
p = √(2m_eE_k) = √(2 × 9.110 × 10⁻³¹ × 8.0 × 10⁻¹⁵) = 1.21 × 10⁻²² kg m s⁻¹	✓ 1	Accept via v = √(2E_k/m_e) and p = m_ev.
Part (c)
λ = h/p = 6.63 × 10⁻³⁴ / 1.21 × 10⁻²² = 5.49 × 10⁻¹² m	✓ 1	Must see the substitution OR an answer to at least 3 s.f. Allow ECF from (b).
Part (d)
v = p/m_e = 1.21 × 10⁻²² / 9.110 × 10⁻³¹ = 1.3 × 10⁸ m s⁻¹	✓ 1	Accept 1.33 × 10⁸ m s⁻¹.
this is about 0.44c, a large fraction of the speed of light, so non-relativistic mechanics is not accurate here and the values in (b) and (c) are only approximate	✓ 1	A comparison with c is needed. Do not accept “the speed is less than c, so it is valid”. Credit a comment that a relativistic calculation gives a momentum about 2 % larger and a wavelength about 2 % shorter «5.36 pm».
Part (e)
θ ≈ 5.49 × 10⁻¹² / 0.20 × 10⁻⁹ ≈ 0.027 rad «≈ 1.6°»	✓ 1	Allow ECF from (c).
for light, λ is about 2500 times larger than the detail «λ/b ≫ 1», so the light diffracts strongly in all directions and carries no information about detail this small	✓ 1	OWTTE. Accept “detail much smaller than the wavelength cannot be resolved”.
the electron wavelength is much smaller than the detail, so the diffraction spreading is small and the detail is resolved	✓ 1
Part (f)(i)
p = √(2m_eeV), so λ = h/√(2m_eeV) ∝ 1/√V	✓ 1
the diffraction angle, and so «for small angles» the ring radius, is proportional to λ, so the hypothesis predicts r ∝ 1/√V	✓ 1
the points lie on a straight line through the origin «within the scatter», so r ∝ 1/√V as predicted	✓ 1	Must refer to both straight line AND origin. Accept a check that r√V is constant «≈ 85 mm kV^½ for every point».
Part (f)(ii)
λ ∝ 1/√V, so λ doubles: λ = 2 × 5.49 × 10⁻¹² = 1.1 × 10⁻¹¹ m	✓ 1	Allow ECF from (c).
r doubles: 2 × 12.0 = 24 mm	✓ 1	Accept 23–25 mm.
Part (g)(i)
photon energy = electron kinetic energy = 8.0 × 10⁻¹⁵ J	✓ 1
λ = hc/E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 8.0 × 10⁻¹⁵ = 2.49 × 10⁻¹¹ m	✓ 1	Must see the substitution OR an answer to at least 3 s.f.
Part (g)(ii)
λ_f = λ_i + (h/m_ec)(1 − cos 60°) = 2.49 × 10⁻¹¹ + 2.43 × 10⁻¹² × 0.500 = 2.61 × 10⁻¹¹ m	✓ 1	Accept 2.62 × 10⁻¹¹ m from λ_i = 25 pm. Award [0] for 2.73 × 10⁻¹¹ m (cos 60° term omitted).
Part (g)(iii)
p_i = h/λ_i = 2.66 × 10⁻²³ kg m s⁻¹ and p_f = h/λ_f = 2.54 × 10⁻²³ kg m s⁻¹	✓ 1	Allow ECF from (g)(i) and (g)(ii).
momentum is conserved as a vector: p_e = p_i − p_f, with 60° between p_i and p_f, so p_e² = p_i² + p_f² − 2p_ip_f cos 60°	✓ 1	May be shown as a vector triangle, or with components: along the beam p_i − p_f cos 60° = 1.39 × 10⁻²³, at right angles p_f sin 60° = 2.20 × 10⁻²³ kg m s⁻¹.
p_e = 2.6 × 10⁻²³ kg m s⁻¹	✓ 1	Award [1 max] for p_i − p_f = 1.2 × 10⁻²⁴ kg m s⁻¹ (vector nature ignored).
Part (h)
electrons show wave behaviour when they are diffracted by the film to form rings «whose size matches λ = h/p», yet show particle behaviour when accelerated by the electric field as charged particles of definite mass and charge	✓ 1	Both aspects of the electron needed.
X-rays are electromagnetic waves, yet in Compton scattering a photon behaves as a particle in a collision, carrying momentum h/λ that is conserved; neither model alone describes either entity	✓ 1	OWTTE.

Answers: (a) 8.0 × 10⁻¹⁵ J · (b) 1.21 × 10⁻²² kg m s⁻¹ · (c) 5.49 × 10⁻¹² m · (d) 1.3 × 10⁸ m s⁻¹ (0.44c) · (e) θ ≈ 0.027 rad · (f)(ii) 1.1 × 10⁻¹¹ m; 24 mm · (g)(i) 2.49 × 10⁻¹¹ m · (g)(ii) 2.61 × 10⁻¹¹ m · (g)(iii) 2.6 × 10⁻²³ kg m s⁻¹

Common error (f)(i) Saying only that “the rings show diffraction”. Rings show wave behaviour, but the hypothesis λ = h/p is tested by the way the ring size changes with V: a straight line through the origin on this graph.

More E.2 questions in the question bank

E.3 Radioactive Decay

+ Add to test✓ In your testCC-IBP-E3-120 marks~25 min

A PET scan tracer: making, transporting and detecting fluorine-18

E.3 Radioactive DecayE.1 Structure of the AtomE.4 FissionSL

Fluorine-18 is a radioactive nuclide used as a tracer in positron emission tomography (PET) scans. It decays by β⁺ emission to oxygen-18. Fluorine-18 is made in a cyclotron at a production centre, and each dose is then driven to a hospital. The journey normally takes 120 minutes.

Each dose has an activity of 800 MBq when it leaves the production centre. Figure 1 shows how the activity of one dose varies with the time t after it leaves.

(1 u = 931.5 MeV c⁻²)

Figure 1: activity of one dose of fluorine-18 against time after it leaves the production centre (drawn to scale)

(a)(i)

State the number of protons and the number of neutrons in a nucleus of fluorine-18, ¹⁸₉F.

RECALLNotes

Show solution

9 protons and 18 − 9 = 9 neutrons

✓ 1

Both needed.

Answer: 9 protons, 9 neutrons

(1)

(a)(ii)

Write down the equation for the decay of fluorine-18 to oxygen-18 (O).

RECALLNotes

Show solution

¹⁸₉F → ¹⁸₈O + …	✓ 1	Both numbers on O needed.
… + ⁰₊₁e + ν	✓ 1	Accept β⁺ for the positron. Do not accept an antineutrino.

Answer: ¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ν

(2)

(b)

In the cyclotron, fluorine-18 is made by bombarding oxygen-18 nuclei with protons:

¹⁸₈O + ¹₁p → ¹⁸₉F + ¹₀n

Particle	Mass / u
oxygen-18 nucleus	17.994 767
fluorine-18 nucleus	17.995 997
proton	1.007 276
neutron	1.008 665

Show that the total rest mass increases in this reaction, and determine the corresponding increase in rest energy, in MeV. Suggest where this energy comes from.

MULTINotes

Show solution

mass before = 17.994 767 + 1.007 276 = 19.002 043 u and mass after = 17.995 997 + 1.008 665 = 19.004 662 u, so the mass increases	✓ 1	Both totals, or the difference with its sign, needed.
Δm = 0.002 619 u, so energy = 0.002 619 × 931.5 = 2.44 MeV	✓ 1	Accept 2.4 MeV.
from the kinetic energy of the incident proton «given to it by the cyclotron»	✓ 1	Do not accept “from the binding energy” alone.

Answer: 2.44 MeV

(3)

(c)

Use Figure 1 to determine the half-life of fluorine-18. Use two successive halvings.

GRAPHNotes

Show solution

800 MBq → 400 MBq at t ≈ 110 min	✓ 1	Lines or readings shown on the graph.
400 MBq → 200 MBq at t ≈ 220 min, so half-life = 110 min	✓ 1	Accept 105–115 min.

Answer: 110 min

(2)

(d)

The patient is to be injected with a dose that has an activity of 370 MBq when it arrives at the hospital. Use Figure 1 to show that the activity of a dose when it leaves the production centre must be about 2.1 times the activity required at the hospital.

DERIVENotes

Show solution

at t = 120 min the activity is ≈ 375 MBq	✓ 1	Accept 370–380 MBq.
ratio = 800 / 375 ≈ 2.1	✓ 1	Accept 2.1–2.2. Accept 2^120/110 = 2.13 as an alternative to the graph reading.

Answer: 800/375 ≈ 2.1

(2)

(e)

On one day, a traffic delay adds 40 minutes to the journey. Determine whether the dose still has the required activity of 370 MBq when it arrives, and the minimum activity it should have had when it left the production centre.

