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Edexcel IGCSE Physics 4PH1 · Concept clusters

Edexcel IGCSE 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. 44 questions · 857 marks · two per subchapter · every part with a worked solution · All courses

Topic

Subchapter

Level

Skill

Topic 1 · Forces and motion

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

Setup	1(b)	1(c)	Other
A 100 m sprint start: timing gates, acceleration and speed	●	○	4(c)
A driverless metro train through a track repair zone: velocity–time graph	●	○	4(c)
An e-scooter emergency stop: reaction time, braking and stopping distance	○	●	4(c)
A tower crane on a building site: moments and the counterweight	○	●	4(c)

1(b) Movement and Position

+ Add to test✓ In your testCC-EIP-1b-120 marks~20 min

A 100 m sprint start: timing gates, acceleration and speed

1(b) Movement and Position1(c) Forces, Movement, Shape and Momentum4(c) Work and PowerBoth papers

A sprinter of mass 72 kg starts a 100 m race from rest. Light gates placed beside the straight track record the time at which she passes each gate. The clock starts as she leaves the blocks.

distance from start / m	0	5.0	10.0	15.0	20.0	30.0	40.0
time / s	0	2.00	2.83	3.46	4.00	5.00	6.00

(a)

Calculate the average speed of the sprinter between the 20.0 m gate and the 40.0 m gate.

CALCNotes

Show solution

average speed = distance moved ÷ time taken = (40.0 − 20.0) ÷ (6.00 − 4.00)	✓ 1	Substitution of 20 (m) and 2.00 (s) earns this mark.
= 10 m/s	✓ 1	Unit required. 6.7 m/s (40 ÷ 6.00) scores 0 for this mark.

Answer: 10 m/s

(2)

(b)

On the grid in Figure 1, plot a distance–time graph of the results in the table.

Draw a suitable line through your points.

Figure 1: grid for the distance–time graph

GRAPHNotes

Show solution

at least five points plotted correctly, to within half a small square	✓ 1
all seven points plotted correctly	✓ 1
single smooth curve from the origin through the points up to 20 m (4.00 s), continuing as a straight line to 40 m (6.00 s)	✓ 1	Do not accept points joined dot to dot with straight segments in the curved region.

(3)

(c)

Describe the motion of the sprinter shown by your graph. Use values from the table in your answer.

EXPLAINNotes

Show solution

0 to 4.00 s / first 20 m: the gradient increases, so the speed is increasing / she accelerates	✓ 1	Accept: the graph curves upwards, so she speeds up.
4.00 s to 6.00 s / 20 m to 40 m: straight line, so she moves at a constant speed of 10 m/s	✓ 1	Allow ECF from (a) for the value of the speed. A value is needed for this mark.

(2)

(d)

The sprinter reaches her top speed as she passes the 20.0 m gate. Assume that her acceleration is uniform until then.

Show that her acceleration during the first 4.00 s is 2.5 m/s².

CALCNotes

Show solution

a = (v − u) ÷ t = (10 − 0) ÷ 4.00	✓ 1	Top speed 10 m/s from (a) must be used with t = 4.00 s. Alternative: average speed for the first 20 m = 20.0 ÷ 4.00 = 5.0 m/s, so top speed = 2 × 5.0 = 10 m/s, then a = 10 ÷ 4.00.
= 2.5 (m/s²)	✓ 1	Answer given: the substitution must be seen.

Answer: 2.5 m/s²

(2)

(e)

Calculate the resultant force on the sprinter during the first 4.00 s.

CALCNotes

Show solution

F = m × a = 72 × 2.5	✓ 1
= 180 N (forwards)	✓ 1	Unit required.

Answer: 180 N

(2)

(f)

Ignore air resistance.

Calculate the kinetic energy of the sprinter at her top speed, and the average useful power she develops during the first 4.00 s.

MULTINotes

Show solution

KE = ½ × m × v² = ½ × 72 × 10²	✓ 1
= 3600 J	✓ 1	Allow ECF from (a) for the speed.
P = W ÷ t = 3600 ÷ 4.00 = 900 W	✓ 1	Allow ECF from the kinetic energy. Accept work done = 180 N × 20 m = 3600 J as an alternative route to the energy.

Answer: 3600 J; 900 W

(3)

(g)(i)

In a training run the sprinter starts without using the blocks. Her acceleration from rest is uniform but is now only 2.0 m/s². She still reaches the same top speed of 10 m/s and then runs at this constant speed.

Calculate the time at which she passes the 40.0 m gate in this run.

MULTINotes

Show solution

distance to reach top speed: v² = u² + 2as ⇒ 10² = 0 + 2 × 2.0 × s, so s = 25 m	✓ 1	Accept distance = area under the velocity–time graph = ½ × 5.0 × 10 = 25 m.
time to reach top speed = (10 − 0) ÷ 2.0 = 5.0 s	✓ 1
time for the remaining 15 m = 15 ÷ 10 = 1.5 s	✓ 1	Allow ECF from an incorrect distance: (40 − s) ÷ 10.
total time = 5.0 + 1.5 = 6.5 s	✓ 1	Using v² = u² + 2as for the whole 40 m scores 0 for the last two marks.

Answer: 6.5 s

(4)

(g)(ii)

Describe how the distance–time graph for this run would differ from the graph you drew in (b).

PREDICTNotes

Show solution

the curved section lasts longer (to 5.0 s and 25 m) and lies below / to the right of the first curve	✓ 1	Accept: it is less steep at the start / takes longer to become straight. Allow ECF from (g)(i).
the straight section has the same gradient (is parallel to the first one) but reaches 40 m later, at 6.5 s	✓ 1	Accept: same steepness, shifted to later times.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
average speed = distance moved ÷ time taken = (40.0 − 20.0) ÷ (6.00 − 4.00)	✓ 1	Substitution of 20 (m) and 2.00 (s) earns this mark.
= 10 m/s	✓ 1	Unit required. 6.7 m/s (40 ÷ 6.00) scores 0 for this mark.
Part (b)
at least five points plotted correctly, to within half a small square	✓ 1
all seven points plotted correctly	✓ 1
single smooth curve from the origin through the points up to 20 m (4.00 s), continuing as a straight line to 40 m (6.00 s)	✓ 1	Do not accept points joined dot to dot with straight segments in the curved region.
Part (c)
0 to 4.00 s / first 20 m: the gradient increases, so the speed is increasing / she accelerates	✓ 1	Accept: the graph curves upwards, so she speeds up.
4.00 s to 6.00 s / 20 m to 40 m: straight line, so she moves at a constant speed of 10 m/s	✓ 1	Allow ECF from (a) for the value of the speed. A value is needed for this mark.
Part (d)
a = (v − u) ÷ t = (10 − 0) ÷ 4.00	✓ 1	Top speed 10 m/s from (a) must be used with t = 4.00 s. Alternative: average speed for the first 20 m = 20.0 ÷ 4.00 = 5.0 m/s, so top speed = 2 × 5.0 = 10 m/s, then a = 10 ÷ 4.00.
= 2.5 (m/s²)	✓ 1	Answer given: the substitution must be seen.
Part (e)
F = m × a = 72 × 2.5	✓ 1
= 180 N (forwards)	✓ 1	Unit required.
Part (f)
KE = ½ × m × v² = ½ × 72 × 10²	✓ 1
= 3600 J	✓ 1	Allow ECF from (a) for the speed.
P = W ÷ t = 3600 ÷ 4.00 = 900 W	✓ 1	Allow ECF from the kinetic energy. Accept work done = 180 N × 20 m = 3600 J as an alternative route to the energy.
Part (g)(i)
distance to reach top speed: v² = u² + 2as ⇒ 10² = 0 + 2 × 2.0 × s, so s = 25 m	✓ 1	Accept distance = area under the velocity–time graph = ½ × 5.0 × 10 = 25 m.
time to reach top speed = (10 − 0) ÷ 2.0 = 5.0 s	✓ 1
time for the remaining 15 m = 15 ÷ 10 = 1.5 s	✓ 1	Allow ECF from an incorrect distance: (40 − s) ÷ 10.
total time = 5.0 + 1.5 = 6.5 s	✓ 1	Using v² = u² + 2as for the whole 40 m scores 0 for the last two marks.
Part (g)(ii)
the curved section lasts longer (to 5.0 s and 25 m) and lies below / to the right of the first curve	✓ 1	Accept: it is less steep at the start / takes longer to become straight. Allow ECF from (g)(i).
the straight section has the same gradient (is parallel to the first one) but reaches 40 m later, at 6.5 s	✓ 1	Accept: same steepness, shifted to later times.

Answers: (a) 10 m/s · (d) 2.5 m/s² · (e) 180 N · (f) 3600 J; 900 W · (g)(i) 6.5 s

Exam tip (b) Join the points with one smooth line of best fit. A curve that bends into a straight line is a single line; do not join neighbouring points with a ruler.

Common error (g)(i) Applying v² = u² + 2as to the whole 40 m. The equation only holds while the acceleration is uniform, so split the run into the accelerating part and the constant-speed part.

More 1(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-1b-219 marks~20 min

A driverless metro train through a track repair zone: velocity–time graph

1(b) Movement and Position1(c) Forces, Movement, Shape and Momentum4(c) Work and PowerBoth papers

A driverless metro train travels along a straight, level track from station A to station B. Part of the track between the stations is being repaired, so the train must pass through this repair zone at a reduced speed.

Figure 1 shows the velocity–time graph of the train from the moment it leaves station A until it stops at station B.

Figure 1: velocity–time graph of the train from station A to station B

(a)

Describe the motion of the train between 50 s and 100 s. Use values from Figure 1 in your answer.

GRAPHNotes

Show solution

50 s to 55 s: uniform deceleration from 12 m/s to 6.0 m/s	✓ 1	Values needed for each mark. “Slows down” alone is not enough.
55 s to 90 s: constant velocity / steady speed of 6.0 m/s	✓ 1
90 s to 100 s: uniform deceleration from 6.0 m/s to rest / stops (at station B) at 100 s	✓ 1

(3)

(b)

Determine the stage of the journey in which the size of the acceleration is greatest, and calculate this acceleration.

CALCNotes

Show solution

50 s to 55 s (the steepest section of the graph)	✓ 1	Accept a comparison of gradients: 0.60, 1.2 and 0.60 m/s².
a = (v − u) ÷ t = (6.0 − 12) ÷ 5	✓ 1	Accept the gradient of this section.
= −1.2 m/s² (a deceleration of 1.2 m/s²)	✓ 1	Unit required. Ignore the sign. Allow this mark for a correct acceleration of a wrongly chosen stage only if its working is shown.

Answer: −1.2 m/s² (deceleration of 1.2 m/s²), from 50 s to 55 s

(3)

(c)

The train enters the repair zone at 55 s.

Show that the train is 525 m from station A when it enters the repair zone.

MULTINotes

Show solution

distance = area under the velocity–time graph (from 0 to 55 s)	✓ 1	May be implied by correct area calculations.
0 to 20 s: ½ × 20 × 12 = 120 m and 20 s to 50 s: 30 × 12 = 360 m	✓ 1
50 s to 55 s: ½ × (12 + 6.0) × 5 = 45 m; total = 120 + 360 + 45 = 525 m	✓ 1	Accept 5 × 6.0 + ½ × 5 × 6.0 = 45 m for the last section. Answer given: working must be seen.

Answer: 525 m

(3)

(d)

The train leaves the repair zone at 90 s. When the repairs are finished, the train will be able to travel through the repair zone at 12 m/s instead of 6.0 m/s.

Calculate the time this will save in crossing the repair zone. Ignore any change in the time taken to slow down and speed up.

MULTINotes

Show solution

length of the repair zone = area from 55 s to 90 s = 35 × 6.0 = 210 m	✓ 1
time at 12 m/s = 210 ÷ 12 = 17.5 s	✓ 1	Allow ECF from an incorrect length.
time saved = 35 − 17.5 = 17.5 s	✓ 1	Accept 18 s. Halving the speed doubles the time, so a correct argument that the time is halved scores the last two marks.

Answer: 17.5 s

(3)

(e)(i)

The mass of the train and its passengers is 1.5 × 10⁵ kg.

Calculate the resultant force on the train between 50 s and 55 s.

CALCNotes

Show solution

F = m × a = 1.5 × 10⁵ × 1.2	✓ 1	Allow ECF from (b) if the same stage was chosen there.
= 1.8 × 10⁵ N (backwards)	✓ 1	Unit required. Ignore a minus sign.

Answer: 1.8 × 10⁵ N

(2)

(e)(ii)

Show that the work done by this resultant force between 50 s and 55 s is equal to the kinetic energy lost by the train.

MULTINotes

Show solution

KE at 12 m/s = ½ × 1.5 × 10⁵ × 12² = 1.08 × 10⁷ J and KE at 6.0 m/s = ½ × 1.5 × 10⁵ × 6.0² = 2.7 × 10⁶ J	✓ 1
kinetic energy lost = 1.08 × 10⁷ − 2.7 × 10⁶ = 8.1 × 10⁶ J	✓ 1	½ × 1.5 × 10⁵ × (12 − 6.0)² = 2.7 × 10⁶ J scores 0 for this mark.
work done = F × d = 1.8 × 10⁵ × 45 = 8.1 × 10⁶ J, the same value	✓ 1	Distance 45 m from (c). Allow ECF from (e)(i).

Answer: both 8.1 × 10⁶ J

(3)

(f)

Between 20 s and 50 s the motors of the train are still producing a forward driving force.

Explain why the speed of the train does not change.

EXPLAINNotes

Show solution

resistive forces (friction / air resistance) act backwards and are equal in size to the driving force	✓ 1	Accept: the forces on the train are balanced.
so the resultant force is zero and the acceleration is zero (F = m × a), so the train keeps a constant speed	✓ 1	“There is no force on the train” scores 0.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
50 s to 55 s: uniform deceleration from 12 m/s to 6.0 m/s	✓ 1	Values needed for each mark. “Slows down” alone is not enough.
55 s to 90 s: constant velocity / steady speed of 6.0 m/s	✓ 1
90 s to 100 s: uniform deceleration from 6.0 m/s to rest / stops (at station B) at 100 s	✓ 1
Part (b)
50 s to 55 s (the steepest section of the graph)	✓ 1	Accept a comparison of gradients: 0.60, 1.2 and 0.60 m/s².
a = (v − u) ÷ t = (6.0 − 12) ÷ 5	✓ 1	Accept the gradient of this section.
= −1.2 m/s² (a deceleration of 1.2 m/s²)	✓ 1	Unit required. Ignore the sign. Allow this mark for a correct acceleration of a wrongly chosen stage only if its working is shown.
Part (c)
distance = area under the velocity–time graph (from 0 to 55 s)	✓ 1	May be implied by correct area calculations.
0 to 20 s: ½ × 20 × 12 = 120 m and 20 s to 50 s: 30 × 12 = 360 m	✓ 1
50 s to 55 s: ½ × (12 + 6.0) × 5 = 45 m; total = 120 + 360 + 45 = 525 m	✓ 1	Accept 5 × 6.0 + ½ × 5 × 6.0 = 45 m for the last section. Answer given: working must be seen.
Part (d)
length of the repair zone = area from 55 s to 90 s = 35 × 6.0 = 210 m	✓ 1
time at 12 m/s = 210 ÷ 12 = 17.5 s	✓ 1	Allow ECF from an incorrect length.
time saved = 35 − 17.5 = 17.5 s	✓ 1	Accept 18 s. Halving the speed doubles the time, so a correct argument that the time is halved scores the last two marks.
Part (e)(i)
F = m × a = 1.5 × 10⁵ × 1.2	✓ 1	Allow ECF from (b) if the same stage was chosen there.
= 1.8 × 10⁵ N (backwards)	✓ 1	Unit required. Ignore a minus sign.
Part (e)(ii)
KE at 12 m/s = ½ × 1.5 × 10⁵ × 12² = 1.08 × 10⁷ J and KE at 6.0 m/s = ½ × 1.5 × 10⁵ × 6.0² = 2.7 × 10⁶ J	✓ 1
kinetic energy lost = 1.08 × 10⁷ − 2.7 × 10⁶ = 8.1 × 10⁶ J	✓ 1	½ × 1.5 × 10⁵ × (12 − 6.0)² = 2.7 × 10⁶ J scores 0 for this mark.
work done = F × d = 1.8 × 10⁵ × 45 = 8.1 × 10⁶ J, the same value	✓ 1	Distance 45 m from (c). Allow ECF from (e)(i).
Part (f)
resistive forces (friction / air resistance) act backwards and are equal in size to the driving force	✓ 1	Accept: the forces on the train are balanced.
so the resultant force is zero and the acceleration is zero (F = m × a), so the train keeps a constant speed	✓ 1	“There is no force on the train” scores 0.

Answers: (b) −1.2 m/s² (deceleration of 1.2 m/s²), from 50 s to 55 s · (c) 525 m · (d) 17.5 s · (e)(i) 1.8 × 10⁵ N · (e)(ii) both 8.1 × 10⁶ J

Common error (b) Choosing the stage that reaches the highest velocity. The size of the acceleration is shown by how steep the line is, not by how high it is.

Exam tip (e)(ii) Kinetic energy depends on speed squared, so work out the kinetic energy at each speed and subtract. Do not square the change in speed.

More 1(b) questions in the question bank

1(c) Forces, Movement, Shape and Momentum

+ Add to test✓ In your testCC-EIP-1c-120 marks~20 min

An e-scooter emergency stop: reaction time, braking and stopping distance

1(c) Forces, Movement, Shape and Momentum1(b) Movement and Position4(c) Work and PowerBoth papers

A rider travels along a straight, level cycle lane on an electric scooter at a constant speed of 6.0 m/s. She sees a pedestrian step into the lane ahead and brakes to a stop.

reaction time of the rider = 0.80 s
once the brakes are applied, rider and scooter decelerate uniformly at 2.5 m/s²
mass of the rider = 65 kg; mass of the scooter = 15 kg

(a)

Calculate the thinking distance.

CALCNotes

Show solution

thinking distance = speed × reaction time = 6.0 × 0.80	✓ 1
= 4.8 m	✓ 1	Unit required.

Answer: 4.8 m

(2)

(b)(i)

On the axes in Figure 1, sketch the velocity–time graph of the scooter from the moment the rider sees the pedestrian until the scooter stops.

Show your working for any time that you calculate.

Figure 1: axes for the velocity–time graph of the stop

SKETCHNotes

Show solution

braking time = (6.0 − 0) ÷ 2.5 = 2.4 s	✓ 1	May be seen on the graph as the scooter stopping at 3.2 s.
horizontal line at 6.0 m/s from 0 to 0.80 s	✓ 1
straight line from (0.80 s, 6.0 m/s) down to zero velocity at 3.2 s	✓ 1	Allow ECF from an incorrect braking time. A curved line scores 0 for this mark.

(3)

(b)(ii)

Use your graph to determine the stopping distance.

GRAPHNotes

Show solution

area under the graph = 6.0 × 0.80 + ½ × 2.4 × 6.0	✓ 1	Allow ECF from (b)(i).
= 4.8 + 7.2 = 12 m	✓ 1	Accept 12.0 m.

Answer: 12 m

(2)

(c)

Show that the braking force on the rider and scooter is about 200 N.

CALCNotes

Show solution

F = m × a with m = 65 + 15 = 80 kg	✓ 1	Using 65 kg or 650 N as the mass scores 0 for this mark.
F = 80 × 2.5 = 200 N	✓ 1	Answer given: substitution must be seen.

Answer: 200 N

(2)

(d)

Calculate the kinetic energy of the rider and scooter just before the brakes are applied.

Use your answer to check the braking distance shown by your graph.

MULTINotes

Show solution

KE = ½ × m × v² = ½ × 80 × 6.0²	✓ 1
= 1440 J	✓ 1	Accept 1400 J.
work done by brakes = KE, so braking distance = 1440 ÷ 200 = 7.2 m, which agrees with the area of the sloping section	✓ 1	Allow ECF from the kinetic energy, from (c) and from (b)(ii).

Answer: 1440 J; braking distance 7.2 m

(3)

(e)(i)

On another day the lane is wet and the rider is looking at her phone. Her speed is again 6.0 m/s.

reaction time = 1.5 s
uniform deceleration when braking = 1.5 m/s²

Calculate the new stopping distance.

MULTINotes

Show solution

thinking distance = 6.0 × 1.5 = 9.0 m	✓ 1
v² = u² + 2as: 0 = 6.0² + 2 × (−1.5) × s, so braking distance = 12 m	✓ 1	Accept braking time 4.0 s and area ½ × 4.0 × 6.0 = 12 m.
stopping distance = 9.0 + 12 = 21 m	✓ 1	Allow ECF from either distance.

Answer: 21 m

(3)

(e)(ii)

The rider says: “At this speed, looking at my phone adds more to my stopping distance than the wet lane does.”

Use your answers to (a), (b)(ii) and (e)(i) to comment on her statement. Explain why each change affects a different part of the stopping distance.

REASONNotes

Show solution

the phone adds 9.0 − 4.8 = 4.2 m to the thinking distance; the wet lane adds 12 − 7.2 = 4.8 m to the braking distance, so the statement is wrong / the wet lane has the (slightly) larger effect	✓ 1	Allow ECF from (a), (b)(ii) and (e)(i). A conclusion without both values scores 0.
the phone increases the reaction time, so she travels further at 6.0 m/s before braking; this changes only the thinking distance	✓ 1
the wet surface reduces the friction between the tyres and the lane, so the braking force / deceleration is smaller; this changes only the braking distance	✓ 1	“Wet lane increases stopping distance” alone is not enough.

(3)

(f)

The rider wears a cycle helmet with a foam lining.

Explain, in terms of momentum, how the foam lining reduces the force on her head if she falls and her head hits the ground.

Paper 2 only onlyEXPLAINNotes

Show solution

the foam crushes / deforms, which increases the time taken for the head to stop	✓ 1
the change in momentum of the head is the same, so the rate of change of momentum is smaller and the force (= change in momentum ÷ time) is smaller	✓ 1	“The helmet absorbs the force” scores 0. Do not accept: the helmet reduces the change in momentum.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
thinking distance = speed × reaction time = 6.0 × 0.80	✓ 1
= 4.8 m	✓ 1	Unit required.
Part (b)(i)
braking time = (6.0 − 0) ÷ 2.5 = 2.4 s	✓ 1	May be seen on the graph as the scooter stopping at 3.2 s.
horizontal line at 6.0 m/s from 0 to 0.80 s	✓ 1
straight line from (0.80 s, 6.0 m/s) down to zero velocity at 3.2 s	✓ 1	Allow ECF from an incorrect braking time. A curved line scores 0 for this mark.
Part (b)(ii)
area under the graph = 6.0 × 0.80 + ½ × 2.4 × 6.0	✓ 1	Allow ECF from (b)(i).
= 4.8 + 7.2 = 12 m	✓ 1	Accept 12.0 m.
Part (c)
F = m × a with m = 65 + 15 = 80 kg	✓ 1	Using 65 kg or 650 N as the mass scores 0 for this mark.
F = 80 × 2.5 = 200 N	✓ 1	Answer given: substitution must be seen.
Part (d)
KE = ½ × m × v² = ½ × 80 × 6.0²	✓ 1
= 1440 J	✓ 1	Accept 1400 J.
work done by brakes = KE, so braking distance = 1440 ÷ 200 = 7.2 m, which agrees with the area of the sloping section	✓ 1	Allow ECF from the kinetic energy, from (c) and from (b)(ii).
Part (e)(i)
thinking distance = 6.0 × 1.5 = 9.0 m	✓ 1
v² = u² + 2as: 0 = 6.0² + 2 × (−1.5) × s, so braking distance = 12 m	✓ 1	Accept braking time 4.0 s and area ½ × 4.0 × 6.0 = 12 m.
stopping distance = 9.0 + 12 = 21 m	✓ 1	Allow ECF from either distance.
Part (e)(ii)
the phone adds 9.0 − 4.8 = 4.2 m to the thinking distance; the wet lane adds 12 − 7.2 = 4.8 m to the braking distance, so the statement is wrong / the wet lane has the (slightly) larger effect	✓ 1	Allow ECF from (a), (b)(ii) and (e)(i). A conclusion without both values scores 0.
the phone increases the reaction time, so she travels further at 6.0 m/s before braking; this changes only the thinking distance	✓ 1
the wet surface reduces the friction between the tyres and the lane, so the braking force / deceleration is smaller; this changes only the braking distance	✓ 1	“Wet lane increases stopping distance” alone is not enough.
Part (f)
the foam crushes / deforms, which increases the time taken for the head to stop	✓ 1
the change in momentum of the head is the same, so the rate of change of momentum is smaller and the force (= change in momentum ÷ time) is smaller	✓ 1	“The helmet absorbs the force” scores 0. Do not accept: the helmet reduces the change in momentum.

Answers: (a) 4.8 m · (b)(ii) 12 m · (c) 200 N · (d) 1440 J; braking distance 7.2 m · (e)(i) 21 m

Exam tip (c) In F = m × a the mass is everything that is being slowed down: here the rider and the scooter together, 80 kg.

More 1(c) questions in the question bank

+ Add to test✓ In your testCC-EIP-1c-219 marks~20 min

A tower crane on a building site: moments and the counterweight

1(c) Forces, Movement, Shape and Momentum4(c) Work and Power1(b) Movement and PositionPaper 2 only

Figure 1 shows a tower crane on a building site. The horizontal jib can turn about the top of the tower. Treat the centre line of the tower as the pivot.

The counterweight has a weight of 60 000 N. The jib, counter-jib, trolley and hook together have a weight of 30 000 N, acting at their centre of gravity G. The crane is lifting a load of mass 2000 kg.

Gravitational field strength g = 10 N/kg.

Figure 1: the tower crane, with the load at 24 m from the centre line of the tower (not to scale)

(a)

Calculate the weight of the load.

CALCNotes

Show solution

W = m × g = 2000 × 10	✓ 1
= 20 000 N	✓ 1	Unit required.

Answer: 20 000 N

(2)

(b)

Calculate the moment of the load about the centre line of the tower when the trolley is 24 m from it.

Paper 2 only onlyCALCNotes

Show solution

moment = force × perpendicular distance from the pivot = 20 000 × 24	✓ 1	Allow ECF from (a).
= 480 000 N m (clockwise)	✓ 1	Accept 4.8 × 10⁵ N m. Unit required.

Answer: 480 000 N m

(2)

(c)

Show that the crane is balanced when the trolley carrying the load is 24 m from the tower.

Paper 2 only onlyMULTINotes

Show solution

anticlockwise moment = 60 000 × 10 = 600 000 N m	✓ 1
clockwise moments = 30 000 × 4.0 + 480 000 = 120 000 + 480 000 = 600 000 N m	✓ 1	Allow ECF from (b). The moment of G must be included.
total clockwise moment = total anticlockwise moment, so the crane is balanced (principle of moments)	✓ 1	The principle must be stated or clearly applied.

Answer: both 600 000 N m

(3)

(d)

Calculate the upward force that the tower exerts on the jib when the crane is balanced.

Paper 2 only onlyCALCNotes

Show solution

upward force = total downward force = 60 000 + 30 000 + 20 000 = 110 000 N

✓ 1

Allow ECF from (a).

Answer: 110 000 N

(1)

(e)(i)

Figure 2 shows how the maximum safe load depends on the distance of the trolley from the tower.

Use Figure 2 to find the maximum safe load when the trolley is 30 m from the tower.

Figure 2: maximum safe load against distance of the trolley from the tower

GRAPHNotes

Show solution

1600 kg

✓ 1

Accept 1580–1620 kg.

Answer: 1600 kg

(1)

(e)(ii)

Between 12 m and 40 m the curve in Figure 2 follows a simple rule.

Use two points on the curve to show what the rule is, and explain the rule using the principle of moments.

Paper 2 only onlyREASONNotes

Show solution

e.g. 4000 × 12 = 48 000 and 1200 × 40 = 48 000: maximum load × distance is constant	✓ 1	Any two correctly read points, e.g. 2000 × 24, 1600 × 30, 3000 × 16.
so the maximum load always has the same moment about the tower (480 000 N m), which is the largest clockwise moment the counterweight can balance (600 000 − 120 000 N m)	✓ 1	Accept: the further out the load, the bigger its moment for each kilogram, so a smaller load gives the same moment.

(2)

(f)

The operator moves the trolley, still carrying the 2000 kg load, out to 30 m from the tower.

Explain why the crane is now in danger of toppling, and state what the operator should do.

Paper 2 only onlyEXPLAINNotes

Show solution

clockwise moment = 30 000 × 4.0 + 20 000 × 30 = 720 000 N m	✓ 1	Accept the moment of the load alone, 600 000 N m, compared with 480 000 N m.
this is greater than the anticlockwise moment of 600 000 N m, so there is a resultant clockwise moment and the crane tips towards the load	✓ 1	Accept: the load is above the maximum safe load of 1600 kg at 30 m (from (e)(i)).
move the trolley back towards the tower (to 24 m or less) / lower the load to the ground before moving out	✓ 1	Do not accept “remove the counterweight”.

(3)

(g)(i)

The hoist lifts the 2000 kg load vertically through 30 m at a constant speed. The lift takes 60 s.

Calculate the speed at which the load is lifted.

CALCNotes

Show solution

speed = 30 ÷ 60 = 0.50 m/s

✓ 1

Unit required.

Answer: 0.50 m/s

(1)

(g)(ii)

Calculate the work done by the hoist on the load.

CALCNotes

Show solution

work done = force × distance = 20 000 × 30	✓ 1	The force is the weight of the load, because the speed is constant. Allow ECF from (a).
= 600 000 J	✓ 1	Accept 6.0 × 10⁵ J. Accept GPE = m × g × h = 2000 × 10 × 30.

Answer: 600 000 J

(2)

(g)(iii)

Calculate the useful power output of the hoist.

CALCNotes

Show solution

P = W ÷ t = 600 000 ÷ 60	✓ 1	Allow ECF from (g)(ii).
= 10 000 W	✓ 1	Accept 10 kW.

Answer: 10 000 W

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
W = m × g = 2000 × 10	✓ 1
= 20 000 N	✓ 1	Unit required.
Part (b)
moment = force × perpendicular distance from the pivot = 20 000 × 24	✓ 1	Allow ECF from (a).
= 480 000 N m (clockwise)	✓ 1	Accept 4.8 × 10⁵ N m. Unit required.
Part (c)
anticlockwise moment = 60 000 × 10 = 600 000 N m	✓ 1
clockwise moments = 30 000 × 4.0 + 480 000 = 120 000 + 480 000 = 600 000 N m	✓ 1	Allow ECF from (b). The moment of G must be included.
total clockwise moment = total anticlockwise moment, so the crane is balanced (principle of moments)	✓ 1	The principle must be stated or clearly applied.
Part (d)
upward force = total downward force = 60 000 + 30 000 + 20 000 = 110 000 N	✓ 1	Allow ECF from (a).
Part (e)(i)
1600 kg	✓ 1	Accept 1580–1620 kg.
Part (e)(ii)
e.g. 4000 × 12 = 48 000 and 1200 × 40 = 48 000: maximum load × distance is constant	✓ 1	Any two correctly read points, e.g. 2000 × 24, 1600 × 30, 3000 × 16.
so the maximum load always has the same moment about the tower (480 000 N m), which is the largest clockwise moment the counterweight can balance (600 000 − 120 000 N m)	✓ 1	Accept: the further out the load, the bigger its moment for each kilogram, so a smaller load gives the same moment.
Part (f)
clockwise moment = 30 000 × 4.0 + 20 000 × 30 = 720 000 N m	✓ 1	Accept the moment of the load alone, 600 000 N m, compared with 480 000 N m.
this is greater than the anticlockwise moment of 600 000 N m, so there is a resultant clockwise moment and the crane tips towards the load	✓ 1	Accept: the load is above the maximum safe load of 1600 kg at 30 m (from (e)(i)).
move the trolley back towards the tower (to 24 m or less) / lower the load to the ground before moving out	✓ 1	Do not accept “remove the counterweight”.
Part (g)(i)
speed = 30 ÷ 60 = 0.50 m/s	✓ 1	Unit required.
Part (g)(ii)
work done = force × distance = 20 000 × 30	✓ 1	The force is the weight of the load, because the speed is constant. Allow ECF from (a).
= 600 000 J	✓ 1	Accept 6.0 × 10⁵ J. Accept GPE = m × g × h = 2000 × 10 × 30.
Part (g)(iii)
P = W ÷ t = 600 000 ÷ 60	✓ 1	Allow ECF from (g)(ii).
= 10 000 W	✓ 1	Accept 10 kW.

Answers: (a) 20 000 N · (b) 480 000 N m · (c) both 600 000 N m · (d) 110 000 N · (e)(i) 1600 kg · (g)(i) 0.50 m/s · (g)(ii) 600 000 J · (g)(iii) 10 000 W

Common error (f) Leaving out the moment of the crane's own weight at G. Every weight that does not act through the pivot has a moment and must be included.

More 1(c) questions in the question bank

Topic 2 · Electricity

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

Setup	2(b)	2(c)	2(d)	Other
An electric kettle: power, fuse choice and safety	●	○		5(c) 4(b)
Charging an electric car at home: power, energy and safety	●	○		4(b) 4(c)
A temperature sensor for a baby incubator: thermistor circuits	○	●		4(b)
Bicycle lights: LEDs, parallel branches and battery charge	○	●		4(b)
Refuelling an aircraft: static charge, sparks and earthing	○	○	●	—
An electrostatic crop sprayer: charged droplets and leaves		○	●	1(c)

2(b) Mains Electricity

+ Add to test✓ In your testCC-EIP-2b-120 marks~20 min

An electric kettle: power, fuse choice and safety

2(b) Mains Electricity2(c) Energy and Voltage in Circuits5(c) Change of State4(b) Energy TransfersBoth papers

An electric kettle is used in a kitchen. The kettle has a stainless steel (metal) body and is connected to the mains by a three-core cable and a plug fitted with a fuse. The label on the base reads:

voltage	230 V
power	2.8 kW

Figure 1 shows how the voltage of two different supplies, X and Y, changes with time. One of them is the mains supply and the other is a battery.

Figure 1: voltage–time graphs for supply X and supply Y, drawn on the same axes

(a)

State which supply, X or Y, is the mains supply. Explain how Figure 1 shows this.

GRAPHNotes

Show solution

X	✓ 1
the voltage keeps changing between positive and negative / goes above and below the time axis, so the current repeatedly reverses direction (alternating current)	✓ 1	Allow: Y is constant / never changes sign, so it is d.c. from the battery. 'X is a wave' alone is not enough.

(2)

(b)(i)

Calculate the current in the kettle when it is working normally.

CALCNotes

Show solution

substitution into I = P ÷ V; e.g. I = 2800 ÷ 230	✓ 1	Power must be converted to 2800 W.
evaluation: (I =) 12 A	✓ 1	Accept 12.2 A or 12.17 A. Do not accept 0.082 A (230 ÷ 2800) or 0.012 A (2.8 ÷ 230).

Answer: 12 A (12.2 A)

(2)

(b)(ii)

Fuses rated 3 A, 5 A and 13 A are available.

Explain which fuse should be fitted in the plug of the kettle.

REASONNotes

Show solution

13 A	✓ 1	ECF from (b)(i): the smallest rating above their current.
13 A is the smallest rating above the normal current (12.2 A) / 3 A and 5 A fuses would melt when the kettle is working normally	✓ 1	Allow: a larger fuse might not melt in a fault, so the cable could overheat.

Answer: 13 A

(2)

(c)

Calculate the resistance of the heating element of the kettle when it is working normally.

CALCNotes

Show solution

substitution into R = V ÷ I; e.g. R = 230 ÷ 12.2	✓ 1	ECF from (b)(i).
evaluation: (R =) 19 Ω	✓ 1	Accept 18.9 Ω to 19.2 Ω (19.2 Ω from I = 12 A).

Answer: 19 Ω

(2)

(d)(i)

The kettle contains 1.0 kg of water at 15 °C. The specific heat capacity of water is 4200 J/kg °C.

Show that about 3.6 × 10⁵ J of energy is needed to bring this water to 100 °C.

Paper 2 only onlyCALCNotes

Show solution

ΔT = 100 − 15 = 85 °C and substitution into ΔQ = m × c × ΔT; e.g. 1.0 × 4200 × 85	✓ 1
evaluation: (ΔQ =) 357 000 J	✓ 1	The answer must be given to more significant figures than 3.6 × 10⁵ J (e.g. 357 000 J or 3.57 × 10⁵ J). Using ΔT = 100 °C (420 000 J) scores 0.

Answer: 357 000 J

(2)

(d)(ii)

The kettle is switched on for 150 s to bring the water to the boil. In this time 357 000 J is transferred to the thermal store of the water.

Calculate the efficiency of the kettle.

MULTINotes

Show solution

energy supplied: E = I × V × t = 12.2 × 230 × 150 (or 2800 × 150)	✓ 1	ECF from (b)(i).
= 420 000 J	✓ 1	Accept 414 000 J (from 12 A) to 420 000 J.
efficiency = 357 000 ÷ 420 000 × 100% = 85%	✓ 1	Accept 0.85. Accept 85% to 86%. ECF from their energy supplied, provided the answer is less than 100%.

Answer: 85%

(3)

(e)

A fault inside the kettle makes the live wire touch the metal body.

Explain how the earth wire and the fuse protect a person who touches the kettle.

EXPLAINNotes

Show solution

the earth wire connects the metal body to earth and gives a low-resistance path	✓ 1	'The earth wire takes the electricity away' does not score.
a (very) large current flows (from the live wire through the metal body and the earth wire), larger than the fuse rating, so the fuse (wire) melts	✓ 1	Accept 'the fuse blows / breaks'.
the live supply is disconnected, so the body is no longer live and the person does not get an electric shock	✓ 1	Must refer to the live wire / supply being cut off.

(3)

(f)(i)

The kettle is taken to a country where the mains voltage is 110 V. Assume that the resistance of the heating element stays the same.

Calculate the power of the kettle on the 110 V supply.

MULTINotes

Show solution

I = V ÷ R = 110 ÷ 18.9	✓ 1	ECF from (c).
= 5.8 A	✓ 1	Accept 5.7 A to 5.9 A.
P = I × V = 5.8 × 110 = 640 W	✓ 1	Accept 630 W to 645 W. Do not accept 2800 W or 1340 W (2800 × 110 ÷ 230: assumes the current is unchanged).

Answer: 640 W

(3)

(f)(ii)

Explain why the kettle takes much longer to boil the same mass of water on the 110 V supply.

REASONNotes

Show solution

the power is lower, so less energy is transferred to the water each second, but the same energy is needed (so it takes longer)

✓ 1

Allow: it takes about 4 to 5 times as long. Allow the idea that, because it takes longer, more energy is wasted to the surroundings.

(1)

Show full solution

Marking point	Mark	Notes
Part (a)
X	✓ 1
the voltage keeps changing between positive and negative / goes above and below the time axis, so the current repeatedly reverses direction (alternating current)	✓ 1	Allow: Y is constant / never changes sign, so it is d.c. from the battery. 'X is a wave' alone is not enough.
Part (b)(i)
substitution into I = P ÷ V; e.g. I = 2800 ÷ 230	✓ 1	Power must be converted to 2800 W.
evaluation: (I =) 12 A	✓ 1	Accept 12.2 A or 12.17 A. Do not accept 0.082 A (230 ÷ 2800) or 0.012 A (2.8 ÷ 230).
Part (b)(ii)
13 A	✓ 1	ECF from (b)(i): the smallest rating above their current.
13 A is the smallest rating above the normal current (12.2 A) / 3 A and 5 A fuses would melt when the kettle is working normally	✓ 1	Allow: a larger fuse might not melt in a fault, so the cable could overheat.
Part (c)
substitution into R = V ÷ I; e.g. R = 230 ÷ 12.2	✓ 1	ECF from (b)(i).
evaluation: (R =) 19 Ω	✓ 1	Accept 18.9 Ω to 19.2 Ω (19.2 Ω from I = 12 A).
Part (d)(i)
ΔT = 100 − 15 = 85 °C and substitution into ΔQ = m × c × ΔT; e.g. 1.0 × 4200 × 85	✓ 1
evaluation: (ΔQ =) 357 000 J	✓ 1	The answer must be given to more significant figures than 3.6 × 10⁵ J (e.g. 357 000 J or 3.57 × 10⁵ J). Using ΔT = 100 °C (420 000 J) scores 0.
Part (d)(ii)
energy supplied: E = I × V × t = 12.2 × 230 × 150 (or 2800 × 150)	✓ 1	ECF from (b)(i).
= 420 000 J	✓ 1	Accept 414 000 J (from 12 A) to 420 000 J.
efficiency = 357 000 ÷ 420 000 × 100% = 85%	✓ 1	Accept 0.85. Accept 85% to 86%. ECF from their energy supplied, provided the answer is less than 100%.
Part (e)
the earth wire connects the metal body to earth and gives a low-resistance path	✓ 1	'The earth wire takes the electricity away' does not score.
a (very) large current flows (from the live wire through the metal body and the earth wire), larger than the fuse rating, so the fuse (wire) melts	✓ 1	Accept 'the fuse blows / breaks'.
the live supply is disconnected, so the body is no longer live and the person does not get an electric shock	✓ 1	Must refer to the live wire / supply being cut off.
Part (f)(i)
I = V ÷ R = 110 ÷ 18.9	✓ 1	ECF from (c).
= 5.8 A	✓ 1	Accept 5.7 A to 5.9 A.
P = I × V = 5.8 × 110 = 640 W	✓ 1	Accept 630 W to 645 W. Do not accept 2800 W or 1340 W (2800 × 110 ÷ 230: assumes the current is unchanged).
Part (f)(ii)
the power is lower, so less energy is transferred to the water each second, but the same energy is needed (so it takes longer)	✓ 1	Allow: it takes about 4 to 5 times as long. Allow the idea that, because it takes longer, more energy is wasted to the surroundings.

Answers: (b)(i) 12 A (12.2 A) · (b)(ii) 13 A · (c) 19 Ω · (d)(i) 357 000 J · (d)(ii) 85% · (f)(i) 640 W

Exam tip (b)(ii) Justify a fuse choice with the calculated current: the rating must be just above it, never below it.

Common error (d)(ii) The useful energy (into the water) goes on top. An efficiency above 100% means the two energies have been swapped.

More 2(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-2b-220 marks~20 min

Charging an electric car at home: power, energy and safety

2(b) Mains Electricity2(c) Energy and Voltage in Circuits4(b) Energy Transfers4(c) Work and PowerBoth papers

A family charges their electric car overnight from a wallbox fixed to the wall of their garage. The wallbox is connected to the 230 V mains. It supplies energy to the car at a constant power of 7.4 kW for the first 4.0 hours. The power then falls steadily to zero as the battery becomes full.

Figure 1 shows how the power supplied by the wallbox changes with time during one charge.

Figure 1: power supplied by the wallbox during one overnight charge

(a)(i)

Calculate the current in the charging cable when the power is 7.4 kW.

CALCNotes

Show solution

substitution into I = P ÷ V; e.g. I = 7400 ÷ 230	✓ 1	Power must be in watts.
evaluation: (I =) 32 A	✓ 1	Accept 32.2 A. Do not accept 0.032 A (7.4 ÷ 230).

Answer: 32 A (32.2 A)

(2)

(a)(ii)

The charging cable is much thicker than the cable of a table lamp.

Explain why a thick cable is needed.

EXPLAINNotes

Show solution

the (large) current transfers energy to the cable / heats the cable	✓ 1	Allow: electrons collide with the ions in the wires.
a thick cable has a lower resistance, so it heats up less and does not overheat (or cause a fire)	✓ 1	Do not accept 'a thick cable lets more electricity through' alone.

(2)

(b)

Calculate the charge that flows through the cable during the first 4.0 hours.

CALCNotes

Show solution

time converted: 4.0 × 60 × 60 = 14 400 s	✓ 1
substitution into Q = I × t; e.g. Q = 32.2 × 14 400	✓ 1	ECF from (a)(i).
evaluation: (Q =) 4.6 × 10⁵ C	✓ 1	Accept 460 800 C to 463 300 C. 129 C (time left in hours) scores 0 for the conversion and the evaluation.

Answer: 4.6 × 10⁵ C

(3)

(c)

Show that the energy transferred by the wallbox in the first 4.0 hours is about 1.1 × 10⁸ J.

CALCNotes

Show solution

substitution into E = Q × V; e.g. E = 4.63 × 10⁵ × 230	✓ 1	Allow E = I × V × t = 32.2 × 230 × 14 400 or P × t = 7400 × 14 400. ECF from (b).
evaluation: (E =) 1.07 × 10⁸ J	✓ 1	Answer must be given to more than 2 significant figures, e.g. 106 560 000 J. Accept 1.06 × 10⁸ J to 1.07 × 10⁸ J.

Answer: 1.07 × 10⁸ J

(2)

(d)

Use Figure 1 to determine the total energy transferred by the wallbox during the whole charge.

GRAPHNotes

Show solution

energy = power × time, so the energy is the area under the power–time graph	✓ 1	May be implied by an area calculation.
energy in the last hour (triangle): ½ × 7400 × 3600 = 1.33 × 10⁷ J	✓ 1	The time must be in seconds and the power in watts. Allow 1 mark for ½ × 7.4 × 1 with no conversion.
total = 1.07 × 10⁸ + 1.33 × 10⁷ = 1.2 × 10⁸ J	✓ 1	Accept 1.19 × 10⁸ J to 1.2 × 10⁸ J. ECF from (c).

Answer: 1.2 × 10⁸ J

(3)

(e)

During the charge, 1.08 × 10⁸ J of energy is stored in the chemical store of the car battery.

Calculate the efficiency of the charging process and state what happens to the rest of the energy.

MULTINotes

Show solution

substitution into efficiency = useful energy output ÷ total energy output (× 100%); e.g. 1.08 × 10⁸ ÷ 1.2 × 10⁸	✓ 1	ECF from (d).
evaluation: 90% (0.90)	✓ 1	Accept 90% to 91%. An answer above 100% scores 0 for this mark.
the rest is transferred (by heating) to the thermal store of the cable / wallbox / battery and the surroundings	✓ 1	Do not accept 'the energy is lost / used up'.

Answer: 90%

(3)

(f)

Away from home, the family uses a portable charger that plugs into an ordinary 230 V socket. The current is limited to 10 A.

Calculate the time this charger takes to transfer the same total energy as in (d). Assume that it transfers energy at a constant rate. Give your answer in hours.

MULTINotes

Show solution

P = I × V = 10 × 230 = 2300 W	✓ 1
t = E ÷ P = 1.2 × 10⁸ ÷ 2300 = 5.2 × 10⁴ s	✓ 1	ECF from (d). Allow t = E ÷ (I × V) in one step.
= 14.5 hours	✓ 1	Accept 14 h to 15 h. ECF for their time in seconds ÷ 3600.

Answer: 14.5 hours (about 3 times as long)

(3)

(g)

The wallbox is protected by a circuit breaker in the live wire instead of a fuse.