PREDICTNotes

Show solution

at t = 160 min the activity is ≈ 290 MBq	✓ 1	Accept 285–300 MBq.
this is less than 370 MBq, so the dose is no longer sufficient	✓ 1	Must follow from a reading at 160 min.
minimum activity on leaving = 370 × 800 / 292 ≈ 1.0 × 10³ MBq «1.0 GBq»	✓ 1	Accept 980–1050 MBq, e.g. 370 × 2^160/110 = 1010 MBq. Allow ECF from their reading.

Answer: ≈ 290 MBq, not enough; ≈ 1.0 × 10³ MBq needed

(3)

(f)

The half-life of fluorine-18 is 110 minutes. Calculate the decay constant of fluorine-18, in s⁻¹, and the number of fluorine-18 nuclei in a dose of activity 370 MBq.

HL onlyMULTINotes

Show solution

λ = ln 2 / (110 × 60) = 1.05 × 10⁻⁴ s⁻¹	✓ 1	Award [0] for this mark if the time is left in minutes and the answer is quoted in s⁻¹.
N = A/λ = 370 × 10⁶ / 1.05 × 10⁻⁴ = 3.5 × 10¹²	✓ 1	Allow ECF from their λ.

Answer: 1.05 × 10⁻⁴ s⁻¹; 3.5 × 10¹² nuclei

(2)

(g)

Inside the patient, each positron meets an electron and the two annihilate, producing two gamma-ray photons that travel in opposite directions to detectors around the patient (Figure 2). Assume that the positron and the electron are at rest when they annihilate.

Show that the energy of each photon is about 0.51 MeV. (mass of an electron = mass of a positron = 0.000 549 u)

Figure 2: annihilation inside the patient and detection of the photons (not to scale)

DERIVENotes

Show solution

total rest mass converted = 2 × 0.000 549 = 0.001 098 u, so energy released = 0.001 098 × 931.5 = 1.02 MeV	✓ 1
the energy is shared equally between the two photons: 1.023 / 2 = 0.511 MeV	✓ 1	Must see the halving OR an answer to at least 3 s.f. Accept 0.000 549 × 931.5 = 0.511 MeV with a statement that each photon carries the energy of one particle.

Answer: 0.511 MeV

(2)

(h)

Discuss why fluorine-18 is a suitable nuclide for this medical use. Refer to its half-life and to the radiation that leaves the patient’s body.

EXPLAINNotes

Show solution

a half-life of about 2 hours is long enough to make the tracer, transport it and carry out the scan	✓ 1
but short enough that the activity in the patient falls quickly «after one day ≈ 13 half-lives, a fraction of about 10⁻⁴ remains», limiting the dose received	✓ 1	Accept “decays to stable oxygen-18, so no further radioactive products remain”.
the radiation that leaves the body is gamma radiation, which is very penetrating, so it reaches the detectors outside the body, and weakly ionising, so it causes little damage on the way out	✓ 1	Both penetration and ionisation needed for this mark.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
9 protons and 18 − 9 = 9 neutrons	✓ 1	Both needed.
Part (a)(ii)
¹⁸₉F → ¹⁸₈O + …	✓ 1	Both numbers on O needed.
… + ⁰₊₁e + ν	✓ 1	Accept β⁺ for the positron. Do not accept an antineutrino.
Part (b)
mass before = 17.994 767 + 1.007 276 = 19.002 043 u and mass after = 17.995 997 + 1.008 665 = 19.004 662 u, so the mass increases	✓ 1	Both totals, or the difference with its sign, needed.
Δm = 0.002 619 u, so energy = 0.002 619 × 931.5 = 2.44 MeV	✓ 1	Accept 2.4 MeV.
from the kinetic energy of the incident proton «given to it by the cyclotron»	✓ 1	Do not accept “from the binding energy” alone.
Part (c)
800 MBq → 400 MBq at t ≈ 110 min	✓ 1	Lines or readings shown on the graph.
400 MBq → 200 MBq at t ≈ 220 min, so half-life = 110 min	✓ 1	Accept 105–115 min.
Part (d)
at t = 120 min the activity is ≈ 375 MBq	✓ 1	Accept 370–380 MBq.
ratio = 800 / 375 ≈ 2.1	✓ 1	Accept 2.1–2.2. Accept 2^120/110 = 2.13 as an alternative to the graph reading.
Part (e)
at t = 160 min the activity is ≈ 290 MBq	✓ 1	Accept 285–300 MBq.
this is less than 370 MBq, so the dose is no longer sufficient	✓ 1	Must follow from a reading at 160 min.
minimum activity on leaving = 370 × 800 / 292 ≈ 1.0 × 10³ MBq «1.0 GBq»	✓ 1	Accept 980–1050 MBq, e.g. 370 × 2^160/110 = 1010 MBq. Allow ECF from their reading.
Part (f)
λ = ln 2 / (110 × 60) = 1.05 × 10⁻⁴ s⁻¹	✓ 1	Award [0] for this mark if the time is left in minutes and the answer is quoted in s⁻¹.
N = A/λ = 370 × 10⁶ / 1.05 × 10⁻⁴ = 3.5 × 10¹²	✓ 1	Allow ECF from their λ.
Part (g)
total rest mass converted = 2 × 0.000 549 = 0.001 098 u, so energy released = 0.001 098 × 931.5 = 1.02 MeV	✓ 1
the energy is shared equally between the two photons: 1.023 / 2 = 0.511 MeV	✓ 1	Must see the halving OR an answer to at least 3 s.f. Accept 0.000 549 × 931.5 = 0.511 MeV with a statement that each photon carries the energy of one particle.
Part (h)
a half-life of about 2 hours is long enough to make the tracer, transport it and carry out the scan	✓ 1
but short enough that the activity in the patient falls quickly «after one day ≈ 13 half-lives, a fraction of about 10⁻⁴ remains», limiting the dose received	✓ 1	Accept “decays to stable oxygen-18, so no further radioactive products remain”.
the radiation that leaves the body is gamma radiation, which is very penetrating, so it reaches the detectors outside the body, and weakly ionising, so it causes little damage on the way out	✓ 1	Both penetration and ionisation needed for this mark.

Answers: (a)(i) 9 protons, 9 neutrons · (a)(ii) ¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ν · (b) 2.44 MeV · (c) 110 min · (d) 800/375 ≈ 2.1 · (e) ≈ 290 MBq, not enough; ≈ 1.0 × 10³ MBq needed · (f) 1.05 × 10⁻⁴ s⁻¹; 3.5 × 10¹² nuclei · (g) 0.511 MeV

Common error (a)(ii) Writing an antineutrino in a β⁺ equation. β⁺ decay emits a positron and a neutrino; β⁻ decay emits an electron and an antineutrino.

More E.3 questions in the question bank

+ Add to test✓ In your testCC-IBP-E3-224 marks~30 min

Americium-241 in a smoke detector: decay, binding energy and the ionisation current

E.3 Radioactive DecayE.1 Structure of the AtomB.5 Current and CircuitsD.2 Electric and Magnetic FieldsSL

A household smoke detector contains an ionisation chamber (Figure 1). Two parallel metal plates 6.0 mm apart, with air between them, are connected in series with a 9.0 V battery and a sensitive ammeter. A source containing 0.30 μg of americium-241, ²⁴¹₉₅Am, is mounted on one plate. Americium-241 is an alpha emitter with a half-life of 432 years; it decays to neptunium (Np).

Each alpha particle is emitted with a kinetic energy of 5.5 MeV and ionises air molecules along its path. On average, 35 eV of energy is needed to create one ion pair (a positive ion and an electron).

(m_p = 1.007 276 u, m_n = 1.008 665 u, 1 u = 931.5 MeV c⁻², e = 1.60 × 10⁻¹⁹ C, N_A = 6.02 × 10²³ mol⁻¹, 1 year = 3.16 × 10⁷ s)

Figure 1: the ionisation chamber (not to scale)

(a)

Write down the equation for the alpha decay of americium-241.

RECALLNotes

Show solution

²⁴¹₉₅Am → ²³⁷₉₃Np + …	✓ 1	Both numbers on Np needed.
… + ⁴₂α	✓ 1	Accept ⁴₂He.

Answer: ²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α

(2)

(b)

Explain, with reference to ionisation and penetration, why an alpha emitter is suitable for use in a smoke detector in a home.

EXPLAINNotes

Show solution

alpha particles are strongly ionising, so they create many ion pairs in the small volume of air between the plates, giving a measurable current	✓ 1
alpha particles have a range of only a few cm in air and are stopped by the casing «or a sheet of paper», so no radiation reaches the people in the room	✓ 1	Do not accept “alpha is not dangerous” without reference to its short range.

(2)

(c)

The mass of an americium-241 nucleus is 241.004 674 u. Show that the binding energy per nucleon of americium-241 is about 7.5 MeV.

DERIVENotes

Show solution

Δm = 95 × 1.007 276 + 146 × 1.008 665 − 241.004 674 = 1.951 636 u	✓ 1	Number of neutrons 241 − 95 = 146 must be used.
binding energy = 1.951 636 × 931.5 = 1818 MeV, so per nucleon = 1818 / 241 = 7.54 MeV	✓ 1	Must see the division by 241 OR an answer to at least 3 s.f.

Answer: 7.54 MeV

(2)

(d)

Figure 2 shows how the binding energy per nucleon varies with nucleon number A.

Explain, with reference to Figure 2, why energy is released when americium-241 decays by alpha emission.

Figure 2: binding energy per nucleon against nucleon number (drawn to scale)

GRAPHNotes

Show solution

americium-241 lies on the slowly falling right-hand part of the curve «at A = 241, about 7.5 MeV», so the daughter Np-237, with a smaller A, has a larger binding energy per nucleon	✓ 1
helium-4 has a high binding energy per nucleon for so light a nucleus «7.07 MeV, well above its light neighbours», so its nucleons are bound almost as tightly as those in americium-241	✓ 1
the total binding energy of the products is greater than that of the americium-241 nucleus, so the products have less mass and the difference is released «as kinetic energy, E = Δmc²»	✓ 1	OWTTE. Do not accept “the products are more stable” alone.

(3)

(e)(i)

With no smoke in the chamber, the current is 80 pA. Calculate the number of ion pairs collected per second.

CALCNotes

Show solution

each ion pair transfers a charge e through the circuit, so the number per second = I/e	✓ 1
= 80 × 10⁻¹² / 1.60 × 10⁻¹⁹ = 5.0 × 10⁸ s⁻¹	✓ 1

Answer: 5.0 × 10⁸ s⁻¹

(2)

(e)(ii)

Estimate the minimum number of alpha particles per second that must enter the air between the plates to produce this current.

MULTINotes

Show solution

ion pairs per alpha particle «if all its energy is used» = 5.5 × 10⁶ / 35 = 1.6 × 10⁵	✓ 1
number of alpha particles per second = 5.0 × 10⁸ / 1.57 × 10⁵ ≈ 3.2 × 10³ s⁻¹	✓ 1	Allow ECF from (e)(i). A minimum because an alpha particle that reaches the far plate has not used all its energy.

Answer: ≈ 3.2 × 10³ s⁻¹

(2)

(f)

Calculate the electric field strength between the plates, and explain why this electric field is needed for the detector to work.

EXPLAINNotes

Show solution

E = V/d = 9.0 / 6.0 × 10⁻³ = 1.5 × 10³ V m⁻¹	✓ 1	Accept N C⁻¹.
the field exerts forces in opposite directions on the positive ions and the electrons, moving them to opposite plates before they can recombine, so there is a current in the circuit	✓ 1	OWTTE.

Answer: 1.5 × 10³ V m⁻¹

(2)

(g)

When smoke enters the chamber, the current falls and the alarm sounds. Suggest why the current falls.

EXPLAINNotes

Show solution

smoke particles absorb alpha particles / shorten their path in the air, so fewer ion pairs are created per second	✓ 1
ions attach to the much more massive smoke particles, which move slowly in the field and are more likely to be neutralised before reaching the plates, so fewer charges reach the plates per second	✓ 1	Either point, or any other valid physics, for each mark.

(2)

(h)

Determine the activity of the americium-241 in the source.

HL onlyMULTINotes

Show solution

N = (0.30 × 10⁻⁶ / 241) × 6.02 × 10²³ = 7.5 × 10¹⁴	✓ 1
λ = ln 2 / (432 × 3.16 × 10⁷) = 5.08 × 10⁻¹¹ s⁻¹	✓ 1
A = λN = 5.08 × 10⁻¹¹ × 7.49 × 10¹⁴ = 3.8 × 10⁴ Bq	✓ 1	Allow ECF.

Answer: 3.8 × 10⁴ Bq

(3)

(i)

Show that the activity of the source falls by less than 2 % during the ten-year working life of the detector.

HL onlyDERIVENotes

Show solution

A/A₀ = e^−λt = e^{−5.08 × 10⁻¹¹ × 10 × 3.16 × 10⁷} = 0.984	✓ 1	Accept (½)^10/432 = 0.984, or λ in y⁻¹: e^{−1.60 × 10⁻³ × 10}.
fall = 1 − 0.984 = 0.016 = 1.6 %, which is less than 2 %	✓ 1	Must see the final comparison.

Answer: 1.6 %

(2)

(j)

A designer suggests replacing the americium-241 source with a beta-minus source of the same activity. Predict, with a reason, the effect on the current with no smoke present, and suggest why this would make the detector less effective.

PREDICTNotes

Show solution

the current would be much smaller, because beta particles are much less ionising: they create far fewer ion pairs per mm of air and cross the 6.0 mm gap having used only a small part of their energy	✓ 1
so the change in current caused by smoke would be very small and hard to detect / beta particles are more penetrating and could escape the casing	✓ 1	Either suggestion.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
²⁴¹₉₅Am → ²³⁷₉₃Np + …	✓ 1	Both numbers on Np needed.
… + ⁴₂α	✓ 1	Accept ⁴₂He.
Part (b)
alpha particles are strongly ionising, so they create many ion pairs in the small volume of air between the plates, giving a measurable current	✓ 1
alpha particles have a range of only a few cm in air and are stopped by the casing «or a sheet of paper», so no radiation reaches the people in the room	✓ 1	Do not accept “alpha is not dangerous” without reference to its short range.
Part (c)
Δm = 95 × 1.007 276 + 146 × 1.008 665 − 241.004 674 = 1.951 636 u	✓ 1	Number of neutrons 241 − 95 = 146 must be used.
binding energy = 1.951 636 × 931.5 = 1818 MeV, so per nucleon = 1818 / 241 = 7.54 MeV	✓ 1	Must see the division by 241 OR an answer to at least 3 s.f.
Part (d)
americium-241 lies on the slowly falling right-hand part of the curve «at A = 241, about 7.5 MeV», so the daughter Np-237, with a smaller A, has a larger binding energy per nucleon	✓ 1
helium-4 has a high binding energy per nucleon for so light a nucleus «7.07 MeV, well above its light neighbours», so its nucleons are bound almost as tightly as those in americium-241	✓ 1
the total binding energy of the products is greater than that of the americium-241 nucleus, so the products have less mass and the difference is released «as kinetic energy, E = Δmc²»	✓ 1	OWTTE. Do not accept “the products are more stable” alone.
Part (e)(i)
each ion pair transfers a charge e through the circuit, so the number per second = I/e	✓ 1
= 80 × 10⁻¹² / 1.60 × 10⁻¹⁹ = 5.0 × 10⁸ s⁻¹	✓ 1
Part (e)(ii)
ion pairs per alpha particle «if all its energy is used» = 5.5 × 10⁶ / 35 = 1.6 × 10⁵	✓ 1
number of alpha particles per second = 5.0 × 10⁸ / 1.57 × 10⁵ ≈ 3.2 × 10³ s⁻¹	✓ 1	Allow ECF from (e)(i). A minimum because an alpha particle that reaches the far plate has not used all its energy.
Part (f)
E = V/d = 9.0 / 6.0 × 10⁻³ = 1.5 × 10³ V m⁻¹	✓ 1	Accept N C⁻¹.
the field exerts forces in opposite directions on the positive ions and the electrons, moving them to opposite plates before they can recombine, so there is a current in the circuit	✓ 1	OWTTE.
Part (g)
smoke particles absorb alpha particles / shorten their path in the air, so fewer ion pairs are created per second	✓ 1
ions attach to the much more massive smoke particles, which move slowly in the field and are more likely to be neutralised before reaching the plates, so fewer charges reach the plates per second	✓ 1	Either point, or any other valid physics, for each mark.
Part (h)
N = (0.30 × 10⁻⁶ / 241) × 6.02 × 10²³ = 7.5 × 10¹⁴	✓ 1
λ = ln 2 / (432 × 3.16 × 10⁷) = 5.08 × 10⁻¹¹ s⁻¹	✓ 1
A = λN = 5.08 × 10⁻¹¹ × 7.49 × 10¹⁴ = 3.8 × 10⁴ Bq	✓ 1	Allow ECF.
Part (i)
A/A₀ = e^−λt = e^{−5.08 × 10⁻¹¹ × 10 × 3.16 × 10⁷} = 0.984	✓ 1	Accept (½)^10/432 = 0.984, or λ in y⁻¹: e^{−1.60 × 10⁻³ × 10}.
fall = 1 − 0.984 = 0.016 = 1.6 %, which is less than 2 %	✓ 1	Must see the final comparison.
Part (j)
the current would be much smaller, because beta particles are much less ionising: they create far fewer ion pairs per mm of air and cross the 6.0 mm gap having used only a small part of their energy	✓ 1
so the change in current caused by smoke would be very small and hard to detect / beta particles are more penetrating and could escape the casing	✓ 1	Either suggestion.

Answers: (a) ²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α · (c) 7.54 MeV · (e)(i) 5.0 × 10⁸ s⁻¹ · (e)(ii) ≈ 3.2 × 10³ s⁻¹ · (f) 1.5 × 10³ V m⁻¹ · (h) 3.8 × 10⁴ Bq · (i) 1.6 %

Exam tip (c) Keep the masses to six decimal places until the subtraction is done: the mass defect is a small difference between two large numbers.