Give two advantages of a circuit breaker compared with a fuse.

EXPLAINNotes

Show solution

it acts (switches off) faster than a fuse	✓ 1
it can be reset (switched back on) and does not need to be replaced after it has tripped	✓ 1	Do not accept 'it is safer' without a reason.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
substitution into I = P ÷ V; e.g. I = 7400 ÷ 230	✓ 1	Power must be in watts.
evaluation: (I =) 32 A	✓ 1	Accept 32.2 A. Do not accept 0.032 A (7.4 ÷ 230).
Part (a)(ii)
the (large) current transfers energy to the cable / heats the cable	✓ 1	Allow: electrons collide with the ions in the wires.
a thick cable has a lower resistance, so it heats up less and does not overheat (or cause a fire)	✓ 1	Do not accept 'a thick cable lets more electricity through' alone.
Part (b)
time converted: 4.0 × 60 × 60 = 14 400 s	✓ 1
substitution into Q = I × t; e.g. Q = 32.2 × 14 400	✓ 1	ECF from (a)(i).
evaluation: (Q =) 4.6 × 10⁵ C	✓ 1	Accept 460 800 C to 463 300 C. 129 C (time left in hours) scores 0 for the conversion and the evaluation.
Part (c)
substitution into E = Q × V; e.g. E = 4.63 × 10⁵ × 230	✓ 1	Allow E = I × V × t = 32.2 × 230 × 14 400 or P × t = 7400 × 14 400. ECF from (b).
evaluation: (E =) 1.07 × 10⁸ J	✓ 1	Answer must be given to more than 2 significant figures, e.g. 106 560 000 J. Accept 1.06 × 10⁸ J to 1.07 × 10⁸ J.
Part (d)
energy = power × time, so the energy is the area under the power–time graph	✓ 1	May be implied by an area calculation.
energy in the last hour (triangle): ½ × 7400 × 3600 = 1.33 × 10⁷ J	✓ 1	The time must be in seconds and the power in watts. Allow 1 mark for ½ × 7.4 × 1 with no conversion.
total = 1.07 × 10⁸ + 1.33 × 10⁷ = 1.2 × 10⁸ J	✓ 1	Accept 1.19 × 10⁸ J to 1.2 × 10⁸ J. ECF from (c).
Part (e)
substitution into efficiency = useful energy output ÷ total energy output (× 100%); e.g. 1.08 × 10⁸ ÷ 1.2 × 10⁸	✓ 1	ECF from (d).
evaluation: 90% (0.90)	✓ 1	Accept 90% to 91%. An answer above 100% scores 0 for this mark.
the rest is transferred (by heating) to the thermal store of the cable / wallbox / battery and the surroundings	✓ 1	Do not accept 'the energy is lost / used up'.
Part (f)
P = I × V = 10 × 230 = 2300 W	✓ 1
t = E ÷ P = 1.2 × 10⁸ ÷ 2300 = 5.2 × 10⁴ s	✓ 1	ECF from (d). Allow t = E ÷ (I × V) in one step.
= 14.5 hours	✓ 1	Accept 14 h to 15 h. ECF for their time in seconds ÷ 3600.
Part (g)
it acts (switches off) faster than a fuse	✓ 1
it can be reset (switched back on) and does not need to be replaced after it has tripped	✓ 1	Do not accept 'it is safer' without a reason.

Answers: (a)(i) 32 A (32.2 A) · (b) 4.6 × 10⁵ C · (c) 1.07 × 10⁸ J · (d) 1.2 × 10⁸ J · (e) 90% · (f) 14.5 hours (about 3 times as long)

Common error (b) Hours must be changed to seconds before using Q = I × t: 4.0 hours is 14 400 s, not 4.0 s or 240 s.

More 2(b) questions in the question bank

2(c) Energy and Voltage in Circuits

+ Add to test✓ In your testCC-EIP-2c-120 marks~20 min

A temperature sensor for a baby incubator: thermistor circuits

2(c) Energy and Voltage in Circuits2(b) Mains Electricity4(b) Energy TransfersBoth papers

A baby incubator keeps the air around a newborn baby warm. The air temperature is monitored by the circuit in Figure 1. A thermistor is connected in series with a 2000 Ω fixed resistor and a 12 V battery. M₁ and M₂ are two meters.

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

Figure 1: the temperature-sensing circuit

Figure 2: resistance of the thermistor against temperature

(a)

One of the meters is an ammeter and the other is a voltmeter.

State which meter, M₁ or M₂, is the voltmeter. Give a reason for your answer.

RECALLNotes

Show solution

M₂	✓ 1
it is connected in parallel / across the fixed resistor	✓ 1	Allow: M₁ is the ammeter because it is in series (in the loop). A reason with no meter identified scores 0.

(2)

(b)

The air in the incubator is at 34 °C.

Use Figure 2 to find the resistance of the thermistor.

GRAPHNotes

Show solution

2800 Ω

✓ 1

Accept 2750 Ω to 2850 Ω.

Answer: 2800 Ω

(1)

(c)(i)

Calculate the current in the circuit at 34 °C.

CALCNotes

Show solution

total resistance R = 2800 + 2000 = 4800 Ω	✓ 1	ECF from (b).
I = V ÷ R = 12 ÷ 4800 = 0.0025 A	✓ 1	Accept 2.5 mA. 0.0043 A (12 ÷ 2800) or 0.006 A (12 ÷ 2000) scores 0.

Answer: 0.0025 A (2.5 mA)

(2)

(c)(ii)

Show that the reading on M₂ is 5.0 V at 34 °C.

CALCNotes

Show solution

V = I × R = 0.0025 × 2000	✓ 1	Allow: voltage shared in proportion to resistance, 12 × 2000 ÷ 4800.
= 5.0 V	✓ 1	Allow: the thermistor has 0.0025 × 2800 = 7.0 V, and 12 − 7.0 = 5.0 V. The given value (5.0 V) with no working does not score.

Answer: 5.0 V

(2)

(d)

The temperature of the air in the incubator rises above 34 °C.

Explain what happens to the current in the circuit and to the reading on M₂.

EXPLAINNotes

Show solution

the resistance of the thermistor decreases	✓ 1	Must be linked to the temperature rise.
the total resistance decreases, so the current increases	✓ 1	Do not accept 'the current increases' with no reason.
the voltage across the fixed resistor (V = I × R) increases, so the reading on M₂ increases	✓ 1	Allow: the thermistor takes a smaller share of the 12 V, so the fixed resistor takes a larger share.

(3)

(e)

A control unit switches the incubator heater off when the reading on M₂ rises to 5.4 V.

Determine the air temperature at which the heater is switched off.

MULTINotes

Show solution

current: I = 5.4 ÷ 2000 = 0.0027 A	✓ 1
voltage across the thermistor = 12 − 5.4 = 6.6 V	✓ 1
resistance of the thermistor = 6.6 ÷ 0.0027 = 2440 Ω	✓ 1	Allow 2400 Ω to 2450 Ω. ECF from their current and voltage.
temperature read from Figure 2: 37 °C	✓ 1	Accept 36.5 °C to 37.5 °C. ECF: the temperature read from Figure 2 for their resistance.

Answer: 37 °C

(4)

(f)

The heater is connected to the 230 V mains. It takes a current of 1.5 A.

Calculate the power of the heater.

CALCNotes

Show solution

substitution into P = I × V; e.g. P = 1.5 × 230	✓ 1
evaluation: (P =) 345 W	✓ 1	Accept 350 W. Do not accept 153 W (230 ÷ 1.5).

Answer: 345 W

(2)

(g)

The walls of the incubator are made of two layers of clear plastic with a layer of air trapped between them.

Explain how this design reduces the energy transferred from the warm air inside the incubator to the room.

EXPLAINNotes

Show solution

air (and plastic) is a poor conductor, so conduction (through the walls) is reduced	✓ 1
the trapped air cannot circulate / move, so convection currents cannot form between the layers	✓ 1	Do not accept 'air is an insulator so it keeps the heat in' alone.

(2)

(h)

A second circuit, identical to Figure 1 except that the thermistor is replaced by an LDR, is used to detect when the cover of the incubator is opened. While the cover is closed the LDR is in the dark. Opening the cover lets light from the room fall on the LDR.

State what happens to the reading on M₂ when the cover is opened. Give a reason for your answer.

PREDICTNotes

Show solution

the resistance of the LDR decreases (when light falls on it)	✓ 1
the current increases, so the reading on M₂ increases	✓ 1	Allow: the fixed resistor takes a larger share of the 12 V.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
M₂	✓ 1
it is connected in parallel / across the fixed resistor	✓ 1	Allow: M₁ is the ammeter because it is in series (in the loop). A reason with no meter identified scores 0.
Part (b)
2800 Ω	✓ 1	Accept 2750 Ω to 2850 Ω.
Part (c)(i)
total resistance R = 2800 + 2000 = 4800 Ω	✓ 1	ECF from (b).
I = V ÷ R = 12 ÷ 4800 = 0.0025 A	✓ 1	Accept 2.5 mA. 0.0043 A (12 ÷ 2800) or 0.006 A (12 ÷ 2000) scores 0.
Part (c)(ii)
V = I × R = 0.0025 × 2000	✓ 1	Allow: voltage shared in proportion to resistance, 12 × 2000 ÷ 4800.
= 5.0 V	✓ 1	Allow: the thermistor has 0.0025 × 2800 = 7.0 V, and 12 − 7.0 = 5.0 V. The given value (5.0 V) with no working does not score.
Part (d)
the resistance of the thermistor decreases	✓ 1	Must be linked to the temperature rise.
the total resistance decreases, so the current increases	✓ 1	Do not accept 'the current increases' with no reason.
the voltage across the fixed resistor (V = I × R) increases, so the reading on M₂ increases	✓ 1	Allow: the thermistor takes a smaller share of the 12 V, so the fixed resistor takes a larger share.
Part (e)
current: I = 5.4 ÷ 2000 = 0.0027 A	✓ 1
voltage across the thermistor = 12 − 5.4 = 6.6 V	✓ 1
resistance of the thermistor = 6.6 ÷ 0.0027 = 2440 Ω	✓ 1	Allow 2400 Ω to 2450 Ω. ECF from their current and voltage.
temperature read from Figure 2: 37 °C	✓ 1	Accept 36.5 °C to 37.5 °C. ECF: the temperature read from Figure 2 for their resistance.
Part (f)
substitution into P = I × V; e.g. P = 1.5 × 230	✓ 1
evaluation: (P =) 345 W	✓ 1	Accept 350 W. Do not accept 153 W (230 ÷ 1.5).
Part (g)
air (and plastic) is a poor conductor, so conduction (through the walls) is reduced	✓ 1
the trapped air cannot circulate / move, so convection currents cannot form between the layers	✓ 1	Do not accept 'air is an insulator so it keeps the heat in' alone.
Part (h)
the resistance of the LDR decreases (when light falls on it)	✓ 1
the current increases, so the reading on M₂ increases	✓ 1	Allow: the fixed resistor takes a larger share of the 12 V.

Answers: (b) 2800 Ω · (c)(i) 0.0025 A (2.5 mA) · (c)(ii) 5.0 V · (e) 37 °C · (f) 345 W

Exam tip (e) Work back one component at a time: the voltmeter reading and the fixed resistance give the current, which is the same in the thermistor. Then use the graph the other way round, from resistance to temperature.

More 2(c) questions in the question bank

+ Add to test✓ In your testCC-EIP-2c-220 marks~20 min

Bicycle lights: LEDs, parallel branches and battery charge

2(c) Energy and Voltage in Circuits2(b) Mains Electricity4(b) Energy TransfersBoth papers

A rechargeable bicycle light set has a white front LED and a red rear LED. Each LED is in series with its own resistor, and the two branches are connected in parallel to a single rechargeable cell of voltage 3.7 V, as shown in Figure 1. The front LED is in series with a 2.5 Ω resistor and the rear LED is in series with an 80 Ω resistor.

When the lights are on, the voltage across the front LED is 3.2 V and the voltage across the rear LED is 2.1 V.

Figure 1: the circuit of the bicycle lights

(a)

Explain why the two LEDs are connected in parallel rather than in series.

REASONNotes

Show solution

each branch has the full (3.7 V) voltage of the cell across it	✓ 1	Allow: in series the 3.7 V would be shared and would not be enough for both LEDs (3.2 V + 2.1 V).
if one LED fails (or breaks), the other stays lit / each branch is a separate complete circuit	✓ 1	Allow: each light could be switched on and off separately.

(2)

(b)

Figure 2 shows current–voltage graphs for a resistor, a filament lamp and an LED.

State which graph, P, Q or R, is for the LED. Describe how its graph shows the way an LED behaves.

Figure 2: current–voltage graphs for three components, P, Q and R

GRAPHNotes

Show solution

R	✓ 1
there is no current when the voltage is reversed (negative), so current flows in one direction only	✓ 1
in the forward direction there is (almost) no current until about 2 V, then the current rises steeply (low resistance)	✓ 1	Allow any voltage from 1.5 V to 2.2 V.

(3)

(c)(i)

Show that the current in the front LED is 0.20 A.

CALCNotes

Show solution

voltage across the 2.5 Ω resistor = 3.7 − 3.2 = 0.5 V	✓ 1
I = V ÷ R = 0.5 ÷ 2.5 = 0.20 A	✓ 1	Using 3.7 V across the resistor (1.48 A) scores 0. The given value (0.20 A) with no working does not score.

Answer: 0.20 A

(2)

(c)(ii)

Calculate the current in the cell.

MULTINotes

Show solution

current in the rear branch = (3.7 − 2.1) ÷ 80 = 0.020 A	✓ 1
current in the cell = 0.20 + 0.020 = 0.22 A	✓ 1	ECF from their rear current. The branch currents add at the junction.

Answer: 0.22 A

(2)

(d)(i)

The cell runs both lights for 2.5 hours before it needs recharging.

Calculate the charge that flows from the cell in this time.

CALCNotes

Show solution

time = 2.5 × 3600 = 9000 s	✓ 1
Q = I × t = 0.22 × 9000 = 1980 C	✓ 1	Accept 2000 C. ECF from (c)(ii).

Answer: 1980 C

(2)

(d)(ii)

Calculate the energy transferred by the cell in this time.

CALCNotes

Show solution

substitution into E = Q × V; e.g. E = 1980 × 3.7	✓ 1	ECF from (d)(i). Allow E = I × V × t.
evaluation: (E =) 7300 J	✓ 1	Accept 7326 J or 7.3 kJ.

Answer: 7300 J

(2)

(e)(i)

Calculate the power supplied by the cell to the front branch (the front LED and its resistor).

CALCNotes

Show solution

substitution into P = I × V; e.g. P = 0.20 × 3.7	✓ 1	Using 3.2 V gives 0.64 W: this is the power of the LED alone and scores 1.
evaluation: (P =) 0.74 W	✓ 1

Answer: 0.74 W

(2)

(e)(ii)

Every second, the front branch transfers 0.20 J of energy usefully as light.

Calculate the efficiency of the front branch.

CALCNotes

Show solution

substitution into efficiency = useful energy output ÷ total energy output (× 100%); e.g. 0.20 ÷ 0.74	✓ 1	ECF from (e)(i).
evaluation: 27% (0.27)	✓ 1	Accept 27% to 31% (31% from 0.64 W).

Answer: 27%

(2)

(f)

The rear LED fails and no longer conducts (it behaves like an open switch).

Explain the effect on the front LED and on the current in the cell.

PREDICTNotes

Show solution

the front branch still has 3.7 V across it (it is still a complete circuit)	✓ 1
so the current in the front LED is still 0.20 A and its brightness does not change	✓ 1	Do not accept 'the front LED gets brighter'.
the current in the cell falls to 0.20 A, because there is no longer a current in the rear branch	✓ 1	ECF from (c). Allow 'falls by 0.020 A'.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
each branch has the full (3.7 V) voltage of the cell across it	✓ 1	Allow: in series the 3.7 V would be shared and would not be enough for both LEDs (3.2 V + 2.1 V).
if one LED fails (or breaks), the other stays lit / each branch is a separate complete circuit	✓ 1	Allow: each light could be switched on and off separately.
Part (b)
R	✓ 1
there is no current when the voltage is reversed (negative), so current flows in one direction only	✓ 1
in the forward direction there is (almost) no current until about 2 V, then the current rises steeply (low resistance)	✓ 1	Allow any voltage from 1.5 V to 2.2 V.
Part (c)(i)
voltage across the 2.5 Ω resistor = 3.7 − 3.2 = 0.5 V	✓ 1
I = V ÷ R = 0.5 ÷ 2.5 = 0.20 A	✓ 1	Using 3.7 V across the resistor (1.48 A) scores 0. The given value (0.20 A) with no working does not score.
Part (c)(ii)
current in the rear branch = (3.7 − 2.1) ÷ 80 = 0.020 A	✓ 1
current in the cell = 0.20 + 0.020 = 0.22 A	✓ 1	ECF from their rear current. The branch currents add at the junction.
Part (d)(i)
time = 2.5 × 3600 = 9000 s	✓ 1
Q = I × t = 0.22 × 9000 = 1980 C	✓ 1	Accept 2000 C. ECF from (c)(ii).
Part (d)(ii)
substitution into E = Q × V; e.g. E = 1980 × 3.7	✓ 1	ECF from (d)(i). Allow E = I × V × t.
evaluation: (E =) 7300 J	✓ 1	Accept 7326 J or 7.3 kJ.
Part (e)(i)
substitution into P = I × V; e.g. P = 0.20 × 3.7	✓ 1	Using 3.2 V gives 0.64 W: this is the power of the LED alone and scores 1.
evaluation: (P =) 0.74 W	✓ 1
Part (e)(ii)
substitution into efficiency = useful energy output ÷ total energy output (× 100%); e.g. 0.20 ÷ 0.74	✓ 1	ECF from (e)(i).
evaluation: 27% (0.27)	✓ 1	Accept 27% to 31% (31% from 0.64 W).
Part (f)
the front branch still has 3.7 V across it (it is still a complete circuit)	✓ 1
so the current in the front LED is still 0.20 A and its brightness does not change	✓ 1	Do not accept 'the front LED gets brighter'.
the current in the cell falls to 0.20 A, because there is no longer a current in the rear branch	✓ 1	ECF from (c). Allow 'falls by 0.020 A'.

Answers: (c)(i) 0.20 A · (c)(ii) 0.22 A · (d)(i) 1980 C · (d)(ii) 7300 J · (e)(i) 0.74 W · (e)(ii) 27%

Common error (c)(i) In each branch the 3.7 V is shared between the LED and its resistor. Only the resistor's share, 3.7 − 3.2 = 0.5 V, goes into I = V ÷ R for the 2.5 Ω resistor.

More 2(c) questions in the question bank

2(d) Electric Charge

+ Add to test✓ In your testCC-EIP-2d-118 marks~20 min

Refuelling an aircraft: static charge, sparks and earthing

2(d) Electric Charge2(c) Energy and Voltage in Circuits2(b) Mains ElectricityPaper 2 only

An aircraft is refuelled from a tanker lorry. Fuel is pumped through a long rubber hose into the fuel tanks in the wings. As the fuel flows quickly along the inside of the hose, it becomes negatively charged and carries this charge into the aircraft.

Before refuelling starts, a metal bonding cable joins the aircraft to the tanker, and an earthing cable joins both of them to a metal spike in the ground.

Figure 1 shows how the voltage between the aircraft and the ground would change during refuelling if the cables were not connected (line A), and how it changes when they are connected (line B). On line A, a spark jumps from the aircraft to a nearby metal ladder each time the voltage falls suddenly.

Without the cables, charge is carried onto the aircraft at a steady rate equivalent to a current of 0.50 μA. The aircraft stands on rubber tyres, and the air on the day of Figure 1 is dry, so no charge leaks away from the aircraft through the air.

Figure 1: voltage of the aircraft during refuelling, without the cables (A) and with the cables connected (B)

(a)

The fuel becomes negatively charged as it rubs against the inside of the hose.

Explain, in terms of electrons, why the hose becomes positively charged and why its charge is equal in size to the charge on the fuel.

Paper 2 only onlyEXPLAINNotes

Show solution

friction transfers electrons from the hose to the fuel, so the hose has lost electrons / has fewer electrons than protons (and is positive)	✓ 1	Electrons must move from hose to fuel. Any mention of positive charges / protons moving: 0 for this mark.
charge is not created; the electrons lost by the hose are the same electrons gained by the fuel (so the charges are equal and opposite)	✓ 1

(2)

(b)

Suggest why, without the cables, the charge stays on the aircraft instead of flowing away to the ground.

Paper 2 only onlyREASONNotes

Show solution

the tyres are (made of) rubber, which is an insulator, so charge cannot flow through them to the ground

✓ 1

Accept: the aircraft is insulated from the ground by its tyres. Accept: dry air is an insulator.

(1)

(c)

Use Figure 1 to find the time at which the first spark occurs on line A.

Calculate the charge carried onto the aircraft in this time.

CALCNotes

Show solution

t = 30 s (read from Figure 1)	✓ 1
Q = I × t = 0.50 × 10⁻⁶ × 30	✓ 1	Conversion of μA to A must be seen or used.
Q = 1.5 × 10⁻⁵ C	✓ 1	Accept 15 μC. Allow ECF from an incorrect time reading.

Answer: 1.5 × 10⁻⁵ C (15 μC)

(3)

(d)

The smallest energy that can ignite fuel vapour in air is 0.25 mJ.

Assume that all the charge on the aircraft flows in the spark, and use the voltage just before the spark. (This simple estimate gives an upper limit: the real energy is smaller, because the voltage falls as the charge flows.)

Use Figure 1 and your answer to (c) to show that this estimate of the energy transferred by the first spark is about 0.2 J, and comment on whether this spark could cause a fire.

MULTINotes

Show solution

V = 15 kV read from Figure 1 and E = Q × V = 1.5 × 10⁻⁵ × 15 000	✓ 1	Allow ECF for Q from (c). 15 kV must be used as 15 000 V.
E = 0.225 J (about 0.2 J)	✓ 1	Answer must be given to more than 1 s.f. or the substitution shown.
0.225 J is far greater than 0.25 mJ (0.000 25 J) / about 900 times the ignition energy, so even if the real energy is a fraction of this estimate the spark could ignite the fuel vapour and cause a fire or explosion	✓ 1	Comparison must use the same units.

Answer: 0.225 J; about 900 times the energy needed, so it could ignite the vapour

(3)

(e)

Describe what happens to the charge on the aircraft at a time of 30 s, and explain the shape of line A just after this time.

Paper 2 only onlyGRAPHNotes

Show solution

electrons flow suddenly from the aircraft through the air (as a spark) to the ladder / to earth, so the aircraft loses its charge	✓ 1	Must refer to electrons (not positive charge) moving off the aircraft.
so the voltage falls almost to zero at once, then rises steadily again (at the same rate) because fuel keeps carrying charge onto the aircraft	✓ 1	Accept: the line rises with the same gradient, so a second spark would occur.

(2)

(f)(i)

Explain why the voltage on line B stays close to zero, and why the bonding and earthing cables must be made of metal.

Paper 2 only onlyEXPLAINNotes

Show solution

electrons flow continuously through the cable(s) to earth as fast as the fuel brings them onto the aircraft	✓ 1	Not ‘the charge is earthed’ without a direction of electron flow.
so charge cannot build up, the voltage stays low and no spark can jump	✓ 1
metals are conductors / contain free electrons, so charge can flow through them; an insulating cable would not let the charge flow away	✓ 1

(3)

(f)(ii)

The metal case of a toaster is connected to the earth wire of the mains supply.

Give one similarity and one difference between the job of the earthing cable on the aircraft and the job of the earth wire of the toaster.

REASONNotes

Show solution

similarity: both give a low-resistance conducting path from a metal body to the ground, so charge flows to earth instead of through a person or as a spark	✓ 1
difference: the toaster's earth wire normally carries no current and only carries a large current in a fault (so that the fuse melts); the aircraft cable carries a tiny current all the time during refuelling	✓ 1	Accept: the earth wire works with a fuse to disconnect the live wire; the aircraft cable stops charge building up.

(2)

(g)

The same aircraft is later refuelled, without the cables, on a day when the air is very humid.

Suggest and explain how line A would be different.

Paper 2 only onlyPREDICTNotes

Show solution

line A would rise less steeply / curve and level off, so the first spark would happen later than 30 s (or not at all)	✓ 1	Accept: the voltage would build up more slowly / reach a lower value.
moisture in the air (and on surfaces) lets some charge leak away from the aircraft, so charge builds up more slowly than it is brought on by the fuel	✓ 1	Must link the change to charge leaking away through the moist air.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
friction transfers electrons from the hose to the fuel, so the hose has lost electrons / has fewer electrons than protons (and is positive)	✓ 1	Electrons must move from hose to fuel. Any mention of positive charges / protons moving: 0 for this mark.
charge is not created; the electrons lost by the hose are the same electrons gained by the fuel (so the charges are equal and opposite)	✓ 1
Part (b)
the tyres are (made of) rubber, which is an insulator, so charge cannot flow through them to the ground	✓ 1	Accept: the aircraft is insulated from the ground by its tyres. Accept: dry air is an insulator.
Part (c)
t = 30 s (read from Figure 1)	✓ 1
Q = I × t = 0.50 × 10⁻⁶ × 30	✓ 1	Conversion of μA to A must be seen or used.
Q = 1.5 × 10⁻⁵ C	✓ 1	Accept 15 μC. Allow ECF from an incorrect time reading.
Part (d)
V = 15 kV read from Figure 1 and E = Q × V = 1.5 × 10⁻⁵ × 15 000	✓ 1	Allow ECF for Q from (c). 15 kV must be used as 15 000 V.
E = 0.225 J (about 0.2 J)	✓ 1	Answer must be given to more than 1 s.f. or the substitution shown.
0.225 J is far greater than 0.25 mJ (0.000 25 J) / about 900 times the ignition energy, so even if the real energy is a fraction of this estimate the spark could ignite the fuel vapour and cause a fire or explosion	✓ 1	Comparison must use the same units.
Part (e)
electrons flow suddenly from the aircraft through the air (as a spark) to the ladder / to earth, so the aircraft loses its charge	✓ 1	Must refer to electrons (not positive charge) moving off the aircraft.
so the voltage falls almost to zero at once, then rises steadily again (at the same rate) because fuel keeps carrying charge onto the aircraft	✓ 1	Accept: the line rises with the same gradient, so a second spark would occur.
Part (f)(i)
electrons flow continuously through the cable(s) to earth as fast as the fuel brings them onto the aircraft	✓ 1	Not ‘the charge is earthed’ without a direction of electron flow.
so charge cannot build up, the voltage stays low and no spark can jump	✓ 1
metals are conductors / contain free electrons, so charge can flow through them; an insulating cable would not let the charge flow away	✓ 1
Part (f)(ii)
similarity: both give a low-resistance conducting path from a metal body to the ground, so charge flows to earth instead of through a person or as a spark	✓ 1
difference: the toaster's earth wire normally carries no current and only carries a large current in a fault (so that the fuse melts); the aircraft cable carries a tiny current all the time during refuelling	✓ 1	Accept: the earth wire works with a fuse to disconnect the live wire; the aircraft cable stops charge building up.
Part (g)
line A would rise less steeply / curve and level off, so the first spark would happen later than 30 s (or not at all)	✓ 1	Accept: the voltage would build up more slowly / reach a lower value.
moisture in the air (and on surfaces) lets some charge leak away from the aircraft, so charge builds up more slowly than it is brought on by the fuel	✓ 1	Must link the change to charge leaking away through the moist air.

Answers: (c) 1.5 × 10⁻⁵ C (15 μC) · (d) 0.225 J; about 900 times the energy needed, so it could ignite the vapour

Common error (a) Saying that positive charge moves onto the hose. Only electrons are transferred; the hose is positive because it has lost electrons.

Exam tip (g) For refuelling questions, link the whole chain: friction → charge builds up → large voltage → spark → ignites fuel vapour. Then say how the precaution stops the first link from leading to the rest.

More 2(d) questions in the question bank

+ Add to test✓ In your testCC-EIP-2d-217 marks~20 min

An electrostatic crop sprayer: charged droplets and leaves

2(d) Electric Charge2(c) Energy and Voltage in Circuits1(c) Forces, Movement, Shape and MomentumPaper 2 only

A farmer sprays a field of vegetable plants with a liquid that protects them from insects. The sprayer gives every droplet a negative charge as it leaves the nozzle, as shown in Figure 1. The plants grow in moist soil, which is connected to earth.

Figure 1: charged droplets from the sprayer moving towards a plant (not to scale)

(a)

Explain why the droplets spread out into an even cloud after they leave the nozzle.

Paper 2 only onlyEXPLAINNotes

Show solution

all the droplets have the same (negative) charge	✓ 1
like charges repel, so each droplet pushes the others away and they spread apart	✓ 1	‘They repel’ alone, without linking to like charges: 1 mark max for the part.

(2)

(b)

As the charged cloud approaches a plant, the surface of each leaf nearest the cloud becomes positively charged.

Explain, in terms of electrons, why this happens and why the droplets then move onto the leaf.

Paper 2 only onlyEXPLAINNotes

Show solution

the negative droplets repel electrons in the leaf	✓ 1	Must be electrons that move, not positive charges.
the electrons move away from the leaf surface, down through the plant to the (earthed) soil, so the surface nearest the cloud is left positive	✓ 1	Accept: electrons move to the far side of the leaf / to earth.
unlike charges attract, so the negative droplets are attracted to the positive leaf surface	✓ 1

(3)

(c)

When the sprayer is working, the charged droplets carry charge away from the nozzle at a steady rate, equivalent to a current of 8.0 μA. The tank that feeds this nozzle lasts 25 minutes.

Show that the total charge carried by the droplets from one full tank is about 0.01 C.

CALCNotes

Show solution

t = 25 × 60 = 1500 s and Q = I × t = 8.0 × 10⁻⁶ × 1500	✓ 1	Both conversions (μA to A and minutes to s) must be seen.
Q = 0.012 C	✓ 1	Answer must be given to 2 s.f. or more.

Answer: 0.012 C

(2)

(d)(i)

One droplet has a mass of 6.0 × 10⁻¹¹ kg. Near a leaf, the electrostatic force on this droplet is 3.0 × 10⁻⁸ N.

Calculate how many times larger the electrostatic force is than the weight of the droplet.

Gravitational field strength g = 10 N/kg

MULTINotes

Show solution

W = m × g = 6.0 × 10⁻¹¹ × 10	✓ 1
W = 6.0 × 10⁻¹⁰ N	✓ 1	Using g = 9.8 N/kg gives 5.9 × 10⁻¹⁰ N: accept.
ratio = 3.0 × 10⁻⁸ ÷ 6.0 × 10⁻¹⁰ = 50	✓ 1	Allow ECF from an incorrect weight. Accept 51 from g = 9.8 N/kg.

Answer: 50 times

(3)

(d)(ii)

Suggest why the result in (d)(i) means that droplets can coat the underside of a leaf.

Paper 2 only onlyREASONNotes

Show solution

the attraction towards the leaf is much larger than the weight, so a droplet can move upwards / curve round onto the underside, against gravity

✓ 1

(1)

(e)(i)

Trials were done with different amounts of charge on the droplets. Figure 2 shows the percentage of the spray that lands on the leaves for different currents carried by the spray.

Use Figure 2 to find the percentage of the spray that lands on the leaves when the droplets are uncharged and when the current is 8.0 μA.

Hence determine how many times more liquid lands on the leaves at 8.0 μA.

Figure 2: percentage of the spray landing on the leaves against the current carried by the spray

Paper 2 only onlyGRAPHNotes

Show solution

30% (uncharged) and 80% (at 8.0 μA)	✓ 1	Both readings needed. Accept 29–31% and 79–81%.
80 ÷ 30 = 2.7 times	✓ 1	Allow ECF from the readings. Accept 2.6–2.8.

Answer: 30% and 80%; about 2.7 times

(2)

(e)(ii)

The farmer suggests doubling the current from 8.0 μA to 16 μA so that much more of the spray lands on the leaves.

Comment on this suggestion. Use Figure 2 in your answer.

Paper 2 only onlyGRAPHNotes

Show solution

the graph levels off: at 16 μA about 84–85% lands on the leaves, only about 5 percentage points more than at 8.0 μA	✓ 1	A reading from the graph at 16 μA (accept 83–86%) or a clear statement that it levels off is needed.
so doubling the current gives very little extra benefit / is not worthwhile (and more charge increases the risk of shocks or sparks)	✓ 1

(2)

(f)

On a very hot day the field is dry and dusty, and the air has almost no moisture in it. The sprayer is pulled by a tractor that has rubber tyres.

Suggest one risk from static charge on such a day and one precaution the farmer could take.

Paper 2 only onlyPREDICTNotes

Show solution

risk: charge builds up on the tractor / sprayer / farmer because it cannot leak away in dry air, so the farmer could get a shock or a spark could ignite the dust (or flammable vapour from the liquid)	✓ 1	Risk must be linked to charge building up.
precaution: connect the metal frame of the sprayer to the ground with a metal (conducting) strap or chain, so electrons can flow to or from earth and charge cannot build up	✓ 1	Accept: earth the sprayer.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
all the droplets have the same (negative) charge	✓ 1
like charges repel, so each droplet pushes the others away and they spread apart	✓ 1	‘They repel’ alone, without linking to like charges: 1 mark max for the part.
Part (b)
the negative droplets repel electrons in the leaf	✓ 1	Must be electrons that move, not positive charges.
the electrons move away from the leaf surface, down through the plant to the (earthed) soil, so the surface nearest the cloud is left positive	✓ 1	Accept: electrons move to the far side of the leaf / to earth.
unlike charges attract, so the negative droplets are attracted to the positive leaf surface	✓ 1
Part (c)
t = 25 × 60 = 1500 s and Q = I × t = 8.0 × 10⁻⁶ × 1500	✓ 1	Both conversions (μA to A and minutes to s) must be seen.
Q = 0.012 C	✓ 1	Answer must be given to 2 s.f. or more.
Part (d)(i)
W = m × g = 6.0 × 10⁻¹¹ × 10	✓ 1
W = 6.0 × 10⁻¹⁰ N	✓ 1	Using g = 9.8 N/kg gives 5.9 × 10⁻¹⁰ N: accept.
ratio = 3.0 × 10⁻⁸ ÷ 6.0 × 10⁻¹⁰ = 50	✓ 1	Allow ECF from an incorrect weight. Accept 51 from g = 9.8 N/kg.
Part (d)(ii)
the attraction towards the leaf is much larger than the weight, so a droplet can move upwards / curve round onto the underside, against gravity	✓ 1
Part (e)(i)
30% (uncharged) and 80% (at 8.0 μA)	✓ 1	Both readings needed. Accept 29–31% and 79–81%.
80 ÷ 30 = 2.7 times	✓ 1	Allow ECF from the readings. Accept 2.6–2.8.
Part (e)(ii)
the graph levels off: at 16 μA about 84–85% lands on the leaves, only about 5 percentage points more than at 8.0 μA	✓ 1	A reading from the graph at 16 μA (accept 83–86%) or a clear statement that it levels off is needed.
so doubling the current gives very little extra benefit / is not worthwhile (and more charge increases the risk of shocks or sparks)	✓ 1
Part (f)
risk: charge builds up on the tractor / sprayer / farmer because it cannot leak away in dry air, so the farmer could get a shock or a spark could ignite the dust (or flammable vapour from the liquid)	✓ 1	Risk must be linked to charge building up.
precaution: connect the metal frame of the sprayer to the ground with a metal (conducting) strap or chain, so electrons can flow to or from earth and charge cannot build up	✓ 1	Accept: earth the sprayer.

Answers: (c) 0.012 C · (d)(i) 50 times · (e)(i) 30% and 80%; about 2.7 times

Common error (b) Writing that positive charge flows up from the ground into the leaf. Only electrons move: the leaf is left positive because electrons have been pushed away from it.

More 2(d) questions in the question bank

Topic 3 · Waves

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

Setup	3(b)	3(c)	3(d)	Other
An ocean data buoy: measuring swell waves	●			4(c) 4(d)
A race car passing a trackside microphone: sound waves and the Doppler effect	●		○	1(b)
Airport security: x-ray baggage scanners and microwave body scanners	○	●		7(b)
Satellite broadband: microwaves to orbit and back	○	●		8(b) 1(b)
A medical endoscope: refractive index, critical angle and optical fibres	○	○	●	—
Bat echolocation: ultrasound echoes and the speed of sound	○		●	1(b)

3(b) Properties of Waves

+ Add to test✓ In your testCC-EIP-3b-119 marks~20 min

An ocean data buoy: measuring swell waves

3(b) Properties of Waves4(c) Work and Power4(d) Energy Resources and Electricity GenerationBoth papers

A data buoy is moored far out at sea. It floats on the surface and records its vertical position every second as swell waves pass it. The buoy rises and falls but stays in the same place.

Figure 1 shows how the vertical displacement of the buoy from its rest position changes with time.

Figure 1: vertical displacement of the buoy against time

(a)

Water waves are transverse waves.

Explain how the motion of the buoy shows that water waves are transverse.

EXPLAINNotes

Show solution

the buoy (water) moves up and down / vibrates vertically, while the wave travels (transfers energy) horizontally past it	✓ 1	Both directions needed.
in a transverse wave the vibrations are perpendicular (at right angles) to the direction of energy transfer	✓ 1	‘Goes up and down’ alone does not score this mark.

(2)

(b)(i)

Use Figure 1 to determine the amplitude of the waves.

GRAPHNotes

Show solution

1.5 m

✓ 1

3.0 m (crest to trough) scores 0.

Answer: 1.5 m

(1)

(b)(ii)

Use Figure 1 to determine the period of the waves, and calculate their frequency.

CALCNotes

Show solution

T = 8.0 s	✓ 1	e.g. crests at 2 s and 10 s, or 3 complete waves in 24 s.
f = 1/T = 1/8.0 = 0.125 Hz	✓ 1	Accept 0.13 Hz. Allow ECF from an incorrect period.

Answer: 8.0 s; 0.125 Hz

(2)

(c)

A satellite measures the distance between neighbouring crests of these waves as 96 m.

Show that the speed of the waves is about 12 m/s.

CALCNotes

Show solution

v = f × λ = 0.125 × 96	✓ 1	Allow ECF for f from (b)(ii); 0.13 × 96 = 12.5 m/s also scores.
v = 12.0 m/s	✓ 1	Answer must be given to 3 s.f. or the substitution shown.

Answer: 12.0 m/s

(2)

(d)(i)

Near the coast the waves pass into shallow water, where their speed falls to 6.0 m/s.

Calculate the wavelength of the waves in the shallow water.

CALCNotes

Show solution

the frequency does not change (0.125 Hz)	✓ 1	Can be implied by correct working.
λ = v ÷ f = 6.0 ÷ 0.125 = 48 m	✓ 1	Allow ECF from (b)(ii). Accept: speed halves so wavelength halves, 96 ÷ 2 = 48 m.

Answer: 48 m

(2)

(d)(ii)

Figure 2 shows the wavefronts in deep water approaching the boundary with the shallow water at an angle.

Complete Figure 2 by drawing three wavefronts in the shallow water.

Figure 2: wavefronts reaching shallow water (not to scale)

SKETCHNotes

Show solution

three straight, parallel wavefronts that join the incident wavefronts at the boundary and are about half the spacing of those in deep water	✓ 1	Spacing judged by eye: clearly smaller than in deep water, about half.
wavefronts turned so that the direction of travel is closer to the normal (wavefronts less steeply angled to the boundary)	✓ 1	Bending away from the normal scores 0 for this mark.

(2)

(d)(iii)

Explain why the direction of travel of the waves changes at the boundary.

EXPLAINNotes

Show solution

one end of each wavefront reaches the shallow water first and slows down, while the rest of the wavefront is still moving at the faster speed	✓ 1
so the wavefront turns and the direction of travel bends towards the normal (refraction)	✓ 1	Accept: the wave refracts because its speed changes when it meets the boundary at an angle. Accept a correct comparison with light slowing down and bending towards the normal as it enters glass.

(2)

(e)(i)

A wave-energy device is tested next to the buoy. It has a float of mass 4000 kg. In a simple model of the device, every wave lifts the float from a trough to a crest, and the useful energy taken from each wave is the gravitational potential energy gained by the float. As the float falls back to a trough, this energy drives a generator. The device is 40% efficient.

Use Figure 1 and this simple model to calculate the electrical power output of the device.

Gravitational field strength g = 10 N/kg

Paper 2 only onlyMULTINotes

Show solution

height from trough to crest h = 2 × 1.5 = 3.0 m	✓ 1	Allow ECF from (b)(i).
GPE = m × g × h = 4000 × 10 × 3.0 = 120 000 J per wave	✓ 1	Using 9.8 N/kg gives 117 600 J: accept.
power input = 120 000 ÷ 8.0 = 15 000 W	✓ 1	One wave every period; allow ECF from (b)(ii).
output = 0.40 × 15 000 = 6000 W (6.0 kW)	✓ 1	Accept 5.9 kW from g = 9.8 N/kg.

Answer: 6000 W (6.0 kW)

(4)

(e)(ii)

During a storm the amplitude of the waves doubles, but their period does not change.

Use the simple model in (e)(i) to state the new electrical power output of the device. Use this to explain why the output of wave-energy devices is described as unreliable.

Paper 2 only onlyPREDICTNotes

Show solution

the float is lifted twice as high (6.0 m) once every 8.0 s, so the output doubles to 12 kW	✓ 1	Allow ECF: twice the answer to (e)(i). Do not penalise a correct comment that in real waves the power rises even more than this (with the square of the amplitude).
the output depends on the size of the waves, which depends on the weather / wind, so it changes from day to day and cannot be controlled to meet demand	✓ 1

Answer: 12 kW

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the buoy (water) moves up and down / vibrates vertically, while the wave travels (transfers energy) horizontally past it	✓ 1	Both directions needed.
in a transverse wave the vibrations are perpendicular (at right angles) to the direction of energy transfer	✓ 1	‘Goes up and down’ alone does not score this mark.
Part (b)(i)
1.5 m	✓ 1	3.0 m (crest to trough) scores 0.
Part (b)(ii)
T = 8.0 s	✓ 1	e.g. crests at 2 s and 10 s, or 3 complete waves in 24 s.
f = 1/T = 1/8.0 = 0.125 Hz	✓ 1	Accept 0.13 Hz. Allow ECF from an incorrect period.
Part (c)
v = f × λ = 0.125 × 96	✓ 1	Allow ECF for f from (b)(ii); 0.13 × 96 = 12.5 m/s also scores.
v = 12.0 m/s	✓ 1	Answer must be given to 3 s.f. or the substitution shown.
Part (d)(i)
the frequency does not change (0.125 Hz)	✓ 1	Can be implied by correct working.
λ = v ÷ f = 6.0 ÷ 0.125 = 48 m	✓ 1	Allow ECF from (b)(ii). Accept: speed halves so wavelength halves, 96 ÷ 2 = 48 m.
Part (d)(ii)
three straight, parallel wavefronts that join the incident wavefronts at the boundary and are about half the spacing of those in deep water	✓ 1	Spacing judged by eye: clearly smaller than in deep water, about half.
wavefronts turned so that the direction of travel is closer to the normal (wavefronts less steeply angled to the boundary)	✓ 1	Bending away from the normal scores 0 for this mark.
Part (d)(iii)
one end of each wavefront reaches the shallow water first and slows down, while the rest of the wavefront is still moving at the faster speed	✓ 1
so the wavefront turns and the direction of travel bends towards the normal (refraction)	✓ 1	Accept: the wave refracts because its speed changes when it meets the boundary at an angle. Accept a correct comparison with light slowing down and bending towards the normal as it enters glass.
Part (e)(i)
height from trough to crest h = 2 × 1.5 = 3.0 m	✓ 1	Allow ECF from (b)(i).
GPE = m × g × h = 4000 × 10 × 3.0 = 120 000 J per wave	✓ 1	Using 9.8 N/kg gives 117 600 J: accept.
power input = 120 000 ÷ 8.0 = 15 000 W	✓ 1	One wave every period; allow ECF from (b)(ii).
output = 0.40 × 15 000 = 6000 W (6.0 kW)	✓ 1	Accept 5.9 kW from g = 9.8 N/kg.
Part (e)(ii)
the float is lifted twice as high (6.0 m) once every 8.0 s, so the output doubles to 12 kW	✓ 1	Allow ECF: twice the answer to (e)(i). Do not penalise a correct comment that in real waves the power rises even more than this (with the square of the amplitude).
the output depends on the size of the waves, which depends on the weather / wind, so it changes from day to day and cannot be controlled to meet demand	✓ 1

Answers: (b)(i) 1.5 m · (b)(ii) 8.0 s; 0.125 Hz · (c) 12.0 m/s · (d)(i) 48 m · (e)(i) 6000 W (6.0 kW) · (e)(ii) 12 kW

Exam tip (a) A transverse/longitudinal answer needs both halves: the direction of the vibration AND what it is compared with (the direction of energy transfer).

Common error (b)(i) Measuring from a trough to a crest gives twice the amplitude. Amplitude is measured from the rest position (zero line) to a crest.

More 3(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-3b-220 marks~20 min

A race car passing a trackside microphone: sound waves and the Doppler effect

3(b) Properties of Waves3(d) Light and Sound1(b) Movement and PositionBoth papers

A microphone M is placed at the edge of a straight section of a race track. A race car travels along the straight at a constant speed of 306 km/h. The driver hears the engine produce a steady note of frequency 510 Hz.

Speed of sound in air = 340 m/s

Figure 1: wavefronts of the engine sound as the car approaches M (not to scale; the car's speed is exaggerated)

(a)

Sound is a longitudinal wave.

Describe how the particles of the air move as the sound passes, and state what is meant by a compression.

RECALLNotes

Show solution

the air particles vibrate backwards and forwards parallel to (along) the direction the sound travels / the direction of energy transfer	✓ 1	Particles moving along with the wave: 0 for this mark.
a compression is a region where the particles are closer together (higher pressure)	✓ 1

(2)

(b)

Show that, when the car is stationary, the wavelength in air of the 510 Hz note produced by the engine is about 0.7 m.

CALCNotes

Show solution

λ = v ÷ f = 340 ÷ 510	✓ 1
λ = 0.67 m	✓ 1	Answer must be given to 2 s.f. or more (0.667 m) or the substitution shown.

Answer: 0.67 m

(2)

(c)

Figure 1 shows wavefronts of the engine sound. Each circle was emitted from the point where the car was at that moment.

Use Figure 1 to explain why the frequency recorded by M while the car approaches is higher than 510 Hz.

EXPLAINNotes

Show solution

the car moves towards M between emitting one wavefront and the next, so the wavefronts in front of the car are closer together (bunched)	✓ 1
so the wavelength reaching M is shorter (than 0.67 m)	✓ 1
the speed of sound is unchanged, so (from f = v/λ) more waves reach M each second: a higher frequency	✓ 1	‘The sound travels faster’ scores 0 for this mark.