More E.3 questions in the question bank

E.4 Fission

+ Add to test✓ In your testCC-IBP-E4-123 marks~30 min

The reactor of a nuclear submarine

E.4 FissionA.3 Work, Energy and PowerA.2 Forces and MomentumE.3 Radioactive DecaySL

A nuclear submarine is powered by a pressurised-water reactor. The fuel is uranium enriched in uranium-235. At full power the reactor core produces thermal energy at a rate of 150 MW. Each fission of a uranium-235 nucleus releases 200 MeV.

Water in the reactor acts as both the moderator and the coolant. The thermal energy is used to make steam, which drives the turbines that propel the submarine.

Figure 1 shows how the binding energy per nucleon varies with nucleon number A.

Figure 1: binding energy per nucleon against nucleon number (dots: light nuclides; line: smooth curve through heavier nuclides)

(a)

Use Figure 1 to explain why energy is released when a uranium-235 nucleus splits into two nuclei of similar mass.

EXPLAINNotes

Show solution

the products (A about 90–150) have a greater binding energy per nucleon than uranium-235 (about 8.5 MeV compared with about 7.6 MeV)	✓ 1	Values read from Figure 1 are not required, but the comparison must refer to binding energy per nucleon.
so the total binding energy of the products is greater / the total mass of the products is less than that of the reactants; the difference (mass decrease × c²) is released as energy	✓ 1	Do not credit "energy is released because bonds are broken" or "binding energy is released" on its own.

(2)

(b)(i)

Show that the number of fissions per second in the core at full power is about 4.7 × 10¹⁸ s⁻¹.

DERIVENotes

Show solution

energy per fission = 200 × 1.60 × 10⁻¹³ = 3.20 × 10⁻¹¹ J	✓ 1
number per second = 150 × 10⁶ ÷ 3.20 × 10⁻¹¹ = 4.69 × 10¹⁸ s⁻¹	✓ 1	The answer must be seen to more significant figures than the given value, or with full working.

Answer: 4.69 × 10¹⁸ s⁻¹

(2)

(b)(ii)

The mass of a uranium-235 atom is 235 u. Determine the mass of uranium-235 that undergoes fission if the reactor runs at full power for 365 days.

MULTINotes

Show solution

number of fissions = 4.69 × 10¹⁸ × 365 × 24 × 3600 = 1.48 × 10²⁶	✓ 1
mass = 1.48 × 10²⁶ × 235 × 1.661 × 10⁻²⁷ = 58 kg	✓ 1	Accept 57–58 kg. Using 4.7 × 10¹⁸ s⁻¹ gives 57.9 kg. Allow ECF from (b)(i).

Answer: 58 kg (57.7 kg)

(2)

(c)

30 % of the thermal power of the core is converted into useful power that propels the submarine. At full power the submarine moves through the water at a constant speed of 15 m s⁻¹.

Determine the resistive force exerted by the water on the submarine at this speed. Explain your reasoning.

MULTINotes

Show solution

useful power = 0.30 × 150 MW = 45 MW	✓ 1
the velocity is constant, so the resultant force is zero: the forward (driving) force equals the resistive force	✓ 1	Accept a reference to Newton's first law.
F = P ÷ v = 45 × 10⁶ ÷ 15 = 3.0 × 10⁶ N	✓ 1	Using 150 MW instead of 45 MW gives 1.0 × 10⁷ N: award 2 max.

Answer: 3.0 × 10⁶ N

(3)

(d)(i)

Fission produces fast neutrons. In the moderator, a neutron of speed v collides head-on with a stationary hydrogen nucleus (a proton). Take the mass of the proton to be equal to the mass m of the neutron. After the collision the neutron is at rest.

Deduce the velocity of the proton after the collision, and show that the collision is elastic.

REASONNotes

Show solution

momentum: mv = 0 + mV, so the proton moves off with speed V = v in the neutron's original direction	✓ 1	Direction needed.
kinetic energy after = ½mv² = kinetic energy before, so kinetic energy is conserved and the collision is elastic	✓ 1

Answer: speed v, in the neutron's original direction

(2)

(d)(ii)

A fast neutron that collides head-on with a stationary uranium-238 nucleus rebounds with almost its original speed.

With reference to your answer to (d)(i), explain why the reactor needs a moderator made of light nuclei such as hydrogen.

EXPLAINNotes

Show solution

in a collision with a heavy nucleus such as uranium-238 the neutron keeps almost all its kinetic energy, so it is hardly slowed down	✓ 1
in a collision with a light nucleus such as a proton the neutron can lose a large fraction (up to all) of its kinetic energy, so far fewer collisions are needed to slow it down	✓ 1
slow (thermal) neutrons are much more likely than fast neutrons to cause fission of uranium-235, so a chain reaction can be sustained	✓ 1	Do not credit "the moderator absorbs neutrons".

(3)

(e)(i)

Figure 2 is a model of how the average kinetic energy of a neutron in the water moderator changes with the number of collisions it has made. The neutron starts with a kinetic energy of 2.0 MeV. A thermal neutron has a kinetic energy of about 0.025 eV.

Use Figure 2 to determine the number of collisions needed for the neutron to reach thermal energy.

Figure 2: model of the average kinetic energy of a neutron in the water moderator against the number of collisions (logarithmic vertical scale)

GRAPHNotes

Show solution

20 collisions

✓ 1

Accept 19–21.

Answer: 20

(1)

(e)(ii)

Hence determine the average fraction of its kinetic energy that the neutron keeps in each collision.

CALCNotes

Show solution

f²⁰ = 0.025 ÷ (2.0 × 10⁶) = 1.25 × 10⁻⁸	✓ 1	Allow ECF from (e)(i): 19 collisions gives 0.38, 21 gives 0.42.
f = (1.25 × 10⁻⁸)^1/20 = 0.40	✓ 1

Answer: 0.40

(2)

(e)(iii)

Suggest why the graph becomes horizontal once the neutron reaches thermal energy.

EXPLAINNotes

Show solution

the neutron is now in thermal equilibrium with the moderator: the moderator nuclei have a similar average kinetic energy, so on average the neutron gains as much energy in collisions as it loses

✓ 1

(1)

(f)(i)

At full power the reactor is critical. An operator withdraws the control rods slightly. As a result, each generation of neutrons is 1.002 times larger than the generation before.

Assuming that the number of fissions per second is proportional to the number of neutrons, calculate the thermal power of the core after 500 more generations.

CALCNotes

Show solution

factor = 1.002⁵⁰⁰ = 2.72	✓ 1
power = 150 × 2.72 = 410 MW	✓ 1	Accept 407 MW.

Answer: 410 MW (407 MW)

(2)

(f)(ii)

State whether the reactor is now critical, subcritical or supercritical, and outline how the operator returns the reactor to a steady thermal power of 150 MW.

EXPLAINNotes

Show solution

supercritical	✓ 1
the control rods are inserted further so that they absorb more neutrons; on average fewer than one neutron per fission then causes another fission and the power falls	✓ 1
when the power has fallen to 150 MW the rods are set so that, on average, exactly one neutron from each fission causes another fission: the reactor is critical again	✓ 1	Accept "the multiplication factor is returned to 1".

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
the products (A about 90–150) have a greater binding energy per nucleon than uranium-235 (about 8.5 MeV compared with about 7.6 MeV)	✓ 1	Values read from Figure 1 are not required, but the comparison must refer to binding energy per nucleon.
so the total binding energy of the products is greater / the total mass of the products is less than that of the reactants; the difference (mass decrease × c²) is released as energy	✓ 1	Do not credit "energy is released because bonds are broken" or "binding energy is released" on its own.
Part (b)(i)
energy per fission = 200 × 1.60 × 10⁻¹³ = 3.20 × 10⁻¹¹ J	✓ 1
number per second = 150 × 10⁶ ÷ 3.20 × 10⁻¹¹ = 4.69 × 10¹⁸ s⁻¹	✓ 1	The answer must be seen to more significant figures than the given value, or with full working.
Part (b)(ii)
number of fissions = 4.69 × 10¹⁸ × 365 × 24 × 3600 = 1.48 × 10²⁶	✓ 1
mass = 1.48 × 10²⁶ × 235 × 1.661 × 10⁻²⁷ = 58 kg	✓ 1	Accept 57–58 kg. Using 4.7 × 10¹⁸ s⁻¹ gives 57.9 kg. Allow ECF from (b)(i).
Part (c)
useful power = 0.30 × 150 MW = 45 MW	✓ 1
the velocity is constant, so the resultant force is zero: the forward (driving) force equals the resistive force	✓ 1	Accept a reference to Newton's first law.
F = P ÷ v = 45 × 10⁶ ÷ 15 = 3.0 × 10⁶ N	✓ 1	Using 150 MW instead of 45 MW gives 1.0 × 10⁷ N: award 2 max.
Part (d)(i)
momentum: mv = 0 + mV, so the proton moves off with speed V = v in the neutron's original direction	✓ 1	Direction needed.
kinetic energy after = ½mv² = kinetic energy before, so kinetic energy is conserved and the collision is elastic	✓ 1
Part (d)(ii)
in a collision with a heavy nucleus such as uranium-238 the neutron keeps almost all its kinetic energy, so it is hardly slowed down	✓ 1
in a collision with a light nucleus such as a proton the neutron can lose a large fraction (up to all) of its kinetic energy, so far fewer collisions are needed to slow it down	✓ 1
slow (thermal) neutrons are much more likely than fast neutrons to cause fission of uranium-235, so a chain reaction can be sustained	✓ 1	Do not credit "the moderator absorbs neutrons".
Part (e)(i)
20 collisions	✓ 1	Accept 19–21.
Part (e)(ii)
f²⁰ = 0.025 ÷ (2.0 × 10⁶) = 1.25 × 10⁻⁸	✓ 1	Allow ECF from (e)(i): 19 collisions gives 0.38, 21 gives 0.42.
f = (1.25 × 10⁻⁸)^1/20 = 0.40	✓ 1
Part (e)(iii)
the neutron is now in thermal equilibrium with the moderator: the moderator nuclei have a similar average kinetic energy, so on average the neutron gains as much energy in collisions as it loses	✓ 1
Part (f)(i)
factor = 1.002⁵⁰⁰ = 2.72	✓ 1
power = 150 × 2.72 = 410 MW	✓ 1	Accept 407 MW.
Part (f)(ii)
supercritical	✓ 1
the control rods are inserted further so that they absorb more neutrons; on average fewer than one neutron per fission then causes another fission and the power falls	✓ 1
when the power has fallen to 150 MW the rods are set so that, on average, exactly one neutron from each fission causes another fission: the reactor is critical again	✓ 1	Accept "the multiplication factor is returned to 1".