(3)

(d)(i)

Figure 2 shows the frequency recorded by M as the car approaches, passes and moves away.

Use Figure 2 to state the frequency recorded while the car is approaching, and the time at which the car passes M.

Figure 2: frequency recorded by M against time

GRAPHNotes

Show solution

680 Hz	✓ 1	Accept 670–690 Hz.
2.0 s (where the frequency falls most steeply / passes through 510 Hz)	✓ 1	Accept 1.9–2.1 s.

Answer: 680 Hz; 2.0 s

(2)

(d)(ii)

Calculate the wavelength of the sound reaching M while the car is approaching, and compare it with your answer to (b).

CALCNotes

Show solution

λ = 340 ÷ 680 = 0.50 m	✓ 1	Allow ECF from (d)(i).
shorter than the 0.67 m emitted / produced by the engine (because the wavefronts are bunched)	✓ 1	Allow ECF from (b).

Answer: 0.50 m, shorter than 0.67 m

(2)

(e)

On a later lap the car brakes while it is still approaching M. The note produced by the engine stays at 510 Hz.

Predict how the frequency recorded by M changes while the car slows down, and explain your answer.

PREDICTNotes

Show solution

the recorded frequency decreases (gets closer to 510 Hz) but stays above 510 Hz while the car is still approaching	✓ 1	‘Decreases’ alone scores this mark; ‘falls below 510 Hz’ does not.
the car moves a smaller distance between wavefronts, so the wavefronts are less bunched and the wavelength reaching M is longer	✓ 1

(2)

(f)

A spectator in the grandstand is 170 m from the car when the car passes M.

Calculate the time for the sound from the car to reach the spectator, and the distance the car travels in this time.

MULTINotes

Show solution

time = distance ÷ speed = 170 ÷ 340 = 0.50 s	✓ 1
car speed = 306 × 1000 ÷ 3600 = 85 m/s	✓ 1	Or 306 ÷ 3.6.
distance = 85 × 0.50 = 42.5 m (43 m)	✓ 1	Allow ECF from the time and the speed.

Answer: 0.50 s; 42.5 m

(3)

(g)(i)

In the pit lane the car's engine runs slowly. M is connected to an oscilloscope, which shows the trace in Figure 3. The timebase is set to 2 ms per division.

Determine the frequency of the sound.

Figure 3: oscilloscope trace of the engine sound in the pit lane (each square is one division)

Paper 2 only onlyGRAPHNotes

Show solution

4 cycles in 10 divisions, so one cycle = 2.5 divisions and T = 2.5 × 2 ms = 5.0 ms (0.0050 s)	✓ 1
f = 1/T = 1 ÷ 0.0050 = 200 Hz	✓ 1	Allow ECF from an incorrect period.

Answer: 200 Hz

(2)

(g)(ii)

The driver then makes the engine run faster. The sound becomes higher in pitch and louder. The oscilloscope settings are not changed.

Describe two changes to the trace in Figure 3.

Paper 2 only onlyREASONNotes

Show solution

more cycles across the screen / each cycle is narrower (shorter period), because the frequency is higher	✓ 1
the trace is taller (greater peak height), because the amplitude is larger	✓ 1

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the air particles vibrate backwards and forwards parallel to (along) the direction the sound travels / the direction of energy transfer	✓ 1	Particles moving along with the wave: 0 for this mark.
a compression is a region where the particles are closer together (higher pressure)	✓ 1
Part (b)
λ = v ÷ f = 340 ÷ 510	✓ 1
λ = 0.67 m	✓ 1	Answer must be given to 2 s.f. or more (0.667 m) or the substitution shown.
Part (c)
the car moves towards M between emitting one wavefront and the next, so the wavefronts in front of the car are closer together (bunched)	✓ 1
so the wavelength reaching M is shorter (than 0.67 m)	✓ 1
the speed of sound is unchanged, so (from f = v/λ) more waves reach M each second: a higher frequency	✓ 1	‘The sound travels faster’ scores 0 for this mark.
Part (d)(i)
680 Hz	✓ 1	Accept 670–690 Hz.
2.0 s (where the frequency falls most steeply / passes through 510 Hz)	✓ 1	Accept 1.9–2.1 s.
Part (d)(ii)
λ = 340 ÷ 680 = 0.50 m	✓ 1	Allow ECF from (d)(i).
shorter than the 0.67 m emitted / produced by the engine (because the wavefronts are bunched)	✓ 1	Allow ECF from (b).
Part (e)
the recorded frequency decreases (gets closer to 510 Hz) but stays above 510 Hz while the car is still approaching	✓ 1	‘Decreases’ alone scores this mark; ‘falls below 510 Hz’ does not.
the car moves a smaller distance between wavefronts, so the wavefronts are less bunched and the wavelength reaching M is longer	✓ 1
Part (f)
time = distance ÷ speed = 170 ÷ 340 = 0.50 s	✓ 1
car speed = 306 × 1000 ÷ 3600 = 85 m/s	✓ 1	Or 306 ÷ 3.6.
distance = 85 × 0.50 = 42.5 m (43 m)	✓ 1	Allow ECF from the time and the speed.
Part (g)(i)
4 cycles in 10 divisions, so one cycle = 2.5 divisions and T = 2.5 × 2 ms = 5.0 ms (0.0050 s)	✓ 1
f = 1/T = 1 ÷ 0.0050 = 200 Hz	✓ 1	Allow ECF from an incorrect period.
Part (g)(ii)
more cycles across the screen / each cycle is narrower (shorter period), because the frequency is higher	✓ 1
the trace is taller (greater peak height), because the amplitude is larger	✓ 1

Answers: (b) 0.67 m · (d)(i) 680 Hz; 2.0 s · (d)(ii) 0.50 m, shorter than 0.67 m · (f) 0.50 s; 42.5 m · (g)(i) 200 Hz

Exam tip (c) A Doppler explanation is a chain: source moves → wavefronts bunched → shorter wavelength → speed of sound unchanged → higher frequency. Leaving out the wavelength step loses marks.

More 3(b) questions in the question bank

3(c) The Electromagnetic Spectrum

+ Add to test✓ In your testCC-EIP-3c-118 marks~20 min

Airport security: x-ray baggage scanners and microwave body scanners

3(c) The Electromagnetic Spectrum3(b) Properties of Waves7(b) RadioactivityBoth papers

At an airport security check, hand luggage passes through an x-ray scanner. Passengers walk through a body scanner that uses microwaves of frequency 80 GHz.

Figure 1 shows the electromagnetic spectrum. Two regions are labelled P and Q.

The speed of electromagnetic waves in free space is 3.0 × 10⁸ m/s.

Figure 1: the electromagnetic spectrum (band widths not to scale)

(a)(i)

Name the regions labelled P and Q in Figure 1.

RECALLNotes

Show solution

P: microwaves	✓ 1	Accept microwave(s).
Q: ultraviolet	✓ 1	Accept UV.

Answer: P: microwaves; Q: ultraviolet

(2)

(a)(ii)

The x-rays and the microwaves have very different frequencies. State one property that they share when they travel through free space.

RECALLNotes

Show solution

same speed (3.0 × 10⁸ m/s) / can travel through a vacuum / transfer energy / are transverse waves

✓ 1

Any one. Do not accept “same frequency” or “same wavelength”.

(1)

(b)(i)

Calculate the wavelength of the microwaves used by the body scanner.

CALCNotes

Show solution

λ = v ÷ f = 3.0 × 10⁸ ÷ 8.0 × 10¹⁰	✓ 1	Rearrangement and substitution, with 80 GHz converted to 8.0 × 10¹⁰ Hz.
λ = 3.75 × 10⁻³ m	✓ 1	Accept 0.00375 m, 3.8 × 10⁻³ m or 3.75 mm. A power-of-ten error (e.g. 3.75 m from using 80 × 10⁶ Hz) scores 1.

Answer: 3.75 × 10⁻³ m (3.75 mm)

(2)

(b)(ii)

The x-rays used in the baggage scanner have a wavelength of 1.5 × 10⁻¹¹ m.

Show that their frequency is about 2 × 10¹⁹ Hz.

DERIVENotes

Show solution

f = v ÷ λ = 3.0 × 10⁸ ÷ 1.5 × 10⁻¹¹	✓ 1	Rearrangement and substitution must be seen.
f = 2.0 × 10¹⁹ Hz	✓ 1	The answer must be stated to at least 2 s.f., with the working above.

Answer: 2.0 × 10¹⁹ Hz

(2)

(c)(i)

Figure 2 shows the percentage of the scanner's x-rays that pass through a 1.0 cm thick layer of different materials.

Use Figure 2 to state the percentage of x-rays that pass through 1.0 cm of aluminium.

Figure 2: percentage of x-rays passing through 1.0 cm of each material

GRAPHNotes

Show solution

58%

✓ 1

Accept 57% to 59%.

Answer: 58%

(1)

(c)(ii)

A suitcase contains folded clothes and a steel pair of scissors.

Use Figure 2 to explain why the scissors show up clearly on the scanner image but the clothes do not.

EXPLAINNotes

Show solution

almost all (95%) of the x-rays pass through the clothes, so they leave little or no shadow on the detector	✓ 1	Must refer to the x-rays passing through the clothes (low-density material).
steel absorbs almost all of the x-rays (only 1% pass through), so the scissors leave a clear shadow / dark outline on the detector	✓ 1	Must refer to absorption by the steel (dense material). Allow the idea of a contrast between the two.

(2)

(d)

X-rays, like gamma rays, are ionising radiation.

Explain why a large dose of x-rays is harmful to the human body.

EXPLAINNotes

Show solution

x-rays knock electrons out of atoms in body cells / ionise atoms or molecules in cells	✓ 1	The meaning of ionising must be given, not just the word.
this damages cells or DNA, which can cause mutations / cancer / kill cells	✓ 1	“It is dangerous” or “it damages you” is not enough: a named effect is needed.

(2)

(e)

The walls of the scanner tunnel are lined with sheets of lead.

Use Figure 2 to explain why lead is used rather than aluminium.

REASONNotes

Show solution

(almost) no x-rays pass through lead, but about 58% pass through the same thickness of aluminium	✓ 1	Values from Figure 2 or a clear comparison needed. Allow ECF from (c)(i).
so lead stops x-rays escaping from the tunnel and protects (reduces the dose to) staff and passengers nearby	✓ 1	Must link to protecting people outside the tunnel.

(2)

(f)

The body scanner sends very low power microwaves towards each passenger for a few seconds.

Comment on why this microwave scanner is considered safe for passengers, while x-rays are not used to scan people at the airport.

REASONNotes

Show solution

microwaves have a much lower frequency than x-rays, so they carry less energy and are not ionising / do not damage DNA	✓ 1	Allow a comparison using the frequencies from (b).
the harm from too much microwave radiation is internal heating of body tissue, and the power is too low and the time too short for any significant heating	✓ 1	Must name internal heating and link it to low power or short exposure.

(2)

(g)

An engineer suggests replacing the body scanner with an infrared (thermal) camera.

Suggest one thing that the infrared camera could detect, and one reason why it could not do the job of the body scanner.

REASONNotes

Show solution

it could detect a passenger with a high body temperature (fever) / warm objects, because warmer objects emit more infrared	✓ 1	Must link the detection to the emission of infrared by warm objects.
infrared is absorbed at the surface (by clothing / skin), so it cannot show objects hidden under clothes	✓ 1	Accept: the camera only shows the temperature of the outer surface of the clothing.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
P: microwaves	✓ 1	Accept microwave(s).
Q: ultraviolet	✓ 1	Accept UV.
Part (a)(ii)
same speed (3.0 × 10⁸ m/s) / can travel through a vacuum / transfer energy / are transverse waves	✓ 1	Any one. Do not accept “same frequency” or “same wavelength”.
Part (b)(i)
λ = v ÷ f = 3.0 × 10⁸ ÷ 8.0 × 10¹⁰	✓ 1	Rearrangement and substitution, with 80 GHz converted to 8.0 × 10¹⁰ Hz.
λ = 3.75 × 10⁻³ m	✓ 1	Accept 0.00375 m, 3.8 × 10⁻³ m or 3.75 mm. A power-of-ten error (e.g. 3.75 m from using 80 × 10⁶ Hz) scores 1.
Part (b)(ii)
f = v ÷ λ = 3.0 × 10⁸ ÷ 1.5 × 10⁻¹¹	✓ 1	Rearrangement and substitution must be seen.
f = 2.0 × 10¹⁹ Hz	✓ 1	The answer must be stated to at least 2 s.f., with the working above.
Part (c)(i)
58%	✓ 1	Accept 57% to 59%.
Part (c)(ii)
almost all (95%) of the x-rays pass through the clothes, so they leave little or no shadow on the detector	✓ 1	Must refer to the x-rays passing through the clothes (low-density material).
steel absorbs almost all of the x-rays (only 1% pass through), so the scissors leave a clear shadow / dark outline on the detector	✓ 1	Must refer to absorption by the steel (dense material). Allow the idea of a contrast between the two.
Part (d)
x-rays knock electrons out of atoms in body cells / ionise atoms or molecules in cells	✓ 1	The meaning of ionising must be given, not just the word.
this damages cells or DNA, which can cause mutations / cancer / kill cells	✓ 1	“It is dangerous” or “it damages you” is not enough: a named effect is needed.
Part (e)
(almost) no x-rays pass through lead, but about 58% pass through the same thickness of aluminium	✓ 1	Values from Figure 2 or a clear comparison needed. Allow ECF from (c)(i).
so lead stops x-rays escaping from the tunnel and protects (reduces the dose to) staff and passengers nearby	✓ 1	Must link to protecting people outside the tunnel.
Part (f)
microwaves have a much lower frequency than x-rays, so they carry less energy and are not ionising / do not damage DNA	✓ 1	Allow a comparison using the frequencies from (b).
the harm from too much microwave radiation is internal heating of body tissue, and the power is too low and the time too short for any significant heating	✓ 1	Must name internal heating and link it to low power or short exposure.
Part (g)
it could detect a passenger with a high body temperature (fever) / warm objects, because warmer objects emit more infrared	✓ 1	Must link the detection to the emission of infrared by warm objects.
infrared is absorbed at the surface (by clothing / skin), so it cannot show objects hidden under clothes	✓ 1	Accept: the camera only shows the temperature of the outer surface of the clothing.

Answers: (a)(i) P: microwaves; Q: ultraviolet · (b)(i) 3.75 × 10⁻³ m (3.75 mm) · (b)(ii) 2.0 × 10¹⁹ Hz · (c)(i) 58%

Exam tip (b)(i) Write 80 GHz = 8.0 × 10¹⁰ Hz on its own line before substituting. 1 GHz is 10⁹ Hz.

Common error (f) Saying microwaves are safe because they “do not go into the body”. Microwaves do penetrate body tissue; the point is that at very low power the heating they cause is negligible and they are not ionising.

More 3(c) questions in the question bank

+ Add to test✓ In your testCC-EIP-3c-218 marks~20 min

Satellite broadband: microwaves to orbit and back

3(c) The Electromagnetic Spectrum3(b) Properties of Waves8(b) Motion in the Universe1(b) Movement and PositionBoth papers

A farmhouse far from any cable network receives its internet connection from satellites. A dish on the house sends microwave signals up to a satellite, which sends them straight back down to a ground station connected to the internet.

The satellites orbit at a height of 550 km above the Earth's surface. The signals sent down from the satellite have a frequency of 12.0 GHz.

The speed of electromagnetic waves in free space is 3.0 × 10⁸ m/s.

Figure 1: the satellite link (not to scale)

(a)

Explain why microwaves are used to send signals between the ground and a satellite.

EXPLAINNotes

Show solution

microwaves pass through the atmosphere with little absorption	✓ 1	Accept: are not absorbed (much) by the atmosphere.
so the signal reaches the satellite in space and can be sent back to the ground	✓ 1	Must link to reaching the satellite / returning to Earth. Do not accept “they travel fast” (all electromagnetic waves have the same speed).

(2)

(b)

Calculate the wavelength of the 12.0 GHz signals.

CALCNotes

Show solution

λ = v ÷ f = 3.0 × 10⁸ ÷ 1.20 × 10¹⁰	✓ 1	Substitution with 12.0 GHz converted to 1.20 × 10¹⁰ Hz.
λ = 0.025 m	✓ 1	Accept 2.5 cm or 2.5 × 10⁻² m. A power-of-ten error scores 1.

Answer: 0.025 m (2.5 cm)

(2)

(c)

The satellite is directly above the house and the ground station.

Show that the time for a signal to travel up from the dish to the satellite and back down to the ground station is about 3.7 ms.

DERIVENotes

Show solution

distance = 2 × 550 km = 1.10 × 10⁶ m	✓ 1	The factor of 2 and the conversion to metres must be seen.
time = distance ÷ speed = 1.10 × 10⁶ ÷ 3.0 × 10⁸ = 3.67 × 10⁻³ s (3.67 ms)	✓ 1	Answer to at least 3 s.f. (or 0.00367 s) needed, since 3.7 ms is given.

Answer: 3.67 ms

(2)

(d)(i)

Figure 2 shows how this travel time (up and back down) depends on the height of the satellite above the Earth's surface.

Television signals are sent through satellites at a height of 36 000 km. Use Figure 2 to find the travel time for these satellites.

Figure 2: travel time (up and back down) against height of the satellite

GRAPHNotes

Show solution

240 ms

✓ 1

Accept 238 ms to 242 ms.

Answer: 240 ms

(1)

(d)(ii)

Explain why the graph in Figure 2 is a straight line through the origin.

REASONNotes

Show solution

the distance travelled by the signal is proportional to the height (distance = 2 × height)	✓ 1	Accept: doubling the height doubles the distance.
all electromagnetic waves travel at the same (constant) speed, so time = distance ÷ speed is proportional to the height	✓ 1	Must refer to the speed being constant / the same.

(2)

(e)

The radius of the Earth is 6370 km. Each satellite moves in a circular orbit at a constant speed of 7.59 km/s.

Calculate the time taken for one complete orbit. Give your answer in minutes.

MULTINotes

Show solution

orbital radius r = 6370 + 550 = 6920 km = 6.92 × 10⁶ m	✓ 1	Using 550 km as the radius scores 0 for this mark, but the later marks can be awarded (giving 7.6 min).
T = 2 × π × r ÷ v = 2 × π × 6.92 × 10⁶ ÷ 7590 (= 5730 s)	✓ 1	Rearrangement and substitution with consistent units (m and m/s, or km and km/s).
T = 95.5 minutes	✓ 1	Accept 95 to 96 minutes. 5730 s left in seconds scores 2.

Answer: 95.5 minutes (5730 s)

(3)

(f)

Explain why the satellite is accelerating, even though it moves at a constant speed.

EXPLAINNotes

Show solution

the direction of motion (velocity) is always changing	✓ 1	Accept: velocity changes because direction changes.
because the gravitational force of the Earth acts on it towards the centre of the Earth / the orbit	✓ 1	Must name gravity (gravitational force) and its direction.

(2)

(g)

Some residents are worried about the microwaves sent up by the dish on the farmhouse roof.

Describe the harm that too much microwave radiation can do to the body, and suggest one reason why the dish is not a risk to people in the house.

REASONNotes

Show solution

internal heating of body tissue (microwaves are absorbed by water in the body)	✓ 1	Do not accept burns to the skin (infrared), cancer or ionisation.
the dish sends a low-power beam / the beam is pointed up at the sky, away from people / people are not close to or in front of the dish	✓ 1	Any one sensible reason.

(2)

(h)

The dish sends its signals up to the satellite at a higher frequency, 14.0 GHz.

Compare the speed and the wavelength of these microwaves with those of the 12.0 GHz signals in (b).

PREDICTNotes

Show solution

the speed is the same (3.0 × 10⁸ m/s)	✓ 1	Do not accept “faster”.
the wavelength is shorter (0.021 m), because λ = v ÷ f and v is unchanged	✓ 1	A statement “shorter” with a reason or the value 0.021 m (2.1 cm). Allow ECF from (b).

Answer: same speed; shorter wavelength (0.021 m)

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
microwaves pass through the atmosphere with little absorption	✓ 1	Accept: are not absorbed (much) by the atmosphere.
so the signal reaches the satellite in space and can be sent back to the ground	✓ 1	Must link to reaching the satellite / returning to Earth. Do not accept “they travel fast” (all electromagnetic waves have the same speed).
Part (b)
λ = v ÷ f = 3.0 × 10⁸ ÷ 1.20 × 10¹⁰	✓ 1	Substitution with 12.0 GHz converted to 1.20 × 10¹⁰ Hz.
λ = 0.025 m	✓ 1	Accept 2.5 cm or 2.5 × 10⁻² m. A power-of-ten error scores 1.
Part (c)
distance = 2 × 550 km = 1.10 × 10⁶ m	✓ 1	The factor of 2 and the conversion to metres must be seen.
time = distance ÷ speed = 1.10 × 10⁶ ÷ 3.0 × 10⁸ = 3.67 × 10⁻³ s (3.67 ms)	✓ 1	Answer to at least 3 s.f. (or 0.00367 s) needed, since 3.7 ms is given.
Part (d)(i)
240 ms	✓ 1	Accept 238 ms to 242 ms.
Part (d)(ii)
the distance travelled by the signal is proportional to the height (distance = 2 × height)	✓ 1	Accept: doubling the height doubles the distance.
all electromagnetic waves travel at the same (constant) speed, so time = distance ÷ speed is proportional to the height	✓ 1	Must refer to the speed being constant / the same.
Part (e)
orbital radius r = 6370 + 550 = 6920 km = 6.92 × 10⁶ m	✓ 1	Using 550 km as the radius scores 0 for this mark, but the later marks can be awarded (giving 7.6 min).
T = 2 × π × r ÷ v = 2 × π × 6.92 × 10⁶ ÷ 7590 (= 5730 s)	✓ 1	Rearrangement and substitution with consistent units (m and m/s, or km and km/s).
T = 95.5 minutes	✓ 1	Accept 95 to 96 minutes. 5730 s left in seconds scores 2.
Part (f)
the direction of motion (velocity) is always changing	✓ 1	Accept: velocity changes because direction changes.
because the gravitational force of the Earth acts on it towards the centre of the Earth / the orbit	✓ 1	Must name gravity (gravitational force) and its direction.
Part (g)
internal heating of body tissue (microwaves are absorbed by water in the body)	✓ 1	Do not accept burns to the skin (infrared), cancer or ionisation.
the dish sends a low-power beam / the beam is pointed up at the sky, away from people / people are not close to or in front of the dish	✓ 1	Any one sensible reason.
Part (h)
the speed is the same (3.0 × 10⁸ m/s)	✓ 1	Do not accept “faster”.
the wavelength is shorter (0.021 m), because λ = v ÷ f and v is unchanged	✓ 1	A statement “shorter” with a reason or the value 0.021 m (2.1 cm). Allow ECF from (b).

Answers: (b) 0.025 m (2.5 cm) · (c) 3.67 ms · (d)(i) 240 ms · (e) 95.5 minutes (5730 s) · (h) same speed; shorter wavelength (0.021 m)

Common error (c) Using 550 km as the whole distance. The signal goes up and comes back down, so it travels twice the height.

More 3(c) questions in the question bank

3(d) Light and Sound

+ Add to test✓ In your testCC-EIP-3d-118 marks~20 min

A medical endoscope: refractive index, critical angle and optical fibres

3(d) Light and Sound3(b) Properties of Waves3(c) The Electromagnetic SpectrumBoth papers

An endoscope lets a doctor see inside a patient's body. Visible light travels along a bundle of thin glass fibres into the body, and a second bundle carries light from the inside of the body back to a camera.

Before the fibres are made, a technician measures the refractive index of the glass. She shines rays of light into a rectangular block of the glass at different angles of incidence i and measures each angle of refraction r. Figure 1 shows her results, plotted as sin i against sin r.

Figure 1: sin i against sin r for the glass block

(a)(i)

Draw a line of best fit on Figure 1 and use it to determine the refractive index of the glass.

GRAPHNotes

Show solution

straight line of best fit through the origin, with points evenly spread about it	✓ 1	The line must pass through (0, 0).
gradient found from a large triangle, e.g. 0.93 ÷ 0.60	✓ 1	Triangle (or two read points) using at least half the line.
refractive index = 1.55	✓ 1	Accept 1.50 to 1.60. No unit. Using a single data point instead of the line scores this mark only if within range, but not the second mark.

Answer: 1.55 (accept 1.50–1.60)

(3)

(a)(ii)

Explain why the technician uses the gradient of the graph rather than calculating the refractive index from a single pair of angles.

REASONNotes

Show solution

the line uses many readings, so it reduces the effect of random errors in measuring each angle

✓ 1

Accept: one reading could be anomalous / have a large error; an average over many readings is more reliable.

(1)

(b)

The glass has a refractive index of 1.55.

Show that the critical angle for this glass is about 40°.

DERIVENotes

Show solution

sin c = 1 ÷ n = 1 ÷ 1.55 (= 0.645)	✓ 1	Equation and substitution must be seen.
c = sin⁻¹(0.645) = 40.2°	✓ 1	The value must be given to at least 3 s.f. Do not accept 0.645° (no inverse sine).

Answer: 40.2°

(2)

(c)

Figure 2 shows a ray of light inside a straight section of one fibre. The ray meets the wall of the fibre at point A. Treat each fibre as glass surrounded by air.

Explain what happens to the ray at A.

Figure 2: a ray inside a straight section of fibre (not to scale)

EXPLAINNotes

Show solution

the ray is travelling from glass towards air (denser to less dense) and its angle of incidence (58°) is greater than the critical angle (40.2°)	✓ 1	Both conditions needed. Allow ECF from (b).
so all of the light is reflected back into the glass: total internal reflection, at 58° to the normal	✓ 1	“It reflects” alone is not enough; total (internal) reflection or “no light leaves” must be stated.

(2)

(d)(i)

Where the fibre is bent sharply, a ray meets the wall with an angle of incidence of only 38°.

Calculate the angle of refraction of this ray in the air outside the fibre.

CALCNotes

Show solution

sin (angle in air) = 1.55 × sin 38° (= 0.954)	✓ 1	Correct use of n = sin i ÷ sin r with the angle in air as i and 38° as the angle in the glass.
angle in air = 72.6°	✓ 1	Accept 73°. 23.4° (from dividing by 1.55) scores 0.

Answer: 72.6°

(2)

(d)(ii)

Suggest why the image seen through an endoscope becomes dimmer if the bundle of fibres is bent too sharply.

REASONNotes

Show solution

at a sharp bend the angle of incidence becomes less than the critical angle, so some light refracts out of the fibre and does not reach the camera

✓ 1

Must link to the angle being smaller than the critical angle (light escaping). Allow ECF from (d)(i).

(1)

(e)

The manufacturer changes to a glass with a refractive index of 1.70.

Calculate the new critical angle, and explain whether the ray in (d) now stays inside the fibre.

PREDICTNotes

Show solution

sin c = 1 ÷ 1.70 = 0.588	✓ 1	Substitution seen.
c = 36.0°	✓ 1	Accept 36°.
38° is now greater than the critical angle, so the ray is totally internally reflected and stays inside the fibre	✓ 1	Allow ECF from their critical angle. A conclusion with no comparison of angles scores 0 for this mark.

Answer: 36.0°; the ray is now totally internally reflected

(3)

(f)

The light in the fibres has a frequency of 5.0 × 10¹⁴ Hz. Light travels through the glass of refractive index 1.55 at 1.94 × 10⁸ m/s.

Calculate the wavelength of this light inside the glass.

CALCNotes

Show solution

λ = v ÷ f = 1.94 × 10⁸ ÷ 5.0 × 10¹⁴	✓ 1	Rearrangement and substitution with the speed in glass.
λ = 3.9 × 10⁻⁷ m	✓ 1	Accept 3.88 × 10⁻⁷ m. Using 3.0 × 10⁸ m/s gives 6.0 × 10⁻⁷ m and scores 1.

Answer: 3.9 × 10⁻⁷ m

(2)

(g)

An x-ray image could also be used to look for an ulcer in the soft lining of a patient's stomach.

Give two reasons why the endoscope is better than an x-ray image for this.

EXPLAINNotes

Show solution

x-rays pass through soft tissue, so the stomach lining (and an ulcer in it) would not show up clearly on an x-ray image	✓ 1	Must link to x-rays passing through soft (low-density) tissue.
x-rays are ionising / can damage cells and cause cancer or mutation, but the visible light of the endoscope is not harmful in this way	✓ 1	A named harm is needed; “x-rays are dangerous” alone is not enough.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
straight line of best fit through the origin, with points evenly spread about it	✓ 1	The line must pass through (0, 0).
gradient found from a large triangle, e.g. 0.93 ÷ 0.60	✓ 1	Triangle (or two read points) using at least half the line.
refractive index = 1.55	✓ 1	Accept 1.50 to 1.60. No unit. Using a single data point instead of the line scores this mark only if within range, but not the second mark.
Part (a)(ii)
the line uses many readings, so it reduces the effect of random errors in measuring each angle	✓ 1	Accept: one reading could be anomalous / have a large error; an average over many readings is more reliable.
Part (b)
sin c = 1 ÷ n = 1 ÷ 1.55 (= 0.645)	✓ 1	Equation and substitution must be seen.
c = sin⁻¹(0.645) = 40.2°	✓ 1	The value must be given to at least 3 s.f. Do not accept 0.645° (no inverse sine).
Part (c)
the ray is travelling from glass towards air (denser to less dense) and its angle of incidence (58°) is greater than the critical angle (40.2°)	✓ 1	Both conditions needed. Allow ECF from (b).
so all of the light is reflected back into the glass: total internal reflection, at 58° to the normal	✓ 1	“It reflects” alone is not enough; total (internal) reflection or “no light leaves” must be stated.
Part (d)(i)
sin (angle in air) = 1.55 × sin 38° (= 0.954)	✓ 1	Correct use of n = sin i ÷ sin r with the angle in air as i and 38° as the angle in the glass.
angle in air = 72.6°	✓ 1	Accept 73°. 23.4° (from dividing by 1.55) scores 0.
Part (d)(ii)
at a sharp bend the angle of incidence becomes less than the critical angle, so some light refracts out of the fibre and does not reach the camera	✓ 1	Must link to the angle being smaller than the critical angle (light escaping). Allow ECF from (d)(i).
Part (e)
sin c = 1 ÷ 1.70 = 0.588	✓ 1	Substitution seen.
c = 36.0°	✓ 1	Accept 36°.
38° is now greater than the critical angle, so the ray is totally internally reflected and stays inside the fibre	✓ 1	Allow ECF from their critical angle. A conclusion with no comparison of angles scores 0 for this mark.
Part (f)
λ = v ÷ f = 1.94 × 10⁸ ÷ 5.0 × 10¹⁴	✓ 1	Rearrangement and substitution with the speed in glass.
λ = 3.9 × 10⁻⁷ m	✓ 1	Accept 3.88 × 10⁻⁷ m. Using 3.0 × 10⁸ m/s gives 6.0 × 10⁻⁷ m and scores 1.
Part (g)
x-rays pass through soft tissue, so the stomach lining (and an ulcer in it) would not show up clearly on an x-ray image	✓ 1	Must link to x-rays passing through soft (low-density) tissue.
x-rays are ionising / can damage cells and cause cancer or mutation, but the visible light of the endoscope is not harmful in this way	✓ 1	A named harm is needed; “x-rays are dangerous” alone is not enough.

Answers: (a)(i) 1.55 (accept 1.50–1.60) · (b) 40.2° · (d)(i) 72.6° · (e) 36.0°; the ray is now totally internally reflected · (f) 3.9 × 10⁻⁷ m

Exam tip (d)(i) In n = sin i ÷ sin r, i is always the angle in the air and r the angle in the glass, whichever way the light is travelling. Light leaving glass bends away from the normal, so the angle in the air must be larger than 38°.

More 3(d) questions in the question bank

+ Add to test✓ In your testCC-EIP-3d-220 marks~20 min

Bat echolocation: ultrasound echoes and the speed of sound

3(d) Light and Sound3(b) Properties of Waves1(b) Movement and PositionBoth papers

A bat hunts moths at night by making short pulses of high-frequency sound and listening for the echoes. A researcher records the bat's calls with a special microphone connected to an oscilloscope.

The speed of sound in air is 340 m/s.

Figure 1: oscilloscope trace of one call (timebase 0.005 ms per division)

(a)

Sound is a longitudinal wave. Describe how the particles of air move as the bat's call passes through the air.

RECALLNotes

Show solution

the particles vibrate (backwards and forwards) about a fixed position	✓ 1	Do not accept “the particles move from the bat to the moth”.
parallel to the direction in which the sound travels / energy is transferred, forming compressions and rarefactions	✓ 1	Accept “along the direction of travel”.

(2)

(b)(i)

Figure 1 shows the trace of one call. The timebase is set to 0.005 ms per division.

Determine the frequency of the call.

Paper 2 only onlyGRAPHNotes

Show solution

one cycle = 5 divisions (2 cycles across 10 divisions)	✓ 1
T = 5 × 0.005 ms = 0.025 ms = 2.5 × 10⁻⁵ s	✓ 1	Conversion to seconds needed. Allow ECF from the number of divisions.
f = 1 ÷ T = 1 ÷ 2.5 × 10⁻⁵ = 40 000 Hz	✓ 1	Accept 40 kHz or 4.0 × 10⁴ Hz. Allow ECF from T.

Answer: 40 000 Hz (40 kHz)

(3)

(b)(ii)

Explain why a person standing next to the bat cannot hear these calls.

Paper 2 only onlyEXPLAINNotes

Show solution

humans can hear frequencies from 20 Hz to 20 000 Hz	✓ 1	Both limits needed.
40 000 Hz is above 20 000 Hz (the upper limit of human hearing)	✓ 1	Allow ECF from (b)(i) if their frequency is above 20 000 Hz.

(2)

(c)

The frequency of the call is 40 kHz.

Calculate the wavelength of the call in air.

CALCNotes

Show solution

λ = v ÷ f = 340 ÷ 40 000	✓ 1	40 kHz converted to 40 000 Hz.
λ = 0.0085 m	✓ 1	Accept 8.5 mm or 8.5 × 10⁻³ m.

Answer: 0.0085 m (8.5 mm)

(2)

(d)

An echo from a moth reaches the bat 17.6 ms after the bat makes a call.

Show that the moth is about 3.0 m from the bat.

DERIVENotes

Show solution

distance travelled by the sound = 340 × 0.0176 = 5.98 m	✓ 1	Conversion of 17.6 ms to 0.0176 s must be seen.
distance to moth = 5.98 ÷ 2 = 2.99 m	✓ 1	Halving for the journey there and back must be shown; answer to at least 3 s.f.

Answer: 2.99 m

(2)

(e)

The bat flies straight towards a moth that is hovering in one place. The researcher uses the echo times to plot Figure 2, which shows the distance between the bat and the moth.

Use Figure 2 to determine the speed of the bat.

Figure 2: distance from the bat to the moth against time

GRAPHNotes

Show solution

gradient from the line, e.g. (3.0 − 0.5) ÷ (0.50 − 0)	✓ 1	Any correct readings from the line over at least 0.2 s.
speed = 5.0 m/s	✓ 1	Accept −5.0 m/s.

Answer: 5.0 m/s

(2)

(f)

Use Figure 2 to calculate the time between a call and its echo at a time of 0.40 s. Assume the bat moves a negligible distance while the call and its echo travel.

MULTINotes

Show solution

distance from Figure 2 = 1.0 m, so the sound travels 2 × 1.0 = 2.0 m	✓ 1	Reading from the graph and the factor of 2 both needed.
time = 2.0 ÷ 340 = 0.0059 s (5.9 ms)	✓ 1	Accept 5.88 ms. 2.9 ms (no factor of 2) scores 1.

Answer: 5.9 ms (0.0059 s)

(2)

(g)

On a warm evening the speed of sound in air is 350 m/s. The researcher measures an echo time of 17.6 ms but still uses 340 m/s to calculate the distance to the moth.

State whether the distance he calculates is larger or smaller than the true distance, and explain your answer.

PREDICTNotes

Show solution

smaller than the true distance	✓ 1	Award this mark only if a reason is also given (see the next mark row).
sound travels faster in warm air, so in 17.6 ms it really covers a greater distance (350 × 0.0176 ÷ 2 = 3.08 m, not 2.99 m)	✓ 1	Must link the higher speed to a greater true distance for the same time.

Answer: smaller (true distance 3.08 m)

(2)

(h)

Describe how two microphones could be used to measure the speed of sound in air in a laboratory.

Paper 2 only onlyEXPTNotes

Show solution

place two microphones a measured distance apart (e.g. 1 m), in line with the sound source, and measure the distance with a metre rule	✓ 1	The distance between the microphones must be measured.
connect them to a fast electronic timer / dual-trace oscilloscope and make a sharp sound (e.g. a clap) in line with them	✓ 1
measure the time delay between the sound reaching the two microphones and calculate speed = distance between microphones ÷ time delay	✓ 1	Speed equation with the microphone separation and the time delay needed.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
the particles vibrate (backwards and forwards) about a fixed position	✓ 1	Do not accept “the particles move from the bat to the moth”.
parallel to the direction in which the sound travels / energy is transferred, forming compressions and rarefactions	✓ 1	Accept “along the direction of travel”.
Part (b)(i)
one cycle = 5 divisions (2 cycles across 10 divisions)	✓ 1
T = 5 × 0.005 ms = 0.025 ms = 2.5 × 10⁻⁵ s	✓ 1	Conversion to seconds needed. Allow ECF from the number of divisions.
f = 1 ÷ T = 1 ÷ 2.5 × 10⁻⁵ = 40 000 Hz	✓ 1	Accept 40 kHz or 4.0 × 10⁴ Hz. Allow ECF from T.
Part (b)(ii)
humans can hear frequencies from 20 Hz to 20 000 Hz	✓ 1	Both limits needed.
40 000 Hz is above 20 000 Hz (the upper limit of human hearing)	✓ 1	Allow ECF from (b)(i) if their frequency is above 20 000 Hz.
Part (c)
λ = v ÷ f = 340 ÷ 40 000	✓ 1	40 kHz converted to 40 000 Hz.
λ = 0.0085 m	✓ 1	Accept 8.5 mm or 8.5 × 10⁻³ m.
Part (d)
distance travelled by the sound = 340 × 0.0176 = 5.98 m	✓ 1	Conversion of 17.6 ms to 0.0176 s must be seen.
distance to moth = 5.98 ÷ 2 = 2.99 m	✓ 1	Halving for the journey there and back must be shown; answer to at least 3 s.f.
Part (e)
gradient from the line, e.g. (3.0 − 0.5) ÷ (0.50 − 0)	✓ 1	Any correct readings from the line over at least 0.2 s.
speed = 5.0 m/s	✓ 1	Accept −5.0 m/s.
Part (f)
distance from Figure 2 = 1.0 m, so the sound travels 2 × 1.0 = 2.0 m	✓ 1	Reading from the graph and the factor of 2 both needed.
time = 2.0 ÷ 340 = 0.0059 s (5.9 ms)	✓ 1	Accept 5.88 ms. 2.9 ms (no factor of 2) scores 1.
Part (g)
smaller than the true distance	✓ 1	Award this mark only if a reason is also given (see the next mark row).
sound travels faster in warm air, so in 17.6 ms it really covers a greater distance (350 × 0.0176 ÷ 2 = 3.08 m, not 2.99 m)	✓ 1	Must link the higher speed to a greater true distance for the same time.
Part (h)
place two microphones a measured distance apart (e.g. 1 m), in line with the sound source, and measure the distance with a metre rule	✓ 1	The distance between the microphones must be measured.
connect them to a fast electronic timer / dual-trace oscilloscope and make a sharp sound (e.g. a clap) in line with them	✓ 1
measure the time delay between the sound reaching the two microphones and calculate speed = distance between microphones ÷ time delay	✓ 1	Speed equation with the microphone separation and the time delay needed.

Answers: (b)(i) 40 000 Hz (40 kHz) · (c) 0.0085 m (8.5 mm) · (d) 2.99 m · (e) 5.0 m/s · (f) 5.9 ms (0.0059 s) · (g) smaller (true distance 3.08 m)

Common error (d) Forgetting that the sound travels to the moth and back again. The time measured is for twice the distance to the moth.

More 3(d) questions in the question bank

Topic 4 · Energy resources and energy transfers

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

Setup	4(b)	4(c)	4(d)	Other
A low-energy house: reducing thermal energy transfer	●	○		2(b) 5(c)
Keeping a gaming laptop cool: heat sinks, fans and efficiency	●			2(b) 2(c)
A launched rollercoaster: work, kinetic energy and power		●		1(b) 1(c)
A bungee jump: gravitational, kinetic and elastic energy stores		●		1(c) 1(b)
A pumped-storage hydroelectric scheme: storing energy in water	○	○	●	—
An offshore wind farm: energy from moving air	○	○	●	6(d)

4(b) Energy Transfers

+ Add to test✓ In your testCC-EIP-4b-120 marks~20 min

A low-energy house: reducing thermal energy transfer

4(b) Energy Transfers2(b) Mains Electricity5(c) Change of State4(c) Work and PowerBoth papers

A family has moved from an old house into a new low-energy house. On a winter day the inside of each house is kept at 21 °C while the temperature outside is 1 °C.

The table shows the rate at which energy is transferred out of each house through each part of the building on that day.

part of the building	old house: rate of energy transfer / W	new house: rate of energy transfer / W
walls	1200	240
roof	900	150
windows	600	210
floor	450	120
draughts through gaps	850	80
total	4000	800

Figure 1 shows some of the features of the new house.

Figure 1: cross-section of the new house (not to scale)

(a)

In the old house, a room is warmed by an electric heater connected to the mains.

Describe the energy transfers that take place, starting with the mains supply and ending with the air outside the house.

RECALLNotes

Show solution

energy is transferred electrically to the thermal store of the heater (element)	✓ 1	Accept 'the current heats the element'. Do not accept 'electrical energy changes into heat' with no store named.
energy is then transferred by heating to the thermal store of the air (and objects) in the room, and through the walls, roof, windows and floor to the thermal store of the air outside / surroundings	✓ 1	Both stages needed: room air and outside surroundings. Accept 'dissipated to the surroundings'.

(2)

(b)

The new house has triple glazing, made of three panes of glass with layers of gas between them. Draught strips are fitted around the doors.

Explain how each of these features reduces the rate of energy transfer out of the house.

EXPLAINNotes

Show solution

triple glazing: the gas between the panes is a poor conductor, so conduction (through the window) is reduced	✓ 1	Accept 'gas/air is an insulator'. Feature and mechanism both needed.
draught strips: they stop warm air escaping (and outside air entering) through the gaps, so energy transfer by convection / moving air is reduced	✓ 1	Do not accept 'they stop the cold getting in'.

(2)

(c)

The heaters in the new house keep the inside temperature constant at 21 °C. Ignore any other sources of energy inside the house.

Explain why the heaters must transfer energy at a rate of 800 W.

REASONNotes

Show solution

energy is transferred out of the house at a total rate of 800 W (from the table)	✓ 1
for the temperature (the thermal store of the house) to stay constant, the energy supplied each second must equal the energy transferred out each second	✓ 1	Accept 'energy in = energy out' or 'if less than 800 W were supplied, the thermal store would decrease and the temperature would fall'.

(2)

(d)(i)

The heaters in the new house are connected to the 230 V mains supply.

Calculate the total current in the heaters when they transfer energy at a rate of 800 W.

CALCNotes

Show solution

I = P ÷ V = 800 ÷ 230	✓ 1	Accept P = I × V rearranged.
I = 3.5 A	✓ 1	Accept 3.48 A. Unit required.

Answer: 3.5 A

(2)

(d)(ii)

Show that the new house needs about 2.8 × 10⁸ J less energy than the old house to keep it at 21 °C for 24 hours.

DERIVENotes

Show solution

difference in power = 4000 − 800 = 3200 W and t = 24 × 60 × 60 = 86 400 s	✓ 1	Both values needed for this mark.
energy = P × t = 3200 × 86 400 = 2.76 × 10⁸ J	✓ 1	Alternative: 4000 × 86 400 = 3.456 × 10⁸ J and 800 × 86 400 = 6.91 × 10⁷ J, then subtracted. A value to at least 3 s.f. must be seen.

Answer: 2.76 × 10⁸ J

(2)

(e)(i)

At 22:00 the heating in both houses is switched off. The temperature outside stays at 1 °C.

Figure 2 shows how the temperature inside each house changes during the next 8 hours.

Use Figure 2 to determine the fall in temperature inside each house during the first 4.0 hours.

Figure 2: inside temperature of each house after the heating is switched off

GRAPHNotes

Show solution

old house: 21 − 13 = 8 °C	✓ 1	Allow 7.5 °C to 8.5 °C.
new house: 21 − 19 = 2 °C	✓ 1	Allow 1.5 °C to 2.5 °C.

Answer: old house 8 °C; new house 2 °C

(2)

(e)(ii)

The graph for the old house becomes less steep as time goes on.

Explain why.

EXPLAINNotes

Show solution

as the inside cools, the temperature difference between the inside and the outside becomes smaller	✓ 1
so the rate of energy transfer out of the house decreases, and the temperature falls more slowly	✓ 1	Do not accept 'less heat is left' or 'the cold slows down'.

(2)

(f)

The family that bought the old house replaces its windows. The new glass has a very thin, shiny, transparent metal coating on its inner surface. This reduces the rate of energy transfer through the windows.

Explain how the shiny coating reduces the rate of energy transfer.

EXPLAINNotes

Show solution

the shiny surface reflects infrared radiation from the room back into the room	✓ 1	Must refer to infrared or radiation.
a shiny surface is a poor emitter (and poor absorber) of infrared, so less energy is radiated away by the glass	✓ 1	Accept 'reduces energy transfer by radiation' for this mark only if linked to poor emission or absorption.

(2)

(g)(i)

One morning the air in the living room of the new house is at 15 °C. A portable 2.0 kW electric heater warms the air to 21 °C.

The mass of air in the room is 60 kg. The specific heat capacity of air is 1000 J/kg °C.

Calculate the energy needed to raise the temperature of the air.

Paper 2 only onlyCALCNotes

Show solution

ΔQ = m × c × ΔT = 60 × 1000 × (21 − 15)	✓ 1	ΔT = 6 °C must be used.
ΔQ = 360 000 J	✓ 1	Accept 3.6 × 10⁵ J or 360 kJ.

Answer: 360 000 J

(2)

(g)(ii)

The heater takes 12 minutes to warm the air from 15 °C to 21 °C.

Calculate the percentage of the energy supplied by the heater that is transferred to the thermal store of the air.

Paper 2 only onlyCALCNotes

Show solution

energy supplied = P × t = 2000 × 12 × 60 = 1 440 000 J	✓ 1	Do not accept 2.0 × 12 = 24.
percentage = 360 000 ÷ 1 440 000 × 100 = 25%	✓ 1	Allow ECF from (g)(i) and from the energy supplied.