Answers: (b)(i) 4.69 × 10¹⁸ s⁻¹ · (b)(ii) 58 kg (57.7 kg) · (c) 3.0 × 10⁶ N · (d)(i) speed v, in the neutron's original direction · (e)(i) 20 · (e)(ii) 0.40 · (f)(i) 410 MW (407 MW)

Common error (d)(ii) Mixing up the two jobs: the moderator slows neutrons down by collisions; control rods absorb neutrons. Write the verb that matches the component.

Exam tip (e)(ii) On a logarithmic scale, equal steps mean equal factors: a straight line on Figure 2 means the energy is multiplied by the same fraction in every collision.

More E.4 questions in the question bank

+ Add to test✓ In your testCC-IBP-E4-222 marks~25 min

Oklo: a natural nuclear reactor two billion years ago

E.4 FissionE.3 Radioactive DecayB.1 Thermal Energy TransfersA.3 Work, Energy and PowerSL

About two billion years ago, groundwater flowed through a rich deposit of uranium ore at Oklo, in present-day Gabon. In several zones of the deposit a self-sustaining chain reaction of uranium-235 fission started and continued, on and off, for hundreds of thousands of years. The groundwater acted as the moderator.

Data:

half-life of uranium-235 = 7.04 × 10⁸ years
half-life of uranium-238 = 4.47 × 10⁹ years
today, 0.720 % of the atoms in natural uranium are uranium-235; the rest are uranium-238

After n half-lives the fraction of the original nuclei that remain undecayed is (½)ⁿ, where n need not be a whole number.

(a)

Define half-life.

RECALLNotes

Show solution

the time taken for the number of undecayed nuclei (or the activity) of a sample of a radioactive nuclide to fall to half of its initial value

✓ 1

Accept "time for half the nuclei to decay". Do not accept "time for half the mass to disappear".

(1)

(b)

A sample of natural uranium today contains 7200 uranium-235 atoms for every 992 800 uranium-238 atoms. Assume that uranium in the deposit has changed only by radioactive decay, and ignore the uranium-235 used up by fission.

Show that, 2.0 × 10⁹ years ago, about 3.7 % of the uranium atoms in the deposit were uranium-235.

DERIVENotes

Show solution

uranium-235: n = 2.0 × 10⁹ ÷ 7.04 × 10⁸ = 2.84, so the number then = 7200 × 2^2.84 = 5.16 × 10⁴	✓ 1
uranium-238: n = 2.0 × 10⁹ ÷ 4.47 × 10⁹ = 0.447, so the number then = 992 800 × 2^0.447 = 1.354 × 10⁶	✓ 1
percentage = 5.16 × 10⁴ ÷ (5.16 × 10⁴ + 1.354 × 10⁶) × 100 = 3.67 % ≈ 3.7 %	✓ 1	Working with the ratio of the two numbers is equally acceptable (0.00725 × 5.25 = 0.0381), but the final step must convert the ratio to a percentage of all uranium atoms.

Answer: 3.67 %

(3)

(c)(i)

Figure 1 shows how the percentage of uranium-235 in natural uranium has changed over the last 3.0 × 10⁹ years.

Assume that, in a deposit like Oklo, a chain reaction moderated by groundwater can be sustained only if at least 3.0 % of the uranium atoms are uranium-235. Use Figure 1 to determine the most recent time at which this condition was met.

Figure 1: percentage of uranium-235 in natural uranium against time before the present

GRAPHNotes

Show solution

1.75 × 10⁹ years ago

✓ 1

Accept 1.7–1.8 × 10⁹ years.

Answer: 1.75 × 10⁹ years ago

(1)

(c)(ii)

Water-moderated power reactors today use fuel enriched to about 3–5 % uranium-235. Explain why a natural reactor like Oklo cannot start in a uranium deposit today.

EXPLAINNotes

Show solution

only 0.72 % of natural uranium is now uranium-235, far below the 3 % needed	✓ 1
too many neutrons are absorbed by uranium-238 (and by the water) without causing fission, so on average fewer than one neutron per fission causes another fission: the reaction is subcritical and cannot be sustained	✓ 1

(2)

(d)(i)

One possible fission reaction in the Oklo reactor is

²³⁵₉₂U + ¹₀n → ¹³⁹₅₄Xe + ⁹⁵_ZSr + x ¹₀n

Deduce the values of Z and x.

REASONNotes

Show solution

Z = 38, since 92 + 0 = 54 + Z	✓ 1
x = 2, since 235 + 1 = 139 + 95 + x	✓ 1

Answer: Z = 38, x = 2

(2)

(d)(ii)

The atomic masses are: uranium-235, 235.043928 u; xenon-139, 138.918792 u; strontium-95, 94.919358 u. The mass of a neutron is 1.008665 u.

Calculate, in MeV, the energy released in this fission.

MULTINotes

Show solution

mass before = 235.043928 + 1.008665 = 236.052593 u and mass after = 138.918792 + 94.919358 + 2 × 1.008665 = 235.855480 u	✓ 1	Allow ECF for the number of neutrons from (d)(i).
Δm = 0.197113 u	✓ 1
E = 0.197113 × 931.5 = 184 MeV	✓ 1	Accept 183.6 MeV. Going via kilograms and E = mc² gives 184 MeV.

Answer: 184 MeV

(3)

(e)(i)

Over the lifetime of the natural reactors, about 5.0 × 10³ kg of uranium-235 underwent fission. Using your answer to (d)(ii) as the energy released per fission, determine the total energy released.

MULTINotes

Show solution

number of nuclei = 5.0 × 10³ ÷ (235 × 1.661 × 10⁻²⁷) = 1.28 × 10²⁸	✓ 1
energy = 1.28 × 10²⁸ × 184 × 1.60 × 10⁻¹³ = 3.8 × 10¹⁷ J	✓ 1	Allow ECF from (d)(ii). Using 200 MeV gives 4.1 × 10¹⁷ J.

Answer: 3.8 × 10¹⁷ J

(2)

(e)(ii)

Coal releases 30 MJ kg⁻¹ when it is burned. Calculate the mass of coal that would release the same energy as in (e)(i), and comment on your answer.

CALCNotes

Show solution

mass = 3.8 × 10¹⁷ ÷ 30 × 10⁶ = 1.3 × 10¹⁰ kg	✓ 1	Allow ECF from (e)(i).
this is about 2.5 × 10⁶ times the mass of uranium-235 used: nuclear fuel releases millions of times more energy per kilogram than a chemical fuel	✓ 1	Any valid comparison of energy per unit mass.

Answer: 1.3 × 10¹⁰ kg

(2)

(f)(i)

During an active phase, which lasted about 30 minutes, the thermal power produced in one reaction zone was about 3.0 × 10⁵ W. Groundwater entered the zone at 30 °C. Assume that the water boils at 100 °C and that all the thermal energy is transferred to the water.

Specific heat capacity of water = 4180 J kg⁻¹ K⁻¹; specific latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹.

Calculate the maximum mass of water that could be turned into steam during one active phase.

MULTINotes

Show solution

energy = 3.0 × 10⁵ × 30 × 60 = 5.4 × 10⁸ J	✓ 1
energy needed per kilogram = 4180 × 70 + 2.26 × 10⁶ = 2.55 × 10⁶ J kg⁻¹	✓ 1
mass = 5.4 × 10⁸ ÷ 2.55 × 10⁶ = 210 kg	✓ 1	Accept 211–212 kg. Omitting the heating from 30 °C to 100 °C gives 239 kg: award 2 max.

Answer: 210 kg

(3)

(f)(ii)

Each active phase ended when the water in the zone boiled away, and a new phase started after about 2.5 hours.