Answer: 25%

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
energy is transferred electrically to the thermal store of the heater (element)	✓ 1	Accept 'the current heats the element'. Do not accept 'electrical energy changes into heat' with no store named.
energy is then transferred by heating to the thermal store of the air (and objects) in the room, and through the walls, roof, windows and floor to the thermal store of the air outside / surroundings	✓ 1	Both stages needed: room air and outside surroundings. Accept 'dissipated to the surroundings'.
Part (b)
triple glazing: the gas between the panes is a poor conductor, so conduction (through the window) is reduced	✓ 1	Accept 'gas/air is an insulator'. Feature and mechanism both needed.
draught strips: they stop warm air escaping (and outside air entering) through the gaps, so energy transfer by convection / moving air is reduced	✓ 1	Do not accept 'they stop the cold getting in'.
Part (c)
energy is transferred out of the house at a total rate of 800 W (from the table)	✓ 1
for the temperature (the thermal store of the house) to stay constant, the energy supplied each second must equal the energy transferred out each second	✓ 1	Accept 'energy in = energy out' or 'if less than 800 W were supplied, the thermal store would decrease and the temperature would fall'.
Part (d)(i)
I = P ÷ V = 800 ÷ 230	✓ 1	Accept P = I × V rearranged.
I = 3.5 A	✓ 1	Accept 3.48 A. Unit required.
Part (d)(ii)
difference in power = 4000 − 800 = 3200 W and t = 24 × 60 × 60 = 86 400 s	✓ 1	Both values needed for this mark.
energy = P × t = 3200 × 86 400 = 2.76 × 10⁸ J	✓ 1	Alternative: 4000 × 86 400 = 3.456 × 10⁸ J and 800 × 86 400 = 6.91 × 10⁷ J, then subtracted. A value to at least 3 s.f. must be seen.
Part (e)(i)
old house: 21 − 13 = 8 °C	✓ 1	Allow 7.5 °C to 8.5 °C.
new house: 21 − 19 = 2 °C	✓ 1	Allow 1.5 °C to 2.5 °C.
Part (e)(ii)
as the inside cools, the temperature difference between the inside and the outside becomes smaller	✓ 1
so the rate of energy transfer out of the house decreases, and the temperature falls more slowly	✓ 1	Do not accept 'less heat is left' or 'the cold slows down'.
Part (f)
the shiny surface reflects infrared radiation from the room back into the room	✓ 1	Must refer to infrared or radiation.
a shiny surface is a poor emitter (and poor absorber) of infrared, so less energy is radiated away by the glass	✓ 1	Accept 'reduces energy transfer by radiation' for this mark only if linked to poor emission or absorption.
Part (g)(i)
ΔQ = m × c × ΔT = 60 × 1000 × (21 − 15)	✓ 1	ΔT = 6 °C must be used.
ΔQ = 360 000 J	✓ 1	Accept 3.6 × 10⁵ J or 360 kJ.
Part (g)(ii)
energy supplied = P × t = 2000 × 12 × 60 = 1 440 000 J	✓ 1	Do not accept 2.0 × 12 = 24.
percentage = 360 000 ÷ 1 440 000 × 100 = 25%	✓ 1	Allow ECF from (g)(i) and from the energy supplied.

Answers: (d)(i) 3.5 A · (d)(ii) 2.76 × 10⁸ J · (e)(i) old house 8 °C; new house 2 °C · (g)(i) 360 000 J · (g)(ii) 25%

Common error (d)(ii) Using 24 for the time. Power in watts is joules per second, so the time must be in seconds: 24 h = 86 400 s.

More 4(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-4b-220 marks~20 min

Keeping a gaming laptop cool: heat sinks, fans and efficiency

4(b) Energy Transfers2(b) Mains Electricity2(c) Energy and Voltage in CircuitsBoth papers

A gaming laptop is powered from an adapter that supplies a steady voltage of 20 V. While a game is running, the current in the laptop is 4.5 A.

Figure 1 shows the cooling system for the processor. A copper base plate sits on top of the processor and carries thin copper fins painted matt black. A fan draws cool air in through vents in the base of the laptop, blows it across the fins and out through a vent in the side.

Figure 1: the cooling system for the processor (side view, not to scale)

(a)

Calculate the power input to the laptop.

CALCNotes

Show solution

P = I × V = 4.5 × 20	✓ 1
P = 90 W	✓ 1	Unit required.

Answer: 90 W

(2)

(b)(i)

The game is played for 1.0 hour.

Calculate the charge that flows through the laptop in this time.

CALCNotes

Show solution

t = 1.0 × 60 × 60 = 3600 s and Q = I × t = 4.5 × 3600	✓ 1
Q = 16 200 C	✓ 1	Accept 1.6 × 10⁴ C. 4.5 C (time not converted) scores 0.

Answer: 16 200 C

(2)

(b)(ii)

Calculate the energy transferred to the laptop in 1.0 hour.

CALCNotes

Show solution

E = Q × V = 16 200 × 20	✓ 1	Allow ECF from (b)(i). Accept E = P × t = 90 × 3600.
E = 324 000 J	✓ 1	Accept 3.2 × 10⁵ J or 324 kJ.

Answer: 324 000 J

(2)

(c)(i)

Figure 2 is a Sankey diagram, drawn to scale, for the energy transferred by the laptop each second. The input arrow represents 90 J.

Use Figure 2 to determine the energy transferred by light each second.

Figure 2: Sankey diagram for the laptop, drawn to scale on a square grid

GRAPHNotes

Show solution

scale: input arrow 30 squares wide, so 90 ÷ 30 = 3 J per square	✓ 1
light arrow 2 squares wide: 2 × 3 = 6 J	✓ 1	Allow ECF from an incorrect scale.

Answer: 6 J

(2)

(c)(ii)

The useful energy outputs of the laptop are light from the screen and sound from the speakers.

Show that the efficiency of the laptop is 10%.

DERIVENotes

Show solution

useful energy output = 6 J (light) + 3 J (sound, 1 square) = 9 J each second	✓ 1	Sound must be included.
efficiency = 9 ÷ 90 × 100% = 10%	✓ 1	Accept 0.10. Working must be shown.

Answer: 10%

(2)

(d)

Explain why the base plate and the fins are made of copper.

EXPLAINNotes

Show solution

copper is a metal, so it is a good (thermal) conductor	✓ 1	Do not accept 'copper absorbs heat'.
free electrons carry energy quickly from the processor to the fins (by conduction)	✓ 1	Accept 'energy is conducted away from the processor quickly'.

(2)

(e)

Explain why the fins are thin and painted matt black.

EXPLAINNotes

Show solution

many thin fins give a large surface area, which increases the rate of energy transfer to the air (and by radiation)	✓ 1
matt black is the best emitter of infrared radiation	✓ 1	Do not accept 'black absorbs heat' or 'black attracts heat'.

(2)

(f)(i)

Figure 3 shows how the temperature of the processor changes after a game is started, with the fan working and with the fan switched off.

Use Figure 3 to state the highest temperature reached with the fan working and to determine the time taken for the processor to reach 80 °C with the fan switched off.

Figure 3: processor temperature after a game is started

GRAPHNotes

Show solution

highest temperature with the fan working = 65 °C	✓ 1	Allow 64 °C to 66 °C.
time to reach 80 °C with the fan off = 160 s	✓ 1	Allow 150 s to 170 s.

Answer: 65 °C; 160 s

(2)

(f)(ii)

With the fan working, the temperature of the processor rises and then becomes constant.

Explain why, in terms of energy transfer.

EXPLAINNotes

Show solution

as the processor and fins get hotter, the temperature difference with the air increases, so energy is transferred away at a greater rate	✓ 1
the temperature stops rising when the rate of energy transfer away equals the rate at which energy is supplied to the processor	✓ 1	Accept 'energy transferred away per second = energy supplied per second'.

(2)

(g)

The laptop is now used on a soft blanket. The blanket blocks the vents in the base of the laptop.

Suggest how this affects the constant temperature reached by the processor. Explain your answer.

PREDICTNotes

Show solution

the constant temperature is higher / the processor gets hotter	✓ 1
cool air cannot be drawn in, so warm air is not replaced and less energy is carried away by the moving air (convection); the blanket traps air, which is a poor conductor	✓ 1	Either reason scores. Do not accept 'the blanket makes heat'.

Answer: higher

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
P = I × V = 4.5 × 20	✓ 1
P = 90 W	✓ 1	Unit required.
Part (b)(i)
t = 1.0 × 60 × 60 = 3600 s and Q = I × t = 4.5 × 3600	✓ 1
Q = 16 200 C	✓ 1	Accept 1.6 × 10⁴ C. 4.5 C (time not converted) scores 0.
Part (b)(ii)
E = Q × V = 16 200 × 20	✓ 1	Allow ECF from (b)(i). Accept E = P × t = 90 × 3600.
E = 324 000 J	✓ 1	Accept 3.2 × 10⁵ J or 324 kJ.
Part (c)(i)
scale: input arrow 30 squares wide, so 90 ÷ 30 = 3 J per square	✓ 1
light arrow 2 squares wide: 2 × 3 = 6 J	✓ 1	Allow ECF from an incorrect scale.
Part (c)(ii)
useful energy output = 6 J (light) + 3 J (sound, 1 square) = 9 J each second	✓ 1	Sound must be included.
efficiency = 9 ÷ 90 × 100% = 10%	✓ 1	Accept 0.10. Working must be shown.
Part (d)
copper is a metal, so it is a good (thermal) conductor	✓ 1	Do not accept 'copper absorbs heat'.
free electrons carry energy quickly from the processor to the fins (by conduction)	✓ 1	Accept 'energy is conducted away from the processor quickly'.
Part (e)
many thin fins give a large surface area, which increases the rate of energy transfer to the air (and by radiation)	✓ 1
matt black is the best emitter of infrared radiation	✓ 1	Do not accept 'black absorbs heat' or 'black attracts heat'.
Part (f)(i)
highest temperature with the fan working = 65 °C	✓ 1	Allow 64 °C to 66 °C.
time to reach 80 °C with the fan off = 160 s	✓ 1	Allow 150 s to 170 s.
Part (f)(ii)
as the processor and fins get hotter, the temperature difference with the air increases, so energy is transferred away at a greater rate	✓ 1
the temperature stops rising when the rate of energy transfer away equals the rate at which energy is supplied to the processor	✓ 1	Accept 'energy transferred away per second = energy supplied per second'.
Part (g)
the constant temperature is higher / the processor gets hotter	✓ 1
cool air cannot be drawn in, so warm air is not replaced and less energy is carried away by the moving air (convection); the blanket traps air, which is a poor conductor	✓ 1	Either reason scores. Do not accept 'the blanket makes heat'.

Answers: (a) 90 W · (b)(i) 16 200 C · (b)(ii) 324 000 J · (c)(i) 6 J · (c)(ii) 10% · (f)(i) 65 °C; 160 s · (g) higher

Exam tip (c)(ii) Count squares across the width of each arrow, not along its length, and work out the scale from the arrow whose value you are given.

More 4(b) questions in the question bank

4(c) Work and Power

+ Add to test✓ In your testCC-EIP-4c-120 marks~20 min

A launched rollercoaster: work, kinetic energy and power

4(c) Work and Power1(b) Movement and Position1(c) Forces, Movement, Shape and MomentumBoth papers

A rollercoaster train of mass 8000 kg starts from rest on a straight, horizontal launch track 36 m long. Electric motors give the train a uniform acceleration, so that it leaves the launch track at 24 m/s, 3.0 s after starting.

The train then travels along a short level section and climbs a hill. The top of the hill is 27 m above the launch track (Figure 1).

Use g = 10 N/kg. Assume that friction and air resistance are negligible unless a part tells you otherwise.

Figure 1: the launch track and the hill (not to scale)

(a)(i)

Calculate the acceleration of the train on the launch track.

CALCNotes

Show solution

a = (v − u) ÷ t = (24 − 0) ÷ 3.0	✓ 1
a = 8.0 m/s²	✓ 1	Unit required.

Answer: 8.0 m/s²

(2)

(a)(ii)

Calculate the resultant force on the train during the launch.

CALCNotes

Show solution

F = m × a = 8000 × 8.0	✓ 1	Allow ECF from (a)(i).
F = 64 000 N	✓ 1	Accept 6.4 × 10⁴ N.

Answer: 64 000 N

(2)

(b)

Calculate the kinetic energy of the train as it leaves the launch track.

CALCNotes

Show solution

KE = ½ × m × v² = ½ × 8000 × 24²	✓ 1	Speed must be squared on its own.
KE = 2 304 000 J	✓ 1	Accept 2.3 × 10⁶ J or 2.3 MJ.

Answer: 2 304 000 J

(2)

(c)

Calculate the work done by the resultant force during the launch. Comment on your answer by comparing it with your answer to (b).

MULTINotes

Show solution

W = F × d = 64 000 × 36	✓ 1	Allow ECF from (a)(ii).
W = 2 304 000 J	✓ 1
the work done is equal to the kinetic energy gained: work done on the train = energy transferred to its kinetic store	✓ 1	Accept 'they are the same, because work done equals energy transferred'. Allow ECF if (b) and (c) differ only through an earlier slip, with a sensible comment.

Answer: 2 304 000 J; equal to the kinetic energy gained

(3)

(d)

Calculate the average power output of the motors during the launch.

CALCNotes

Show solution

P = W ÷ t = 2 304 000 ÷ 3.0	✓ 1	Allow ECF from (c).
P = 768 000 W	✓ 1	Accept 768 kW or 7.7 × 10⁵ W.

Answer: 768 000 W

(2)

(e)

Show that the speed of the train at the top of the hill is 6.0 m/s.

DERIVENotes

Show solution

GPE gained = m × g × h = 8000 × 10 × 27 = 2 160 000 J	✓ 1
KE at the top = 2 304 000 − 2 160 000 = 144 000 J	✓ 1	Allow ECF from (b).
v = √(2 × 144 000 ÷ 8000) = √36 = 6.0 m/s	✓ 1	Full working needed, including the square root.

Answer: 6.0 m/s

(3)

(f)

Figure 2 shows how the gravitational potential energy of the train changes as it travels over the hill.

On Figure 2, sketch a graph to show how the kinetic energy of the train changes over the same distance.

Figure 2: gravitational potential energy of the train over the hill

SKETCHNotes

Show solution

kinetic energy starts at 2.3 MJ at 0 m and returns to 2.3 MJ at 60 m	✓ 1	Allow ECF from (b).
curve falls to a minimum of about 0.14 MJ at 30 m, so that KE + GPE = 2.3 MJ at every point (mirror image of the GPE curve)	✓ 1	Minimum must be above zero, at the top of the hill. Allow ECF from (e).

(2)

(g)

In practice, the speed of the train at the top of the hill is less than your answer to (e).

Explain why.

EXPLAINNotes

Show solution

work is done against friction and air resistance	✓ 1
energy is transferred to the thermal store of the train, track and surroundings (and by sound), so less energy is left in the kinetic store at the top	✓ 1	Do not accept 'energy is lost' without saying where it goes.

(2)

(h)

On a busy day the train is full and its mass is 9600 kg. The motors are controlled so that the train still leaves the launch track at 24 m/s.

Ignoring friction and air resistance, explain whether the speed of the train at the top of the hill is different from your answer to (e).

PREDICTNotes

Show solution

the speed is the same (6.0 m/s)	✓ 1	No mark for 'same' with no reason or a wrong reason.
both the kinetic energy and the gravitational potential energy are proportional to the mass, so the mass cancels (½mv² lost = mgh gained)	✓ 1	Accept a calculation with 9600 kg giving 6.0 m/s.

Answer: the same, 6.0 m/s

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
a = (v − u) ÷ t = (24 − 0) ÷ 3.0	✓ 1
a = 8.0 m/s²	✓ 1	Unit required.
Part (a)(ii)
F = m × a = 8000 × 8.0	✓ 1	Allow ECF from (a)(i).
F = 64 000 N	✓ 1	Accept 6.4 × 10⁴ N.
Part (b)
KE = ½ × m × v² = ½ × 8000 × 24²	✓ 1	Speed must be squared on its own.
KE = 2 304 000 J	✓ 1	Accept 2.3 × 10⁶ J or 2.3 MJ.
Part (c)
W = F × d = 64 000 × 36	✓ 1	Allow ECF from (a)(ii).
W = 2 304 000 J	✓ 1
the work done is equal to the kinetic energy gained: work done on the train = energy transferred to its kinetic store	✓ 1	Accept 'they are the same, because work done equals energy transferred'. Allow ECF if (b) and (c) differ only through an earlier slip, with a sensible comment.
Part (d)
P = W ÷ t = 2 304 000 ÷ 3.0	✓ 1	Allow ECF from (c).
P = 768 000 W	✓ 1	Accept 768 kW or 7.7 × 10⁵ W.
Part (e)
GPE gained = m × g × h = 8000 × 10 × 27 = 2 160 000 J	✓ 1
KE at the top = 2 304 000 − 2 160 000 = 144 000 J	✓ 1	Allow ECF from (b).
v = √(2 × 144 000 ÷ 8000) = √36 = 6.0 m/s	✓ 1	Full working needed, including the square root.
Part (f)
kinetic energy starts at 2.3 MJ at 0 m and returns to 2.3 MJ at 60 m	✓ 1	Allow ECF from (b).
curve falls to a minimum of about 0.14 MJ at 30 m, so that KE + GPE = 2.3 MJ at every point (mirror image of the GPE curve)	✓ 1	Minimum must be above zero, at the top of the hill. Allow ECF from (e).
Part (g)
work is done against friction and air resistance	✓ 1
energy is transferred to the thermal store of the train, track and surroundings (and by sound), so less energy is left in the kinetic store at the top	✓ 1	Do not accept 'energy is lost' without saying where it goes.
Part (h)
the speed is the same (6.0 m/s)	✓ 1	No mark for 'same' with no reason or a wrong reason.
both the kinetic energy and the gravitational potential energy are proportional to the mass, so the mass cancels (½mv² lost = mgh gained)	✓ 1	Accept a calculation with 9600 kg giving 6.0 m/s.

Answers: (a)(i) 8.0 m/s² · (a)(ii) 64 000 N · (b) 2 304 000 J · (c) 2 304 000 J; equal to the kinetic energy gained · (d) 768 000 W · (e) 6.0 m/s · (h) the same, 6.0 m/s

Common error (e) Forgetting the square root at the end, or using the 2 304 000 J at the bottom as the kinetic energy at the top. The kinetic store at the top is only what is left after the gravitational store has gained mgh. The track is curved, so the acceleration is not uniform: use energy, not v² = u² + 2as.

More 4(c) questions in the question bank

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A bungee jump: gravitational, kinetic and elastic energy stores

4(c) Work and Power1(c) Forces, Movement, Shape and Momentum1(b) Movement and PositionBoth papers

A bungee jumper of mass 70 kg steps off a high bridge. One end of an elastic cord is fixed to the bridge and the other end to the jumper's ankles. The unstretched length of the cord is 20 m.

Figure 1 shows the jumper at the lowest point B. At point A, 20 m below the bridge, the cord becomes taut and starts to stretch.

Use g = 10 N/kg. Ignore air resistance and the mass of the cord.

Figure 1: the bungee jump (not to scale)

(a)

Calculate the gravitational potential energy lost by the jumper in falling from the bridge to point A.

CALCNotes

Show solution

GPE = m × g × h = 70 × 10 × 20	✓ 1
GPE = 14 000 J	✓ 1	Accept 13 720 J using g = 9.8 N/kg.

Answer: 14 000 J

(2)

(b)

Calculate the speed of the jumper at point A.

CALCNotes

Show solution

KE gained = GPE lost, so ½ × 70 × v² = 14 000	✓ 1	Allow ECF from (a).
v = √400 = 20 m/s	✓ 1	400 m/s (no square root) scores this mark 0.

Answer: 20 m/s

(2)

(c)

Between the bridge and point A the jumper has a uniform acceleration of 10 m/s².

Use the equation v² = u² + 2as to check your answer to (b).

CALCNotes

Show solution

v² = 0² + 2 × 10 × 20 = 400	✓ 1	u = 0 must be used.
v = 20 m/s, which agrees with (b)	✓ 1	Allow a correct comparison with an ECF value from (b).

Answer: 20 m/s

(2)

(d)

Describe how the energy stores change between point A and the lowest point B.

RECALLNotes

Show solution

the gravitational store of the jumper decreases and the elastic store of the cord increases	✓ 1	Both needed.
the kinetic store of the jumper increases at first and then decreases to zero at B	✓ 1	Accept 'the kinetic store decreases to zero at B'.

(2)

(e)

Figure 2 is the force–extension graph for the cord.

Explain how Figure 2 shows that the cord obeys Hooke's law.

Figure 2: force–extension graph for the cord

GRAPHNotes

Show solution

the graph is a straight line through the origin	✓ 1	Both 'straight' and 'through the origin' needed.
so the extension is directly proportional to the force	✓ 1	Do not accept 'the extension increases as the force increases'.

(2)

(f)

At B the jumper is momentarily at rest. The energy in the elastic store of the cord is equal to the area under the force–extension graph.

Show that the extension of the cord at B is 40 m.

DERIVENotes

Show solution

GPE lost from the bridge to B = 70 × 10 × (20 + 40) = 42 000 J	✓ 1	The fall must be 60 m, not 40 m.
force at 40 m extension = 2100 N (from Figure 2), so elastic store = ½ × 40 × 2100 = 42 000 J	✓ 1
the two energies are equal, so all the GPE lost is in the elastic store and the kinetic store is zero at an extension of 20 m	✓ 1	A conclusion linking the equal values to conservation of energy is needed.

Answer: 40 m

(3)

(g)

Determine the size and direction of the resultant force on the jumper at B. Calculate the acceleration of the jumper at B.

MULTINotes

Show solution

weight = m × g = 70 × 10 = 700 N	✓ 1
resultant force = 2100 − 700 = 1400 N, upwards	✓ 1	Direction required. 2100 N is the pull of the cord at 40 m extension (Figure 2). Allow ECF from the force used in (f).
a = F ÷ m = 1400 ÷ 70 = 20 m/s² (upwards)	✓ 1	Allow ECF from the resultant force.

Answer: 1400 N upwards; 20 m/s² upwards

(3)

(h)

When the jumper stops bouncing, the jumper hangs at rest 33 m below the bridge. A winch then lifts the jumper back up to the bridge in 55 s.

Calculate the useful power output of the winch.

CALCNotes

Show solution

work done = m × g × h = 70 × 10 × 33 = 23 100 J	✓ 1
P = W ÷ t = 23 100 ÷ 55 = 420 W	✓ 1	Allow ECF from the work done.

Answer: 420 W

(2)

(i)

A jumper of mass 90 kg uses the same cord from the same bridge.

Explain how the lowest point reached by this jumper compares with point B.

PREDICTNotes

Show solution

the lowest point is lower than B / the cord stretches further	✓ 1
the heavier jumper loses more gravitational potential energy for each metre fallen, so the cord must store more elastic energy (and pull with a larger force) before the jumper stops, which needs a larger extension	✓ 1	Do not accept 'heavier objects fall faster'.

Answer: lower than B

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
GPE = m × g × h = 70 × 10 × 20	✓ 1
GPE = 14 000 J	✓ 1	Accept 13 720 J using g = 9.8 N/kg.
Part (b)
KE gained = GPE lost, so ½ × 70 × v² = 14 000	✓ 1	Allow ECF from (a).
v = √400 = 20 m/s	✓ 1	400 m/s (no square root) scores this mark 0.
Part (c)
v² = 0² + 2 × 10 × 20 = 400	✓ 1	u = 0 must be used.
v = 20 m/s, which agrees with (b)	✓ 1	Allow a correct comparison with an ECF value from (b).
Part (d)
the gravitational store of the jumper decreases and the elastic store of the cord increases	✓ 1	Both needed.
the kinetic store of the jumper increases at first and then decreases to zero at B	✓ 1	Accept 'the kinetic store decreases to zero at B'.
Part (e)
the graph is a straight line through the origin	✓ 1	Both 'straight' and 'through the origin' needed.
so the extension is directly proportional to the force	✓ 1	Do not accept 'the extension increases as the force increases'.
Part (f)
GPE lost from the bridge to B = 70 × 10 × (20 + 40) = 42 000 J	✓ 1	The fall must be 60 m, not 40 m.
force at 40 m extension = 2100 N (from Figure 2), so elastic store = ½ × 40 × 2100 = 42 000 J	✓ 1
the two energies are equal, so all the GPE lost is in the elastic store and the kinetic store is zero at an extension of 20 m	✓ 1	A conclusion linking the equal values to conservation of energy is needed.
Part (g)
weight = m × g = 70 × 10 = 700 N	✓ 1
resultant force = 2100 − 700 = 1400 N, upwards	✓ 1	Direction required. 2100 N is the pull of the cord at 40 m extension (Figure 2). Allow ECF from the force used in (f).
a = F ÷ m = 1400 ÷ 70 = 20 m/s² (upwards)	✓ 1	Allow ECF from the resultant force.
Part (h)
work done = m × g × h = 70 × 10 × 33 = 23 100 J	✓ 1
P = W ÷ t = 23 100 ÷ 55 = 420 W	✓ 1	Allow ECF from the work done.
Part (i)
the lowest point is lower than B / the cord stretches further	✓ 1
the heavier jumper loses more gravitational potential energy for each metre fallen, so the cord must store more elastic energy (and pull with a larger force) before the jumper stops, which needs a larger extension	✓ 1	Do not accept 'heavier objects fall faster'.

Answers: (a) 14 000 J · (b) 20 m/s · (c) 20 m/s · (f) 40 m · (g) 1400 N upwards; 20 m/s² upwards · (h) 420 W · (i) lower than B

Common error (f) Using only the 40 m extension for the GPE lost. The jumper falls the 20 m length of the cord and then the extension, 60 m in total.

More 4(c) questions in the question bank

4(d) Energy Resources and Electricity Generation

+ Add to test✓ In your testCC-EIP-4d-120 marks~20 min

A pumped-storage hydroelectric scheme: storing energy in water

4(d) Energy Resources and Electricity Generation4(b) Energy Transfers4(c) Work and PowerPaper 2 only

In a pumped-storage hydroelectric scheme, water flows from an upper reservoir on a mountain, down a pipe, to a power station at the foot of the mountain and then into a lower reservoir (Figure 1). The machines in the power station can work in two ways: as turbines that drive generators, or as electric pumps that lift water from the lower reservoir back up to the upper reservoir.

When the scheme is generating, 100 m³ of water flows through the turbines each second. The turbines are 500 m below the water surface of the upper reservoir. Each cubic metre of water has a mass of 1000 kg.

Assume g = 10 N/kg.

Figure 1: the pumped-storage scheme (not to scale)

(a)

Describe the energy transfers that take place when the scheme is pumping water from the lower reservoir up to the upper reservoir.

Paper 2 only onlyRECALLNotes

Show solution

energy is transferred electrically to the pumps (motors), which transfer it (mechanically) to the kinetic store of the moving water	✓ 1	Accept: electrical → kinetic energy of the pump / water. “Electricity is turned into movement” is not enough: a store or a transfer must be named.
→ gravitational (potential energy) store of the water in the upper reservoir	✓ 1	Accept GPE of the water. Ignore correct mention of energy wasted to the thermal store of the surroundings.

(2)

(b)(i)

Show that the gravitational potential energy lost by the water each second is 5.0 × 10⁸ J.

Paper 2 only onlyMULTINotes

Show solution

mass of water each second = 100 × 1000 = 100 000 kg	✓ 1
GPE = m × g × h = 100 000 × 10 × 500	✓ 1	Use of 500 m as the height.
= 500 000 000 J (5.0 × 10⁸ J)	✓ 1	The answer is given: the full substitution and the value 500 000 000 J (or 5.0 × 10⁸ J) must be seen. Using g = 9.8 N/kg gives 4.9 × 10⁸ J: allow.

Answer: 5.0 × 10⁸ J each second

(3)

(b)(ii)

When it is generating, the scheme has an efficiency of 80%.

Calculate the electrical power output of the scheme. Give your answer in MW.

CALCNotes

Show solution

useful energy output each second = 0.80 × 5.0 × 10⁸ = 4.0 × 10⁸ J	✓ 1	Allow ECF from (b)(i).
power = 4.0 × 10⁸ W = 400 MW	✓ 1	Energy per second in J is power in W. Unit must be MW for this mark (or 4.0 × 10⁸ W converted correctly).

Answer: 400 MW

(2)

(c)

The power station has four identical turbines, and each one takes an equal share of the flow. One turbine is shut down for repairs, so only 75 m³ of water flows through the turbines each second. The height and the efficiency do not change.

Calculate the new electrical power output of the scheme.

CALCNotes

Show solution

mass each second = 75 000 kg, so GPE lost each second = 75 000 × 10 × 500 = 3.75 × 10⁸ J	✓ 1
output = 0.80 × 3.75 × 10⁸ = 3.0 × 10⁸ W = 300 MW	✓ 1	Alternative: the flow is ¾ of the original, so output = ¾ × 400 = 300 MW. Allow ECF from (b)(ii).

Answer: 300 MW

(2)

(d)(i)

One night, the pumps are supplied with 1.0 × 10¹³ J electrically to refill the upper reservoir. The next day, all of this water flows back down through the turbines and the scheme supplies 7.2 × 10¹² J electrically.

Calculate the overall efficiency of storing energy in this way.

CALCNotes

Show solution

efficiency = (7.2 × 10¹² ÷ 1.0 × 10¹³) × 100%	✓ 1	Useful output divided by the total energy supplied.
= 72%	✓ 1	Accept 0.72. An answer greater than 100% scores 0.

Answer: 72%

(2)

(d)(ii)

Calculate the time for which the scheme can generate at the power output in (b)(ii) using this stored water. Give your answer in hours.

CALCNotes

Show solution

t = W ÷ P = 7.2 × 10¹² ÷ 4.0 × 10⁸ = 18 000 s	✓ 1	Allow ECF from (b)(ii).
18 000 ÷ 3600 = 5.0 hours	✓ 1	Allow ECF from an incorrect time in s.

Answer: 18 000 s = 5.0 hours

(2)

(e)(i)

Figure 2 shows how the demand for electricity in the country changes during one day. The other power stations in the country can supply a maximum total power of 2300 MW (the dashed line). The pumped-storage scheme generates whenever the demand is greater than this.

Use Figure 2 to find

the time interval during which the scheme must generate
the greatest power that the scheme must supply.

Figure 2: demand for electricity during one day

Paper 2 only onlyGRAPHNotes

Show solution

from about 16:40 to 20:00	✓ 1	Accept a start time from 16:30 to 17:00 and an end time from 19:45 to 20:15.
2600 − 2300 = 300 MW	✓ 1	Peak demand of 2600 MW must be read from the graph.

Answer: about 16:40 to 20:00; 300 MW

(2)

(e)(ii)

Use your answers to (b)(ii), (d)(ii) and (e)(i) to explain whether the scheme can meet this extra demand.

Paper 2 only onlyREASONNotes

Show solution

the greatest extra power needed (300 MW) is less than the power output of the scheme (400 MW)	✓ 1	Allow ECF from earlier parts; the comparison must be stated.
the extra demand lasts about 3.4 hours, which is less than the 5.0 hours for which the stored water can generate, so the scheme can meet the demand	✓ 1	Accept a duration from 3 to 3.75 hours consistent with (e)(i). Allow ECF. Both comparisons are needed for the conclusion.

(2)

(f)

The government considers building a new gas-fired power station, instead of using the pumped-storage scheme, to meet the evening peak in demand.

Discuss the advantages and disadvantages of using the pumped-storage scheme rather than the gas-fired power station for this purpose. Give a conclusion.

Paper 2 only onlyEXPLAINNotes

Show solution

an advantage of the pumped-storage scheme, e.g. no carbon dioxide / greenhouse gas is released while it generates, whereas burning gas releases carbon dioxide, which contributes to climate change; it can start up very quickly; no fuel cost	✓ 1	Point must be specific. “Better for the environment” alone is not enough.
a disadvantage of the pumped-storage scheme, e.g. it is not a source of energy: it uses electricity from other power stations and wastes 28% of it; it can generate for only about 5 hours before the upper reservoir is empty; it needs a suitable mountain site and building reservoirs floods land and destroys habitats; whereas a gas station can run for as long as needed	✓ 1	Accept any correct, specific disadvantage, or a correct advantage of the gas station.
a conclusion consistent with the points made, e.g. the pumped-storage scheme is the better choice because it meets the short evening peak quickly without releasing carbon dioxide	✓ 1	A conclusion with no supporting points scores 0.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
energy is transferred electrically to the pumps (motors), which transfer it (mechanically) to the kinetic store of the moving water	✓ 1	Accept: electrical → kinetic energy of the pump / water. “Electricity is turned into movement” is not enough: a store or a transfer must be named.
→ gravitational (potential energy) store of the water in the upper reservoir	✓ 1	Accept GPE of the water. Ignore correct mention of energy wasted to the thermal store of the surroundings.
Part (b)(i)
mass of water each second = 100 × 1000 = 100 000 kg	✓ 1
GPE = m × g × h = 100 000 × 10 × 500	✓ 1	Use of 500 m as the height.
= 500 000 000 J (5.0 × 10⁸ J)	✓ 1	The answer is given: the full substitution and the value 500 000 000 J (or 5.0 × 10⁸ J) must be seen. Using g = 9.8 N/kg gives 4.9 × 10⁸ J: allow.
Part (b)(ii)
useful energy output each second = 0.80 × 5.0 × 10⁸ = 4.0 × 10⁸ J	✓ 1	Allow ECF from (b)(i).
power = 4.0 × 10⁸ W = 400 MW	✓ 1	Energy per second in J is power in W. Unit must be MW for this mark (or 4.0 × 10⁸ W converted correctly).
Part (c)
mass each second = 75 000 kg, so GPE lost each second = 75 000 × 10 × 500 = 3.75 × 10⁸ J	✓ 1
output = 0.80 × 3.75 × 10⁸ = 3.0 × 10⁸ W = 300 MW	✓ 1	Alternative: the flow is ¾ of the original, so output = ¾ × 400 = 300 MW. Allow ECF from (b)(ii).
Part (d)(i)
efficiency = (7.2 × 10¹² ÷ 1.0 × 10¹³) × 100%	✓ 1	Useful output divided by the total energy supplied.
= 72%	✓ 1	Accept 0.72. An answer greater than 100% scores 0.
Part (d)(ii)
t = W ÷ P = 7.2 × 10¹² ÷ 4.0 × 10⁸ = 18 000 s	✓ 1	Allow ECF from (b)(ii).
18 000 ÷ 3600 = 5.0 hours	✓ 1	Allow ECF from an incorrect time in s.
Part (e)(i)
from about 16:40 to 20:00	✓ 1	Accept a start time from 16:30 to 17:00 and an end time from 19:45 to 20:15.
2600 − 2300 = 300 MW	✓ 1	Peak demand of 2600 MW must be read from the graph.
Part (e)(ii)
the greatest extra power needed (300 MW) is less than the power output of the scheme (400 MW)	✓ 1	Allow ECF from earlier parts; the comparison must be stated.
the extra demand lasts about 3.4 hours, which is less than the 5.0 hours for which the stored water can generate, so the scheme can meet the demand	✓ 1	Accept a duration from 3 to 3.75 hours consistent with (e)(i). Allow ECF. Both comparisons are needed for the conclusion.
Part (f)
an advantage of the pumped-storage scheme, e.g. no carbon dioxide / greenhouse gas is released while it generates, whereas burning gas releases carbon dioxide, which contributes to climate change; it can start up very quickly; no fuel cost	✓ 1	Point must be specific. “Better for the environment” alone is not enough.
a disadvantage of the pumped-storage scheme, e.g. it is not a source of energy: it uses electricity from other power stations and wastes 28% of it; it can generate for only about 5 hours before the upper reservoir is empty; it needs a suitable mountain site and building reservoirs floods land and destroys habitats; whereas a gas station can run for as long as needed	✓ 1	Accept any correct, specific disadvantage, or a correct advantage of the gas station.
a conclusion consistent with the points made, e.g. the pumped-storage scheme is the better choice because it meets the short evening peak quickly without releasing carbon dioxide	✓ 1	A conclusion with no supporting points scores 0.

Answers: (b)(i) 5.0 × 10⁸ J each second · (b)(ii) 400 MW · (c) 300 MW · (d)(i) 72% · (d)(ii) 18 000 s = 5.0 hours · (e)(i) about 16:40 to 20:00; 300 MW

Common error (b)(i) Using 100 kg as the mass. The flow rate is a volume (100 m³ each second); each cubic metre has a mass of 1000 kg, so the mass is 100 000 kg.

Exam tip (f) When you discuss two resources, give at least one point on each side, say which resource each point applies to, and finish with a conclusion that follows from your points.

More 4(d) questions in the question bank

+ Add to test✓ In your testCC-EIP-4d-220 marks~20 min

An offshore wind farm: energy from moving air

4(d) Energy Resources and Electricity Generation4(c) Work and Power4(b) Energy Transfers6(d) Electromagnetic InductionPaper 2 only

An offshore wind farm is made up of many identical wind turbines. Figure 1 shows one of the turbines. As the blades turn, they sweep out a circle of area 8000 m².

On one day, a steady wind blows straight towards the blades at 10 m/s. Each cubic metre of air has a mass of 1.2 kg.

Each turbine has a control system that limits its electrical power output to a maximum of 3.0 MW. The turbine is stopped when the wind speed is greater than 25 m/s.

Figure 1: one turbine of the wind farm (not to scale)

(a)

Describe the energy transfers that take place when the wind turbine generates electricity.

Paper 2 only onlyRECALLNotes

Show solution

kinetic store of the moving air → kinetic store of the (turning) blades / turbine	✓ 1	“Movement energy” is not enough: a store must be named.
the turbine turns the generator, which transfers the energy electrically	✓ 1	Ignore correct mention of energy wasted to the thermal store of the surroundings or as sound.

(2)

(b)(i)

In one second, the air that passes through the blades would fill a cylinder with a cross-sectional area of 8000 m² and a length of 10 m.

Calculate the mass of air that passes through the blades each second.

Paper 2 only onlyCALCNotes

Show solution

volume each second = 8000 × 10 = 80 000 m³	✓ 1
mass each second = 80 000 × 1.2 = 96 000 kg	✓ 1	Allow ECF from an incorrect volume.

Answer: 96 000 kg

(2)

(b)(ii)

Show that the kinetic energy of the air that passes through the blades each second is about 5 × 10⁶ J.

Paper 2 only onlyMULTINotes

Show solution

KE = ½ × m × v² = ½ × 96 000 × 10²	✓ 1	Allow ECF for the mass from (b)(i).
= 4 800 000 J (4.8 × 10⁶ J)	✓ 1	The answer is given: 4.8 × 10⁶ J or the full substitution must be seen. Squaring the speed is essential: ½ × 96 000 × 10 = 480 000 J scores 0 for this mark.

Answer: 4.8 × 10⁶ J each second

(2)

(c)

At this wind speed, the turbine transfers 45% of the kinetic energy of the air that passes through its blades to electrical energy.

Calculate the electrical power output of the turbine. Give your answer in MW.

Paper 2 only onlyCALCNotes

Show solution

useful energy output each second = 0.45 × 4.8 × 10⁶ = 2.16 × 10⁶ J	✓ 1	Allow ECF from (b)(ii). Using 5 × 10⁶ J gives 2.25 MW: allow.
power = 2.16 × 10⁶ W = 2.16 MW	✓ 1	Accept 2.2 MW. Unit must be MW for this mark (or 2.16 × 10⁶ W converted correctly).

Answer: 2.16 MW (about 2.2 MW)

(2)

(d)

Later, the wind speed doubles to 20 m/s.

Explain why the kinetic energy of the air that passes through the blades each second becomes eight times as large.

REASONNotes

Show solution

the cylinder of air passing through the blades each second is twice as long, so twice the mass of air passes each second	✓ 1
kinetic energy depends on speed squared, so each kilogram of air has 2² = 4 times as much kinetic energy	✓ 1	Accept: KE = ½mv², so doubling v multiplies KE by 4 (for the same mass).
2 × 4 = 8 (times as much kinetic energy each second)	✓ 1	Both factors must be combined for this mark.

(3)

(e)(i)

Figure 2 shows how the electrical power output of the turbine depends on the wind speed.

Use Figure 2 to explain why the electrical power output at 20 m/s is not eight times the output at 10 m/s.

Figure 2: electrical power output of one turbine against wind speed

Paper 2 only onlyGRAPHNotes

Show solution

output at 20 m/s is 3.0 MW (from Figure 2), which is much less than 8 × 2.16 = 17 MW	✓ 1	The value 3.0 MW must be read from the graph. Allow ECF from (c).
above about 11 m/s the control system limits the output to its maximum of 3.0 MW, so the extra energy of the wind is not transferred	✓ 1	Accept: the graph is flat / constant above about 11 m/s because the output cannot exceed 3.0 MW.

(2)

(e)(ii)

Suggest why the turbine is stopped when the wind speed is greater than 25 m/s.

Paper 2 only onlyREASONNotes

Show solution

to prevent damage to the blades / turbine / generator in very strong winds

✓ 1

Accept: for safety, so that the turbine is not damaged.

(1)

(f)

The electricity from the wind farm is sent to the shore along a cable laid on the seabed. A transformer at the wind farm increases the voltage to 132 kV before the electricity is sent along the cable.

Explain why the voltage is increased.

Paper 2 only onlyEXPLAINNotes

Show solution

for the same power, a higher voltage means a smaller current in the cable	✓ 1	Accept reference to P = I × V, or V_pI_p = V_sI_s.
a smaller current causes less heating of the cable (because the cable has resistance)	✓ 1	“Reduces the resistance of the cable” scores 0.
so less energy is wasted as thermal energy / transmission is more efficient	✓ 1

(3)

(g)

Discuss the advantages and disadvantages of generating electricity with this offshore wind farm rather than with a coal-fired power station. Give a conclusion.

Paper 2 only onlyEXPLAINNotes

Show solution

an advantage of the wind farm, e.g. no carbon dioxide / greenhouse gas released while operating, whereas burning coal releases carbon dioxide (and sulfur dioxide, causing acid rain); no fuel cost; wind is renewable, coal will run out	✓ 1	Point must be specific and say which resource it applies to.
a disadvantage of the wind farm, e.g. output depends on the wind speed, so it is unreliable: no output in calm weather or above 25 m/s, whereas a coal station gives a reliable output whatever the weather; many turbines needed for the output of one coal station	✓ 1	Accept any correct, specific disadvantage.
a justified conclusion, e.g. the wind farm is better for reducing carbon dioxide emissions but needs a reliable back-up supply for calm days	✓ 1	A conclusion with no supporting points scores 0.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
kinetic store of the moving air → kinetic store of the (turning) blades / turbine	✓ 1	“Movement energy” is not enough: a store must be named.
the turbine turns the generator, which transfers the energy electrically	✓ 1	Ignore correct mention of energy wasted to the thermal store of the surroundings or as sound.
Part (b)(i)
volume each second = 8000 × 10 = 80 000 m³	✓ 1
mass each second = 80 000 × 1.2 = 96 000 kg	✓ 1	Allow ECF from an incorrect volume.
Part (b)(ii)
KE = ½ × m × v² = ½ × 96 000 × 10²	✓ 1	Allow ECF for the mass from (b)(i).
= 4 800 000 J (4.8 × 10⁶ J)	✓ 1	The answer is given: 4.8 × 10⁶ J or the full substitution must be seen. Squaring the speed is essential: ½ × 96 000 × 10 = 480 000 J scores 0 for this mark.
Part (c)
useful energy output each second = 0.45 × 4.8 × 10⁶ = 2.16 × 10⁶ J	✓ 1	Allow ECF from (b)(ii). Using 5 × 10⁶ J gives 2.25 MW: allow.
power = 2.16 × 10⁶ W = 2.16 MW	✓ 1	Accept 2.2 MW. Unit must be MW for this mark (or 2.16 × 10⁶ W converted correctly).
Part (d)
the cylinder of air passing through the blades each second is twice as long, so twice the mass of air passes each second	✓ 1
kinetic energy depends on speed squared, so each kilogram of air has 2² = 4 times as much kinetic energy	✓ 1	Accept: KE = ½mv², so doubling v multiplies KE by 4 (for the same mass).
2 × 4 = 8 (times as much kinetic energy each second)	✓ 1	Both factors must be combined for this mark.
Part (e)(i)
output at 20 m/s is 3.0 MW (from Figure 2), which is much less than 8 × 2.16 = 17 MW	✓ 1	The value 3.0 MW must be read from the graph. Allow ECF from (c).
above about 11 m/s the control system limits the output to its maximum of 3.0 MW, so the extra energy of the wind is not transferred	✓ 1	Accept: the graph is flat / constant above about 11 m/s because the output cannot exceed 3.0 MW.
Part (e)(ii)
to prevent damage to the blades / turbine / generator in very strong winds	✓ 1	Accept: for safety, so that the turbine is not damaged.
Part (f)
for the same power, a higher voltage means a smaller current in the cable	✓ 1	Accept reference to P = I × V, or V_pI_p = V_sI_s.
a smaller current causes less heating of the cable (because the cable has resistance)	✓ 1	“Reduces the resistance of the cable” scores 0.
so less energy is wasted as thermal energy / transmission is more efficient	✓ 1
Part (g)
an advantage of the wind farm, e.g. no carbon dioxide / greenhouse gas released while operating, whereas burning coal releases carbon dioxide (and sulfur dioxide, causing acid rain); no fuel cost; wind is renewable, coal will run out	✓ 1	Point must be specific and say which resource it applies to.
a disadvantage of the wind farm, e.g. output depends on the wind speed, so it is unreliable: no output in calm weather or above 25 m/s, whereas a coal station gives a reliable output whatever the weather; many turbines needed for the output of one coal station	✓ 1	Accept any correct, specific disadvantage.
a justified conclusion, e.g. the wind farm is better for reducing carbon dioxide emissions but needs a reliable back-up supply for calm days	✓ 1	A conclusion with no supporting points scores 0.

Answers: (b)(i) 96 000 kg · (b)(ii) 4.8 × 10⁶ J each second · (c) 2.16 MW (about 2.2 MW)

Common error (c) Expecting a wind turbine to be close to 100% efficient. The air must still be moving after it passes the blades, so only part of its kinetic energy can be transferred.

More 4(d) questions in the question bank

Topic 5 · Solids, liquids and gases

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

Setup	5(b)	5(c)	5(d)	Other
A scuba dive: pressure under water	●		○	1(c)
Rescue on thin ice: pressure and area	●	○		1(c)
Tempering chocolate: a heating curve, specific heat capacity and wasted energy		●		2(b) 4(b)
A marathon runner keeping cool: specific heat capacity and evaporation		●		4(b) 4(c)
Pumping up a road-bike tyre: gas pressure, volume and temperature	○		●	1(c)
A compressed-air can left in a hot car: pressure, temperature and absolute zero	○		●	4(b)

5(b) Density and Pressure

+ Add to test✓ In your testCC-EIP-5b-120 marks~20 min

A scuba dive: pressure under water

5(b) Density and Pressure5(d) Ideal Gas Molecules1(c) Forces, Movement, Shape and MomentumBoth papers

A scuba diver descends to the flat seabed, 25.0 m below the surface of the sea (Figure 1). The density of the seawater is 1030 kg/m³. Atmospheric pressure at the sea surface is 1.0 × 10⁵ Pa.

Assume g = 10 N/kg.

Figure 1: the diver on the seabed (not to scale)

(a)(i)

Figure 1 also shows a column of seawater that stands on an area of 0.40 m² of the seabed and reaches up to the surface.

Calculate the mass of seawater in this column.

CALCNotes

Show solution

volume of column = 0.40 × 25.0 = 10 m³	✓ 1
m = ρ × V = 1030 × 10 = 10 300 kg	✓ 1	Allow ECF from an incorrect volume.

Answer: 10 300 kg

(2)

(a)(ii)

Calculate the weight of this column of seawater, and hence show that the pressure on the seabed due to the seawater is about 2.6 × 10⁵ Pa.

MULTINotes

Show solution

W = m × g = 10 300 × 10 = 103 000 N	✓ 1	Allow ECF from (a)(i).
p = F ÷ A = 103 000 ÷ 0.40 = 257 500 Pa	✓ 1	The answer is given: 257 500 Pa (or 2.58 × 10⁵ Pa) must be seen. Using p = h × ρ × g alone does not answer the question and scores this mark only if the weight has been calculated.

Answer: 257 500 Pa (about 2.6 × 10⁵ Pa)

(2)

(a)(iii)

Calculate the total pressure on the diver at the seabed.

CALCNotes

Show solution

1.0 × 10⁵ + 2.575 × 10⁵ = 3.6 × 10⁵ Pa

✓ 1

Accept 357 500 Pa or 3.58 × 10⁵ Pa. Allow ECF from (a)(ii).

Answer: 3.6 × 10⁵ Pa

(1)

(b)

On the way down, at a depth of 3.0 m, the diver feels discomfort in her ears. The air behind her eardrum is still at atmospheric pressure. The area of her eardrum is 0.50 cm².

Calculate the size of the resultant force on her eardrum.

MULTINotes

Show solution

pressure difference p = h × ρ × g = 3.0 × 1030 × 10 = 30 900 Pa	✓ 1	Atmospheric pressure acts on both sides and is not added.
A = 0.50 ÷ 10 000 = 5.0 × 10⁻⁵ m²	✓ 1
F = p × A = 30 900 × 5.0 × 10⁻⁵ = 1.5 N	✓ 1	Accept 1.55 N. Dividing the area by 100 (155 N): max 2. Adding atmospheric pressure (6.5 N): max 2.

Answer: 1.5 N

(3)

(c)

Later, on the seabed, the diver turns her head so that her ear faces sideways instead of upwards. Her ear stays at the same depth.

Explain why the pressure of the water on her ear does not change.

EXPLAINNotes

Show solution

at any point in a liquid at rest, the pressure acts equally in all directions	✓ 1	“The water pushes everywhere” is not enough.
the ear is still at the same depth (in the same liquid), so the pressure is the same; the force on the eardrum still acts at right angles to it	✓ 1	Accept: pressure depends only on the depth (and density), not on the direction the surface faces.

(2)

(d)(i)

The diver also dives in a freshwater lake high in the mountains. Figure 2 shows how the total pressure changes with depth in the sea (line A) and in the mountain lake (line B).

Use line B to determine the density of the water in the lake.

Figure 2: total pressure against depth in the sea (A) and in a mountain lake (B)

GRAPHNotes

Show solution

pressure difference read from line B, e.g. 375 − 75 = 300 kPa over a depth of 30 m	✓ 1	Any correct pair of readings from line B, or a gradient of 10 kPa per m.
ρ = p ÷ (h × g) = 300 000 ÷ (30 × 10)	✓ 1	Pressure must be in Pa (or kPa converted). Allow ECF from the readings.
ρ = 1000 kg/m³	✓ 1	Accept 980 to 1020 kg/m³.

Answer: 1000 kg/m³

(3)

(d)(ii)

Use Figure 2 to give two reasons why, at the same depth, the total pressure on the diver is smaller in the lake than in the sea.

REASONNotes

Show solution

the atmospheric pressure at the lake surface is smaller (75 kPa compared with 100 kPa: line B starts lower)	✓ 1	Reference to the values at zero depth, or to line B starting lower, is needed.
the lake water is less dense than seawater, so the pressure increases less for each metre of depth (line B has a smaller gradient)	✓ 1

(2)

(e)

The diver's air cylinder has a fixed volume. Just after it was filled, the air in the cylinder was at a temperature of 47 °C and a pressure of 20 000 kPa. Before any air is used, the cylinder is carried into the water and the air cools to the temperature of the water, 7 °C.

Calculate the new pressure of the air in the cylinder.

CALCNotes

Show solution

T₁ = 47 + 273 = 320 K and T₂ = 7 + 273 = 280 K	✓ 1	Both conversions needed.
p₂ = p₁ × T₂ ÷ T₁ = 20 000 × 280 ÷ 320	✓ 1	From p₁/T₁ = p₂/T₂.
p₂ = 17 500 kPa	✓ 1	Accept 1.75 × 10⁷ Pa. Using °C (about 2980 kPa) scores 0 for the first and third marks.

Answer: 17 500 kPa

(3)

(f)

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

EXPLAINNotes

Show solution

the molecules have less kinetic energy, so they move more slowly on average	✓ 1
they hit the walls less often and with a smaller force, so the force on each unit area of wall (the pressure) is smaller	✓ 1	Both “less often” and “smaller force” are needed for this mark.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
volume of column = 0.40 × 25.0 = 10 m³	✓ 1
m = ρ × V = 1030 × 10 = 10 300 kg	✓ 1	Allow ECF from an incorrect volume.
Part (a)(ii)
W = m × g = 10 300 × 10 = 103 000 N	✓ 1	Allow ECF from (a)(i).
p = F ÷ A = 103 000 ÷ 0.40 = 257 500 Pa	✓ 1	The answer is given: 257 500 Pa (or 2.58 × 10⁵ Pa) must be seen. Using p = h × ρ × g alone does not answer the question and scores this mark only if the weight has been calculated.
Part (a)(iii)
1.0 × 10⁵ + 2.575 × 10⁵ = 3.6 × 10⁵ Pa	✓ 1	Accept 357 500 Pa or 3.58 × 10⁵ Pa. Allow ECF from (a)(ii).
Part (b)
pressure difference p = h × ρ × g = 3.0 × 1030 × 10 = 30 900 Pa	✓ 1	Atmospheric pressure acts on both sides and is not added.
A = 0.50 ÷ 10 000 = 5.0 × 10⁻⁵ m²	✓ 1
F = p × A = 30 900 × 5.0 × 10⁻⁵ = 1.5 N	✓ 1	Accept 1.55 N. Dividing the area by 100 (155 N): max 2. Adding atmospheric pressure (6.5 N): max 2.
Part (c)
at any point in a liquid at rest, the pressure acts equally in all directions	✓ 1	“The water pushes everywhere” is not enough.
the ear is still at the same depth (in the same liquid), so the pressure is the same; the force on the eardrum still acts at right angles to it	✓ 1	Accept: pressure depends only on the depth (and density), not on the direction the surface faces.
Part (d)(i)
pressure difference read from line B, e.g. 375 − 75 = 300 kPa over a depth of 30 m	✓ 1	Any correct pair of readings from line B, or a gradient of 10 kPa per m.
ρ = p ÷ (h × g) = 300 000 ÷ (30 × 10)	✓ 1	Pressure must be in Pa (or kPa converted). Allow ECF from the readings.
ρ = 1000 kg/m³	✓ 1	Accept 980 to 1020 kg/m³.
Part (d)(ii)
the atmospheric pressure at the lake surface is smaller (75 kPa compared with 100 kPa: line B starts lower)	✓ 1	Reference to the values at zero depth, or to line B starting lower, is needed.
the lake water is less dense than seawater, so the pressure increases less for each metre of depth (line B has a smaller gradient)	✓ 1
Part (e)
T₁ = 47 + 273 = 320 K and T₂ = 7 + 273 = 280 K	✓ 1	Both conversions needed.
p₂ = p₁ × T₂ ÷ T₁ = 20 000 × 280 ÷ 320	✓ 1	From p₁/T₁ = p₂/T₂.
p₂ = 17 500 kPa	✓ 1	Accept 1.75 × 10⁷ Pa. Using °C (about 2980 kPa) scores 0 for the first and third marks.
Part (f)
the molecules have less kinetic energy, so they move more slowly on average	✓ 1
they hit the walls less often and with a smaller force, so the force on each unit area of wall (the pressure) is smaller	✓ 1	Both “less often” and “smaller force” are needed for this mark.

Answers: (a)(i) 10 300 kg · (a)(ii) 257 500 Pa (about 2.6 × 10⁵ Pa) · (a)(iii) 3.6 × 10⁵ Pa · (b) 1.5 N · (d)(i) 1000 kg/m³ · (e) 17 500 kPa

Exam tip (b) Convert cm² to m² by dividing by 10 000 (100 × 100), not by 100: 0.50 cm² = 5.0 × 10⁻⁵ m².

Common error (e) Substituting temperatures in °C. The pressure of a gas at constant volume is proportional to its kelvin temperature, so always add 273 first.

More 5(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-5b-220 marks~20 min

Rescue on thin ice: pressure and area

5(b) Density and Pressure1(c) Forces, Movement, Shape and Momentum5(c) Change of StateBoth papers

A person has fallen through the ice on a frozen lake. A rescuer of mass 75 kg goes to help. The sole of each of the rescuer's boots has an area of 250 cm².

The rescuer can either walk across the ice (Figure 1, A) or lie flat on a rescue board and push it across the ice (Figure 1, B). The rescue board has a mass of 15 kg, and the area of the board in contact with the ice is 0.60 m².

Assume g = 10 N/kg.

Figure 1: two ways of crossing the ice (not to scale)

(a)

Calculate the weight of the rescuer.

CALCNotes

Show solution

W = m × g = 75 × 10	✓ 1
= 750 N	✓ 1	Unit required. Using g = 9.8 N/kg gives 735 N: allow.

Answer: 750 N

(2)

(b)(i)

Calculate the pressure on the ice when the rescuer stands still on both feet.

CALCNotes

Show solution

area = 2 × 250 = 500 cm² = 0.050 m²	✓ 1	Converting by dividing by 100 (5.0 m²): max 2.
p = F ÷ A = 750 ÷ 0.050	✓ 1	Allow ECF from (a).
= 15 000 Pa	✓ 1	Accept 15 kPa. Using one boot only (30 000 Pa): max 2.

Answer: 15 000 Pa (15 kPa)

(3)

(b)(ii)

Show that the pressure on the ice when the rescuer lies on the board is about 1.5 kPa.

MULTINotes

Show solution

total weight on the ice = (75 + 15) × 10 = 900 N	✓ 1	The weight of the board must be included.
p = 900 ÷ 0.60	✓ 1
= 1500 Pa = 1.5 kPa	✓ 1	The answer is given: 1500 Pa must be seen. Leaving out the board (1250 Pa): max 1.

Answer: 1500 Pa (1.5 kPa)

(3)

(c)

A member of the rescue team drills out a cylinder of ice (a core) that reaches from the top surface of the ice to the water below. The core has a mass of 0.46 kg and a cross-sectional area of 50 cm². The density of the ice is 920 kg/m³.

Calculate the thickness of the ice.

MULTINotes

Show solution

V = m ÷ ρ = 0.46 ÷ 920 = 5.0 × 10⁻⁴ m³	✓ 1	Alternative in lab units: 0.92 g/cm³ and V = 460 ÷ 0.92 = 500 cm³.
A = 50 cm² = 5.0 × 10⁻³ m²	✓ 1	Or work in cm³ and cm² throughout.
thickness = V ÷ A = 5.0 × 10⁻⁴ ÷ 5.0 × 10⁻³ = 0.10 m (10 cm)	✓ 1	Allow ECF from the volume.

Answer: 0.10 m (10 cm)

(3)

(d)(i)

Figure 2 shows the greatest pressure that ice of different thicknesses can safely support, according to the rescue team's safety guidance.

Use Figure 2 and your answers to (b) and (c) to explain why the rescuer should lie on the board rather than walk.

Figure 2: greatest safe pressure against thickness of ice (rescue team's safety guidance)

REASONNotes

Show solution

ice 10 cm thick can safely support up to 5.0 kPa (read from Figure 2)	✓ 1	Allow ECF for the thickness from (c); the reading must match it.
standing on both feet gives 15 kPa (walking, with one foot at a time, gives even more), more than 5.0 kPa, so the ice may break; on the board the pressure is only 1.5 kPa, less than 5.0 kPa, so the ice is safe (the same weight is spread over a larger area)	✓ 1	Both comparisons needed. Allow ECF from (b).

(2)

(d)(ii)

Use Figure 2 to find the smallest thickness of ice on which the rescuer could safely stand still on both feet.

GRAPHNotes

Show solution

reads the graph at the standing pressure of 15 kPa	✓ 1	Allow ECF from (b)(i).
thickness = 17 cm	✓ 1	Accept 17 to 18 cm.

Answer: 17 cm

(2)

(e)

A second rescuer, of mass 65 kg, crawls onto the board beside the first rescuer.

Determine whether the ice, 10 cm thick, can still safely support the board with both rescuers on it.

PREDICTNotes

Show solution

total weight = 900 + 650 = 1550 N	✓ 1	Allow (75 + 15 + 65) × 10. Allow ECF from (b)(ii).
p = 1550 ÷ 0.60 = 2600 Pa (2.6 kPa)	✓ 1	Accept 2583 Pa.
2.6 kPa is less than the 5.0 kPa that 10 cm of ice can safely support, so the ice is still safe	✓ 1	Accept: Figure 2 shows that only about 7 cm of ice is needed for 2.6 kPa. Conclusion must follow from the values.

Answer: about 2600 Pa, which is below 5.0 kPa, so still safe

(3)

(f)

The ice and the water just beneath it are both at 0 °C.

Describe how the arrangement and the motion of the particles in the ice differ from those in the water.

Paper 2 only onlyRECALLNotes

Show solution

arrangement: in ice the particles are in a regular pattern (lattice); in water they are in an irregular arrangement	✓ 1	Both states needed. “Ice particles are closer together” alone is not enough.
motion: in ice the particles vibrate about fixed positions; in water they move around and past each other	✓ 1	Both states needed.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
W = m × g = 75 × 10	✓ 1
= 750 N	✓ 1	Unit required. Using g = 9.8 N/kg gives 735 N: allow.
Part (b)(i)
area = 2 × 250 = 500 cm² = 0.050 m²	✓ 1	Converting by dividing by 100 (5.0 m²): max 2.
p = F ÷ A = 750 ÷ 0.050	✓ 1	Allow ECF from (a).
= 15 000 Pa	✓ 1	Accept 15 kPa. Using one boot only (30 000 Pa): max 2.
Part (b)(ii)
total weight on the ice = (75 + 15) × 10 = 900 N	✓ 1	The weight of the board must be included.
p = 900 ÷ 0.60	✓ 1
= 1500 Pa = 1.5 kPa	✓ 1	The answer is given: 1500 Pa must be seen. Leaving out the board (1250 Pa): max 1.
Part (c)
V = m ÷ ρ = 0.46 ÷ 920 = 5.0 × 10⁻⁴ m³	✓ 1	Alternative in lab units: 0.92 g/cm³ and V = 460 ÷ 0.92 = 500 cm³.
A = 50 cm² = 5.0 × 10⁻³ m²	✓ 1	Or work in cm³ and cm² throughout.
thickness = V ÷ A = 5.0 × 10⁻⁴ ÷ 5.0 × 10⁻³ = 0.10 m (10 cm)	✓ 1	Allow ECF from the volume.
Part (d)(i)
ice 10 cm thick can safely support up to 5.0 kPa (read from Figure 2)	✓ 1	Allow ECF for the thickness from (c); the reading must match it.
standing on both feet gives 15 kPa (walking, with one foot at a time, gives even more), more than 5.0 kPa, so the ice may break; on the board the pressure is only 1.5 kPa, less than 5.0 kPa, so the ice is safe (the same weight is spread over a larger area)	✓ 1	Both comparisons needed. Allow ECF from (b).
Part (d)(ii)
reads the graph at the standing pressure of 15 kPa	✓ 1	Allow ECF from (b)(i).
thickness = 17 cm	✓ 1	Accept 17 to 18 cm.
Part (e)
total weight = 900 + 650 = 1550 N	✓ 1	Allow (75 + 15 + 65) × 10. Allow ECF from (b)(ii).
p = 1550 ÷ 0.60 = 2600 Pa (2.6 kPa)	✓ 1	Accept 2583 Pa.
2.6 kPa is less than the 5.0 kPa that 10 cm of ice can safely support, so the ice is still safe	✓ 1	Accept: Figure 2 shows that only about 7 cm of ice is needed for 2.6 kPa. Conclusion must follow from the values.
Part (f)
arrangement: in ice the particles are in a regular pattern (lattice); in water they are in an irregular arrangement	✓ 1	Both states needed. “Ice particles are closer together” alone is not enough.
motion: in ice the particles vibrate about fixed positions; in water they move around and past each other	✓ 1	Both states needed.

Answers: (a) 750 N · (b)(i) 15 000 Pa (15 kPa) · (b)(ii) 1500 Pa (1.5 kPa) · (c) 0.10 m (10 cm) · (d)(ii) 17 cm · (e) about 2600 Pa, which is below 5.0 kPa, so still safe

Common error (b)(ii) Forgetting the weight of the board. Everything resting on the ice pushes down on it, so the force is the weight of the rescuer and the board together.

More 5(b) questions in the question bank

5(c) Change of State

+ Add to test✓ In your testCC-EIP-5c-120 marks~20 min

Tempering chocolate: a heating curve, specific heat capacity and wasted energy

5(c) Change of State2(b) Mains Electricity4(b) Energy TransfersPaper 2 only

A chocolatier melts 0.80 kg of solid dark chocolate in an electric melting pot. The heating element in the base of the pot transfers energy at a steady rate of 40 W. A temperature probe in the chocolate records its temperature, and the chocolate is stirred gently throughout.

Figure 1 shows how the temperature of the chocolate changes from the moment the heater is switched on (A) until it is switched off (D).

Figure 1: temperature of the chocolate against time while the heater is switched on

(a)

Describe the arrangement and the motion of the particles in the chocolate in section AB and in section CD of the graph.

Paper 2 only onlyRECALLNotes

Show solution

AB (solid): particles close together / touching in a regular pattern, vibrating about fixed positions	✓ 1	Both arrangement and motion needed for the mark.
CD (liquid): particles close together / touching but in an irregular arrangement, moving around (and past) each other	✓ 1	Do not accept 'far apart' for the liquid.

(2)

(b)

Calculate the energy transferred by the heater during section BC.

CALCNotes

Show solution

time for BC read from the graph = 21 − 7 = 14 min = 840 s	✓ 1	The conversion to seconds must be seen for this mark.
energy = power × time = 40 × 840 = 33 600 J	✓ 1	Accept 3.4 × 10⁴ J. ECF for an incorrect time in seconds. 560 J (40 × 14) scores 0.

Answer: 33 600 J (3.4 × 10⁴ J)

(2)

(c)

Explain why the temperature of the chocolate stays the same during section BC, even though the heater is still switched on.

Paper 2 only onlyEXPLAINNotes

Show solution

the chocolate is melting / changing state from solid to liquid	✓ 1
the energy supplied is used to break the bonds / overcome the forces holding the particles in their fixed positions	✓ 1	Allow: it increases the energy stored in the arrangement of the particles.
so the kinetic energy (speed) of the particles does not increase, so the temperature does not rise	✓ 1	'The energy is used to change state' alone scores the first mark only.

(3)

(d)

Use information from Figure 1 to show that the specific heat capacity of liquid chocolate is about 2000 J/kg °C.

Assume that all the energy from the heater is transferred to the chocolate.

Paper 2 only onlyDERIVENotes

Show solution

energy supplied in CD = 40 × (8 × 60) = 19 200 J	✓ 1	Time for CD read from the graph: 29 − 21 = 8 min = 480 s.
temperature change = 46 − 34 = 12 °C	✓ 1
c = ΔQ ÷ (m × ΔT) = 19 200 ÷ (0.80 × 12) = 2000 (J/kg °C)	✓ 1	A 'show that': the substitution and an answer of 2000 (or more figures) must be seen.

Answer: 2000 J/kg °C

(3)

(e)(i)

A data book gives the specific heat capacity of liquid dark chocolate as 1600 J/kg °C.

Calculate the efficiency of the melting pot in heating the liquid chocolate during section CD.

Paper 2 only onlyMULTINotes

Show solution

useful energy = m × c × ΔT = 0.80 × 1600 × 12 = 15 360 J	✓ 1	Allow ECF for ΔT from (d).
efficiency = useful energy ÷ total energy = 15 360 ÷ 19 200	✓ 1	Allow ECF for the total energy from (d).
evaluation: 0.80 or 80%	✓ 1	Alternative: 1600 ÷ 2000 = 0.80 scores full marks with a reason (same m, ΔT and energy). An answer above 100% scores 0 for the last mark.

Answer: 0.80 (80%)

(3)

(e)(ii)

Explain why the value in (d) is higher than the data-book value, and suggest one change to the melting pot that would make the value closer to 1600 J/kg °C.

Paper 2 only onlyEXPLAINNotes

Show solution

some of the energy from the heater is transferred to the pot / to the surroundings (thermal store of the surroundings)	✓ 1
so less than 19 200 J heats the chocolate: its temperature rise is smaller than it should be, so the calculated c is too large	✓ 1	Must link the wasted energy to the value being too high.
suggestion with how it works, e.g. fit a lid, reducing energy transfer by convection / evaporation from the surface; or wrap the pot in an insulating layer with trapped air, reducing conduction and convection	✓ 1	Naming the change without how it reduces energy transfer does not score.

(3)

(f)

The chocolatier now melts 1.60 kg of the same chocolate in the same pot, again starting at 20 °C and heating until the temperature reaches 46 °C.

Describe how the graph for 1.60 kg would differ from Figure 1.

Paper 2 only onlyPREDICTNotes

Show solution

the flat section is at the same temperature, 34 °C	✓ 1
each section takes longer / the sloping sections are less steep and the flat section is longer (about twice as long for each)	✓ 1	Accept 'D is reached at about 58 min'.

(2)

(g)

At D the heater is switched off. The chocolate is left in the pot in a room at 20 °C until it is completely solid and has cooled to room temperature.

On Figure 1, sketch a graph to show how the temperature of the chocolate changes after the heater is switched off.

Paper 2 only onlySKETCHNotes

Show solution

line starting at D and falling, then a horizontal section at 34 °C	✓ 1	The flat section must be at 34 °C (± 2 °C).
after the flat section the temperature falls again, levelling off at 20 °C, with no extra flat sections and never below 20 °C	✓ 1	Accept straight-line or curved falling sections.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
AB (solid): particles close together / touching in a regular pattern, vibrating about fixed positions	✓ 1	Both arrangement and motion needed for the mark.
CD (liquid): particles close together / touching but in an irregular arrangement, moving around (and past) each other	✓ 1	Do not accept 'far apart' for the liquid.
Part (b)
time for BC read from the graph = 21 − 7 = 14 min = 840 s	✓ 1	The conversion to seconds must be seen for this mark.
energy = power × time = 40 × 840 = 33 600 J	✓ 1	Accept 3.4 × 10⁴ J. ECF for an incorrect time in seconds. 560 J (40 × 14) scores 0.
Part (c)
the chocolate is melting / changing state from solid to liquid	✓ 1
the energy supplied is used to break the bonds / overcome the forces holding the particles in their fixed positions	✓ 1	Allow: it increases the energy stored in the arrangement of the particles.
so the kinetic energy (speed) of the particles does not increase, so the temperature does not rise	✓ 1	'The energy is used to change state' alone scores the first mark only.
Part (d)
energy supplied in CD = 40 × (8 × 60) = 19 200 J	✓ 1	Time for CD read from the graph: 29 − 21 = 8 min = 480 s.
temperature change = 46 − 34 = 12 °C	✓ 1
c = ΔQ ÷ (m × ΔT) = 19 200 ÷ (0.80 × 12) = 2000 (J/kg °C)	✓ 1	A 'show that': the substitution and an answer of 2000 (or more figures) must be seen.
Part (e)(i)
useful energy = m × c × ΔT = 0.80 × 1600 × 12 = 15 360 J	✓ 1	Allow ECF for ΔT from (d).
efficiency = useful energy ÷ total energy = 15 360 ÷ 19 200	✓ 1	Allow ECF for the total energy from (d).
evaluation: 0.80 or 80%	✓ 1	Alternative: 1600 ÷ 2000 = 0.80 scores full marks with a reason (same m, ΔT and energy). An answer above 100% scores 0 for the last mark.
Part (e)(ii)
some of the energy from the heater is transferred to the pot / to the surroundings (thermal store of the surroundings)	✓ 1
so less than 19 200 J heats the chocolate: its temperature rise is smaller than it should be, so the calculated c is too large	✓ 1	Must link the wasted energy to the value being too high.
suggestion with how it works, e.g. fit a lid, reducing energy transfer by convection / evaporation from the surface; or wrap the pot in an insulating layer with trapped air, reducing conduction and convection	✓ 1	Naming the change without how it reduces energy transfer does not score.
Part (f)
the flat section is at the same temperature, 34 °C	✓ 1
each section takes longer / the sloping sections are less steep and the flat section is longer (about twice as long for each)	✓ 1	Accept 'D is reached at about 58 min'.
Part (g)
line starting at D and falling, then a horizontal section at 34 °C	✓ 1	The flat section must be at 34 °C (± 2 °C).
after the flat section the temperature falls again, levelling off at 20 °C, with no extra flat sections and never below 20 °C	✓ 1	Accept straight-line or curved falling sections.

Answers: (b) 33 600 J (3.4 × 10⁴ J) · (d) 2000 J/kg °C · (e)(i) 0.80 (80%)

Common error (b) Using the time in minutes: energy = power × time needs the time in seconds, so 14 min must become 840 s.

Exam tip (e)(ii) For an insulation mark, give the change AND the transfer it reduces, e.g. 'a lid reduces convection from the surface'.

More 5(c) questions in the question bank

+ Add to test✓ In your testCC-EIP-5c-219 marks~20 min

A marathon runner keeping cool: specific heat capacity and evaporation

5(c) Change of State4(b) Energy Transfers4(c) Work and PowerPaper 2 only

A runner of mass 60 kg takes part in a marathon. While she runs at a steady pace, her muscles transfer 840 W of power to the thermal store of her body.

Average specific heat capacity of the human body = 3500 J/kg °C

(a)

Calculate the thermal energy produced in her body in the first 10.0 minutes of the run.

CALCNotes

Show solution

substitution into energy = power × time with t = 600 s; e.g. 840 × 600	✓ 1	Time must be converted to seconds.
evaluation: 504 000 J	✓ 1	Accept 5.0 × 10⁵ J. 8400 J (840 × 10) scores 0.

Answer: 504 000 J (5.0 × 10⁵ J)

(2)

(b)

Show that, if none of this energy left her body, her body temperature would rise by 2.4 °C in the first 10.0 minutes.

Paper 2 only onlyDERIVENotes

Show solution

rearrangement and substitution: ΔT = ΔQ ÷ (m × c) = 504 000 ÷ (60 × 3500)	✓ 1	Allow ECF from (a) for the substitution mark only.
evaluation: 2.4 (°C)	✓ 1	A 'show that': working must be seen. Reverse calculation (60 × 3500 × 2.4 = 504 000 J) is acceptable.

Answer: 2.4 °C

(2)

(c)(i)

Figure 1 shows how her body temperature actually changes during the first 40 minutes of the run.

Use Figure 1 to find her actual rise in temperature in the first 10.0 minutes. Hence calculate the thermal energy that stayed in her body during this time.

Figure 1: the runner's body temperature during the first 40 minutes of the run

Paper 2 only onlyGRAPHNotes

Show solution

temperature rise read from graph = 37.6 − 37.0 = 0.6 °C	✓ 1	Accept 0.55–0.65 °C.
substitution: ΔQ = 60 × 3500 × 0.6	✓ 1	ECF for their temperature rise.
evaluation: 126 000 J	✓ 1	Accept 1.3 × 10⁵ J; allow values from 0.55–0.65 °C (115 500 J to 136 500 J).

Answer: 0.6 °C; 126 000 J

(3)

(c)(ii)

Calculate the energy transferred from her body to the surroundings in the first 10.0 minutes.

Paper 2 only onlyCALCNotes

Show solution

energy transferred = energy produced − energy stored = 504 000 − 126 000	✓ 1	ECF from (a) and (c)(i).
evaluation: 378 000 J	✓ 1	Accept 3.8 × 10⁵ J.

Answer: 378 000 J (3.8 × 10⁵ J)

(2)

(c)(iii)

After about 30 minutes her body temperature stays almost constant even though she keeps running at the same pace.

Explain why.

REASONNotes

Show solution

energy is now transferred from her body to the surroundings at the same rate as it is produced / at about 840 W	✓ 1	Accept 'energy lost per second = energy produced per second'.
so there is no (net) increase in the thermal store of her body, so its temperature does not change	✓ 1

(2)

(d)

Much of the energy leaves her body when sweat evaporates from her skin.

Explain, in terms of particles, why the evaporation of sweat cools her skin.

Paper 2 only onlyEXPLAINNotes

Show solution

the fastest-moving particles / particles with the most kinetic energy escape from the surface of the sweat	✓ 1
so the average kinetic energy (speed) of the particles left behind falls, so the temperature of the sweat falls	✓ 1
energy is then transferred from the (warmer) skin to the cooler sweat	✓ 1	Allow: energy from the skin is used to evaporate more sweat.

(3)

(e)

On another day there is a steady breeze.

Explain two ways in which the moving air helps to keep her cool.

Paper 2 only onlyEXPLAINNotes

Show solution

air moving across the skin increases the rate of evaporation of sweat (evaporated particles are carried away)	✓ 1	Must refer to faster evaporation.
air warmed by her skin is carried away and replaced by cooler air, so energy is transferred from her skin (by convection) at a greater rate	✓ 1	Allow: keeps a larger temperature difference between skin and air.

(2)

(f)

In a later race, in the same sunny weather and at the same pace as the race in Figure 1, she wears a tight-fitting matt black top instead of the loose white top she wore before.

Suggest why her body temperature levels off at a higher value than in Figure 1.

Paper 2 only onlyPREDICTNotes

Show solution

matt black is a better absorber of (infrared) radiation (from the Sun) than white	✓ 1	Do not accept 'black attracts heat'.
so energy is transferred to her body at a greater rate	✓ 1
a tight top reduces the air moving over her skin, so sweat evaporates more slowly / less energy is transferred away; so a higher temperature is needed before the rate of energy loss equals the rate of energy gain	✓ 1	Award for either the evaporation idea or the balance-of-rates idea.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
substitution into energy = power × time with t = 600 s; e.g. 840 × 600	✓ 1	Time must be converted to seconds.
evaluation: 504 000 J	✓ 1	Accept 5.0 × 10⁵ J. 8400 J (840 × 10) scores 0.
Part (b)
rearrangement and substitution: ΔT = ΔQ ÷ (m × c) = 504 000 ÷ (60 × 3500)	✓ 1	Allow ECF from (a) for the substitution mark only.
evaluation: 2.4 (°C)	✓ 1	A 'show that': working must be seen. Reverse calculation (60 × 3500 × 2.4 = 504 000 J) is acceptable.
Part (c)(i)
temperature rise read from graph = 37.6 − 37.0 = 0.6 °C	✓ 1	Accept 0.55–0.65 °C.
substitution: ΔQ = 60 × 3500 × 0.6	✓ 1	ECF for their temperature rise.
evaluation: 126 000 J	✓ 1	Accept 1.3 × 10⁵ J; allow values from 0.55–0.65 °C (115 500 J to 136 500 J).
Part (c)(ii)
energy transferred = energy produced − energy stored = 504 000 − 126 000	✓ 1	ECF from (a) and (c)(i).
evaluation: 378 000 J	✓ 1	Accept 3.8 × 10⁵ J.
Part (c)(iii)
energy is now transferred from her body to the surroundings at the same rate as it is produced / at about 840 W	✓ 1	Accept 'energy lost per second = energy produced per second'.
so there is no (net) increase in the thermal store of her body, so its temperature does not change	✓ 1
Part (d)
the fastest-moving particles / particles with the most kinetic energy escape from the surface of the sweat	✓ 1
so the average kinetic energy (speed) of the particles left behind falls, so the temperature of the sweat falls	✓ 1
energy is then transferred from the (warmer) skin to the cooler sweat	✓ 1	Allow: energy from the skin is used to evaporate more sweat.
Part (e)
air moving across the skin increases the rate of evaporation of sweat (evaporated particles are carried away)	✓ 1	Must refer to faster evaporation.
air warmed by her skin is carried away and replaced by cooler air, so energy is transferred from her skin (by convection) at a greater rate	✓ 1	Allow: keeps a larger temperature difference between skin and air.
Part (f)
matt black is a better absorber of (infrared) radiation (from the Sun) than white	✓ 1	Do not accept 'black attracts heat'.
so energy is transferred to her body at a greater rate	✓ 1
a tight top reduces the air moving over her skin, so sweat evaporates more slowly / less energy is transferred away; so a higher temperature is needed before the rate of energy loss equals the rate of energy gain	✓ 1	Award for either the evaporation idea or the balance-of-rates idea.

Answers: (a) 504 000 J (5.0 × 10⁵ J) · (b) 2.4 °C · (c)(i) 0.6 °C; 126 000 J · (c)(ii) 378 000 J (3.8 × 10⁵ J)

Common error (d) Writing that 'the sweat takes the heat away' without particles. The key idea is that the fastest particles leave, so the average kinetic energy of those left behind falls.

More 5(c) questions in the question bank

5(d) Ideal Gas Molecules

+ Add to test✓ In your testCC-EIP-5d-119 marks~20 min

Pumping up a road-bike tyre: gas pressure, volume and temperature

5(d) Ideal Gas Molecules5(b) Density and Pressure1(c) Forces, Movement, Shape and MomentumBoth papers

A cyclist uses a track pump to inflate a road-bike tyre (Figure 1). The tyre already contains air at a pressure of 700 kPa.

At the start of each stroke the piston is at the top of the barrel, and the barrel contains 350 cm³ of air at atmospheric pressure, 100 kPa. The cyclist pushes the piston down slowly, so the temperature of the air stays constant. The valve stays closed until the pressure of the air in the barrel reaches 700 kPa; air then flows into the tyre.

Figure 1: the track pump connected to the tyre (not to scale)

(a)

Explain how the air molecules in the barrel exert a pressure on the piston.

EXPLAINNotes

Show solution

the molecules move randomly / in all directions at high speed	✓ 1
they collide with the piston (and the walls) and each collision exerts a force on it	✓ 1	Collisions between molecules do not score.
pressure = total force (from all the collisions) ÷ area of the piston	✓ 1	Accept 'force per unit area'.

(3)

(b)

Show that the volume of the air in the barrel is 50 cm³ at the moment the valve opens.

DERIVENotes

Show solution

substitution into p₁ × V₁ = p₂ × V₂; e.g. 100 × 350 = 700 × V₂	✓ 1	Units need not be converted: kPa and cm³ on both sides.
V₂ = 35 000 ÷ 700 = 50 (cm³)	✓ 1	A 'show that': the division must be seen.

Answer: 50 cm³

(2)

(c)(i)

Figure 2 shows how the pressure of the air in the barrel changes with its volume during one slow stroke, until the valve opens.

Use Figure 2 to find the pressure of the air when its volume is 140 cm³.

Figure 2: pressure against volume for the air in the barrel during one slow stroke

GRAPHNotes

Show solution

250 kPa

✓ 1

Accept 240–260 kPa.

Answer: 250 kPa

(1)

(c)(ii)

Explain, in terms of molecules, why the pressure of the air increases as its volume decreases during the stroke.

EXPLAINNotes

Show solution

the temperature is constant, so the (average) speed of the molecules does not change	✓ 1	Do not credit 'the molecules move faster'.
in a smaller volume the molecules hit the walls / piston more often (more collisions per second on each unit area), so the force per unit area is greater	✓ 1	'More collisions' without a rate or area idea is not enough.

(2)

(d)(i)

When the valve opens, the pressure difference between the air in the barrel and the atmosphere above the piston is 600 kPa. The area of the piston is 7.0 cm².

Calculate the force the cyclist must apply to the piston to keep it still at this moment. Ignore friction and the weight of the piston.

CALCNotes

Show solution

area = 7.0 ÷ 10 000 = 7.0 × 10⁻⁴ m² and substitution into F = p × A; e.g. 600 000 × 7.0 × 10⁻⁴	✓ 1
evaluation: 420 N	✓ 1	4.2 N or 42 000 N (power-of-ten errors in the area) score 1 mark if the method is clear.

Answer: 420 N

(2)

(d)(ii)

The cyclist has a mass of 45 kg. She leans on the handle so that her whole weight acts downwards on the piston.

Explain whether this is enough to push air into the tyre. Use g = 10 N/kg.

REASONNotes

Show solution

weight = m × g = 45 × 10 = 450 N	✓ 1
450 N is greater than 420 N, so yes, the piston keeps moving down and air is pushed into the tyre	✓ 1	ECF from (d)(i): conclusion must match their force.

Answer: yes (450 N > 420 N)

(2)

(e)

A different cyclist pushes the piston down very quickly, and the air in the barrel becomes warm.

Explain why the valve now opens before the volume of the air in the barrel has fallen to 50 cm³.

REASONNotes

Show solution

the temperature of the air is higher, so its molecules have more kinetic energy / a greater average speed	✓ 1
so they hit the walls / piston more often and harder (with more force per collision), giving a greater force per unit area	✓ 1	Both 'more often' and 'harder' needed for this mark.
so at any volume the pressure is greater than in the slow stroke, and 700 kPa is reached at a volume greater than 50 cm³	✓ 1	Accept: the pressure reaches 700 kPa earlier in the stroke.

(3)

(f)(i)

The cyclist finishes inflating the tyre to 700 kPa at 17 °C. She then rides down a long mountain road on a hot day. Braking heats the wheel rims, and the air in the tyre warms to 46 °C. Assume the volume of the tyre does not change.

Convert 17 °C and 46 °C to kelvin.

RECALLNotes

Show solution

290 K and 319 K

✓ 1

Both needed.

Answer: 290 K; 319 K

(1)

(f)(ii)

Calculate the new pressure of the air in the tyre.

CALCNotes

Show solution

substitution into p₁ ÷ T₁ = p₂ ÷ T₂; e.g. p₂ = 700 × 319 ÷ 290	✓ 1	ECF from (f)(i). Temperatures must be in kelvin.
evaluation: 770 kPa	✓ 1	Using Celsius (700 × 46 ÷ 17 = 1890 kPa) scores 0.

Answer: 770 kPa

(2)

(f)(iii)

The tyre is designed for a maximum pressure of 800 kPa.

Comment on whether the cyclist should inflate the tyre to more than 700 kPa before a long, hot descent.

REASONNotes

Show solution

no: 770 kPa is already close to 800 kPa, and a higher starting pressure (or a hotter tyre) would make the pressure exceed the maximum, so the tyre could burst

✓ 1

ECF from (f)(ii). Must use the numbers.

(1)

Show full solution

Marking point	Mark	Notes
Part (a)
the molecules move randomly / in all directions at high speed	✓ 1
they collide with the piston (and the walls) and each collision exerts a force on it	✓ 1	Collisions between molecules do not score.
pressure = total force (from all the collisions) ÷ area of the piston	✓ 1	Accept 'force per unit area'.
Part (b)
substitution into p₁ × V₁ = p₂ × V₂; e.g. 100 × 350 = 700 × V₂	✓ 1	Units need not be converted: kPa and cm³ on both sides.
V₂ = 35 000 ÷ 700 = 50 (cm³)	✓ 1	A 'show that': the division must be seen.
Part (c)(i)
250 kPa	✓ 1	Accept 240–260 kPa.
Part (c)(ii)
the temperature is constant, so the (average) speed of the molecules does not change	✓ 1	Do not credit 'the molecules move faster'.
in a smaller volume the molecules hit the walls / piston more often (more collisions per second on each unit area), so the force per unit area is greater	✓ 1	'More collisions' without a rate or area idea is not enough.
Part (d)(i)
area = 7.0 ÷ 10 000 = 7.0 × 10⁻⁴ m² and substitution into F = p × A; e.g. 600 000 × 7.0 × 10⁻⁴	✓ 1
evaluation: 420 N	✓ 1	4.2 N or 42 000 N (power-of-ten errors in the area) score 1 mark if the method is clear.
Part (d)(ii)
weight = m × g = 45 × 10 = 450 N	✓ 1
450 N is greater than 420 N, so yes, the piston keeps moving down and air is pushed into the tyre	✓ 1	ECF from (d)(i): conclusion must match their force.
Part (e)
the temperature of the air is higher, so its molecules have more kinetic energy / a greater average speed	✓ 1
so they hit the walls / piston more often and harder (with more force per collision), giving a greater force per unit area	✓ 1	Both 'more often' and 'harder' needed for this mark.
so at any volume the pressure is greater than in the slow stroke, and 700 kPa is reached at a volume greater than 50 cm³	✓ 1	Accept: the pressure reaches 700 kPa earlier in the stroke.
Part (f)(i)
290 K and 319 K	✓ 1	Both needed.
Part (f)(ii)
substitution into p₁ ÷ T₁ = p₂ ÷ T₂; e.g. p₂ = 700 × 319 ÷ 290	✓ 1	ECF from (f)(i). Temperatures must be in kelvin.
evaluation: 770 kPa	✓ 1	Using Celsius (700 × 46 ÷ 17 = 1890 kPa) scores 0.
Part (f)(iii)
no: 770 kPa is already close to 800 kPa, and a higher starting pressure (or a hotter tyre) would make the pressure exceed the maximum, so the tyre could burst	✓ 1	ECF from (f)(ii). Must use the numbers.

Answers: (b) 50 cm³ · (c)(i) 250 kPa · (d)(i) 420 N · (d)(ii) yes (450 N > 420 N) · (f)(i) 290 K; 319 K · (f)(ii) 770 kPa

Exam tip (c)(ii) For a squeeze at constant temperature, only 'more frequent collisions' applies. The molecules are not faster, so do not write 'harder collisions'.

Common error (d)(i) Dividing 7.0 cm² by 100 instead of 10 000. 1 m² = 100 × 100 = 10 000 cm².

More 5(d) questions in the question bank

+ Add to test✓ In your testCC-EIP-5d-218 marks~20 min

A compressed-air can left in a hot car: pressure, temperature and absolute zero

5(d) Ideal Gas Molecules5(b) Density and Pressure4(b) Energy TransfersBoth papers

A rigid steel can contains only compressed air. It is used to blow dust off camera lenses. On a warm morning the air in the can is at 27 °C and its pressure is 400 kPa.

The can is left on the back seat of a car parked in direct sunshine with the windows closed. By the afternoon the air in the can has reached 72 °C. Assume the volume of the can does not change and no air escapes.

(a)

Convert 27 °C and 72 °C to kelvin.

RECALLNotes

Show solution

300 K and 345 K

✓ 1

Both needed. Do not accept °K.

Answer: 300 K; 345 K

(1)

(b)

Explain, in terms of molecules, why the pressure of the air in the can increases as its temperature rises.

EXPLAINNotes

Show solution

the molecules gain kinetic energy / move faster on average	✓ 1	Do not accept 'the molecules expand' or 'get bigger'.
they collide with the walls of the can more often / more collisions per second	✓ 1
and each collision exerts a greater force, so the force per unit area (pressure) increases	✓ 1

(3)

(c)

Show that the pressure of the air in the can at 72 °C is 460 kPa.

DERIVENotes

Show solution

substitution into p₁ ÷ T₁ = p₂ ÷ T₂ with kelvin temperatures; e.g. p₂ = 400 × 345 ÷ 300	✓ 1	ECF from (a).
evaluation: 460 (kPa)	✓ 1	A 'show that': working must be seen.

Answer: 460 kPa

(2)

(d)

The flat base of the can has an area of 30 cm².

Calculate the force exerted by the air in the can on its base at 72 °C.

CALCNotes

Show solution

area = 30 ÷ 10 000 = 3.0 × 10⁻³ m² and substitution into F = p × A; e.g. 460 000 × 3.0 × 10⁻³	✓ 1	Pressure must be in Pa for a force in N.
evaluation: 1380 N	✓ 1	Accept 1400 N. ECF from (c).

Answer: 1380 N (1400 N)

(2)

(e)(i)

Before sale, the can was tested at several temperatures. Figure 1 shows the results.

Use Figure 1 to find the temperature at which the pressure of the air in the can would be zero. Show on Figure 1 how you obtained your answer.

Figure 1: pressure of the air in the can at different temperatures, with the line of best fit

GRAPHNotes

Show solution

straight line extended back (with a ruler) to meet the temperature axis	✓ 1
−273 °C	✓ 1	Accept −260 °C to −285 °C.

Answer: −273 °C

(2)

(e)(ii)

Explain why it is not possible to cool the air to a temperature lower than the value in (e)(i).

EXPLAINNotes

Show solution

at this temperature (absolute zero) the molecules have the least possible kinetic energy / stop moving (so exert no pressure)	✓ 1
kinetic energy cannot be less than zero / no more energy can be removed, so the temperature cannot fall further	✓ 1

(2)

(f)

Explain why the air inside a car parked in direct sunshine with its windows closed becomes much hotter than the air outside.

EXPLAINNotes

Show solution

(infrared) radiation from the Sun passes through the windows and is absorbed by the seats and dashboard	✓ 1
dark / matt surfaces are good absorbers, so they become hot and transfer energy to the air next to them (by conduction and convection)	✓ 1
the windows are closed, so the warm air cannot escape and be replaced by cooler air (convection to the outside is prevented)	✓ 1

(3)

(g)

The next day the driver fits a shiny silver sun shade inside the windscreen. This time the air in the can reaches only 42 °C.

Explain how the shade reduces the temperature, and calculate the pressure of the air in the can at 42 °C. The can again starts at 400 kPa and 27 °C.

MULTINotes

Show solution

shiny silver surface is a poor absorber / good reflector, so it reflects most of the Sun's radiation back out	✓ 1
T₂ = 42 + 273 = 315 K and p₂ = 400 × 315 ÷ 300	✓ 1
evaluation: 420 kPa	✓ 1	Using Celsius (400 × 42 ÷ 27 = 622 kPa) scores 0 for the calculation.