Explain how the loss of liquid water stopped the chain reaction, and why the reaction later started again.

EXPLAINNotes

Show solution

without liquid water (steam has far fewer nuclei per unit volume) the fast neutrons are no longer slowed down / moderated	✓ 1
fast neutrons are much less likely to cause fission of uranium-235 (they escape or are absorbed by uranium-238), so on average fewer than one neutron per fission causes another fission and the chain reaction dies away	✓ 1
when the zone cools, groundwater flows back in / the steam condenses, so the neutrons are moderated again and the reaction becomes critical again	✓ 1	The idea of a self-regulating cycle is needed for this mark.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
the time taken for the number of undecayed nuclei (or the activity) of a sample of a radioactive nuclide to fall to half of its initial value	✓ 1	Accept "time for half the nuclei to decay". Do not accept "time for half the mass to disappear".
Part (b)
uranium-235: n = 2.0 × 10⁹ ÷ 7.04 × 10⁸ = 2.84, so the number then = 7200 × 2^2.84 = 5.16 × 10⁴	✓ 1
uranium-238: n = 2.0 × 10⁹ ÷ 4.47 × 10⁹ = 0.447, so the number then = 992 800 × 2^0.447 = 1.354 × 10⁶	✓ 1
percentage = 5.16 × 10⁴ ÷ (5.16 × 10⁴ + 1.354 × 10⁶) × 100 = 3.67 % ≈ 3.7 %	✓ 1	Working with the ratio of the two numbers is equally acceptable (0.00725 × 5.25 = 0.0381), but the final step must convert the ratio to a percentage of all uranium atoms.
Part (c)(i)
1.75 × 10⁹ years ago	✓ 1	Accept 1.7–1.8 × 10⁹ years.
Part (c)(ii)
only 0.72 % of natural uranium is now uranium-235, far below the 3 % needed	✓ 1
too many neutrons are absorbed by uranium-238 (and by the water) without causing fission, so on average fewer than one neutron per fission causes another fission: the reaction is subcritical and cannot be sustained	✓ 1
Part (d)(i)
Z = 38, since 92 + 0 = 54 + Z	✓ 1
x = 2, since 235 + 1 = 139 + 95 + x	✓ 1
Part (d)(ii)
mass before = 235.043928 + 1.008665 = 236.052593 u and mass after = 138.918792 + 94.919358 + 2 × 1.008665 = 235.855480 u	✓ 1	Allow ECF for the number of neutrons from (d)(i).
Δm = 0.197113 u	✓ 1
E = 0.197113 × 931.5 = 184 MeV	✓ 1	Accept 183.6 MeV. Going via kilograms and E = mc² gives 184 MeV.
Part (e)(i)
number of nuclei = 5.0 × 10³ ÷ (235 × 1.661 × 10⁻²⁷) = 1.28 × 10²⁸	✓ 1
energy = 1.28 × 10²⁸ × 184 × 1.60 × 10⁻¹³ = 3.8 × 10¹⁷ J	✓ 1	Allow ECF from (d)(ii). Using 200 MeV gives 4.1 × 10¹⁷ J.
Part (e)(ii)
mass = 3.8 × 10¹⁷ ÷ 30 × 10⁶ = 1.3 × 10¹⁰ kg	✓ 1	Allow ECF from (e)(i).
this is about 2.5 × 10⁶ times the mass of uranium-235 used: nuclear fuel releases millions of times more energy per kilogram than a chemical fuel	✓ 1	Any valid comparison of energy per unit mass.
Part (f)(i)
energy = 3.0 × 10⁵ × 30 × 60 = 5.4 × 10⁸ J	✓ 1
energy needed per kilogram = 4180 × 70 + 2.26 × 10⁶ = 2.55 × 10⁶ J kg⁻¹	✓ 1
mass = 5.4 × 10⁸ ÷ 2.55 × 10⁶ = 210 kg	✓ 1	Accept 211–212 kg. Omitting the heating from 30 °C to 100 °C gives 239 kg: award 2 max.
Part (f)(ii)
without liquid water (steam has far fewer nuclei per unit volume) the fast neutrons are no longer slowed down / moderated	✓ 1
fast neutrons are much less likely to cause fission of uranium-235 (they escape or are absorbed by uranium-238), so on average fewer than one neutron per fission causes another fission and the chain reaction dies away	✓ 1
when the zone cools, groundwater flows back in / the steam condenses, so the neutrons are moderated again and the reaction becomes critical again	✓ 1	The idea of a self-regulating cycle is needed for this mark.

Answers: (b) 3.67 % · (c)(i) 1.75 × 10⁹ years ago · (d)(i) Z = 38, x = 2 · (d)(ii) 184 MeV · (e)(i) 3.8 × 10¹⁷ J · (e)(ii) 1.3 × 10¹⁰ kg · (f)(i) 210 kg

Common error (b) Stopping at the ratio of uranium-235 to uranium-238 atoms (3.81 %). The question asks for the percentage of all uranium atoms, so divide by the total.

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E.5 Fusion and Stars

+ Add to test✓ In your testCC-IBP-E5-124 marks~30 min

Fusion in the Sun's core

E.5 Fusion and StarsB.1 Thermal Energy TransfersE.3 Radioactive DecaySL

The Sun is a main-sequence star. In its core, hydrogen nuclei fuse to form helium.

Data for the Sun:

luminosity L_☉ = 3.83 × 10²⁶ W
radius R_☉ = 6.96 × 10⁸ m
mass M_☉ = 2.0 × 10³⁰ kg

(a)

Explain why the Sun neither collapses nor expands.

EXPLAINNotes

Show solution

an outward force due to radiation pressure (and gas pressure), maintained by the energy released by fusion in the core	✓ 1	The word "radiation pressure" or "gas pressure" and the outward direction are needed.
is equal in magnitude and opposite in direction to the inward gravitational force on each layer, so the resultant force is zero	✓ 1	"Pressure balances gravity" on its own scores this mark only if both directions are clear.

(2)

(b)(i)

Overall, four protons form one helium-4 nucleus. Complete the equation for this overall change by identifying the two missing particles.

4 ¹₁H → ⁴₂He + 2 ____ + 2 ____

RECALLNotes

Show solution

2 positrons (⁰₊₁e) and 2 (electron) neutrinos (ν_e)

✓ 1

Both particles needed. Do not accept electrons.

Answer: 2 ⁰₊₁e + 2 ν_e

(1)

(b)(ii)

The following masses are given: proton = 1.007276 u; helium-4 nucleus = 4.001506 u; positron = 0.000549 u. The mass of a neutrino is negligible.

Calculate, in MeV, the energy released in the overall change in (b)(i).

CALCNotes

Show solution

Δm = 4 × 1.007276 − 4.001506 − 2 × 0.000549 = 0.026500 u	✓ 1	Omitting the positrons gives 0.027598 u (25.7 MeV): award 1 max.
E = 0.026500 × 931.5 = 24.7 MeV	✓ 1

Answer: 24.7 MeV

(2)

(b)(iii)

Each positron annihilates with an electron in the core. The mass of an electron is 0.000549 u.

Show that the total energy released for each helium-4 nucleus formed is about 26.7 MeV.

DERIVENotes

Show solution

two annihilations convert 4 × 0.000549 = 0.002196 u, releasing 0.002196 × 931.5 = 2.05 MeV	✓ 1
total = 24.68 + 2.05 = 26.73 MeV ≈ 26.7 MeV	✓ 1	Allow ECF from (b)(ii) only if the final value is shown to round to 26.7 MeV.

Answer: 26.7 MeV

(2)

(c)

Determine the mass of hydrogen that is converted into helium in the Sun each second. Take the mass of a hydrogen nucleus to be 1.007276 u.

MULTINotes

Show solution

26.7 MeV = 26.7 × 1.60 × 10⁻¹³ = 4.27 × 10⁻¹² J	✓ 1
number of helium nuclei formed per second = 3.83 × 10²⁶ ÷ 4.27 × 10⁻¹² = 8.96 × 10³⁷ s⁻¹	✓ 1
mass = 8.96 × 10³⁷ × 4 × 1.007276 × 1.661 × 10⁻²⁷ = 6.0 × 10¹¹ kg	✓ 1	Forgetting the factor of 4 gives 1.5 × 10¹¹ kg: award 2 max.

Answer: 6.0 × 10¹¹ kg

(3)

(d)

Assume that the Sun leaves the main sequence when a mass of hydrogen equal to 10 % of its total mass has been fused into helium, and that its luminosity stays constant until then.

Estimate the main-sequence lifetime of the Sun in years.

CALCNotes

Show solution

mass fused = 0.10 × 2.0 × 10³⁰ = 2.0 × 10²⁹ kg	✓ 1
time = 2.0 × 10²⁹ ÷ 6.0 × 10¹¹ = 3.3 × 10¹⁷ s = 1.1 × 10¹⁰ years	✓ 1	Allow ECF from (c).

Answer: 1.1 × 10¹⁰ years

(2)

(e)

Show that the surface temperature of the Sun is about 5800 K.