Answer: 420 kPa

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
300 K and 345 K	✓ 1	Both needed. Do not accept °K.
Part (b)
the molecules gain kinetic energy / move faster on average	✓ 1	Do not accept 'the molecules expand' or 'get bigger'.
they collide with the walls of the can more often / more collisions per second	✓ 1
and each collision exerts a greater force, so the force per unit area (pressure) increases	✓ 1
Part (c)
substitution into p₁ ÷ T₁ = p₂ ÷ T₂ with kelvin temperatures; e.g. p₂ = 400 × 345 ÷ 300	✓ 1	ECF from (a).
evaluation: 460 (kPa)	✓ 1	A 'show that': working must be seen.
Part (d)
area = 30 ÷ 10 000 = 3.0 × 10⁻³ m² and substitution into F = p × A; e.g. 460 000 × 3.0 × 10⁻³	✓ 1	Pressure must be in Pa for a force in N.
evaluation: 1380 N	✓ 1	Accept 1400 N. ECF from (c).
Part (e)(i)
straight line extended back (with a ruler) to meet the temperature axis	✓ 1
−273 °C	✓ 1	Accept −260 °C to −285 °C.
Part (e)(ii)
at this temperature (absolute zero) the molecules have the least possible kinetic energy / stop moving (so exert no pressure)	✓ 1
kinetic energy cannot be less than zero / no more energy can be removed, so the temperature cannot fall further	✓ 1
Part (f)
(infrared) radiation from the Sun passes through the windows and is absorbed by the seats and dashboard	✓ 1
dark / matt surfaces are good absorbers, so they become hot and transfer energy to the air next to them (by conduction and convection)	✓ 1
the windows are closed, so the warm air cannot escape and be replaced by cooler air (convection to the outside is prevented)	✓ 1
Part (g)
shiny silver surface is a poor absorber / good reflector, so it reflects most of the Sun's radiation back out	✓ 1
T₂ = 42 + 273 = 315 K and p₂ = 400 × 315 ÷ 300	✓ 1
evaluation: 420 kPa	✓ 1	Using Celsius (400 × 42 ÷ 27 = 622 kPa) scores 0 for the calculation.

Answers: (a) 300 K; 345 K · (c) 460 kPa · (d) 1380 N (1400 N) · (e)(i) −273 °C · (g) 420 kPa

Common error (g) Putting Celsius temperatures into p₁ ÷ T₁ = p₂ ÷ T₂. Add 273 to every temperature first.

More 5(d) questions in the question bank

Topic 6 · Magnetism and electromagnetism

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

Setup	6(b)	6(c)	6(d)	Other
A recycling plant: separating steel cans with magnets	●			1(c) 1(b)
An MRI scanner room: strong magnetic fields and safety	●	○		3(c)
An electric skateboard: the d.c. motor		●		2(c) 2(b) 4(c)
Inside a pair of headphones: the loudspeaker principle		●		3(b) 3(d)
A bicycle hub dynamo: generating a voltage			●	2(b) 2(c) 3(b)
A garden lighting transformer: stepping down the mains			●	2(b) 4(b)

6(b) Magnetism

+ Add to test✓ In your testCC-EIP-6b-120 marks~20 min

A recycling plant: separating steel cans with magnets

6(b) Magnetism1(c) Forces, Movement, Shape and Momentum1(b) Movement and PositionBoth papers

At a recycling plant, a conveyor belt carries a mixture of empty steel cans and aluminium cans. The belt passes under a large permanent magnet, as shown in Figure 1. Steel cans are lifted up to the magnet and removed; aluminium cans stay on the belt.

Each steel can has a mass of 45 g. The lower face of the magnet is a north pole and is 0.60 m long in the direction the belt moves. The belt moves at a constant speed of 1.5 m/s.

Take the gravitational field strength g to be 10 N/kg.

Figure 1: the magnetic separator above the conveyor belt (not to scale)

(a)

Explain why the magnet lifts the steel cans but not the aluminium cans.

EXPLAINNotes

Show solution

steel is a magnetic material (so it is attracted by a magnet)	✓ 1	Accept “steel contains iron, which is magnetic”.
aluminium is not a magnetic material (so there is no magnetic force on it)	✓ 1	Do not accept “aluminium is not a metal”. “Steel is a metal” alone scores 0.

(2)

(b)

A steel can is not a magnet before it reaches the separator. Explain, in terms of induced magnetism, why the steel can is attracted to the north pole of the magnet.

EXPLAINNotes

Show solution

the can becomes an (induced) magnet when it is in the magnet’s field	✓ 1	Accept “the can is magnetised by the magnet”.
the end of the can nearest the magnet becomes a south pole, and unlike poles attract	✓ 1	Must name the S pole (or “opposite pole”) and attraction.

(2)

(c)

An engineer uses a small bar magnet as a model of the separator magnet. Describe how she could use a plotting compass to draw one field line around the bar magnet, including its direction.

EXPTNotes

Show solution

place the magnet on paper and put the compass next to the N pole; draw a dot at the end the compass needle points to	✓ 1	Accept “at the N end of the needle”.
move the compass so the tail of the needle is over the dot, draw a new dot, and repeat (until the S pole is reached)	✓ 1
join the dots with a smooth line and add an arrow in the direction the needle points (N to S)	✓ 1	Allow iron filings for the shape only if a compass is still used for the direction.

(3)

(d)(i)

Show that the weight of one steel can is about 0.45 N.

DERIVENotes

Show solution

m = 45 g = 0.045 kg	✓ 1	Conversion seen.
W = m × g = 0.045 × 10 = 0.45 N	✓ 1	Substitution and answer. Using g = 9.8 N/kg gives 0.44 N: accept.

Answer: 0.45 N

(2)

(d)(ii)

Figure 2 shows how the magnetic force on a steel can depends on the gap between the magnet face and the top of the can.

The plant manager wants the gap to be as large as possible, so that large items do not jam under the magnet. Use Figure 2 and your answer to (d)(i) to find the largest gap at which the magnet can still lift a steel can. Explain your answer.

Figure 2: magnetic force on a steel can against the size of the gap

GRAPHNotes

Show solution

the can is lifted only if the magnetic force (upwards) is greater than its weight / greater than 0.45 N	✓ 1	Accept “resultant force must be upwards”.
largest gap = 120 mm	✓ 1	Accept 115–125 mm. Allow ECF from (d)(i) for a reading at a different weight.

Answer: 120 mm

(2)

(d)(iii)

A steel can passes under the magnet with a gap of 60 mm.

Calculate the acceleration of the can at the moment it starts to rise.

MULTINotes

Show solution

magnetic force read from Figure 2 = 0.90 N	✓ 1	Accept 0.88–0.92 N.
resultant force = 0.90 − 0.45 = 0.45 N (upwards)	✓ 1	Allow ECF from the reading and from (d)(i).
a = F ÷ m = 0.45 ÷ 0.045 = 10 m/s²	✓ 1	Using the magnetic force alone (0.90 ÷ 0.045 = 20 m/s²) scores the reading mark only. Mass must be in kg.

Answer: 10 m/s² (upwards)

(3)

(e)(i)

Calculate the time a can spends directly under the magnet face as it is carried along by the belt.

CALCNotes

Show solution

time = distance ÷ speed = 0.60 ÷ 1.5	✓ 1
= 0.40 s	✓ 1

Answer: 0.40 s

(2)

(e)(ii)

To process more cans, the belt speed is increased to 3.0 m/s. The magnet and its height above the belt are not changed. Some cans lie on their side, so the gap above them is close to the largest gap found in (d)(ii).

Suggest why some steel cans might now stay on the belt.

REASONNotes

Show solution

each can spends less time under the magnet (0.20 s, half the time)	✓ 1	Accept a correct time calculated using 3.0 m/s.
near the largest gap the resultant upward force (and so the acceleration) is very small, so in the shorter time such a can may not rise to the magnet before it passes the end of the magnet face	✓ 1	Do not accept “the magnetic force is smaller”: the force at a given gap does not depend on the speed. Allow ECF from (d)(ii).

(2)

(f)

The separator magnet is made of a magnetically hard material. Explain why a magnetically soft material such as iron would not be suitable.

EXPLAINNotes

Show solution

a hard material keeps its magnetism, so it is a permanent magnet	✓ 1
a soft material (iron) loses its magnetism (once the magnetising field is removed), so it would not keep attracting the cans	✓ 1	The comparison must refer to keeping or losing magnetism. “Iron is too soft/weak” scores 0.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
steel is a magnetic material (so it is attracted by a magnet)	✓ 1	Accept “steel contains iron, which is magnetic”.
aluminium is not a magnetic material (so there is no magnetic force on it)	✓ 1	Do not accept “aluminium is not a metal”. “Steel is a metal” alone scores 0.
Part (b)
the can becomes an (induced) magnet when it is in the magnet’s field	✓ 1	Accept “the can is magnetised by the magnet”.
the end of the can nearest the magnet becomes a south pole, and unlike poles attract	✓ 1	Must name the S pole (or “opposite pole”) and attraction.
Part (c)
place the magnet on paper and put the compass next to the N pole; draw a dot at the end the compass needle points to	✓ 1	Accept “at the N end of the needle”.
move the compass so the tail of the needle is over the dot, draw a new dot, and repeat (until the S pole is reached)	✓ 1
join the dots with a smooth line and add an arrow in the direction the needle points (N to S)	✓ 1	Allow iron filings for the shape only if a compass is still used for the direction.
Part (d)(i)
m = 45 g = 0.045 kg	✓ 1	Conversion seen.
W = m × g = 0.045 × 10 = 0.45 N	✓ 1	Substitution and answer. Using g = 9.8 N/kg gives 0.44 N: accept.
Part (d)(ii)
the can is lifted only if the magnetic force (upwards) is greater than its weight / greater than 0.45 N	✓ 1	Accept “resultant force must be upwards”.
largest gap = 120 mm	✓ 1	Accept 115–125 mm. Allow ECF from (d)(i) for a reading at a different weight.
Part (d)(iii)
magnetic force read from Figure 2 = 0.90 N	✓ 1	Accept 0.88–0.92 N.
resultant force = 0.90 − 0.45 = 0.45 N (upwards)	✓ 1	Allow ECF from the reading and from (d)(i).
a = F ÷ m = 0.45 ÷ 0.045 = 10 m/s²	✓ 1	Using the magnetic force alone (0.90 ÷ 0.045 = 20 m/s²) scores the reading mark only. Mass must be in kg.
Part (e)(i)
time = distance ÷ speed = 0.60 ÷ 1.5	✓ 1
= 0.40 s	✓ 1
Part (e)(ii)
each can spends less time under the magnet (0.20 s, half the time)	✓ 1	Accept a correct time calculated using 3.0 m/s.
near the largest gap the resultant upward force (and so the acceleration) is very small, so in the shorter time such a can may not rise to the magnet before it passes the end of the magnet face	✓ 1	Do not accept “the magnetic force is smaller”: the force at a given gap does not depend on the speed. Allow ECF from (d)(ii).
Part (f)
a hard material keeps its magnetism, so it is a permanent magnet	✓ 1
a soft material (iron) loses its magnetism (once the magnetising field is removed), so it would not keep attracting the cans	✓ 1	The comparison must refer to keeping or losing magnetism. “Iron is too soft/weak” scores 0.

Answers: (d)(i) 0.45 N · (d)(ii) 120 mm · (d)(iii) 10 m/s² (upwards) · (e)(i) 0.40 s

Exam tip (a) Say “magnetic material”, not “metal”: aluminium is a metal but it is not attracted by a magnet.

Common error (d)(iii) Dividing the magnetic force by the mass. F = m × a uses the resultant force: subtract the weight first.

More 6(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-6b-220 marks~20 min

An MRI scanner room: strong magnetic fields and safety

6(b) Magnetism6(c) Electromagnetism3(c) The Electromagnetic SpectrumBoth papers

A hospital MRI scanner produces a very strong magnetic field inside the tunnel (the bore) where the patient lies. Figure 1 shows some of the field lines inside and outside the scanner.

Steel objects brought into the room can be pulled towards the scanner with a large force, so staff follow strict safety rules.

Figure 1: field lines of the MRI scanner, side view (not to scale)

(a)

The field in region X is uniform. Give two features of the field lines in region X that show this.

RECALLNotes

Show solution

any one of: (lines are) straight / parallel / equally spaced	✓ 1	Any two of straight / parallel / equally spaced, one mark each.
a second, different feature from the list	✓ 1	“Same direction everywhere” on its own is not a feature of the lines: do not accept.

(2)

(b)

A steel oxygen cylinder must never be taken into the scanner room, but an oxygen cylinder made of aluminium can be used safely.

Explain this difference.

EXPLAINNotes

Show solution

steel is a magnetic material; it becomes an induced magnet in the scanner’s field	✓ 1	Accept “is magnetised by the field”.
the end nearest the scanner becomes the opposite pole, so the cylinder is attracted / pulled towards the scanner	✓ 1
aluminium is not a magnetic material, so there is no (magnetic) force on it	✓ 1	Not “aluminium is not a metal”.

(3)

(c)

A technician accidentally carries a steel screwdriver and an iron nail close to the scanner. Later, the screwdriver picks up small steel pins, but the iron nail does not.

Explain these observations.

REASONNotes

Show solution

steel is magnetically hard: it keeps (some of) the magnetism induced by the scanner and is now a (permanent) magnet	✓ 1
iron is magnetically soft: it loses its induced magnetism once it is taken out of the field	✓ 1	Each mark needs the material’s behaviour (keeps / loses its magnetism) linked to the observation (picks up pins / does not).

(2)

(d)(i)

The field of the scanner is produced by a current in a long coil (a solenoid) wound round the bore.

Explain why the coil is made from insulated wire.

Paper 2 only onlyEXPLAINNotes

Show solution

insulation stops the current passing (short-circuiting) between turns that touch	✓ 1
so the current flows around every turn and the field is produced	✓ 1	Accept “so the field is as strong as possible”.

(2)

(d)(ii)

State two changes to the coil that would make the field in the bore stronger.

Paper 2 only onlyRECALLNotes

Show solution

increase the current	✓ 1	Any two, one mark each.
increase the number of turns (on the coil)	✓ 1	Do not accept “add an iron core”: it is not a change to the coil, and the bore must stay clear for the patient. Do not accept “use thicker wire” or “bigger voltage” without reference to the current.

(2)

(e)

The current in the coil is reversed.

State what happens to the field in the bore, and explain why a steel object is still pulled towards the scanner.

PREDICTNotes

Show solution

the field reverses direction (same pattern and strength)	✓ 1	Accept “the arrows on the field lines point the other way”.
the induced poles in the steel also reverse, so the nearest end is still the opposite pole and the object is still attracted	✓ 1	Accept “induced magnetism always gives attraction, whichever pole is nearest”.

(2)

(f)(i)

The scanner sends radio waves of frequency 64 MHz into the patient’s body. The speed of electromagnetic waves is 3.0 × 10⁸ m/s.

Calculate the wavelength of the radio waves.

CALCNotes

Show solution

λ = v ÷ f = 3.0 × 10⁸ ÷ 6.4 × 10⁷	✓ 1	64 MHz converted to 6.4 × 10⁷ Hz.
λ = 4.7 m	✓ 1	Accept 4.69 m or 4.6875 m. A power-of-ten error scores 1.

Answer: 4.7 m

(2)

(f)(ii)

Some patients have an x-ray scan instead. Explain why the radio waves used in the MRI scanner are much less likely to damage body cells than x-rays.

EXPLAINNotes

Show solution

radio waves have a much lower frequency (longer wavelength) than x-rays	✓ 1
lower-frequency radiation carries less energy, so it is less able to damage cells / DNA	✓ 1	Accept “x-rays carry more energy and can cause cancer or mutation”.

(2)

(g)(i)

Figure 2 shows how the strength of the field outside the scanner changes with distance from the end of the scanner. Field strength is measured in millitesla (mT).

Steel equipment trolleys must be kept further from the scanner than the point where the field strength is 3 mT.

Use Figure 2 to find the smallest distance from the end of the scanner at which a trolley may be parked.

Figure 2: strength of the magnetic field outside the scanner against distance from the end of the scanner

GRAPHNotes

Show solution

2.4 m

✓ 1

Accept 2.3–2.5 m.

Answer: 2.4 m

(1)

(g)(ii)

Use readings from Figure 2 to show that, for this scanner, when the distance from the end of the scanner doubles from 1.0 m to 2.0 m, the field strength falls to about one-eighth of its value.

DATANotes

Show solution

readings: 40 mT at 1.0 m and 5 mT at 2.0 m	✓ 1	Accept 39–41 mT and 4.5–5.5 mT.
40 ÷ 5 = 8, so the field at 2.0 m is 1/8 of the field at 1.0 m	✓ 1	Ratio must be calculated from the readings.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
any one of: (lines are) straight / parallel / equally spaced	✓ 1	Any two of straight / parallel / equally spaced, one mark each.
a second, different feature from the list	✓ 1	“Same direction everywhere” on its own is not a feature of the lines: do not accept.
Part (b)
steel is a magnetic material; it becomes an induced magnet in the scanner’s field	✓ 1	Accept “is magnetised by the field”.
the end nearest the scanner becomes the opposite pole, so the cylinder is attracted / pulled towards the scanner	✓ 1
aluminium is not a magnetic material, so there is no (magnetic) force on it	✓ 1	Not “aluminium is not a metal”.
Part (c)
steel is magnetically hard: it keeps (some of) the magnetism induced by the scanner and is now a (permanent) magnet	✓ 1
iron is magnetically soft: it loses its induced magnetism once it is taken out of the field	✓ 1	Each mark needs the material’s behaviour (keeps / loses its magnetism) linked to the observation (picks up pins / does not).
Part (d)(i)
insulation stops the current passing (short-circuiting) between turns that touch	✓ 1
so the current flows around every turn and the field is produced	✓ 1	Accept “so the field is as strong as possible”.
Part (d)(ii)
increase the current	✓ 1	Any two, one mark each.
increase the number of turns (on the coil)	✓ 1	Do not accept “add an iron core”: it is not a change to the coil, and the bore must stay clear for the patient. Do not accept “use thicker wire” or “bigger voltage” without reference to the current.
Part (e)
the field reverses direction (same pattern and strength)	✓ 1	Accept “the arrows on the field lines point the other way”.
the induced poles in the steel also reverse, so the nearest end is still the opposite pole and the object is still attracted	✓ 1	Accept “induced magnetism always gives attraction, whichever pole is nearest”.
Part (f)(i)
λ = v ÷ f = 3.0 × 10⁸ ÷ 6.4 × 10⁷	✓ 1	64 MHz converted to 6.4 × 10⁷ Hz.
λ = 4.7 m	✓ 1	Accept 4.69 m or 4.6875 m. A power-of-ten error scores 1.
Part (f)(ii)
radio waves have a much lower frequency (longer wavelength) than x-rays	✓ 1
lower-frequency radiation carries less energy, so it is less able to damage cells / DNA	✓ 1	Accept “x-rays carry more energy and can cause cancer or mutation”.
Part (g)(i)
2.4 m	✓ 1	Accept 2.3–2.5 m.
Part (g)(ii)
readings: 40 mT at 1.0 m and 5 mT at 2.0 m	✓ 1	Accept 39–41 mT and 4.5–5.5 mT.
40 ÷ 5 = 8, so the field at 2.0 m is 1/8 of the field at 1.0 m	✓ 1	Ratio must be calculated from the readings.

Answers: (f)(i) 4.7 m · (g)(i) 2.4 m

More 6(b) questions in the question bank

6(c) Electromagnetism

+ Add to test✓ In your testCC-EIP-6c-120 marks~20 min

An electric skateboard: the d.c. motor

6(c) Electromagnetism2(c) Energy and Voltage in Circuits2(b) Mains Electricity4(c) Work and PowerBoth papers

An electric skateboard is driven by a d.c. motor that runs from a 36 V battery. Figure 1 shows a simplified view, from above, of a motor of this type. The coil ABCD lies flat between the poles of a magnet and can turn about the axle.

Figure 1: a simple d.c. motor seen from above (not to scale)

(a)

Explain why there is a force on side AB of the coil when there is a current in it.

EXPLAINNotes

Show solution

the current (in AB) produces its own magnetic field (around the wire)	✓ 1
this field interacts with the field of the (permanent) magnet, producing a force	✓ 1	Quoting Fleming’s left-hand rule alone scores 0.

(2)

(b)

Use Fleming’s left-hand rule to determine the direction of the force on side AB and on side CD at the instant shown in Figure 1. Hence explain why the coil turns.

REASONNotes

Show solution

force on AB is into the page	✓ 1	Accept “downwards, into the paper”.
force on CD is out of the page	✓ 1	Accept “upwards, out of the paper”. Both reversed: allow 1 mark out of the first two if the forces are opposite.
the forces on the two sides are in opposite directions, so they produce a turning effect (about the axle)	✓ 1

(3)

(c)

Explain the purpose of the split-ring commutator.

EXPLAINNotes

Show solution

it reverses the current in the coil every half turn	✓ 1	Accept “when the coil passes the vertical”.
so the forces keep turning the coil in the same direction (continuous rotation)	✓ 1	“Stops the wires tangling” or “changes a.c. to d.c.” scores 0.

(2)

(d)(i)

When the skateboard travels on level ground at a steady speed, the electrical power input to the motor is 360 W.

Calculate the current in the motor.

CALCNotes

Show solution

I = P ÷ V = 360 ÷ 36	✓ 1
= 10 A	✓ 1

Answer: 10 A

(2)

(d)(ii)

Calculate the charge that flows through the motor in 25 s.

CALCNotes

Show solution

Q = I × t = 10 × 25	✓ 1	Allow ECF from (d)(i).
= 250 C	✓ 1	Unit required.

Answer: 250 C

(2)

(e)(i)

In 25 s the skateboard travels 150 m along the level ground at a steady speed. The forward driving force on the skateboard is 48 N.

Show that the useful output power of the motor is about 290 W.

DERIVENotes

Show solution

work done = force × distance = 48 × 150	✓ 1
= 7200 J	✓ 1
P = W ÷ t = 7200 ÷ 25 = 288 W (≈ 290 W)	✓ 1	The unrounded value 288 W must be seen. Alternative: speed = 150 ÷ 25 = 6.0 m/s (✓), power = force × speed = 48 × 6.0 (✓) = 288 W (✓).

Answer: 288 W

(3)

(e)(ii)

Calculate the power wasted in the motor.

CALCNotes

Show solution

360 − 288 = 72 W

✓ 1

Accept 70 W if 290 W is used. Allow ECF from (e)(i).

Answer: 72 W

(1)

(f)(i)

Figure 2 shows how the driving force produced by the motor depends on the current in it.

The rider now goes up a steep hill at a steady speed. This needs a driving force of 84 N.

Use Figure 2 to find the current in the motor on the hill.

Figure 2: driving force produced by the motor against the current in the motor

GRAPHNotes

Show solution

17.5 A

✓ 1

Accept 17.2–17.8 A.

Answer: 17.5 A

(1)

(f)(ii)

The skateboard’s instructions warn that the motor may overheat if the rider climbs long, steep hills. Suggest why.

EXPLAINNotes

Show solution

a larger force needs a larger current (17.5 A instead of 10 A)	✓ 1	Allow ECF from (f)(i).
a larger current transfers more energy per second as heat in the (resistance of the) coil, so its temperature rises (more)	✓ 1	Accept a description in terms of electrons colliding more often with ions in the wire.

(2)

(f)(iii)

A designer wants the motor to produce the same driving force with a smaller current. State two changes she could make to the motor.

REASONNotes

Show solution

use a stronger magnet	✓ 1	Any two, one mark each. Accept “move the poles closer together”.
use more turns on the coil	✓ 1	Do not accept “increase the current” or “increase the voltage”: the current must be smaller.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the current (in AB) produces its own magnetic field (around the wire)	✓ 1
this field interacts with the field of the (permanent) magnet, producing a force	✓ 1	Quoting Fleming’s left-hand rule alone scores 0.
Part (b)
force on AB is into the page	✓ 1	Accept “downwards, into the paper”.
force on CD is out of the page	✓ 1	Accept “upwards, out of the paper”. Both reversed: allow 1 mark out of the first two if the forces are opposite.
the forces on the two sides are in opposite directions, so they produce a turning effect (about the axle)	✓ 1
Part (c)
it reverses the current in the coil every half turn	✓ 1	Accept “when the coil passes the vertical”.
so the forces keep turning the coil in the same direction (continuous rotation)	✓ 1	“Stops the wires tangling” or “changes a.c. to d.c.” scores 0.
Part (d)(i)
I = P ÷ V = 360 ÷ 36	✓ 1
= 10 A	✓ 1
Part (d)(ii)
Q = I × t = 10 × 25	✓ 1	Allow ECF from (d)(i).
= 250 C	✓ 1	Unit required.
Part (e)(i)
work done = force × distance = 48 × 150	✓ 1
= 7200 J	✓ 1
P = W ÷ t = 7200 ÷ 25 = 288 W (≈ 290 W)	✓ 1	The unrounded value 288 W must be seen. Alternative: speed = 150 ÷ 25 = 6.0 m/s (✓), power = force × speed = 48 × 6.0 (✓) = 288 W (✓).
Part (e)(ii)
360 − 288 = 72 W	✓ 1	Accept 70 W if 290 W is used. Allow ECF from (e)(i).
Part (f)(i)
17.5 A	✓ 1	Accept 17.2–17.8 A.
Part (f)(ii)
a larger force needs a larger current (17.5 A instead of 10 A)	✓ 1	Allow ECF from (f)(i).
a larger current transfers more energy per second as heat in the (resistance of the) coil, so its temperature rises (more)	✓ 1	Accept a description in terms of electrons colliding more often with ions in the wire.
Part (f)(iii)
use a stronger magnet	✓ 1	Any two, one mark each. Accept “move the poles closer together”.
use more turns on the coil	✓ 1	Do not accept “increase the current” or “increase the voltage”: the current must be smaller.

Answers: (d)(i) 10 A · (d)(ii) 250 C · (e)(i) 288 W · (e)(ii) 72 W · (f)(i) 17.5 A

Exam tip (a) “Explain why there is a force” needs the two fields: the field of the current and the field of the magnet interact. The left-hand rule only gives the direction.

More 6(c) questions in the question bank

+ Add to test✓ In your testCC-EIP-6c-219 marks~20 min

Inside a pair of headphones: the loudspeaker principle

6(c) Electromagnetism3(b) Properties of Waves3(d) Light and SoundBoth papers

Each earpiece of a pair of headphones contains a small driver that works like a loudspeaker. Figure 1 shows a cross-section through the driver. A coil fixed to a thin cone sits in the gap between the poles of a permanent magnet. The music player sends an alternating current through the coil.

The speed of sound in air is 340 m/s.

Figure 1: cross-section of a headphone driver at one instant (not to scale)

(a)

Describe how the alternating current in the coil makes the cone produce a sound wave.

EXPLAINNotes

Show solution

the current in the coil is in a magnetic field, so there is a force on the coil (motor effect)	✓ 1
the current keeps changing direction, so the force keeps reversing direction	✓ 1	The idea of the force changing direction is essential for this mark.
the coil and cone move in and out / vibrate, making the air vibrate (and a sound wave travels to the ear)	✓ 1	Accept “cone vibrates at the frequency of the current”.

(3)

(b)

Use Fleming’s left-hand rule to determine the direction of the force on the coil at the instant shown in Figure 1.

REASONNotes

Show solution

to the left / towards the magnet / away from the ear

✓ 1

Both wires give a force to the left. Accept “the cone is pulled in”.

Answer: to the left (away from the ear)

(1)

(c)(i)

The headphones play a note of frequency 500 Hz.

Calculate the wavelength of this sound in air.

CALCNotes

Show solution

λ = v ÷ f = 340 ÷ 500	✓ 1
= 0.68 m	✓ 1

Answer: 0.68 m

(2)

(c)(ii)

Show that the time period of the 500 Hz note is 2.0 ms.

DERIVENotes

Show solution

T = 1 ÷ f = 1 ÷ 500 = 0.0020 s	✓ 1
0.0020 s × 1000 = 2.0 ms	✓ 1	The conversion from s to ms must be shown.

Answer: 2.0 ms

(2)

(d)(i)

Figure 2 shows, on the same axes, how the current in the coil changes with time for two different notes, P and Q. The headphones play the notes one after the other.

Use Figure 2 to determine the frequency of note P.

Figure 2: current in the coil against time for two notes, P and Q

GRAPHNotes

Show solution

period of P read from the graph = 4.0 ms (0.0040 s)	✓ 1	Accept the time for two cycles (8.0 ms) divided by 2.
f = 1 ÷ 0.0040 = 250 Hz	✓ 1	Allow ECF from the period. Using 4.0 without converting (0.25 Hz) scores 1.

Answer: 250 Hz

(2)

(d)(ii)

Compare the sounds of notes P and Q heard by the listener. Explain your answers using Figure 2.

Paper 2 only onlyEXPLAINNotes

Show solution

Q has the higher pitch because it has the higher frequency (shorter period: 2.0 ms, 500 Hz)	✓ 1
P is louder	✓ 1
because P has the larger current amplitude, so a larger force and a larger amplitude of vibration of the cone	✓ 1	Must link current amplitude to the amplitude of vibration (or of the sound wave).

(3)

(e)

During a test, the driver is fed an alternating current of frequency 25 kHz. The cone vibrates, but the tester hears nothing. Explain why.

Paper 2 only onlyREASONNotes

Show solution

the human hearing range is 20 Hz to 20 000 Hz (20 kHz)	✓ 1
25 kHz is above the upper limit, so the sound (ultrasound) cannot be heard	✓ 1	Must compare 25 kHz with the upper limit.

(2)

(f)

The manufacturer replaces the magnet with a stronger magnet of the same size. The current from the music player is not changed.

Predict how the sound changes. Explain your answer.

PREDICTNotes

Show solution

stronger field gives a larger force on the coil for the same current	✓ 1
so the cone vibrates with a larger amplitude and the sound is louder (the pitch is unchanged)	✓ 1	Do not accept “higher pitch”.

(2)

(g)

A student repairing the headphones connects the two wires to the driver the other way round.

Explain why the music sounds the same to the listener.

REASONNotes

Show solution

the current (at each instant) is reversed, so the force on the coil is reversed at each instant	✓ 1
the cone still vibrates at the same frequency and with the same amplitude, so the pitch and loudness are unchanged	✓ 1	Accept “the force still alternates in the same way, only the timing of in and out is swapped”.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the current in the coil is in a magnetic field, so there is a force on the coil (motor effect)	✓ 1
the current keeps changing direction, so the force keeps reversing direction	✓ 1	The idea of the force changing direction is essential for this mark.
the coil and cone move in and out / vibrate, making the air vibrate (and a sound wave travels to the ear)	✓ 1	Accept “cone vibrates at the frequency of the current”.
Part (b)
to the left / towards the magnet / away from the ear	✓ 1	Both wires give a force to the left. Accept “the cone is pulled in”.
Part (c)(i)
λ = v ÷ f = 340 ÷ 500	✓ 1
= 0.68 m	✓ 1
Part (c)(ii)
T = 1 ÷ f = 1 ÷ 500 = 0.0020 s	✓ 1
0.0020 s × 1000 = 2.0 ms	✓ 1	The conversion from s to ms must be shown.
Part (d)(i)
period of P read from the graph = 4.0 ms (0.0040 s)	✓ 1	Accept the time for two cycles (8.0 ms) divided by 2.
f = 1 ÷ 0.0040 = 250 Hz	✓ 1	Allow ECF from the period. Using 4.0 without converting (0.25 Hz) scores 1.
Part (d)(ii)
Q has the higher pitch because it has the higher frequency (shorter period: 2.0 ms, 500 Hz)	✓ 1
P is louder	✓ 1
because P has the larger current amplitude, so a larger force and a larger amplitude of vibration of the cone	✓ 1	Must link current amplitude to the amplitude of vibration (or of the sound wave).
Part (e)
the human hearing range is 20 Hz to 20 000 Hz (20 kHz)	✓ 1
25 kHz is above the upper limit, so the sound (ultrasound) cannot be heard	✓ 1	Must compare 25 kHz with the upper limit.
Part (f)
stronger field gives a larger force on the coil for the same current	✓ 1
so the cone vibrates with a larger amplitude and the sound is louder (the pitch is unchanged)	✓ 1	Do not accept “higher pitch”.
Part (g)
the current (at each instant) is reversed, so the force on the coil is reversed at each instant	✓ 1
the cone still vibrates at the same frequency and with the same amplitude, so the pitch and loudness are unchanged	✓ 1	Accept “the force still alternates in the same way, only the timing of in and out is swapped”.

Answers: (b) to the left (away from the ear) · (c)(i) 0.68 m · (c)(ii) 2.0 ms · (d)(i) 250 Hz

Exam tip (a) Build the chain: alternating current → force that reverses → cone vibrates → air vibrates → sound.

More 6(c) questions in the question bank

6(d) Electromagnetic Induction

+ Add to test✓ In your testCC-EIP-6d-120 marks~20 min

A bicycle hub dynamo: generating a voltage

6(d) Electromagnetic Induction2(b) Mains Electricity2(c) Energy and Voltage in Circuits3(b) Properties of WavesBoth papers

Some bicycles have a dynamo built into the hub of the front wheel to power the lights. A ring-shaped magnet is fixed inside the hub, so it turns with the wheel. The magnet has 20 poles, arranged alternately N, S, N, S, … around the ring. A coil of insulated wire, wound on an iron core, is fixed to the axle inside the ring and does not turn. The ends of the coil are connected to the front lamp of the bicycle.

Figure 1 shows the arrangement.

Figure 1: the hub dynamo seen along the axle (simplified, not to scale)

(a)

Explain why an alternating voltage is induced across the coil while the wheel is turning.

EXPLAINNotes

Show solution

as the magnet turns, the magnetic field through the coil changes / field lines are cut by the coil	✓ 1	Do not accept 'the magnet is near the coil' or 'the coil is in a magnetic field' on its own.
a changing field (relative movement between magnet and coil) induces a voltage across the coil	✓ 1
N and S poles pass the coil in turn, so the field through the coil keeps reversing and the induced voltage reverses direction	✓ 1	Accept 'the poles alternate, so the voltage keeps changing direction'. Must link the reversal to the alternating poles.

(3)

(b)(i)

The lamp is disconnected and an oscilloscope is connected across the coil. Figure 2 shows how the voltage across the coil varies with time while the bicycle moves at a steady speed.

Write down the peak voltage.

Figure 2: voltage across the coil against time, lamp disconnected

GRAPHNotes

Show solution

8.0 V

✓ 1

Accept 8 V. Do not accept 16 V (that is the peak-to-peak value).

Answer: 8.0 V

(1)

(b)(ii)

Use Figure 2 to determine the frequency of the alternating voltage.

CALCNotes

Show solution

time period T = 25 ms = 0.025 s	✓ 1	Any complete cycle, e.g. 0 to 25 ms or 25 to 50 ms.
f = 1 ÷ T = 1 ÷ 0.025 = 40 Hz	✓ 1	Allow ECF from an incorrect period read from the graph.

Answer: 40 Hz

(2)

(b)(iii)

The voltage goes through 10 complete cycles for each turn of the wheel.

Calculate the number of turns the wheel makes each second.

CALCNotes

Show solution

40 ÷ 10 = 4.0 turns per second

✓ 1

Allow ECF from (b)(ii).

Answer: 4.0 turns per second

(1)

(c)

The cyclist slows down until the wheel makes 2.0 turns each second.

Describe and explain how the trace in Figure 2 changes. Include the new frequency in your answer.

PREDICTNotes

Show solution

new frequency = 10 × 2.0 = 20 Hz / the period doubles to 0.050 s	✓ 1	Allow ECF from (b)(iii): frequency halves.
the peak voltage is smaller (halved, to about 4 V)	✓ 1	Accept 'the amplitude decreases'.
because the poles pass the coil less often, so the field through the coil changes more slowly / field lines are cut at a lower rate	✓ 1	Do not accept 'fewer field lines are cut' without a reference to rate or time.

(3)

(d)(i)

The lamp is connected to the coil again. When the lamp works normally, the voltage across it is 6.0 V and its power is 3.0 W.

Calculate the current in the lamp.

CALCNotes

Show solution

I = P ÷ V = 3.0 ÷ 6.0	✓ 1	Accept P = I × V rearranged.
I = 0.50 A	✓ 1	Unit required.

Answer: 0.50 A

(2)

(d)(ii)

Calculate the resistance of the lamp filament when the lamp works normally.

CALCNotes

Show solution

R = V ÷ I = 6.0 ÷ 0.50	✓ 1	Allow ECF from (d)(i).
R = 12 Ω	✓ 1	Unit required.

Answer: 12 Ω

(2)

(e)

The cyclist rides home with the lamp working normally. The journey takes 24 minutes.

Show that about 700 C of charge passes through the lamp during the journey.

CALCNotes

Show solution

t = 24 × 60 = 1440 s	✓ 1
Q = I × t = 0.50 × 1440 = 720 C	✓ 1	The value 720 C (or 7.2 × 10² C) must be seen. Using 24 for the time (giving 12 C) loses the first mark.

Answer: 720 C

(2)

(f)

At very low speeds the lamp is dim.

Suggest two changes to the design of the dynamo that would give a larger voltage at the same low speed.

RECALLNotes

Show solution

use a stronger magnet	✓ 1	Accept 'move the magnet closer to the coil'. Accept 'more poles on the ring magnet' (the field through the coil then changes more quickly).
use more turns of wire on the coil	✓ 1	Do not accept 'add an iron core' (the coil already has one) or 'turn the wheel faster' (the speed is fixed).

(2)

(g)

The cyclist switches the lamp off, which breaks the lamp circuit. Pedalling then becomes slightly easier.

Explain why.

REASONNotes

Show solution

a voltage is still induced across the coil, but with the circuit broken there is no current	✓ 1	Do not accept 'no voltage is induced'.
so no energy is transferred electrically to the lamp (P = I × V = 0), so the cyclist needs to transfer less energy each second to turn the wheel	✓ 1	Accept a conservation of energy argument: less energy output, so less energy input needed.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
as the magnet turns, the magnetic field through the coil changes / field lines are cut by the coil	✓ 1	Do not accept 'the magnet is near the coil' or 'the coil is in a magnetic field' on its own.
a changing field (relative movement between magnet and coil) induces a voltage across the coil	✓ 1
N and S poles pass the coil in turn, so the field through the coil keeps reversing and the induced voltage reverses direction	✓ 1	Accept 'the poles alternate, so the voltage keeps changing direction'. Must link the reversal to the alternating poles.
Part (b)(i)
8.0 V	✓ 1	Accept 8 V. Do not accept 16 V (that is the peak-to-peak value).
Part (b)(ii)
time period T = 25 ms = 0.025 s	✓ 1	Any complete cycle, e.g. 0 to 25 ms or 25 to 50 ms.
f = 1 ÷ T = 1 ÷ 0.025 = 40 Hz	✓ 1	Allow ECF from an incorrect period read from the graph.
Part (b)(iii)
40 ÷ 10 = 4.0 turns per second	✓ 1	Allow ECF from (b)(ii).
Part (c)
new frequency = 10 × 2.0 = 20 Hz / the period doubles to 0.050 s	✓ 1	Allow ECF from (b)(iii): frequency halves.
the peak voltage is smaller (halved, to about 4 V)	✓ 1	Accept 'the amplitude decreases'.
because the poles pass the coil less often, so the field through the coil changes more slowly / field lines are cut at a lower rate	✓ 1	Do not accept 'fewer field lines are cut' without a reference to rate or time.
Part (d)(i)
I = P ÷ V = 3.0 ÷ 6.0	✓ 1	Accept P = I × V rearranged.
I = 0.50 A	✓ 1	Unit required.
Part (d)(ii)
R = V ÷ I = 6.0 ÷ 0.50	✓ 1	Allow ECF from (d)(i).
R = 12 Ω	✓ 1	Unit required.
Part (e)
t = 24 × 60 = 1440 s	✓ 1
Q = I × t = 0.50 × 1440 = 720 C	✓ 1	The value 720 C (or 7.2 × 10² C) must be seen. Using 24 for the time (giving 12 C) loses the first mark.
Part (f)
use a stronger magnet	✓ 1	Accept 'move the magnet closer to the coil'. Accept 'more poles on the ring magnet' (the field through the coil then changes more quickly).
use more turns of wire on the coil	✓ 1	Do not accept 'add an iron core' (the coil already has one) or 'turn the wheel faster' (the speed is fixed).
Part (g)
a voltage is still induced across the coil, but with the circuit broken there is no current	✓ 1	Do not accept 'no voltage is induced'.
so no energy is transferred electrically to the lamp (P = I × V = 0), so the cyclist needs to transfer less energy each second to turn the wheel	✓ 1	Accept a conservation of energy argument: less energy output, so less energy input needed.

Answers: (b)(i) 8.0 V · (b)(ii) 40 Hz · (b)(iii) 4.0 turns per second · (d)(i) 0.50 A · (d)(ii) 12 Ω · (e) 720 C

Exam tip (a) Write the chain in order: the magnet turns → the field through the coil changes → a voltage is induced. The word 'changing' (or 'cutting') must appear.

Common error (b)(ii) Using 25 for the period gives 0.04 Hz. Convert 25 ms to 0.025 s before using f = 1/T.

More 6(d) questions in the question bank

+ Add to test✓ In your testCC-EIP-6d-220 marks~20 min

A garden lighting transformer: stepping down the mains

6(d) Electromagnetic Induction2(b) Mains Electricity4(b) Energy TransfersPaper 2 only

A set of low-voltage garden lights is supplied from the 230 V a.c. mains through a transformer. The transformer is in a weatherproof box by the house and steps the voltage down to 24 V a.c. A cable carries the 24 V supply to the lamps along the garden path.

Figure 1 shows the transformer. The primary coil has 2070 turns.

Figure 1: the garden lighting transformer (not to scale)

(a)

State why the core of the transformer is made of soft iron.

Paper 2 only onlyRECALLNotes

Show solution

soft iron is magnetised and demagnetised easily (as the field keeps reversing)

✓ 1

Accept 'it links the magnetic field of the primary coil to the secondary coil'.

(1)

(b)

Explain how the transformer produces an alternating voltage across the secondary coil.

Paper 2 only onlyEXPLAINNotes

Show solution

the alternating current in the primary coil produces a magnetic field in the core	✓ 1
this field is constantly changing in size and direction	✓ 1	Accept 'alternating magnetic field'.
the changing field passes through the secondary coil and induces an alternating voltage across it	✓ 1	Do not accept 'the current passes through the core to the secondary coil'.

(3)

(c)

Calculate the number of turns on the secondary coil.

Paper 2 only onlyCALCNotes

Show solution

N_s = N_p × V_s ÷ V_p = 2070 × 24 ÷ 230	✓ 1	Accept V_p/V_s = N_p/N_s with correct substitution.
N_s = 216 (turns)	✓ 1

Answer: 216 turns

(2)

(d)

When all the lamps are on, the current in the secondary coil is 2.5 A. Assume that the transformer is 100% efficient.

Show that the current in the primary coil is about 0.26 A.

Paper 2 only onlyCALCNotes

Show solution

output power = V_s × I_s = 24 × 2.5 = 60 W	✓ 1
V_p × I_p = V_s × I_s, so 230 × I_p = 60	✓ 1	Alternative: current ratio = turns ratio inverted, I_p = 2.5 × 216 ÷ 2070 (allow ECF from (c)), scores both marks.
I_p = 0.261 A	✓ 1	Answer to at least 3 s.f. needed (0.261 A), since 0.26 A is given.

Answer: 0.261 A

(3)

(e)(i)

In fact, the current in the primary coil is 0.29 A when the output is 24 V and 2.5 A.

Calculate the efficiency of the transformer.

CALCNotes

Show solution

input power = 230 × 0.29 = 66.7 W	✓ 1	Accept 67 W.
efficiency = useful power output ÷ total power input (× 100%) = 60 ÷ 66.7	✓ 1	Allow ECF for the output power from (d).
efficiency = 0.90 or 90%	✓ 1	Accept 0.9 or 89.96%. Do not accept a value greater than 1 (100%).

Answer: 0.90 (90%)

(3)

(e)(ii)

The transformer becomes warm while the lamps are on.

Explain why.

EXPLAINNotes

Show solution

the current in the coils heats them: electrons collide with the metal ions, which vibrate more	✓ 1	Accept 'the coils (and core) have resistance so they heat up'.
the energy not transferred to the output (66.7 − 60 = 6.7 J each second) goes to the thermal store of the transformer and is dissipated to the surroundings	✓ 1	Accept 'wasted energy is transferred by heating to the surroundings'. Allow ECF from (e)(i).

(2)

(f)(i)

A student builds a model transformer. She connects its primary coil to a 5.0 V a.c. supply and changes the number of turns on the secondary coil, measuring the output voltage each time. Figure 2 shows her results.

Determine the gradient of the line in Figure 2.

Figure 2: output voltage of the model transformer against the number of turns on its secondary coil (primary supply 5.0 V a.c.)

Paper 2 only onlyGRAPHNotes

Show solution

uses a large triangle, e.g. 12.0 V ÷ 300 turns	✓ 1	Values read from the line, not from one data point near the origin.
gradient = 0.040 V per turn	✓ 1	Accept 0.038–0.042. A unit is not required.

Answer: 0.040 V per turn

(2)

(f)(ii)

Use your answer to (f)(i) to determine the number of turns on the primary coil of the model transformer.

Paper 2 only onlyMULTINotes

Show solution

the gradient is V_s ÷ N_s, which equals V_p ÷ N_p	✓ 1	Accept 'voltage per turn is the same for both coils' or use of a point on the line in the turns ratio equation.
N_p = 5.0 ÷ 0.040 = 125 (turns)	✓ 1	Allow ECF from (f)(i).

Answer: 125 turns

(2)

(g)

The owner takes the transformer and lamps to a holiday home in a country where the mains supply is 115 V a.c.

Calculate the output voltage of the transformer in that country, and suggest why the lamps are much dimmer.

Paper 2 only onlyPREDICTNotes

Show solution

V_s = 115 × 216 ÷ 2070 = 12 V	✓ 1	Allow ECF from (c). Accept 12.0 V or the idea that halving the input halves the output.
the lamps get only half of the 24 V they are designed for, so the current in them is smaller and their power is much less	✓ 1	Accept 'the power of each lamp is much less than its rated power'.

Answer: 12 V

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
soft iron is magnetised and demagnetised easily (as the field keeps reversing)	✓ 1	Accept 'it links the magnetic field of the primary coil to the secondary coil'.
Part (b)
the alternating current in the primary coil produces a magnetic field in the core	✓ 1
this field is constantly changing in size and direction	✓ 1	Accept 'alternating magnetic field'.
the changing field passes through the secondary coil and induces an alternating voltage across it	✓ 1	Do not accept 'the current passes through the core to the secondary coil'.
Part (c)
N_s = N_p × V_s ÷ V_p = 2070 × 24 ÷ 230	✓ 1	Accept V_p/V_s = N_p/N_s with correct substitution.
N_s = 216 (turns)	✓ 1
Part (d)
output power = V_s × I_s = 24 × 2.5 = 60 W	✓ 1
V_p × I_p = V_s × I_s, so 230 × I_p = 60	✓ 1	Alternative: current ratio = turns ratio inverted, I_p = 2.5 × 216 ÷ 2070 (allow ECF from (c)), scores both marks.
I_p = 0.261 A	✓ 1	Answer to at least 3 s.f. needed (0.261 A), since 0.26 A is given.
Part (e)(i)
input power = 230 × 0.29 = 66.7 W	✓ 1	Accept 67 W.
efficiency = useful power output ÷ total power input (× 100%) = 60 ÷ 66.7	✓ 1	Allow ECF for the output power from (d).
efficiency = 0.90 or 90%	✓ 1	Accept 0.9 or 89.96%. Do not accept a value greater than 1 (100%).
Part (e)(ii)
the current in the coils heats them: electrons collide with the metal ions, which vibrate more	✓ 1	Accept 'the coils (and core) have resistance so they heat up'.
the energy not transferred to the output (66.7 − 60 = 6.7 J each second) goes to the thermal store of the transformer and is dissipated to the surroundings	✓ 1	Accept 'wasted energy is transferred by heating to the surroundings'. Allow ECF from (e)(i).
Part (f)(i)
uses a large triangle, e.g. 12.0 V ÷ 300 turns	✓ 1	Values read from the line, not from one data point near the origin.
gradient = 0.040 V per turn	✓ 1	Accept 0.038–0.042. A unit is not required.
Part (f)(ii)
the gradient is V_s ÷ N_s, which equals V_p ÷ N_p	✓ 1	Accept 'voltage per turn is the same for both coils' or use of a point on the line in the turns ratio equation.
N_p = 5.0 ÷ 0.040 = 125 (turns)	✓ 1	Allow ECF from (f)(i).
Part (g)
V_s = 115 × 216 ÷ 2070 = 12 V	✓ 1	Allow ECF from (c). Accept 12.0 V or the idea that halving the input halves the output.
the lamps get only half of the 24 V they are designed for, so the current in them is smaller and their power is much less	✓ 1	Accept 'the power of each lamp is much less than its rated power'.

Answers: (c) 216 turns · (d) 0.261 A · (e)(i) 0.90 (90%) · (f)(i) 0.040 V per turn · (f)(ii) 125 turns · (g) 12 V

Common error (b) Saying that current flows through the core from one coil to the other. The coils are not electrically connected: only the changing magnetic field links them.

More 6(d) questions in the question bank

Topic 7 · Radioactivity and particles

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

Setup	7(b)	7(c)	Other
A paper mill thickness gauge: beta radiation in industry	●		3(c) 1(b)
A kidney scan with a gamma tracer: half-life and safety	●	○	3(c)
A nuclear-powered icebreaker: fission, control rods and the moderator	○	●	4(d) 4(b)
A tokamak fusion experiment: plasma temperature and fusion power	○	●	8(c) 5(d)

7(b) Radioactivity

+ Add to test✓ In your testCC-EIP-7b-120 marks~20 min

A paper mill thickness gauge: beta radiation in industry

7(b) Radioactivity3(c) The Electromagnetic Spectrum1(b) Movement and PositionBoth papers

A paper mill makes paper as one long, continuous sheet. To keep the thickness of the paper correct, the sheet passes between a radioactive source and a Geiger–Müller tube, as shown in Figure 1. The tube is connected to a counter and a control unit, which adjusts the gap between the rollers.

The source contains krypton-85, ⁸⁵₃₆Kr, which emits beta particles. Krypton-85 has a half-life of 10.8 years.

Figure 1: the thickness gauge (not to scale)

(a)(i)

State the number of protons and the number of neutrons in a nucleus of krypton-85.

RECALLNotes

Show solution

36 protons	✓ 1
85 − 36 = 49 neutrons	✓ 1

Answer: 36 protons, 49 neutrons

(2)

(a)(ii)

Krypton-85 decays by beta emission to an isotope of rubidium (Rb).

Complete the nuclear equation for this decay.

⁸⁵₃₆Kr → ^…_…Rb + ^…_…e

REASONNotes

Show solution

rubidium: ⁸⁵₃₇Rb	✓ 1	Both numbers needed.
beta particle: ⁰₋₁e	✓ 1	Accept β in place of e. The −1 must be shown.

Answer: ⁸⁵₃₆Kr → ⁸⁵₃₇Rb + ⁰₋₁e

(2)

(b)(i)

A gamma source was also considered for the gauge. Gamma rays are part of the electromagnetic spectrum.

Give one harmful effect on the human body of too much exposure to gamma rays.

RECALLNotes

Show solution

damage to cells inside the body, causing cancer / mutation of DNA

✓ 1

Accept 'kills cells' or 'radiation sickness'. Do not accept 'burns the skin' or 'it is harmful' on its own.

(1)

(b)(ii)

Explain why a source that emits only alpha particles, or only gamma rays, would not be suitable for this gauge.

EXPLAINNotes

Show solution

alpha would be absorbed completely by the paper, so only the background count would be detected whatever the thickness	✓ 1	Accept 'alpha cannot pass through paper'.
gamma would pass through the paper almost unchanged, so the count rate would hardly change when the thickness changes	✓ 1	Must link to the count rate (not) changing.

(2)

(c)

With no source present, the counter records a background count rate of 30 counts per minute. With the source in place and paper of the correct thickness passing through, the counter records 1230 counts per minute.

Figure 2 shows how the corrected count rate depends on the thickness of the paper.

Use Figure 2 to determine the correct thickness of the paper.

Figure 2: corrected count rate against thickness of paper for the krypton-85 source

GRAPHNotes

Show solution

corrected count rate = 1230 − 30 = 1200 counts per minute	✓ 1
thickness = 0.25 mm	✓ 1	Accept 0.24–0.26 mm. Allow ECF from an incorrect corrected count rate read correctly from the graph.

Answer: 0.25 mm

(2)

(d)

During production, the corrected count rate suddenly falls to 900 counts per minute.

Use Figure 2 to determine the new thickness of the paper, and explain how the control unit should adjust the rollers.

REASONNotes

Show solution

thickness = 0.35 mm	✓ 1	Accept 0.34–0.37 mm.
the paper is now too thick, so fewer beta particles pass through it	✓ 1	Accept 'thicker than 0.25 mm' (allow ECF from (c)).
the rollers are moved closer together / the gap is reduced (so the paper is made thinner)	✓ 1	Do not accept 'the rollers are moved apart'.

Answer: 0.35 mm; the gap between the rollers is reduced

(3)

(e)(i)

An engineer suggests replacing the krypton-85 with a beta emitter that has a half-life of 20 days.

Show that, 60 days after it is installed, the corrected count rate through paper of the correct thickness would have fallen to 150 counts per minute.

CALCNotes

Show solution

60 ÷ 20 = 3 half-lives	✓ 1
1200 → 600 → 300 → 150 counts per minute	✓ 1	Accept 1200 × ⅛ = 150. Allow 1200 ÷ 2³.

Answer: 150 counts per minute

(2)

(e)(ii)

Explain why this source would make the gauge set the wrong thickness, and why a half-life of 10.8 years avoids this problem.

EXPLAINNotes

Show solution

the count rate falls because of decay, not because of the thickness; the control unit would treat the fall as paper that is too thick and make the paper thinner	✓ 1
with a half-life of 10.8 years the activity hardly changes over weeks or months, so any change in count rate is caused by a change in thickness	✓ 1

(2)

(f)

The mill changes to making a thicker paper, 0.50 mm thick. The same krypton-85 source is used.

Use Figure 2 to determine the count rate that the counter should record when this paper has the correct thickness.

GRAPHNotes

Show solution

corrected count rate = 600 counts per minute (from the graph)	✓ 1	Accept 580–620.
recorded count rate = 600 + 30 = 630 counts per minute	✓ 1	Allow ECF from the graph reading. The background must be added back.

Answer: 630 counts per minute

(2)

(g)

The paper moves through the gauge at a constant speed of 12.5 m/s. When the thickness changes, the control unit takes 3.0 s to correct the gap between the rollers.

Calculate the length of paper that passes through the gauge in this time.

CALCNotes

Show solution

distance = speed × time = 12.5 × 3.0	✓ 1
distance = 37.5 m	✓ 1	Accept 38 m. Unit required.

Answer: 37.5 m

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
36 protons	✓ 1
85 − 36 = 49 neutrons	✓ 1
Part (a)(ii)
rubidium: ⁸⁵₃₇Rb	✓ 1	Both numbers needed.
beta particle: ⁰₋₁e	✓ 1	Accept β in place of e. The −1 must be shown.
Part (b)(i)
damage to cells inside the body, causing cancer / mutation of DNA	✓ 1	Accept 'kills cells' or 'radiation sickness'. Do not accept 'burns the skin' or 'it is harmful' on its own.
Part (b)(ii)
alpha would be absorbed completely by the paper, so only the background count would be detected whatever the thickness	✓ 1	Accept 'alpha cannot pass through paper'.
gamma would pass through the paper almost unchanged, so the count rate would hardly change when the thickness changes	✓ 1	Must link to the count rate (not) changing.
Part (c)
corrected count rate = 1230 − 30 = 1200 counts per minute	✓ 1
thickness = 0.25 mm	✓ 1	Accept 0.24–0.26 mm. Allow ECF from an incorrect corrected count rate read correctly from the graph.
Part (d)
thickness = 0.35 mm	✓ 1	Accept 0.34–0.37 mm.
the paper is now too thick, so fewer beta particles pass through it	✓ 1	Accept 'thicker than 0.25 mm' (allow ECF from (c)).
the rollers are moved closer together / the gap is reduced (so the paper is made thinner)	✓ 1	Do not accept 'the rollers are moved apart'.
Part (e)(i)
60 ÷ 20 = 3 half-lives	✓ 1
1200 → 600 → 300 → 150 counts per minute	✓ 1	Accept 1200 × ⅛ = 150. Allow 1200 ÷ 2³.
Part (e)(ii)
the count rate falls because of decay, not because of the thickness; the control unit would treat the fall as paper that is too thick and make the paper thinner	✓ 1
with a half-life of 10.8 years the activity hardly changes over weeks or months, so any change in count rate is caused by a change in thickness	✓ 1
Part (f)
corrected count rate = 600 counts per minute (from the graph)	✓ 1	Accept 580–620.
recorded count rate = 600 + 30 = 630 counts per minute	✓ 1	Allow ECF from the graph reading. The background must be added back.
Part (g)
distance = speed × time = 12.5 × 3.0	✓ 1
distance = 37.5 m	✓ 1	Accept 38 m. Unit required.

Answers: (a)(i) 36 protons, 49 neutrons · (a)(ii) ⁸⁵₃₆Kr → ⁸⁵₃₇Rb + ⁰₋₁e · (c) 0.25 mm · (d) 0.35 mm; the gap between the rollers is reduced · (e)(i) 150 counts per minute · (f) 630 counts per minute · (g) 37.5 m

Exam tip (b)(ii) For each rejected radiation, say what the paper does to it and what that means for the count rate. 'Alpha is too weak' does not explain anything.

Common error (f) Giving 600 counts per minute. The graph shows the corrected count rate; the counter also records the background, so add the 30 counts per minute back on.

More 7(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-7b-220 marks~20 min

A kidney scan with a gamma tracer: half-life and safety

7(b) Radioactivity3(c) The Electromagnetic Spectrum7(c) Fission and FusionBoth papers

A hospital uses a gamma camera to check how well a patient's kidneys are working. A small amount of a compound containing technetium-99m is injected into the patient's blood. The kidneys remove the compound from the blood and it leaves the body in the urine. Technetium-99m emits gamma rays only. The gamma camera above the patient detects the gamma rays and builds up an image of the kidneys, as shown in Figure 1.

1 MBq = 1 000 000 Bq.

Figure 1: a kidney scan with a gamma camera (not to scale)

(a)

Explain why a source that emits gamma rays, rather than alpha or beta particles, is used for this scan.

EXPLAINNotes

Show solution

gamma rays are very penetrating, so they pass out of the body and reach the camera; alpha or beta would be absorbed inside the body	✓ 1
gamma rays are weakly ionising, so they cause less damage to the patient's cells as they pass through	✓ 1	Do not accept 'gamma is not harmful'.

(2)

(b)(i)

Figure 2 shows how the activity of the injected technetium-99m changes with time. The tracer leaving the body in the urine is ignored.

Use Figure 2 to determine the half-life of technetium-99m. Show your working.

Figure 2: activity of the injected technetium-99m against time

GRAPHNotes

Show solution

reads the time for the activity to halve, e.g. 400 MBq → 200 MBq	✓ 1	Any pair of values where the second is half the first.
half-life = 6.0 h	✓ 1	Accept 5.8–6.2 h.

Answer: 6.0 h

(2)

(b)(ii)

Show that, 24 hours after the injection, the activity of the technetium-99m is less than 7% of its activity at the time of injection.

CALCNotes

Show solution

24 ÷ 6.0 = 4 half-lives	✓ 1	Allow ECF from (b)(i).
(½)⁴ = 1/16 = 0.0625 = 6.25% (or 400 → 200 → 100 → 50 → 25 MBq and 25/400 = 6.25%), which is less than 7%	✓ 1	The comparison with 7% must be made.

Answer: 1/16 (6.25%)

(2)

(c)

Explain why a half-life of a few hours makes technetium-99m suitable for this scan.

EXPLAINNotes

Show solution

long enough for the tracer to reach the kidneys and for the scan to be completed while the activity is still high enough to detect	✓ 1
short enough that the activity falls quickly after the scan, so the patient receives only a small dose of radiation	✓ 1

(2)

(d)

The gamma rays emitted by technetium-99m have a wavelength of 8.8 × 10⁻¹² m. The speed of electromagnetic waves is 3.0 × 10⁸ m/s.

Calculate the frequency of the gamma rays.

CALCNotes

Show solution

f = v ÷ λ = 3.0 × 10⁸ ÷ 8.8 × 10⁻¹²	✓ 1	Accept v = f × λ rearranged.
f = 3.4 × 10¹⁹ Hz	✓ 1	Accept 3.41 × 10¹⁹ Hz. Unit required.

Answer: 3.4 × 10¹⁹ Hz

(2)

(e)

After the scan, the patient is asked to drink plenty of water and to empty their bladder often for the rest of the day. The nurse who prepares the injection wears gloves and handles the syringe behind a lead screen.

Explain these precautions. Refer to contamination and irradiation in your answer.

EXPLAINNotes

Show solution

the patient is contaminated (the tracer is inside the body); drinking water and emptying the bladder remove the radioactive material sooner, so the patient's dose is smaller	✓ 1	Accept 'the patient is exposed for less time'.
gloves stop the tracer getting onto the nurse's skin, so the nurse is not contaminated	✓ 1
the lead screen absorbs most of the gamma rays from the syringe, so the nurse is irradiated less	✓ 1	Do not accept 'the screen stops all of the radiation'.

(3)

(f)

Another gamma-emitting tracer that would also collect in the kidneys has a half-life of 2 days.

Discuss whether this tracer would be better or worse than technetium-99m for the scan.

PREDICTNotes

Show solution

24 hours is less than one half-life, so more than half of its activity would remain the next day (compared with 1/16 for technetium-99m)	✓ 1
so the patient stays radioactive for longer and receives a larger dose (and irradiates other people for longer)	✓ 1	Accept 'more risk of cell damage / mutation'.
conclusion: worse, because the larger dose outweighs any advantage (such as it decaying less before it is used)	✓ 1	The conclusion must follow from at least one comparison.

(3)

(g)(i)

Technetium-99m is made at the hospital from molybdenum-99. Molybdenum-99 is one of the products of the fission of uranium-235 in a nuclear reactor. One possible fission reaction is

¹₀n + ²³⁵₉₂U → ⁹⁹₄₂Mo + ^A_ZSn + 3 ¹₀n

Determine the values of A and Z.

REASONNotes

Show solution

A: 1 + 235 = 99 + A + 3, so A = 134	✓ 1	Using 235 instead of 236 on the left (giving 133) scores 0 for this mark.
Z: 0 + 92 = 42 + Z + 0, so Z = 50	✓ 1

Answer: A = 134, Z = 50

(2)

(g)(ii)

Molybdenum-99 has a half-life of 66 hours. The hospital receives a new supply of molybdenum-99 every week.

Suggest why a new supply is needed so often.

REASONNotes

Show solution

one week (168 h) is more than two half-lives (2 × 66 = 132 h), so less than a quarter of the molybdenum-99 is left after a week	✓ 1	Accept 'about 2.5 half-lives'.
so it can no longer produce enough technetium-99m for scans; it cannot be stored for long and must be made again in a reactor	✓ 1

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
gamma rays are very penetrating, so they pass out of the body and reach the camera; alpha or beta would be absorbed inside the body	✓ 1
gamma rays are weakly ionising, so they cause less damage to the patient's cells as they pass through	✓ 1	Do not accept 'gamma is not harmful'.
Part (b)(i)
reads the time for the activity to halve, e.g. 400 MBq → 200 MBq	✓ 1	Any pair of values where the second is half the first.
half-life = 6.0 h	✓ 1	Accept 5.8–6.2 h.
Part (b)(ii)
24 ÷ 6.0 = 4 half-lives	✓ 1	Allow ECF from (b)(i).
(½)⁴ = 1/16 = 0.0625 = 6.25% (or 400 → 200 → 100 → 50 → 25 MBq and 25/400 = 6.25%), which is less than 7%	✓ 1	The comparison with 7% must be made.
Part (c)
long enough for the tracer to reach the kidneys and for the scan to be completed while the activity is still high enough to detect	✓ 1
short enough that the activity falls quickly after the scan, so the patient receives only a small dose of radiation	✓ 1
Part (d)
f = v ÷ λ = 3.0 × 10⁸ ÷ 8.8 × 10⁻¹²	✓ 1	Accept v = f × λ rearranged.
f = 3.4 × 10¹⁹ Hz	✓ 1	Accept 3.41 × 10¹⁹ Hz. Unit required.
Part (e)
the patient is contaminated (the tracer is inside the body); drinking water and emptying the bladder remove the radioactive material sooner, so the patient's dose is smaller	✓ 1	Accept 'the patient is exposed for less time'.
gloves stop the tracer getting onto the nurse's skin, so the nurse is not contaminated	✓ 1
the lead screen absorbs most of the gamma rays from the syringe, so the nurse is irradiated less	✓ 1	Do not accept 'the screen stops all of the radiation'.
Part (f)
24 hours is less than one half-life, so more than half of its activity would remain the next day (compared with 1/16 for technetium-99m)	✓ 1
so the patient stays radioactive for longer and receives a larger dose (and irradiates other people for longer)	✓ 1	Accept 'more risk of cell damage / mutation'.
conclusion: worse, because the larger dose outweighs any advantage (such as it decaying less before it is used)	✓ 1	The conclusion must follow from at least one comparison.
Part (g)(i)
A: 1 + 235 = 99 + A + 3, so A = 134	✓ 1	Using 235 instead of 236 on the left (giving 133) scores 0 for this mark.
Z: 0 + 92 = 42 + Z + 0, so Z = 50	✓ 1
Part (g)(ii)
one week (168 h) is more than two half-lives (2 × 66 = 132 h), so less than a quarter of the molybdenum-99 is left after a week	✓ 1	Accept 'about 2.5 half-lives'.
so it can no longer produce enough technetium-99m for scans; it cannot be stored for long and must be made again in a reactor	✓ 1

Answers: (b)(i) 6.0 h · (b)(ii) 1/16 (6.25%) · (d) 3.4 × 10¹⁹ Hz · (g)(i) A = 134, Z = 50

Exam tip (c) Give both sides of a half-life argument: why it is not too short (the scan can be done) and why it is not too long (a small dose).

More 7(b) questions in the question bank

7(c) Fission and Fusion

+ Add to test✓ In your testCC-EIP-7c-120 marks~20 min

A nuclear-powered icebreaker: fission, control rods and the moderator

7(c) Fission and Fusion7(b) Radioactivity4(d) Energy Resources and Electricity Generation4(b) Energy TransfersBoth papers

A nuclear-powered icebreaker ship has a pressurised water reactor. The fuel rods contain uranium, some of which is uranium-235, ²³⁵₉₂U. Inside a thick steel vessel, water surrounds the fuel rods. The water is the moderator and also the coolant: it is kept at high pressure so that it does not boil, and it carries thermal energy to a steam generator. Control rods containing boron can be raised or lowered between the fuel rods.

Figure 1 shows the reactor.

At full power, fission in the reactor produces 160 MW of thermal power. Each fission of a uranium-235 nucleus releases 3.2 × 10⁻¹¹ J of energy.

Figure 1: the reactor and its shielding (not to scale)

(a)

Describe what happens when a nucleus of uranium-235 absorbs a neutron.

RECALLNotes

Show solution

the nucleus becomes unstable and splits into two smaller (daughter) nuclei	✓ 1	Accept 'becomes uranium-236, which splits in two'. Do not accept 'the atom splits into two atoms'.
two or three neutrons are released, and energy is released as kinetic energy of the products	✓ 1	Both neutrons and energy needed. Accept 'the daughter nuclei and neutrons fly apart at high speed'.

(2)

(b)

In one fission, a uranium-235 nucleus absorbs a neutron and splits into caesium-140, ¹⁴⁰₅₅Cs, a nucleus X and three neutrons.

element	krypton	rubidium	strontium	yttrium	zirconium
proton number	36	37	38	39	40

Determine the nucleon number and the proton number of X. Use the table to state which element X is.

CALCNotes

Show solution

nucleon number: 1 + 235 = 140 + A + 3 × 1, so A = 93	✓ 1	Using 235 on the left (giving 92) scores 0 for this row.
proton number: 0 + 92 = 55 + Z + 3 × 0, so Z = 37	✓ 1
X is rubidium	✓ 1	Allow ECF from an incorrect proton number that is in the table.

Answer: nucleon number 93, proton number 37: rubidium (⁹³₃₇Rb)

(3)

(c)

The crew live and work close to the reactor compartment.

Explain why the reactor is surrounded by very thick concrete shielding.

EXPLAINNotes

Show solution

the core emits (large amounts of) ionising radiation, such as gamma rays and neutrons from fission and radiation from the radioactive daughter nuclei	✓ 1	Accept 'radiation' only if it is linked to damage in the next row.
the concrete absorbs the radiation, so very little reaches the crew, whose cells it would damage	✓ 1	Accept 'protects the crew from cell damage / cancer'.

(2)

(d)

Show that about 5 × 10¹⁸ uranium-235 nuclei undergo fission each second when the reactor is at full power.

DERIVENotes

Show solution

160 MW = 1.6 × 10⁸ W, so 1.6 × 10⁸ J of energy is released each second	✓ 1
number of fissions each second = 1.6 × 10⁸ ÷ 3.2 × 10⁻¹¹ = 5.0 × 10¹⁸	✓ 1	Working must be shown; the given value alone scores 0.

Answer: 5.0 × 10¹⁸ fissions each second

(2)

(e)(i)

As the ship approaches a harbour, the operators reduce the power of the reactor. Figure 2 shows how the thermal power changes after the control rods are lowered further into the core at time 2 min.

Use Figure 2 to determine the number of fissions each second once the power has become steady again.

Figure 2: thermal power of the reactor; the control rods are lowered further at time 2 min

GRAPHNotes

Show solution

steady power read from Figure 2 = 60 MW	✓ 1
60 × 10⁶ ÷ 3.2 × 10⁻¹¹ = 1.9 × 10¹⁸ (fissions each second)	✓ 1	Accept 1.875 × 10¹⁸. Allow (60 ÷ 160) × 5.0 × 10¹⁸. Allow ECF from the reading of the graph.

Answer: 1.9 × 10¹⁸ fissions each second

(2)

(e)(ii)

Explain why lowering the control rods further into the core reduces the thermal power.

EXPLAINNotes

Show solution

the control rods absorb (more) neutrons	✓ 1	Must name neutrons: 'absorb radiation / particles / energy' does not score.
so fewer neutrons are absorbed by uranium-235 nuclei, fewer fissions happen each second and less energy is released each second	✓ 1

(2)

(f)

The reactor is designed so that the chain reaction stops if the water is lost from the core, even if the control rods do not move.

Explain why losing the water stops the chain reaction.

PREDICTNotes

Show solution

the water is the moderator: it slows down the neutrons released by fission	✓ 1	Do not accept 'the water slows down the reaction'.
without the water the neutrons stay fast	✓ 1
fast neutrons are much less likely to be absorbed by uranium-235 nuclei, so (on average fewer than one neutron from each fission causes another fission and) the number of fissions falls until the chain reaction stops	✓ 1	Accept 'fewer fissions, so the chain reaction cannot continue'.

(3)

(g)(i)

The steam from the steam generator turns turbines that drive generators. The generators supply electricity to motors that turn the propellers.

Describe the energy transfers that take place from the steam to the electrical output of the generators.

Paper 2 only onlyRECALLNotes

Show solution

the steam pushes on the turbine blades, transferring energy mechanically to the kinetic store of the turbine	✓ 1	Accept 'thermal store of the steam to kinetic store of the turbine'.
the turbine turns the generator, which transfers the energy electrically (to the motors)	✓ 1	Do not accept 'the generator makes energy'.

(2)

(g)(ii)

Figure 3 is a Sankey diagram for the reactor plant at full power.

Calculate the rate at which energy is transferred to the sea water in the condenser, and the efficiency of the plant.

Figure 3: Sankey diagram for the reactor plant at full power (drawn to scale)

CALCNotes

Show solution

160 − 48 − 7 = 105 MW	✓ 1
efficiency = 48 ÷ 160 (× 100%) = 0.30 or 30%	✓ 1	Accept either form. An answer greater than 1 (or 100%) scores 0 for this row.

Answer: 105 MW; efficiency 0.30 (30%)

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the nucleus becomes unstable and splits into two smaller (daughter) nuclei	✓ 1	Accept 'becomes uranium-236, which splits in two'. Do not accept 'the atom splits into two atoms'.
two or three neutrons are released, and energy is released as kinetic energy of the products	✓ 1	Both neutrons and energy needed. Accept 'the daughter nuclei and neutrons fly apart at high speed'.
Part (b)
nucleon number: 1 + 235 = 140 + A + 3 × 1, so A = 93	✓ 1	Using 235 on the left (giving 92) scores 0 for this row.
proton number: 0 + 92 = 55 + Z + 3 × 0, so Z = 37	✓ 1
X is rubidium	✓ 1	Allow ECF from an incorrect proton number that is in the table.
Part (c)
the core emits (large amounts of) ionising radiation, such as gamma rays and neutrons from fission and radiation from the radioactive daughter nuclei	✓ 1	Accept 'radiation' only if it is linked to damage in the next row.
the concrete absorbs the radiation, so very little reaches the crew, whose cells it would damage	✓ 1	Accept 'protects the crew from cell damage / cancer'.
Part (d)
160 MW = 1.6 × 10⁸ W, so 1.6 × 10⁸ J of energy is released each second	✓ 1
number of fissions each second = 1.6 × 10⁸ ÷ 3.2 × 10⁻¹¹ = 5.0 × 10¹⁸	✓ 1	Working must be shown; the given value alone scores 0.
Part (e)(i)
steady power read from Figure 2 = 60 MW	✓ 1
60 × 10⁶ ÷ 3.2 × 10⁻¹¹ = 1.9 × 10¹⁸ (fissions each second)	✓ 1	Accept 1.875 × 10¹⁸. Allow (60 ÷ 160) × 5.0 × 10¹⁸. Allow ECF from the reading of the graph.
Part (e)(ii)
the control rods absorb (more) neutrons	✓ 1	Must name neutrons: 'absorb radiation / particles / energy' does not score.
so fewer neutrons are absorbed by uranium-235 nuclei, fewer fissions happen each second and less energy is released each second	✓ 1
Part (f)
the water is the moderator: it slows down the neutrons released by fission	✓ 1	Do not accept 'the water slows down the reaction'.
without the water the neutrons stay fast	✓ 1
fast neutrons are much less likely to be absorbed by uranium-235 nuclei, so (on average fewer than one neutron from each fission causes another fission and) the number of fissions falls until the chain reaction stops	✓ 1	Accept 'fewer fissions, so the chain reaction cannot continue'.
Part (g)(i)
the steam pushes on the turbine blades, transferring energy mechanically to the kinetic store of the turbine	✓ 1	Accept 'thermal store of the steam to kinetic store of the turbine'.
the turbine turns the generator, which transfers the energy electrically (to the motors)	✓ 1	Do not accept 'the generator makes energy'.
Part (g)(ii)
160 − 48 − 7 = 105 MW	✓ 1
efficiency = 48 ÷ 160 (× 100%) = 0.30 or 30%	✓ 1	Accept either form. An answer greater than 1 (or 100%) scores 0 for this row.

Answers: (b) nucleon number 93, proton number 37: rubidium (⁹³₃₇Rb) · (d) 5.0 × 10¹⁸ fissions each second · (e)(i) 1.9 × 10¹⁸ fissions each second · (g)(ii) 105 MW; efficiency 0.30 (30%)

Common error (b) A common error is to forget the neutron that starts the fission. The mass numbers on the left add up to 1 + 235 = 236, not 235.

Exam tip (e)(ii) Say exactly what the control rods do: they absorb neutrons. 'They control the reaction' does not explain anything.

More 7(c) questions in the question bank

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A tokamak fusion experiment: plasma temperature and fusion power

7(c) Fission and Fusion8(c) Stellar Evolution7(b) Radioactivity5(d) Ideal Gas MoleculesBoth papers

In a tokamak, a ring-shaped cloud of extremely hot, ionised gas (a plasma) is held away from the walls of a vacuum chamber by strong magnetic fields. The plasma contains deuterium nuclei, ²₁H, and tritium nuclei, ³₁H, which fuse in the reaction

²₁H + ³₁H → ⁴₂He + ¹₀n

Each of these reactions releases 2.82 × 10⁻¹² J of energy.

(a)

State what is meant by nuclear fusion, and state why energy is released in the reaction above.

RECALLNotes

Show solution

two small (light) nuclei join together to form a larger (heavier) nucleus	✓ 1	Do not accept 'atoms join' or 'molecules join'.
the mass of the products is less than the mass of the nuclei that joined, and this lost mass is released as energy	✓ 1	Accept 'mass is lost / converted to energy'.

(2)

(b)

Two deuterium nuclei can also fuse with each other:

²₁H + ²₁H → ³₁H + X

Determine the nucleon number and the proton number of X, and give the name of particle X.

CALCNotes

Show solution

nucleon number = 2 + 2 − 3 = 1 and proton number = 1 + 1 − 1 = 1	✓ 1	Both needed.
X is a proton	✓ 1	Accept ¹₁H, ¹₁p or 'hydrogen-1 nucleus'. Allow ECF from the numbers.

Answer: nucleon number 1, proton number 1: a proton

(2)

(c)

Explain why the nuclei only fuse if the plasma is at a very high temperature.

EXPLAINNotes

Show solution

nuclei are positively charged (they contain protons), so they repel each other	✓ 1	Accept 'electrostatic repulsion'.
at a very high temperature the nuclei have a lot of kinetic energy / move very fast	✓ 1
so they can overcome the repulsion and get close enough to fuse	✓ 1	Do not accept 'so they collide more often' alone.

(3)

(d)

The plasma is made from gas at 27 °C, which is then heated to an operating temperature of 1.8 × 10⁸ K.

Calculate how many times greater the average kinetic energy of the particles is at the operating temperature than at 27 °C.

CALCNotes

Show solution

27 °C = 27 + 273 = 300 K	✓ 1
average kinetic energy is proportional to kelvin temperature: 1.8 × 10⁸ ÷ 300 = 6.0 × 10⁵	✓ 1	Using 27 instead of 300 (6.7 × 10⁶) scores 0 for this row.

Answer: 6.0 × 10⁵ times greater

(2)

(e)

The fuel is a mixture of deuterium and tritium. 1.0 g of the fuel contains 1.2 × 10²³ deuterium nuclei and 1.2 × 10²³ tritium nuclei.

Show that the energy released when all the nuclei in 1.0 g of fuel fuse is about 3 × 10¹¹ J.

DERIVENotes

Show solution

each reaction uses one deuterium and one tritium nucleus, so there are 1.2 × 10²³ reactions	✓ 1	Using 2.4 × 10²³ reactions scores 0 for this row.
energy = 1.2 × 10²³ × 2.82 × 10⁻¹² = 3.4 × 10¹¹ J	✓ 1	Accept 3.38 × 10¹¹ J. The answer must be given to more than 1 s.f.

Answer: 3.4 × 10¹¹ J

(2)

(f)(i)

Figure 1 shows how the fusion power produced in the tokamak depends on the temperature of the plasma, when the amount of fuel in the plasma stays the same.

Describe how the fusion power changes as the temperature increases from 0 to 250 × 10⁶ K.

Figure 1: fusion power produced in the tokamak against the temperature of the plasma (same amount of fuel)

GRAPHNotes

Show solution

almost no fusion power below about 60 × 10⁶ K, then the power increases very rapidly (steeply) up to about 200 × 10⁶ K	✓ 1	Accept any temperatures in the ranges 40–80 × 10⁶ K and 160–210 × 10⁶ K.
at higher temperatures the power still increases but less steeply / starts to level off (reaching about 820 MW at 250 × 10⁶ K)	✓ 1	Do not accept 'the power is proportional to temperature'.

(2)

(f)(ii)

The tokamak normally operates at 180 × 10⁶ K. During one test, the plasma cools until its kelvin temperature has fallen by one third.

Use Figure 1 to determine the fusion power at the new temperature. Compare the decrease in fusion power with the decrease in temperature, and explain the difference.

PREDICTNotes

Show solution

new temperature = 120 × 10⁶ K, and the fusion power falls from 600 MW to 240 MW	✓ 1	Accept 230–250 MW and 590–610 MW.
the power falls by 60% (to 40% of its value), much more than the one-third fall in temperature	✓ 1	Allow ECF from the readings.
the average kinetic energy of the nuclei falls (by one third), so far fewer nuclei are moving fast enough to overcome the repulsion, and the number of fusions each second falls by much more	✓ 1	Accept 'fewer collisions have enough energy to fuse'.

(3)

(g)(i)

The core of the Sun has a temperature of about 1.5 × 10⁷ K, less than one tenth of the operating temperature of the tokamak. Yet hydrogen nuclei fuse there all the time.

Suggest why fusion can continue in the core of the Sun at this lower temperature.

REASONNotes

Show solution

the pressure / density in the core of the Sun is extremely high (because gravity pulls the Sun's huge mass inwards)	✓ 1	Accept 'much higher pressure than in the tokamak'.
so the nuclei are packed closely together and collide very often, so enough fusions take place each second	✓ 1	Do not accept 'the Sun is bigger' without reference to pressure or density.

(2)

(g)(ii)

The Sun is a main sequence star.

Describe what will happen to the Sun when the hydrogen in its core runs out.

RECALLNotes

Show solution

it will expand (and its surface cools) to become a red giant	✓ 1	Red supergiant does not score.
it loses its outer layers and its core becomes a white dwarf, which slowly cools	✓ 1	Supernova, neutron star or black hole scores 0 for this row.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
two small (light) nuclei join together to form a larger (heavier) nucleus	✓ 1	Do not accept 'atoms join' or 'molecules join'.
the mass of the products is less than the mass of the nuclei that joined, and this lost mass is released as energy	✓ 1	Accept 'mass is lost / converted to energy'.
Part (b)
nucleon number = 2 + 2 − 3 = 1 and proton number = 1 + 1 − 1 = 1	✓ 1	Both needed.
X is a proton	✓ 1	Accept ¹₁H, ¹₁p or 'hydrogen-1 nucleus'. Allow ECF from the numbers.
Part (c)
nuclei are positively charged (they contain protons), so they repel each other	✓ 1	Accept 'electrostatic repulsion'.
at a very high temperature the nuclei have a lot of kinetic energy / move very fast	✓ 1
so they can overcome the repulsion and get close enough to fuse	✓ 1	Do not accept 'so they collide more often' alone.
Part (d)
27 °C = 27 + 273 = 300 K	✓ 1
average kinetic energy is proportional to kelvin temperature: 1.8 × 10⁸ ÷ 300 = 6.0 × 10⁵	✓ 1	Using 27 instead of 300 (6.7 × 10⁶) scores 0 for this row.
Part (e)
each reaction uses one deuterium and one tritium nucleus, so there are 1.2 × 10²³ reactions	✓ 1	Using 2.4 × 10²³ reactions scores 0 for this row.
energy = 1.2 × 10²³ × 2.82 × 10⁻¹² = 3.4 × 10¹¹ J	✓ 1	Accept 3.38 × 10¹¹ J. The answer must be given to more than 1 s.f.
Part (f)(i)
almost no fusion power below about 60 × 10⁶ K, then the power increases very rapidly (steeply) up to about 200 × 10⁶ K	✓ 1	Accept any temperatures in the ranges 40–80 × 10⁶ K and 160–210 × 10⁶ K.
at higher temperatures the power still increases but less steeply / starts to level off (reaching about 820 MW at 250 × 10⁶ K)	✓ 1	Do not accept 'the power is proportional to temperature'.
Part (f)(ii)
new temperature = 120 × 10⁶ K, and the fusion power falls from 600 MW to 240 MW	✓ 1	Accept 230–250 MW and 590–610 MW.
the power falls by 60% (to 40% of its value), much more than the one-third fall in temperature	✓ 1	Allow ECF from the readings.
the average kinetic energy of the nuclei falls (by one third), so far fewer nuclei are moving fast enough to overcome the repulsion, and the number of fusions each second falls by much more	✓ 1	Accept 'fewer collisions have enough energy to fuse'.
Part (g)(i)
the pressure / density in the core of the Sun is extremely high (because gravity pulls the Sun's huge mass inwards)	✓ 1	Accept 'much higher pressure than in the tokamak'.
so the nuclei are packed closely together and collide very often, so enough fusions take place each second	✓ 1	Do not accept 'the Sun is bigger' without reference to pressure or density.
Part (g)(ii)
it will expand (and its surface cools) to become a red giant	✓ 1	Red supergiant does not score.
it loses its outer layers and its core becomes a white dwarf, which slowly cools	✓ 1	Supernova, neutron star or black hole scores 0 for this row.

Answers: (b) nucleon number 1, proton number 1: a proton · (d) 6.0 × 10⁵ times greater · (e) 3.4 × 10¹¹ J

Exam tip (c) Link each step: positive nuclei repel, high temperature means high kinetic energy, so the nuclei get close enough to fuse.

Common error (d) A common error is to divide by the Celsius temperature. Kinetic energy is proportional to temperature only on the kelvin scale.

More 7(c) questions in the question bank

Topic 8 · Astrophysics

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

Setup	8(b)	8(c)	8(d)	Other
Satellite navigation: orbits, speed and signal timing	●			3(c) 1(b)
A helicopter on Mars: weight, energy and the orbit of Phobos	●			1(c) 4(c)
Betelgeuse: the life of a massive star	○	●		7(c)
Surveying an open star cluster: the Hertzsprung–Russell diagram	○	●		7(c)
A galaxy red-shift survey: evidence for an expanding universe			●	3(b) 3(c)
Detecting the cosmic microwave background with a radio telescope			●	3(c) 5(d) 3(b)

8(b) Motion in the Universe

+ Add to test✓ In your testCC-EIP-8b-119 marks~20 min

Satellite navigation: orbits, speed and signal timing

8(b) Motion in the Universe3(c) The Electromagnetic Spectrum1(b) Movement and PositionBoth papers

A satellite navigation system uses satellites in circular orbits around the Earth. Each satellite sends out signals that carry the exact time at which they left the satellite. A receiver on the ground uses the time each signal takes to arrive to work out its distance from that satellite.

One navigation satellite has an orbital radius of 26 600 km and a time period of 12.0 hours. The radius of the Earth is 6370 km. Figure 1 shows the satellite at a moment when it is directly above a receiver.

The speed of electromagnetic waves is 3.0 × 10⁸ m/s.

Figure 1: the navigation satellite directly above a receiver (drawn to scale)

(a)

State the force that keeps the satellite in its orbit, and the direction in which this force acts.

RECALLNotes

Show solution

gravitational force (gravity) of the Earth on the satellite	✓ 1	Do not accept 'centrifugal force' or 'magnetic force'.
towards the centre of the Earth	✓ 1	Accept 'towards the Earth'. Do not accept 'downwards' alone.

(2)

(b)

Show that the orbital speed of the satellite is about 3.9 km/s.

DERIVENotes

Show solution

T = 12.0 × 3600 = 43 200 s and r = 2.66 × 10⁷ m	✓ 1	Both conversions needed.
v = 2 × π × r ÷ T = 2 × π × 2.66 × 10⁷ ÷ 43 200	✓ 1
v = 3870 m/s (= 3.87 km/s)	✓ 1	The answer must be given to at least 3 s.f., or with the full substitution seen; 3.9 km/s alone does not score this row.

Answer: 3870 m/s (3.87 km/s)

(3)

(c)(i)

The signal from the satellite has a frequency of 1.58 × 10⁹ Hz.

Calculate the wavelength of the signal, and state the region of the electromagnetic spectrum it belongs to.

CALCNotes

Show solution

λ = v ÷ f = 3.0 × 10⁸ ÷ 1.58 × 10⁹ = 0.19 m	✓ 1	Accept 0.190 m.
microwaves	✓ 1	Do not accept 'radio waves'.

Answer: 0.19 m; microwaves

(2)

(c)(ii)

Calculate the time taken for a signal to travel from the satellite to the receiver in Figure 1.

MULTINotes

Show solution

distance = 26 600 − 6370 = 20 230 km = 2.023 × 10⁷ m	✓ 1
time = distance ÷ speed = 2.023 × 10⁷ ÷ 3.0 × 10⁸	✓ 1	Allow ECF for an incorrect distance, including 2.66 × 10⁷ m.
time = 0.067 s	✓ 1	Using the orbital radius (0.089 s) scores the second row only.

Answer: 0.067 s

(3)

(d)

An error of only 1.0 microsecond (1.0 × 10⁻⁶ s) in the time measured by the receiver makes the position it calculates useless for finding a street.

Explain why, using a calculation.

EXPLAINNotes

Show solution

distance = speed × time = 3.0 × 10⁸ × 1.0 × 10⁻⁶ = 300 m	✓ 1
the signals travel so fast that the calculated distance to the satellite (and so the position) is wrong by 300 m, which is far larger than the width of a street	✓ 1	Accept any sensible comparison showing that 300 m is a large error on the ground.

Answer: the distance is wrong by 300 m

(2)

(e)(i)

Figure 2 shows the orbital speed and orbital radius of four satellites, A, B, C and D, with a curve drawn through the points. Satellite C is the navigation satellite.

Describe how the orbital speed of a satellite depends on its orbital radius.

Figure 2: orbital speed against orbital radius for satellites A, B, C and D in circular orbits around the Earth

GRAPHNotes

Show solution

the orbital speed decreases as the orbital radius increases	✓ 1
the curve gets less steep: the speed falls by smaller amounts for equal increases in radius at larger radii	✓ 1	Accept a correct numerical comment, e.g. doubling the radius from 2 × 10⁷ m to 4 × 10⁷ m reduces the speed from about 4.5 km/s to about 3.2 km/s, which is not half, so the speed is not inversely proportional to the radius.

(2)

(e)(ii)

Another navigation satellite is moved into a lower circular orbit, of radius 2.0 × 10⁷ m.

Use Figure 2 to find its new orbital speed. Calculate its new time period in hours.

MULTINotes

Show solution

v = 4.5 km/s from Figure 2	✓ 1	Accept 4.4–4.6 km/s.
T = 2 × π × r ÷ v = 2 × π × 2.0 × 10⁷ ÷ 4500 = 2.8 × 10⁴ s	✓ 1	Allow ECF from the graph reading.
T = 2.8 × 10⁴ ÷ 3600 = 7.8 hours	✓ 1	Accept 7.6–7.9 hours. Allow ECF.

Answer: 4.5 km/s; 7.8 hours

(3)

(f)

The navigation satellite moves at a constant speed. A comet also moves in an orbit, around the Sun.

Describe the shape of a comet's orbit, and how the speed of the comet changes as it moves around its orbit.

RECALLNotes

Show solution

the orbit is highly elliptical (very elongated), with the Sun near one end	✓ 1
the comet moves fastest when it is closest to the Sun and slowest when it is furthest from the Sun	✓ 1	Both ends needed.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
gravitational force (gravity) of the Earth on the satellite	✓ 1	Do not accept 'centrifugal force' or 'magnetic force'.
towards the centre of the Earth	✓ 1	Accept 'towards the Earth'. Do not accept 'downwards' alone.
Part (b)
T = 12.0 × 3600 = 43 200 s and r = 2.66 × 10⁷ m	✓ 1	Both conversions needed.
v = 2 × π × r ÷ T = 2 × π × 2.66 × 10⁷ ÷ 43 200	✓ 1
v = 3870 m/s (= 3.87 km/s)	✓ 1	The answer must be given to at least 3 s.f., or with the full substitution seen; 3.9 km/s alone does not score this row.
Part (c)(i)
λ = v ÷ f = 3.0 × 10⁸ ÷ 1.58 × 10⁹ = 0.19 m	✓ 1	Accept 0.190 m.
microwaves	✓ 1	Do not accept 'radio waves'.
Part (c)(ii)
distance = 26 600 − 6370 = 20 230 km = 2.023 × 10⁷ m	✓ 1
time = distance ÷ speed = 2.023 × 10⁷ ÷ 3.0 × 10⁸	✓ 1	Allow ECF for an incorrect distance, including 2.66 × 10⁷ m.
time = 0.067 s	✓ 1	Using the orbital radius (0.089 s) scores the second row only.
Part (d)
distance = speed × time = 3.0 × 10⁸ × 1.0 × 10⁻⁶ = 300 m	✓ 1
the signals travel so fast that the calculated distance to the satellite (and so the position) is wrong by 300 m, which is far larger than the width of a street	✓ 1	Accept any sensible comparison showing that 300 m is a large error on the ground.
Part (e)(i)
the orbital speed decreases as the orbital radius increases	✓ 1
the curve gets less steep: the speed falls by smaller amounts for equal increases in radius at larger radii	✓ 1	Accept a correct numerical comment, e.g. doubling the radius from 2 × 10⁷ m to 4 × 10⁷ m reduces the speed from about 4.5 km/s to about 3.2 km/s, which is not half, so the speed is not inversely proportional to the radius.
Part (e)(ii)
v = 4.5 km/s from Figure 2	✓ 1	Accept 4.4–4.6 km/s.
T = 2 × π × r ÷ v = 2 × π × 2.0 × 10⁷ ÷ 4500 = 2.8 × 10⁴ s	✓ 1	Allow ECF from the graph reading.
T = 2.8 × 10⁴ ÷ 3600 = 7.8 hours	✓ 1	Accept 7.6–7.9 hours. Allow ECF.
Part (f)
the orbit is highly elliptical (very elongated), with the Sun near one end	✓ 1
the comet moves fastest when it is closest to the Sun and slowest when it is furthest from the Sun	✓ 1	Both ends needed.