DERIVENotes

Show solution

L = σ4πR²T⁴, so T⁴ = L ÷ (σ4πR²)	✓ 1
T⁴ = 3.83 × 10²⁶ ÷ (5.67 × 10⁻⁸ × 4π × (6.96 × 10⁸)²) = 1.11 × 10¹⁵ K⁴	✓ 1
T = 5.77 × 10³ K	✓ 1	The answer must be given to at least 3 s.f. Using πR² instead of 4πR² scores 0 for the last two marks.

Answer: 5770 K

(3)

(f)

Calculate the wavelength at which the Sun's emission spectrum peaks. Use a surface temperature of 5800 K.

CALCNotes

Show solution

λ_max = 2.9 × 10⁻³ ÷ 5800 = 5.0 × 10⁻⁷ m

✓ 1

Accept 500 nm.

Answer: 5.0 × 10⁻⁷ m

(1)

(g)

The temperature in the Sun's core is about 1.5 × 10⁷ K. Explain why fusion needs both a very high temperature and a very high density, and why it takes place only in the core of the Sun.

EXPLAINNotes

Show solution

nuclei are all positively charged, so they repel each other electrically	✓ 1
only at a very high temperature do the nuclei have enough kinetic energy to get close enough (about 10⁻¹⁵ m) for the strong nuclear force to bind them	✓ 1	Accept a reference to Ē_k = 3/2 k_BT.
a very high density makes collisions frequent enough for the rate of fusion to supply the Sun's luminosity; only the core is hot and dense enough (the surface, at about 5800 K, is far too cool)	✓ 1

(3)

(h)(i)

Figure 1 is a Hertzsprung–Russell diagram on which five stars, A to E, are marked.

Identify the star that represents the Sun today, and the star that represents the Sun at the end of its evolution. Give a reason for each choice.

Figure 1: Hertzsprung–Russell diagram (logarithmic scales; L☉ is the luminosity of the Sun)

GRAPHNotes

Show solution

A: on the main sequence with a surface temperature of about 5800 K and a luminosity of 1 L_☉	✓ 1
D: a white dwarf (hot but very dim, so very small); a star of the Sun's mass (below about 8 M_☉) ends as a white dwarf	✓ 1	Do not accept C: a red giant is a stage the Sun passes through, not its final state.

Answer: A (Sun today); D (white dwarf)

(2)

(h)(ii)

A main-sequence star has a mass of 4.0 M_☉. Assume that, for main-sequence stars, the luminosity L is proportional to M^3.5, and that the star fuses the same fraction of its mass during the main sequence as the Sun.

Determine the luminosity of this star in terms of L_☉ and its main-sequence lifetime. Hence identify the star in Figure 1 that it could be.

PREDICTNotes

Show solution

L = 4.0^3.5 L_☉ = 128 L_☉	✓ 1
lifetime ∝ M/L: lifetime = 1.1 × 10¹⁰ × 4.0 ÷ 128 = 3.4 × 10⁸ years	✓ 1	Allow ECF from (d). Accept 3.3–3.4 × 10⁸ years.
B: the main-sequence star with a luminosity of about 130 L_☉	✓ 1

Answer: 128 L_☉; 3.4 × 10⁸ years; star B

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
an outward force due to radiation pressure (and gas pressure), maintained by the energy released by fusion in the core	✓ 1	The word "radiation pressure" or "gas pressure" and the outward direction are needed.
is equal in magnitude and opposite in direction to the inward gravitational force on each layer, so the resultant force is zero	✓ 1	"Pressure balances gravity" on its own scores this mark only if both directions are clear.
Part (b)(i)
2 positrons (⁰₊₁e) and 2 (electron) neutrinos (ν_e)	✓ 1	Both particles needed. Do not accept electrons.
Part (b)(ii)
Δm = 4 × 1.007276 − 4.001506 − 2 × 0.000549 = 0.026500 u	✓ 1	Omitting the positrons gives 0.027598 u (25.7 MeV): award 1 max.
E = 0.026500 × 931.5 = 24.7 MeV	✓ 1
Part (b)(iii)
two annihilations convert 4 × 0.000549 = 0.002196 u, releasing 0.002196 × 931.5 = 2.05 MeV	✓ 1
total = 24.68 + 2.05 = 26.73 MeV ≈ 26.7 MeV	✓ 1	Allow ECF from (b)(ii) only if the final value is shown to round to 26.7 MeV.
Part (c)
26.7 MeV = 26.7 × 1.60 × 10⁻¹³ = 4.27 × 10⁻¹² J	✓ 1
number of helium nuclei formed per second = 3.83 × 10²⁶ ÷ 4.27 × 10⁻¹² = 8.96 × 10³⁷ s⁻¹	✓ 1
mass = 8.96 × 10³⁷ × 4 × 1.007276 × 1.661 × 10⁻²⁷ = 6.0 × 10¹¹ kg	✓ 1	Forgetting the factor of 4 gives 1.5 × 10¹¹ kg: award 2 max.
Part (d)
mass fused = 0.10 × 2.0 × 10³⁰ = 2.0 × 10²⁹ kg	✓ 1
time = 2.0 × 10²⁹ ÷ 6.0 × 10¹¹ = 3.3 × 10¹⁷ s = 1.1 × 10¹⁰ years	✓ 1	Allow ECF from (c).
Part (e)
L = σ4πR²T⁴, so T⁴ = L ÷ (σ4πR²)	✓ 1
T⁴ = 3.83 × 10²⁶ ÷ (5.67 × 10⁻⁸ × 4π × (6.96 × 10⁸)²) = 1.11 × 10¹⁵ K⁴	✓ 1
T = 5.77 × 10³ K	✓ 1	The answer must be given to at least 3 s.f. Using πR² instead of 4πR² scores 0 for the last two marks.
Part (f)
λ_max = 2.9 × 10⁻³ ÷ 5800 = 5.0 × 10⁻⁷ m	✓ 1	Accept 500 nm.
Part (g)
nuclei are all positively charged, so they repel each other electrically	✓ 1
only at a very high temperature do the nuclei have enough kinetic energy to get close enough (about 10⁻¹⁵ m) for the strong nuclear force to bind them	✓ 1	Accept a reference to Ē_k = 3/2 k_BT.
a very high density makes collisions frequent enough for the rate of fusion to supply the Sun's luminosity; only the core is hot and dense enough (the surface, at about 5800 K, is far too cool)	✓ 1
Part (h)(i)
A: on the main sequence with a surface temperature of about 5800 K and a luminosity of 1 L_☉	✓ 1
D: a white dwarf (hot but very dim, so very small); a star of the Sun's mass (below about 8 M_☉) ends as a white dwarf	✓ 1	Do not accept C: a red giant is a stage the Sun passes through, not its final state.
Part (h)(ii)
L = 4.0^3.5 L_☉ = 128 L_☉	✓ 1
lifetime ∝ M/L: lifetime = 1.1 × 10¹⁰ × 4.0 ÷ 128 = 3.4 × 10⁸ years	✓ 1	Allow ECF from (d). Accept 3.3–3.4 × 10⁸ years.
B: the main-sequence star with a luminosity of about 130 L_☉	✓ 1

Answers: (b)(i) 2 ⁰₊₁e + 2 ν_e · (b)(ii) 24.7 MeV · (b)(iii) 26.7 MeV · (c) 6.0 × 10¹¹ kg · (d) 1.1 × 10¹⁰ years · (e) 5770 K · (f) 5.0 × 10⁻⁷ m · (h)(i) A (Sun today); D (white dwarf) · (h)(ii) 128 L_☉; 3.4 × 10⁸ years; star B

Common error (c) Confusing the mass of hydrogen fused (6.0 × 10¹¹ kg each second) with the mass converted to energy (L/c² = 4.3 × 10⁹ kg each second). Only about 0.7 % of the hydrogen's mass is released as energy.

More E.5 questions in the question bank

+ Add to test✓ In your testCC-IBP-E5-222 marks~25 min

Betelgeuse: measuring a red supergiant

E.5 Fusion and StarsB.1 Thermal Energy TransfersC.5 Doppler EffectSL

Betelgeuse is a red supergiant in the constellation Orion. Data for Betelgeuse:

parallax angle = 0.0059 arc-second
apparent brightness = 1.3 × 10⁻⁷ W m⁻²
surface temperature = 3600 K

Data for the Sun: luminosity L_☉ = 3.83 × 10²⁶ W; radius R_☉ = 6.96 × 10⁸ m; surface temperature = 5800 K.

(a)

Define the parsec.

RECALLNotes

Show solution

the distance at which a length of 1 AU subtends an angle of 1 arc-second

✓ 1

Accept "the distance to a star whose parallax angle is 1 arc-second".

(1)

(b)

Calculate the distance from the Earth to Betelgeuse in metres.

CALCNotes

Show solution

d = 1 ÷ 0.0059 = 169 pc	✓ 1
d = 169 × 3.26 × 9.46 × 10¹⁵ = 5.2 × 10¹⁸ m	✓ 1	Accept 5.21–5.23 × 10¹⁸ m.

Answer: 5.2 × 10¹⁸ m

(2)

(c)

Determine the luminosity of Betelgeuse, in W and in terms of L_☉.