Answers: (b) 3870 m/s (3.87 km/s) · (c)(i) 0.19 m; microwaves · (c)(ii) 0.067 s · (d) the distance is wrong by 300 m · (e)(ii) 4.5 km/s; 7.8 hours

Common error (c)(ii) A common error is to use the orbital radius as the distance to the receiver. The orbital radius is measured from the centre of the Earth, so the radius of the Earth must be subtracted.

More 8(b) questions in the question bank

+ Add to test✓ In your testCC-EIP-8b-220 marks~20 min

A helicopter on Mars: weight, energy and the orbit of Phobos

8(b) Motion in the Universe1(c) Forces, Movement, Shape and Momentum4(c) Work and PowerBoth papers

A rover exploring the surface of Mars carries a small helicopter of mass 1.8 kg. The helicopter flies by spinning its rotor blades very fast in the thin atmosphere of Mars.

gravitational field strength at the surface of the Earth = 10 N/kg
gravitational field strength at the surface of Mars = 3.7 N/kg

(a)

Calculate the weight of the helicopter on the Earth and on Mars.

CALCNotes

Show solution

on the Earth: W = m × g = 1.8 × 10 = 18 N	✓ 1	Accept 17.6 N (using 9.8 N/kg).
on Mars: W = 1.8 × 3.7 = 6.7 N	✓ 1	Accept 6.66 N.

Answer: 18 N on the Earth; 6.7 N on Mars

(2)

(b)

Mars has about one tenth of the mass of the Earth, and its radius is about half the radius of the Earth.

Explain why the gravitational field strength at the surface of Mars is less than at the surface of the Earth.

EXPLAINNotes

Show solution

Mars has a much smaller mass, so it pulls less strongly on each kilogram of mass near it	✓ 1	Do not accept 'Mars is smaller' without reference to mass.
the smaller radius (surface closer to the centre) on its own would make g larger, but the effect of the much smaller mass is greater	✓ 1	Accept 'g depends on both the mass and the distance from the centre, and the mass effect wins'.

(2)

(c)

On one flight, the helicopter rises vertically at a constant speed. Ignore air resistance.

State the size of the upward force on the helicopter from its rotors during this part of the flight. Explain your answer.

REASONNotes

Show solution

6.7 N	✓ 1	Allow ECF from (a).
the speed is constant, so the resultant force is zero and the upward force balances (equals) the weight	✓ 1	Do not accept 'the upward force must be bigger than the weight to rise'.

Answer: 6.7 N

(2)

(d)

Figure 1 shows how the gravitational potential energy (GPE) of the helicopter changes with its height above the ground on Mars.

The helicopter climbs from the ground to a height of 10 m in 4.0 s. Use Figure 1 to find the GPE gained, and calculate the useful power output of the rotors during the climb.

Figure 1: gravitational potential energy (GPE) of the helicopter against its height above the ground on Mars

CALCNotes

Show solution

GPE gained = 67 J (from Figure 1)	✓ 1	Accept 66–68 J.
P = W ÷ t = 67 ÷ 4.0	✓ 1
P = 17 W	✓ 1	Accept 16–17 W. Allow ECF from the reading.

Answer: 67 J; 17 W

(3)

(e)

Explain why the gradient of the line in Figure 1 is equal to the weight of the helicopter on Mars.

GRAPHNotes

Show solution

GPE = m × g × h, so the gradient GPE ÷ h = m × g	✓ 1
m × g is the weight (W = m × g); the gradient is 67 J ÷ 10 m = 6.7 N, the same as the weight on Mars	✓ 1	Accept 'the GPE gained equals the work done against the weight, weight × height'.

(2)

(f)(i)

Mars has a small moon, Phobos. Figure 2 shows the orbit of Phobos around Mars. The orbital radius of Phobos is 9380 km and its time period is 7.65 hours.

Show that the orbital speed of Phobos is about 2.1 km/s.

Figure 2: the orbit of Phobos around Mars (drawn to scale)

DERIVENotes

Show solution

r = 9.38 × 10⁶ m and T = 7.65 × 3600 = 27 540 s	✓ 1	Both conversions needed.
v = 2 × π × r ÷ T = 2 × π × 9.38 × 10⁶ ÷ 27 540	✓ 1
v = 2140 m/s (2.14 km/s)	✓ 1	The answer must be given to at least 3 s.f., or with the full substitution seen.

Answer: 2140 m/s (2.14 km/s)

(3)

(f)(ii)

Phobos orbits Mars, and Mars orbits the Sun.

Describe how the Sun, the Solar System and the Milky Way galaxy are related.

RECALLNotes

Show solution

the Sun is a star; the Solar System is the Sun and everything that orbits it (planets such as Mars, their moons such as Phobos, comets)	✓ 1
the Solar System is a tiny part of the Milky Way galaxy, which is a collection of billions of stars	✓ 1	Do not accept 'the Milky Way is a solar system'.

(2)

(g)(i)

The same helicopter is taken to the Moon, where the gravitational field strength is 1.6 N/kg.

Calculate the GPE the helicopter would gain if it were raised through 10 m on the Moon. Describe how a graph of GPE against height on the Moon would differ from Figure 1.

PREDICTNotes

Show solution

GPE = 1.8 × 1.6 × 10 = 29 J	✓ 1	Accept 28.8 J.
it would still be a straight line through the origin, but less steep (gradient 2.9 J/m instead of 6.7 J/m)	✓ 1	Accept 'gradient less than half as steep'.

Answer: 29 J; a straight line through the origin but less steep

(2)

(g)(ii)

Even though its weight would be smaller, the helicopter could not fly on the Moon.

Explain why.

Paper 2 only onlyEXPLAINNotes

Show solution

the rotor blades push air downwards, and (by Newton's third law) the air pushes the blades upwards with an equal and opposite force	✓ 1
the Moon has no atmosphere, so there is no air for the blades to push on and no upward force is produced	✓ 1	Do not accept 'there is no gravity on the Moon'.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
on the Earth: W = m × g = 1.8 × 10 = 18 N	✓ 1	Accept 17.6 N (using 9.8 N/kg).
on Mars: W = 1.8 × 3.7 = 6.7 N	✓ 1	Accept 6.66 N.
Part (b)
Mars has a much smaller mass, so it pulls less strongly on each kilogram of mass near it	✓ 1	Do not accept 'Mars is smaller' without reference to mass.
the smaller radius (surface closer to the centre) on its own would make g larger, but the effect of the much smaller mass is greater	✓ 1	Accept 'g depends on both the mass and the distance from the centre, and the mass effect wins'.
Part (c)
6.7 N	✓ 1	Allow ECF from (a).
the speed is constant, so the resultant force is zero and the upward force balances (equals) the weight	✓ 1	Do not accept 'the upward force must be bigger than the weight to rise'.
Part (d)
GPE gained = 67 J (from Figure 1)	✓ 1	Accept 66–68 J.
P = W ÷ t = 67 ÷ 4.0	✓ 1
P = 17 W	✓ 1	Accept 16–17 W. Allow ECF from the reading.
Part (e)
GPE = m × g × h, so the gradient GPE ÷ h = m × g	✓ 1
m × g is the weight (W = m × g); the gradient is 67 J ÷ 10 m = 6.7 N, the same as the weight on Mars	✓ 1	Accept 'the GPE gained equals the work done against the weight, weight × height'.
Part (f)(i)
r = 9.38 × 10⁶ m and T = 7.65 × 3600 = 27 540 s	✓ 1	Both conversions needed.
v = 2 × π × r ÷ T = 2 × π × 9.38 × 10⁶ ÷ 27 540	✓ 1
v = 2140 m/s (2.14 km/s)	✓ 1	The answer must be given to at least 3 s.f., or with the full substitution seen.
Part (f)(ii)
the Sun is a star; the Solar System is the Sun and everything that orbits it (planets such as Mars, their moons such as Phobos, comets)	✓ 1
the Solar System is a tiny part of the Milky Way galaxy, which is a collection of billions of stars	✓ 1	Do not accept 'the Milky Way is a solar system'.
Part (g)(i)
GPE = 1.8 × 1.6 × 10 = 29 J	✓ 1	Accept 28.8 J.
it would still be a straight line through the origin, but less steep (gradient 2.9 J/m instead of 6.7 J/m)	✓ 1	Accept 'gradient less than half as steep'.
Part (g)(ii)
the rotor blades push air downwards, and (by Newton's third law) the air pushes the blades upwards with an equal and opposite force	✓ 1
the Moon has no atmosphere, so there is no air for the blades to push on and no upward force is produced	✓ 1	Do not accept 'there is no gravity on the Moon'.

Answers: (a) 18 N on the Earth; 6.7 N on Mars · (c) 6.7 N · (d) 67 J; 17 W · (f)(i) 2140 m/s (2.14 km/s) · (g)(i) 29 J; a straight line through the origin but less steep

Exam tip (b) When explaining a value of g, refer to both the mass of the planet and the distance from its centre.

More 8(b) questions in the question bank

8(c) Stellar Evolution

+ Add to test✓ In your testCC-EIP-8c-120 marks~20 min

Betelgeuse: the life of a massive star

8(c) Stellar Evolution7(c) Fission and Fusion8(b) Motion in the UniverseBoth papers

Betelgeuse and Rigel are two of the brightest stars in the constellation of Orion.

Betelgeuse is a red supergiant. Its mass is about 18 times the mass of the Sun and its surface temperature is about 3500 K.
Rigel is a blue-white star with a surface temperature of about 12 000 K.

Both stars, like the Sun, are in the Milky Way galaxy.

(a)

Explain what the colours of Betelgeuse and Rigel tell you about their surface temperatures.

RECALLNotes

Show solution

the colour of a star depends on its surface temperature / red stars are the coolest and blue stars the hottest	✓ 1	Accept the order red → orange → yellow → white → blue as the order of increasing surface temperature.
Rigel has a higher surface temperature than Betelgeuse / Betelgeuse is cooler than Rigel	✓ 1	A direct comparison of the two stars is needed. “Red means hot” scores 0 for this row.

Answer: Rigel (blue-white) is hotter than Betelgeuse (red)

(2)

(b)(i)

Before it became a red supergiant, Betelgeuse was a main sequence star.

Describe how energy was released in the core of Betelgeuse while it was a main sequence star.

RECALLNotes

Show solution

(nuclear) fusion: hydrogen nuclei join together to make helium (nuclei)	✓ 1	“Hydrogen fuses into helium” is enough. Do not accept fission, burning or a chemical reaction.
the mass of the product is less than the mass of the nuclei that joined / mass is lost and released as energy	✓ 1	Accept “the lost mass is converted into energy”.

(2)

(b)(ii)

Explain why this process can only take place where the temperature and the pressure are extremely high, such as in the core of a star.

EXPLAINNotes

Show solution

nuclei are positively charged (contain protons), so they repel each other	✓ 1	Accept electrostatic repulsion between the nuclei.
at a very high temperature the nuclei move very fast / have enough kinetic energy to overcome the repulsion and get close enough to fuse	✓ 1	The link between high temperature and speed (or kinetic energy) of the nuclei is needed.
at a very high pressure (density) the nuclei are close together, so collisions happen often enough for fusion to continue	✓ 1	Accept “more frequent collisions”.

(3)

(c)

Betelgeuse has used up the hydrogen in its core.

Describe what is expected to happen to Betelgeuse in the future.

PREDICTNotes

Show solution

when fusion in the core can no longer continue, the core collapses (suddenly)	✓ 1	Accept “it runs out of fuel and collapses”.
the star explodes as a supernova	✓ 1	Accept the material thrown out forms a new nebula as an extra detail, not instead of this row.
the core left behind becomes a neutron star or (if massive enough) a black hole	✓ 1	Either answer, or both. Do not accept white dwarf.

Answer: core collapses → supernova → neutron star or black hole

(3)

(d)

A student says:

“In about 5 billion years the Sun will use up the hydrogen in its core. Then it will also explode as a supernova, just like Betelgeuse.”

Comment on the student’s statement.

REASONNotes

Show solution

the student is wrong: the Sun does not have enough mass to become a supernova / only stars much more massive than the Sun explode as supernovae	✓ 1	A reference to the Sun’s (smaller) mass is needed for this mark.
the Sun will become a red giant, then lose its outer layers gently and leave a white dwarf	✓ 1	Both red giant and white dwarf needed. Do not accept red supergiant or neutron star.

(2)

(e)(i)

Figure 1 is a Hertzsprung–Russell (HR) diagram on which the Sun and Betelgeuse have been plotted.

State what is meant by the absolute magnitude of a star.

Figure 1: HR diagram showing the Sun and Betelgeuse (the temperature scale halves at each labelled step)

Paper 2 only onlyRECALLNotes

Show solution

how bright the star would appear if it were at a standard distance from the Earth

✓ 1

The idea of every star being compared at the same (standard) distance is needed. “How bright it looks” alone scores 0.

(1)

(e)(ii)

Use Figure 1 to determine the absolute magnitude of Betelgeuse.

Paper 2 only onlyGRAPHNotes

Show solution

−6

✓ 1

Accept −5.5 to −6.5. The minus sign is needed.

Answer: −6

(1)

(e)(iii)

Betelgeuse has a lower surface temperature than the Sun, but Figure 1 shows that it gives out far more light.

Explain how this is possible.

Paper 2 only onlyEXPLAINNotes

Show solution

Betelgeuse is very much larger than the Sun (it is a supergiant)	✓ 1	Accept “huge”, “enormous radius / surface”.
so although each part of its surface is cooler / redder, its total light output is much greater (smaller, more negative absolute magnitude)	✓ 1	Must link size to the total light given out. A reference to its distance from the Earth scores 0: absolute magnitude already removes the effect of distance.

(2)

(f)(i)

Betelgeuse is 5.2 × 10¹⁸ m from the Earth.

speed of light = 3.0 × 10⁸ m/s; 1 year = 3.15 × 10⁷ s

Show that light from Betelgeuse takes about 550 years to reach the Earth.

DERIVENotes

Show solution

time = distance ÷ speed	✓ 1	Accept the equation in symbols or with numbers substituted.
time = 5.2 × 10¹⁸ ÷ 3.0 × 10⁸ = 1.73 × 10¹⁰ s	✓ 1	Accept 1.7 × 10¹⁰ s.
1.73 × 10¹⁰ ÷ 3.15 × 10⁷ = 550.3 years (≈ 550 years)	✓ 1	The conversion to years must be shown, with an answer to more significant figures than the given 550 (e.g. 550.3 or 550.26).

Answer: 1.73 × 10¹⁰ s ≈ 550 years

(3)

(f)(ii)

When Betelgeuse explodes, for a short time it will be one of the brightest objects in the night sky.

Explain why astronomers say that Betelgeuse might already have exploded, even though it still looks the same tonight.

REASONNotes

Show solution

the light we see now left Betelgeuse about 550 years ago, so light from an explosion in the last 550 years has not reached the Earth yet

✓ 1

Both ideas (light takes about 550 years / we see the star as it was then) in one statement for the mark.

(1)

Show full solution

Marking point	Mark	Notes
Part (a)
the colour of a star depends on its surface temperature / red stars are the coolest and blue stars the hottest	✓ 1	Accept the order red → orange → yellow → white → blue as the order of increasing surface temperature.
Rigel has a higher surface temperature than Betelgeuse / Betelgeuse is cooler than Rigel	✓ 1	A direct comparison of the two stars is needed. “Red means hot” scores 0 for this row.
Part (b)(i)
(nuclear) fusion: hydrogen nuclei join together to make helium (nuclei)	✓ 1	“Hydrogen fuses into helium” is enough. Do not accept fission, burning or a chemical reaction.
the mass of the product is less than the mass of the nuclei that joined / mass is lost and released as energy	✓ 1	Accept “the lost mass is converted into energy”.
Part (b)(ii)
nuclei are positively charged (contain protons), so they repel each other	✓ 1	Accept electrostatic repulsion between the nuclei.
at a very high temperature the nuclei move very fast / have enough kinetic energy to overcome the repulsion and get close enough to fuse	✓ 1	The link between high temperature and speed (or kinetic energy) of the nuclei is needed.
at a very high pressure (density) the nuclei are close together, so collisions happen often enough for fusion to continue	✓ 1	Accept “more frequent collisions”.
Part (c)
when fusion in the core can no longer continue, the core collapses (suddenly)	✓ 1	Accept “it runs out of fuel and collapses”.
the star explodes as a supernova	✓ 1	Accept the material thrown out forms a new nebula as an extra detail, not instead of this row.
the core left behind becomes a neutron star or (if massive enough) a black hole	✓ 1	Either answer, or both. Do not accept white dwarf.
Part (d)
the student is wrong: the Sun does not have enough mass to become a supernova / only stars much more massive than the Sun explode as supernovae	✓ 1	A reference to the Sun’s (smaller) mass is needed for this mark.
the Sun will become a red giant, then lose its outer layers gently and leave a white dwarf	✓ 1	Both red giant and white dwarf needed. Do not accept red supergiant or neutron star.
Part (e)(i)
how bright the star would appear if it were at a standard distance from the Earth	✓ 1	The idea of every star being compared at the same (standard) distance is needed. “How bright it looks” alone scores 0.
Part (e)(ii)
−6	✓ 1	Accept −5.5 to −6.5. The minus sign is needed.
Part (e)(iii)
Betelgeuse is very much larger than the Sun (it is a supergiant)	✓ 1	Accept “huge”, “enormous radius / surface”.
so although each part of its surface is cooler / redder, its total light output is much greater (smaller, more negative absolute magnitude)	✓ 1	Must link size to the total light given out. A reference to its distance from the Earth scores 0: absolute magnitude already removes the effect of distance.
Part (f)(i)
time = distance ÷ speed	✓ 1	Accept the equation in symbols or with numbers substituted.
time = 5.2 × 10¹⁸ ÷ 3.0 × 10⁸ = 1.73 × 10¹⁰ s	✓ 1	Accept 1.7 × 10¹⁰ s.
1.73 × 10¹⁰ ÷ 3.15 × 10⁷ = 550.3 years (≈ 550 years)	✓ 1	The conversion to years must be shown, with an answer to more significant figures than the given 550 (e.g. 550.3 or 550.26).
Part (f)(ii)
the light we see now left Betelgeuse about 550 years ago, so light from an explosion in the last 550 years has not reached the Earth yet	✓ 1	Both ideas (light takes about 550 years / we see the star as it was then) in one statement for the mark.

Answers: (a) Rigel (blue-white) is hotter than Betelgeuse (red) · (c) core collapses → supernova → neutron star or black hole · (e)(ii) −6 · (f)(i) 1.73 × 10¹⁰ s ≈ 550 years

Exam tip (a) Give the direction of the link: “blue stars have a higher surface temperature than red stars”. “Colour depends on temperature” on its own is not enough for both marks.

Common error (d) Mixing up the two life cycles. A star of similar mass to the Sun ends as a white dwarf; supernova, neutron star and black hole are only for stars much more massive than the Sun.

More 8(c) questions in the question bank

+ Add to test✓ In your testCC-EIP-8c-219 marks~20 min

Surveying an open star cluster: the Hertzsprung–Russell diagram

8(c) Stellar Evolution7(c) Fission and Fusion8(b) Motion in the UniversePaper 2 only

An open star cluster is a group of a few hundred stars inside the Milky Way. The stars formed at about the same time from the same nebula, and they are all at about the same distance from the Earth.

Astronomers measured the surface temperature and the absolute magnitude of stars in one open cluster. Six of the stars, A to F, have been plotted on the Hertzsprung–Russell (HR) diagram in Figure 1. Table 1 gives the data for three more stars, P, Q and R.

Star	Surface temperature / K	Absolute magnitude
P	2500	+14
Q	4000	−1
R	20 000	+11

Table 1

Figure 1: HR diagram for the open cluster, stars A to F (temperature decreases to the right)

(a)

State what is meant by the absolute magnitude of a star.

Paper 2 only onlyRECALLNotes

Show solution

how bright the star would appear if it were at a standard distance from the Earth

✓ 1

Accept “if all stars were placed at the same distance”. “How bright the star is” alone scores 0.

(1)

(b)

For most stars in the sky, the order of how bright they appear from the Earth is different from the order of their absolute magnitudes.

Explain why, for the stars in this cluster, the two orders are the same.

Paper 2 only onlyEXPLAINNotes

Show solution

how bright a star appears depends on how much light it gives out and on its distance from the Earth	✓ 1	Both factors needed.
the stars in the cluster are all at about the same distance, so differences in how bright they appear are caused only by differences in the light they give out (which is what absolute magnitude compares)	✓ 1	Must use the same-distance idea from the stem.

(2)

(c)

Plot the stars P, Q and R from Table 1 on Figure 1.

Paper 2 only onlyGRAPHNotes

Show solution

P plotted at 2500 K and +14, and R plotted at 20 000 K and +11 (on the 2500 K and 20 000 K grid lines)	✓ 1	Allow ± 0.5 magnitude (half a small square vertically). Both points for the mark.
Q plotted at 4000 K and −1	✓ 1	Allow ± 0.5 magnitude, on the 4000 K grid line. Note that magnitudes become more negative upwards.

Answer: P: bottom right, at the lower end of the band through A–F; Q: right, well above E; R: bottom left

(2)

(d)

Stars A to F and P all lie on the main sequence.

State the type of star Q and the type of star R. Give a reason for each answer.

Paper 2 only onlyDATANotes

Show solution

Q: red giant – it is cool (red) but much brighter than main sequence stars of the same temperature / it lies above the main sequence on the right	✓ 1	Type and reason needed. Accept giant. Do not accept supergiant (supergiants are at the very top, around −8).
R: white dwarf – it is hot (white) but very dim / it lies at the bottom left, below the main sequence	✓ 1	Type and reason needed. Do not accept main sequence: R is far below the band through A to F.

Answer: Q: red giant; R: white dwarf

(2)

(e)

Star C has a mass similar to the mass of the Sun.

Describe how star C will change after it leaves the main sequence, and how its position on Figure 1 will change.

Paper 2 only onlyPREDICTNotes

Show solution

C becomes a red giant (when the hydrogen in its core runs out)	✓ 1	Do not accept red supergiant.
its point moves up and to the right (cooler but brighter), towards the position of Q	✓ 1	Both directions needed.
it loses its outer layers and the core that is left becomes a white dwarf	✓ 1	Do not accept supernova, neutron star or black hole.
its point ends up at the bottom left (near R) and the white dwarf slowly cools / fades	✓ 1	Accept “below and to the left of the main sequence”.

(4)

(f)

Every main sequence star in the cluster releases energy by nuclear fusion.

Give two differences between the nuclear reactions in the core of a main sequence star and the nuclear reactions in a nuclear power station.

RECALLNotes

Show solution

in a star small (hydrogen) nuclei join to make a larger nucleus; in a power station a large (uranium-235) nucleus splits into smaller nuclei	✓ 1	Each difference must give both sides.
fusion needs a very high temperature and pressure but fission does not / fission is started by absorbing a neutron but fusion is not / fission products are radioactive but fusion products usually are not	✓ 1	Any one of these for the second mark, with both sides stated.

(2)

(g)

Astronomers observe a second open cluster that is several billion years older. In its HR diagram, there are no main sequence stars as hot and as bright as stars A and B.

Explain this observation. Use star A as an example in your answer.

Paper 2 only onlyREASONNotes

Show solution

stars like A (the hottest, brightest, most massive stars on the main sequence) use up their hydrogen much faster than less massive stars	✓ 1	Accept “shorter main sequence lifetime” as an equivalent for this row only if linked to greater mass.
so after billions of years they have already left the main sequence (to become red giants or supergiants), while cooler, less massive stars such as D, E, F and P are still on it	✓ 1	Must state that these stars have left the main sequence.

(2)

(h)(i)

A planet has been discovered orbiting star D. The planet moves in a circular orbit of radius 7.5 × 10⁹ m and takes 4.1 days to complete one orbit.

Show that the orbital speed of the planet is about 1.3 × 10⁵ m/s.

DERIVENotes

Show solution

T = 4.1 × 86 400 = 354 240 s (3.54 × 10⁵ s)	✓ 1	Conversion of days to seconds.
v = 2πr ÷ T = 2 × π × 7.5 × 10⁹ ÷ 354 240	✓ 1	Correct substitution.
v = 1.33 × 10⁵ m/s	✓ 1	Answer to at least 3 significant figures (133 000 m/s), since the value to show is given to 2 s.f.

Answer: 1.33 × 10⁵ m/s

(3)

(h)(ii)

State the force that keeps the planet in its orbit around star D, and the direction in which this force acts.

RECALLNotes

Show solution

gravitational force (of star D on the planet), acting towards the centre of star D

✓ 1

Both name and direction for the mark. Do not accept centrifugal force.

Answer: gravity, towards star D

(1)

Show full solution

Marking point	Mark	Notes
Part (a)
how bright the star would appear if it were at a standard distance from the Earth	✓ 1	Accept “if all stars were placed at the same distance”. “How bright the star is” alone scores 0.
Part (b)
how bright a star appears depends on how much light it gives out and on its distance from the Earth	✓ 1	Both factors needed.
the stars in the cluster are all at about the same distance, so differences in how bright they appear are caused only by differences in the light they give out (which is what absolute magnitude compares)	✓ 1	Must use the same-distance idea from the stem.
Part (c)
P plotted at 2500 K and +14, and R plotted at 20 000 K and +11 (on the 2500 K and 20 000 K grid lines)	✓ 1	Allow ± 0.5 magnitude (half a small square vertically). Both points for the mark.
Q plotted at 4000 K and −1	✓ 1	Allow ± 0.5 magnitude, on the 4000 K grid line. Note that magnitudes become more negative upwards.
Part (d)
Q: red giant – it is cool (red) but much brighter than main sequence stars of the same temperature / it lies above the main sequence on the right	✓ 1	Type and reason needed. Accept giant. Do not accept supergiant (supergiants are at the very top, around −8).
R: white dwarf – it is hot (white) but very dim / it lies at the bottom left, below the main sequence	✓ 1	Type and reason needed. Do not accept main sequence: R is far below the band through A to F.
Part (e)
C becomes a red giant (when the hydrogen in its core runs out)	✓ 1	Do not accept red supergiant.
its point moves up and to the right (cooler but brighter), towards the position of Q	✓ 1	Both directions needed.
it loses its outer layers and the core that is left becomes a white dwarf	✓ 1	Do not accept supernova, neutron star or black hole.
its point ends up at the bottom left (near R) and the white dwarf slowly cools / fades	✓ 1	Accept “below and to the left of the main sequence”.
Part (f)
in a star small (hydrogen) nuclei join to make a larger nucleus; in a power station a large (uranium-235) nucleus splits into smaller nuclei	✓ 1	Each difference must give both sides.
fusion needs a very high temperature and pressure but fission does not / fission is started by absorbing a neutron but fusion is not / fission products are radioactive but fusion products usually are not	✓ 1	Any one of these for the second mark, with both sides stated.
Part (g)
stars like A (the hottest, brightest, most massive stars on the main sequence) use up their hydrogen much faster than less massive stars	✓ 1	Accept “shorter main sequence lifetime” as an equivalent for this row only if linked to greater mass.
so after billions of years they have already left the main sequence (to become red giants or supergiants), while cooler, less massive stars such as D, E, F and P are still on it	✓ 1	Must state that these stars have left the main sequence.
Part (h)(i)
T = 4.1 × 86 400 = 354 240 s (3.54 × 10⁵ s)	✓ 1	Conversion of days to seconds.
v = 2πr ÷ T = 2 × π × 7.5 × 10⁹ ÷ 354 240	✓ 1	Correct substitution.
v = 1.33 × 10⁵ m/s	✓ 1	Answer to at least 3 significant figures (133 000 m/s), since the value to show is given to 2 s.f.
Part (h)(ii)
gravitational force (of star D on the planet), acting towards the centre of star D	✓ 1	Both name and direction for the mark. Do not accept centrifugal force.

Answers: (c) P: bottom right, at the lower end of the band through A–F; Q: right, well above E; R: bottom left · (d) Q: red giant; R: white dwarf · (h)(i) 1.33 × 10⁵ m/s · (h)(ii) gravity, towards star D

Exam tip (c) On an HR diagram both axes run the “wrong” way: temperature falls to the right and brighter (more negative) magnitudes are higher up. Check the direction before plotting.

Common error (g) Thinking massive stars last longer because they contain more hydrogen. They use it very much faster, so their main sequence stage is much shorter.

More 8(c) questions in the question bank

8(d) Cosmology

+ Add to test✓ In your testCC-EIP-8d-120 marks~20 min

A galaxy red-shift survey: evidence for an expanding universe

8(d) Cosmology3(b) Properties of Waves3(c) The Electromagnetic SpectrumPaper 2 only

Astronomers use a telescope fitted with a spectrometer to study light from five distant galaxies, A to E. In the spectrum of each galaxy there is a dark line produced by calcium. Measured in a laboratory on the Earth, this line has a wavelength of 396.8 nm.

Table 1 gives the distance of each galaxy from the Earth and the wavelength at which the calcium line is observed in its light.

Galaxy	Distance / million light-years	Observed wavelength / nm
A	140	400.6
B	230	403.5
C	330	406.1
D	400	408.3
E	520	411.4

Table 1

speed of light c = 3.0 × 10⁸ m/s

(a)

In the light from every galaxy in Table 1, the calcium line is at a longer wavelength than in the laboratory.

Explain, in terms of wavefronts, why light from a galaxy that is moving away from the Earth is received with a longer wavelength.

Paper 2 only onlyEXPLAINNotes

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the galaxy moves away from the Earth in the time between emitting one wavefront and the next	✓ 1	Accept “each wavefront is emitted from a point further from the Earth”.
so the wavefronts reaching the Earth are further apart / spread out	✓ 1
the speed of light does not change (it does not depend on the source), so the wavelength received is longer (and the frequency lower)	✓ 1	The unchanged wave speed is needed for this mark.

(3)

(b)(i)

Calculate the frequency of the calcium line measured in the laboratory.

CALCNotes

Show solution

f = c ÷ λ = 3.0 × 10⁸ ÷ 396.8 × 10⁻⁹	✓ 1	Rearrangement of v = f × λ with the wavelength converted to metres.
f = 7.6 × 10¹⁴ Hz	✓ 1	Accept 7.56 × 10¹⁴ Hz. A power-of-ten error (e.g. 7.6 × 10⁵ Hz from using 396.8 m) scores 1.

Answer: 7.6 × 10¹⁴ Hz

(2)

(b)(ii)

State whether the frequency of this line in the light received from galaxy D is higher or lower than your answer to (b)(i). Give a reason.

REASONNotes

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lower, because the wavelength is longer and the speed of light is the same (f = c ÷ λ)

✓ 1

Answer and reason for the mark.

Answer: lower

(1)

(c)(i)

Show that galaxy D is moving away from the Earth at a speed of about 8.7 × 10⁶ m/s.

Paper 2 only onlyDERIVENotes

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Δλ = 408.3 − 396.8 = 11.5 nm	✓ 1
v = (Δλ ÷ λ₀) × c = (11.5 ÷ 396.8) × 3.0 × 10⁸	✓ 1	Must divide by the laboratory wavelength 396.8 nm. Dividing by 408.3 nm gives 8.45 × 10⁶ m/s: max 1 mark.
v = 8.69 × 10⁶ m/s	✓ 1	At least 3 significant figures needed, since 8.7 × 10⁶ m/s is given.

Answer: 8.69 × 10⁶ m/s

(3)

(c)(ii)

Figure 1 shows the speeds of galaxies A, B, C and E plotted against their distances from the Earth.

Plot the point for galaxy D on Figure 1 and draw a line of best fit.

Figure 1: speed away from the Earth against distance for galaxies A, B, C and E

Paper 2 only onlyGRAPHNotes

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D plotted at 400 million light-years and 8.7 × 10⁶ m/s	✓ 1	Allow ± half a small square. Allow ECF from (c)(i).
straight line of best fit passing through (or very close to) the origin, with points evenly spread about it	✓ 1	A dot-to-dot line scores 0 for this row.

Answer: D at (400, 8.7); straight line through the origin

(2)

(c)(iii)

Describe the relationship between the speed of a galaxy and its distance from the Earth.

Paper 2 only onlyDATANotes

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the further away the galaxy, the faster it is moving away	✓ 1	Accept “the greater the distance, the greater the red-shift and speed”.
the speed is (approximately) directly proportional to the distance – straight line through the origin	✓ 1	Accept a numerical check, e.g. E is about 3.7 times as far as A and moving about 3.8 times as fast.

(2)

(d)

Galaxy F is in the same direction as galaxy D but twice as far away from the Earth.

Assuming the relationship you described in (c)(iii) still applies, determine the wavelength at which the calcium line is observed in light from galaxy F.

Paper 2 only onlyPREDICTNotes

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speed of F = 2 × 8.69 × 10⁶ = 1.74 × 10⁷ m/s	✓ 1	Accept 1.70–1.75 × 10⁷ m/s (e.g. from the gradient of the line of best fit × 800 million light-years). Allow ECF from (c)(i).
Δλ = λ₀ × v ÷ c = 396.8 × 1.74 × 10⁷ ÷ 3.0 × 10⁸ = 23 nm	✓ 1	Alternative: red-shift doubles, so Δλ = 2 × 11.5 = 23.0 nm scores both of the first two marks.
observed wavelength = 396.8 + 23.0 = 419.8 nm (about 420 nm)	✓ 1	Accept 419–420 nm. Giving only Δλ (23 nm) as the final answer loses this mark.

Answer: about 420 nm (419.8 nm)

(3)

(e)

In light from a nearby galaxy G, the calcium line is observed at 396.5 nm.

State what this tells you about the motion of galaxy G.

Paper 2 only onlyREASONNotes

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G is moving towards the Earth (the line is blue-shifted / shorter wavelength) – and only slowly, as the shift is small

✓ 1

“Towards the Earth” is needed. Accept a calculated speed of about 2.3 × 10⁵ m/s towards the Earth.

Answer: moving towards the Earth (blue-shift)

(1)

(f)

Explain how the results in Table 1 and Figure 1 support the Big Bang theory of the origin of the universe.

Paper 2 only onlyEXPLAINNotes

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light from the galaxies is red-shifted, so they are all moving away from the Earth	✓ 1
more distant galaxies have greater red-shifts / move away faster, so the universe is expanding	✓ 1	The distance link is needed: red-shift alone does not show expansion.
so in the past the galaxies were closer together, and the universe began from a very small, hot, dense region (about 13.8 billion years ago)	✓ 1	Accept “all the matter started from one very small region”.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
the galaxy moves away from the Earth in the time between emitting one wavefront and the next	✓ 1	Accept “each wavefront is emitted from a point further from the Earth”.
so the wavefronts reaching the Earth are further apart / spread out	✓ 1
the speed of light does not change (it does not depend on the source), so the wavelength received is longer (and the frequency lower)	✓ 1	The unchanged wave speed is needed for this mark.
Part (b)(i)
f = c ÷ λ = 3.0 × 10⁸ ÷ 396.8 × 10⁻⁹	✓ 1	Rearrangement of v = f × λ with the wavelength converted to metres.
f = 7.6 × 10¹⁴ Hz	✓ 1	Accept 7.56 × 10¹⁴ Hz. A power-of-ten error (e.g. 7.6 × 10⁵ Hz from using 396.8 m) scores 1.
Part (b)(ii)
lower, because the wavelength is longer and the speed of light is the same (f = c ÷ λ)	✓ 1	Answer and reason for the mark.
Part (c)(i)
Δλ = 408.3 − 396.8 = 11.5 nm	✓ 1
v = (Δλ ÷ λ₀) × c = (11.5 ÷ 396.8) × 3.0 × 10⁸	✓ 1	Must divide by the laboratory wavelength 396.8 nm. Dividing by 408.3 nm gives 8.45 × 10⁶ m/s: max 1 mark.
v = 8.69 × 10⁶ m/s	✓ 1	At least 3 significant figures needed, since 8.7 × 10⁶ m/s is given.
Part (c)(ii)
D plotted at 400 million light-years and 8.7 × 10⁶ m/s	✓ 1	Allow ± half a small square. Allow ECF from (c)(i).
straight line of best fit passing through (or very close to) the origin, with points evenly spread about it	✓ 1	A dot-to-dot line scores 0 for this row.
Part (c)(iii)
the further away the galaxy, the faster it is moving away	✓ 1	Accept “the greater the distance, the greater the red-shift and speed”.
the speed is (approximately) directly proportional to the distance – straight line through the origin	✓ 1	Accept a numerical check, e.g. E is about 3.7 times as far as A and moving about 3.8 times as fast.
Part (d)
speed of F = 2 × 8.69 × 10⁶ = 1.74 × 10⁷ m/s	✓ 1	Accept 1.70–1.75 × 10⁷ m/s (e.g. from the gradient of the line of best fit × 800 million light-years). Allow ECF from (c)(i).
Δλ = λ₀ × v ÷ c = 396.8 × 1.74 × 10⁷ ÷ 3.0 × 10⁸ = 23 nm	✓ 1	Alternative: red-shift doubles, so Δλ = 2 × 11.5 = 23.0 nm scores both of the first two marks.
observed wavelength = 396.8 + 23.0 = 419.8 nm (about 420 nm)	✓ 1	Accept 419–420 nm. Giving only Δλ (23 nm) as the final answer loses this mark.
Part (e)
G is moving towards the Earth (the line is blue-shifted / shorter wavelength) – and only slowly, as the shift is small	✓ 1	“Towards the Earth” is needed. Accept a calculated speed of about 2.3 × 10⁵ m/s towards the Earth.
Part (f)
light from the galaxies is red-shifted, so they are all moving away from the Earth	✓ 1
more distant galaxies have greater red-shifts / move away faster, so the universe is expanding	✓ 1	The distance link is needed: red-shift alone does not show expansion.
so in the past the galaxies were closer together, and the universe began from a very small, hot, dense region (about 13.8 billion years ago)	✓ 1	Accept “all the matter started from one very small region”.

Answers: (b)(i) 7.6 × 10¹⁴ Hz · (b)(ii) lower · (c)(i) 8.69 × 10⁶ m/s · (c)(ii) D at (400, 8.7); straight line through the origin · (d) about 420 nm (419.8 nm) · (e) moving towards the Earth (blue-shift)

Exam tip (b)(i) Convert nm to m first: 1 nm = 10⁻⁹ m. The ratio Δλ/λ₀ does not need this, but v = f × λ does.

Common error (d) Stopping at Δλ. The equation gives the change in wavelength; the observed wavelength is λ₀ + Δλ for a red-shift.

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Detecting the cosmic microwave background with a radio telescope

8(d) Cosmology3(c) The Electromagnetic Spectrum5(d) Ideal Gas Molecules3(b) Properties of WavesPaper 2 only

A radio telescope detects faint radiation arriving from every direction in the sky, with almost exactly the same intensity everywhere, even from regions where there are no visible stars or galaxies. This is the cosmic microwave background (CMB) radiation.

Figure 1 shows how the intensity of the CMB radiation varies with wavelength.

speed of electromagnetic waves in free space = 3.0 × 10⁸ m/s

Figure 1: relative intensity of the CMB radiation against wavelength

(a)(i)

Use Figure 1 to find the wavelength at which the intensity of the CMB radiation is greatest.

GRAPHNotes

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1.1 mm

✓ 1

Accept 1.0 mm to 1.1 mm.

Answer: 1.1 mm

(1)

(a)(ii)

Calculate the frequency of radiation with this wavelength.

CALCNotes

Show solution

f = v ÷ λ = 3.0 × 10⁸ ÷ 1.1 × 10⁻³	✓ 1	Wavelength must be converted to metres. Allow ECF from (a)(i).
f = 2.7 × 10¹¹ Hz	✓ 1	Accept 2.73 × 10¹¹ Hz; 3.0 × 10¹¹ Hz from 1.0 mm. A power-of-ten error scores 1.

Answer: 2.7 × 10¹¹ Hz

(2)

(a)(iii)

Name the region of the electromagnetic spectrum that contains this radiation, and the region with the next longer wavelengths.

RECALLNotes

Show solution

microwaves; radio waves have the next longer wavelengths

✓ 1

Both needed for the mark.

Answer: microwaves; radio waves

(1)

(b)

Explain how the Big Bang theory accounts for the CMB radiation, and why this radiation is now detected as microwaves.

Paper 2 only onlyEXPLAINNotes

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the CMB is radiation left over from the early universe, when the universe was extremely hot (and dense)	✓ 1	“Radiation from stars” or “background noise” scores 0.
it fills the universe, so it arrives from every direction with almost the same intensity	✓ 1	Accept “not from any particular object”.
as the universe has expanded, the wavelength of the radiation has been stretched, so it is now in the microwave region	✓ 1	Stretching of the wavelength by the expansion is needed.

(3)

(c)(i)

The CMB radiation has the same pattern of wavelengths as the radiation from an object at a temperature of 2.7 K.

Convert 2.7 K to degrees Celsius.

CALCNotes

Show solution

2.7 − 273 = −270 °C

✓ 1

Accept −270.3 °C. The minus sign is needed.

Answer: −270 °C

(1)

(c)(ii)

This temperature is only a little above absolute zero.

Explain why no object can have a temperature below absolute zero.

EXPLAINNotes

Show solution

temperature is a measure of the average kinetic energy of the particles (molecules)	✓ 1
at absolute zero (0 K, −273 °C) the particles have the least possible kinetic energy, so no more energy can be removed / the temperature cannot fall further	✓ 1	Accept “the particles stop moving”.

(2)

(d)(i)

The radio telescope is also used to observe a line emitted by hydrogen. Measured on the Earth, this line has a wavelength of 21.10 cm. In the radiation from a distant galaxy, the same line is observed at 21.52 cm.

Show that the galaxy is moving away from the Earth at about 6.0 × 10⁶ m/s.

Paper 2 only onlyDERIVENotes

Show solution

Δλ = 21.52 − 21.10 = 0.42 cm	✓ 1	Both wavelengths may stay in cm.
v = (Δλ ÷ λ₀) × c = (0.42 ÷ 21.10) × 3.0 × 10⁸	✓ 1	Dividing by 21.52 cm gives 5.86 × 10⁶ m/s: max 1 mark.
v = 5.97 × 10⁶ m/s	✓ 1	At least 3 significant figures needed.

Answer: 5.97 × 10⁶ m/s

(3)

(d)(ii)

The frequency of the hydrogen line measured on the Earth is 1.42 × 10⁹ Hz.

Calculate the frequency of this line in the radiation received from the galaxy, and comment on your answer.

CALCNotes

Show solution

f = 3.0 × 10⁸ ÷ 0.2152 = 1.39 × 10⁹ Hz	✓ 1	Wavelength must be in metres. Accept 1.394 × 10⁹ Hz.
lower than the emitted frequency, as expected for a source moving away (longer wavelength, same wave speed)	✓ 1	Comment must refer to the galaxy moving away (or to the Doppler effect / red-shift).

Answer: 1.39 × 10⁹ Hz, lower than emitted

(2)

(e)

According to the Big Bang theory, the universe will continue to expand.

Predict how the wavelength at the peak of Figure 1 and the temperature of the CMB radiation will be different in the far future. Explain your answer.

Paper 2 only onlyPREDICTNotes

Show solution

the peak will be at a longer wavelength (the curve shifts to the right)	✓ 1
because further expansion stretches the wavelength of the radiation even more	✓ 1	Allow ECF from (b).
the temperature will be lower (the universe continues to cool), closer to absolute zero	✓ 1	Do not accept a temperature below 0 K.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
1.1 mm	✓ 1	Accept 1.0 mm to 1.1 mm.
Part (a)(ii)
f = v ÷ λ = 3.0 × 10⁸ ÷ 1.1 × 10⁻³	✓ 1	Wavelength must be converted to metres. Allow ECF from (a)(i).
f = 2.7 × 10¹¹ Hz	✓ 1	Accept 2.73 × 10¹¹ Hz; 3.0 × 10¹¹ Hz from 1.0 mm. A power-of-ten error scores 1.
Part (a)(iii)
microwaves; radio waves have the next longer wavelengths	✓ 1	Both needed for the mark.
Part (b)
the CMB is radiation left over from the early universe, when the universe was extremely hot (and dense)	✓ 1	“Radiation from stars” or “background noise” scores 0.
it fills the universe, so it arrives from every direction with almost the same intensity	✓ 1	Accept “not from any particular object”.
as the universe has expanded, the wavelength of the radiation has been stretched, so it is now in the microwave region	✓ 1	Stretching of the wavelength by the expansion is needed.
Part (c)(i)
2.7 − 273 = −270 °C	✓ 1	Accept −270.3 °C. The minus sign is needed.
Part (c)(ii)
temperature is a measure of the average kinetic energy of the particles (molecules)	✓ 1
at absolute zero (0 K, −273 °C) the particles have the least possible kinetic energy, so no more energy can be removed / the temperature cannot fall further	✓ 1	Accept “the particles stop moving”.
Part (d)(i)
Δλ = 21.52 − 21.10 = 0.42 cm	✓ 1	Both wavelengths may stay in cm.
v = (Δλ ÷ λ₀) × c = (0.42 ÷ 21.10) × 3.0 × 10⁸	✓ 1	Dividing by 21.52 cm gives 5.86 × 10⁶ m/s: max 1 mark.
v = 5.97 × 10⁶ m/s	✓ 1	At least 3 significant figures needed.
Part (d)(ii)
f = 3.0 × 10⁸ ÷ 0.2152 = 1.39 × 10⁹ Hz	✓ 1	Wavelength must be in metres. Accept 1.394 × 10⁹ Hz.
lower than the emitted frequency, as expected for a source moving away (longer wavelength, same wave speed)	✓ 1	Comment must refer to the galaxy moving away (or to the Doppler effect / red-shift).
Part (e)
the peak will be at a longer wavelength (the curve shifts to the right)	✓ 1
because further expansion stretches the wavelength of the radiation even more	✓ 1	Allow ECF from (b).
the temperature will be lower (the universe continues to cool), closer to absolute zero	✓ 1	Do not accept a temperature below 0 K.

Answers: (a)(i) 1.1 mm · (a)(ii) 2.7 × 10¹¹ Hz · (a)(iii) microwaves; radio waves · (c)(i) −270 °C · (d)(i) 5.97 × 10⁶ m/s · (d)(ii) 1.39 × 10⁹ Hz, lower than emitted

Exam tip (d)(i) The red-shift equation works for any electromagnetic wave, not only visible light. Because Δλ ÷ λ₀ is a ratio, the wavelengths can stay in cm as long as both use the same unit.

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