CALCNotes

Show solution

L = 4πd²b = 4π × (5.23 × 10¹⁸)² × 1.3 × 10⁻⁷ = 4.5 × 10³¹ W	✓ 1	Allow ECF from (b).
L = 4.46 × 10³¹ ÷ 3.83 × 10²⁶ = 1.2 × 10⁵ L_☉	✓ 1

Answer: 4.5 × 10³¹ W = 1.2 × 10⁵ L_☉

(2)

(d)(i)

Calculate the wavelength at which the emission spectrum of Betelgeuse peaks, and state the region of the electromagnetic spectrum in which this wavelength lies.

CALCNotes

Show solution

λ_max = 2.9 × 10⁻³ ÷ 3600 = 8.1 × 10⁻⁷ m	✓ 1
infrared (near infrared)	✓ 1	Allow ECF from an incorrect wavelength.

Answer: 8.1 × 10⁻⁷ m, infrared

(2)

(d)(ii)

Show that the radius of Betelgeuse is about 900 R_☉.

DERIVENotes

Show solution

R = √(L ÷ 4πσT⁴)	✓ 1	Or the ratio method: R/R_☉ = √(L/L_☉) × (T_☉/T)².
R = √(4.46 × 10³¹ ÷ (4π × 5.67 × 10⁻⁸ × 3600⁴)) = 6.1 × 10¹¹ m	✓ 1	Ratio method: √(1.17 × 10⁵) × (5800/3600)² = 342 × 2.60.
R = 6.1 × 10¹¹ ÷ 6.96 × 10⁸ = 880 R_☉ ≈ 900 R_☉	✓ 1	Accept 870–890 R_☉. The value must be seen to more than 1 s.f.

Answer: 880 R_☉

(3)

(e)

Figure 1 is a Hertzsprung–Russell diagram showing the Sun (S), Betelgeuse (B) and lines of constant radius.

Using Figure 1, explain why Betelgeuse must be very large even though its surface is cooler than the Sun's surface.

Figure 1: Hertzsprung–Russell diagram showing the Sun (S), Betelgeuse (B) and lines of constant radius (dashed); logarithmic scales

GRAPHNotes

Show solution

B has a lower surface temperature than S, so each square metre of its surface emits less power, yet its luminosity is about 10⁵ times greater, so its surface area must be enormous	✓ 1
B lies just below the 1000 R_☉ line (between the 100 R_☉ and 1000 R_☉ lines, close to 1000 R_☉), at the top right of the diagram, consistent with about 900 R_☉	✓ 1

(2)

(f)

The mass of Betelgeuse is estimated to be about 18 M_☉, where M_☉ is the mass of the Sun. Outline the likely future evolution of Betelgeuse, including its final state.

EXPLAINNotes

Show solution

the core fuses successively heavier nuclei (helium → carbon → … → silicon → iron), in shells around the core	✓ 1
fusing iron releases no energy, so the iron core collapses and the star explodes as a supernova	✓ 1
the remnant core becomes a neutron star, or a black hole if its mass is greater than about 3 M_☉	✓ 1	Do not accept white dwarf.

(3)

(g)

In the laboratory, a sodium absorption line has a wavelength of 589.592 nm. In the spectrum of Betelgeuse the same line is observed at 589.635 nm.

Determine the speed of Betelgeuse along the line of sight, and state its direction of motion relative to the Earth.

CALCNotes

Show solution

v = cΔλ/λ = 3.00 × 10⁸ × 0.043 ÷ 589.592 = 2.2 × 10⁴ m s⁻¹	✓ 1
the wavelength is longer (red-shift), so Betelgeuse is moving away from the Earth	✓ 1

Answer: 2.2 × 10⁴ m s⁻¹, away from the Earth

(2)

(h)

The parallax angle of Betelgeuse has an absolute uncertainty of ±0.0008 arc-second. Assume that the uncertainties in the apparent brightness and the surface temperature are negligible.

Determine the percentage uncertainty in the radius found in (d)(ii). Suggest why the parallax uncertainty is so significant for Betelgeuse.

DATANotes

Show solution

percentage uncertainty in d = percentage uncertainty in p = 0.0008 ÷ 0.0059 × 100 = 14 %	✓ 1
L ∝ d² and, at fixed T, R ∝ √L, so R ∝ d: the percentage uncertainty in R is also 14 %	✓ 1	28 % is the uncertainty in L, not in R: stopping at 28 % does not score this mark. Accept R ∝ d stated directly.
Betelgeuse is far away, so its parallax angle is very small, comparable to the smallest angles that can be measured; the absolute uncertainty is a large fraction of the angle	✓ 1

Answer: 14 %

(3)

(i)

Between late 2019 and early 2020 Betelgeuse dimmed noticeably. Suppose that its luminosity decreased by 20 % while its radius stayed the same.

Predict the new surface temperature of Betelgeuse.

PREDICTNotes

Show solution

at constant radius L ∝ T⁴, so T ∝ L^1/4	✓ 1
T = 3600 × 0.80^1/4 = 3400 K	✓ 1	Accept 3405 K. Using a factor of 0.20 instead of 0.80 gives 2400 K: award 1 max.

Answer: 3400 K

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the distance at which a length of 1 AU subtends an angle of 1 arc-second	✓ 1	Accept "the distance to a star whose parallax angle is 1 arc-second".
Part (b)
d = 1 ÷ 0.0059 = 169 pc	✓ 1
d = 169 × 3.26 × 9.46 × 10¹⁵ = 5.2 × 10¹⁸ m	✓ 1	Accept 5.21–5.23 × 10¹⁸ m.
Part (c)
L = 4πd²b = 4π × (5.23 × 10¹⁸)² × 1.3 × 10⁻⁷ = 4.5 × 10³¹ W	✓ 1	Allow ECF from (b).
L = 4.46 × 10³¹ ÷ 3.83 × 10²⁶ = 1.2 × 10⁵ L_☉	✓ 1
Part (d)(i)
λ_max = 2.9 × 10⁻³ ÷ 3600 = 8.1 × 10⁻⁷ m	✓ 1
infrared (near infrared)	✓ 1	Allow ECF from an incorrect wavelength.
Part (d)(ii)
R = √(L ÷ 4πσT⁴)	✓ 1	Or the ratio method: R/R_☉ = √(L/L_☉) × (T_☉/T)².
R = √(4.46 × 10³¹ ÷ (4π × 5.67 × 10⁻⁸ × 3600⁴)) = 6.1 × 10¹¹ m	✓ 1	Ratio method: √(1.17 × 10⁵) × (5800/3600)² = 342 × 2.60.
R = 6.1 × 10¹¹ ÷ 6.96 × 10⁸ = 880 R_☉ ≈ 900 R_☉	✓ 1	Accept 870–890 R_☉. The value must be seen to more than 1 s.f.
Part (e)
B has a lower surface temperature than S, so each square metre of its surface emits less power, yet its luminosity is about 10⁵ times greater, so its surface area must be enormous	✓ 1
B lies just below the 1000 R_☉ line (between the 100 R_☉ and 1000 R_☉ lines, close to 1000 R_☉), at the top right of the diagram, consistent with about 900 R_☉	✓ 1
Part (f)
the core fuses successively heavier nuclei (helium → carbon → … → silicon → iron), in shells around the core	✓ 1
fusing iron releases no energy, so the iron core collapses and the star explodes as a supernova	✓ 1
the remnant core becomes a neutron star, or a black hole if its mass is greater than about 3 M_☉	✓ 1	Do not accept white dwarf.
Part (g)
v = cΔλ/λ = 3.00 × 10⁸ × 0.043 ÷ 589.592 = 2.2 × 10⁴ m s⁻¹	✓ 1
the wavelength is longer (red-shift), so Betelgeuse is moving away from the Earth	✓ 1
Part (h)
percentage uncertainty in d = percentage uncertainty in p = 0.0008 ÷ 0.0059 × 100 = 14 %	✓ 1
L ∝ d² and, at fixed T, R ∝ √L, so R ∝ d: the percentage uncertainty in R is also 14 %	✓ 1	28 % is the uncertainty in L, not in R: stopping at 28 % does not score this mark. Accept R ∝ d stated directly.
Betelgeuse is far away, so its parallax angle is very small, comparable to the smallest angles that can be measured; the absolute uncertainty is a large fraction of the angle	✓ 1
Part (i)
at constant radius L ∝ T⁴, so T ∝ L^1/4	✓ 1
T = 3600 × 0.80^1/4 = 3400 K	✓ 1	Accept 3405 K. Using a factor of 0.20 instead of 0.80 gives 2400 K: award 1 max.

Answers: (b) 5.2 × 10¹⁸ m · (c) 4.5 × 10³¹ W = 1.2 × 10⁵ L_☉ · (d)(i) 8.1 × 10⁻⁷ m, infrared · (d)(ii) 880 R_☉ · (g) 2.2 × 10⁴ m s⁻¹, away from the Earth · (h) 14 % · (i) 3400 K

Exam tip (d)(ii) In the ratio method the luminosity ratio is square-rooted but the temperature ratio is squared, because L ∝ R²T⁴.

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