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Cambridge IGCSE Physics 0625 · Concept clusters

Cambridge 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. 48 questions · 895 marks · two per subchapter · every part with a worked solution · All courses

Topic

Subchapter

Level

Skill

Topic 1 · Motion, forces and energy

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

Setup	1.1	1.2	1.3	1.4	1.5	1.6	1.7	1.8	Other
Checking 3D-printer filament: micrometers, layers and feed speed	●	○	○						—
A rowing crew on a river: scalars, vectors and timing	●	○		○					—
A skydiver jumping from a balloon: free fall, terminal speed and landing		●		○			○		—
A driverless metro train between two stations	○	●		○					—
Is the gold bar real? Density in a jeweller's lab	○		●					○	—
Why a hot-air balloon floats: density of hot and cold air			●					○	2.1
Testing a bungee cord: extension, limit of proportionality and safety		○		●			○		—
An electric racing car: thrust, drag and cornering		○		●		○			—
A tower crane on a building site: moments and counterweights			○	○	●				—
Designing a playground seesaw for two children and a parent			○		●			○	—
Curling stones: collisions on the ice		○				●	○		—
Crash-test dummy: airbags, crumple zones and impulse		○		○		●			—
A pumped-storage hydroelectric plant: energy, power and efficiency				○			●		4.2
An electric scooter on a hill: energy changes, power and braking		○		○			●		—
A deep-sea submersible: pressure on its window and hatch			○		○			●	—
Snowboards, stilettos and tractors: pressure on soft ground			○	○				●	—

1.1 Physical quantities and measurement

+ Add to test✓ In your testCC-CIP-11-117 marks~20 min

Checking 3D-printer filament: micrometers, layers and feed speed

1.1 Physical quantities and measurement1.3 Mass, weight and density1.2 Speed, velocity and motion graphsCore

A student who runs a 3D printer checks a new spool of plastic filament. The label on the spool states that the diameter of the filament is 1.75 mm ± 0.03 mm.

The student first closes the jaws of a micrometer screw gauge with nothing between them. The gauge reads +0.03 mm. The filament is then placed between the jaws. Figure 1 shows the scales of the micrometer.

Figure 1: micrometer scales with the filament between the jaws (enlarged, not to scale)

(a)

Determine the diameter of the filament.

CALCNotes

Show solution

reading = 1.5 mm + 26 × 0.01 mm = 1.76 mm	✓ 1	Sleeve 1.5 mm and thimble 0.26 mm both needed.
diameter = 1.76 − 0.03 = 1.73 mm	✓ 1	Allow ECF from an incorrect reading. Unit required. 1.79 mm (zero error added) scores this mark 0.

Answer: 1.73 mm

(2)

(b)

State, with a reason, whether the filament is within the range given on the label.

REASONNotes

Show solution

yes: 1.73 mm lies within the range 1.72 mm to 1.78 mm

✓ 1

Allow ECF from (a). The range or the lower limit 1.72 mm must be seen.

Answer: Yes, 1.73 mm is between 1.72 mm and 1.78 mm

(1)

(c)(i)

The student prints a test block 40 layers tall and measures its height with a 30 cm ruler. The height is 8.0 mm.

Calculate the thickness of one layer.

CALCNotes

Show solution

8.0 ÷ 40 = 0.20 mm

✓ 1

Accept 0.020 cm or 2.0 × 10⁻⁴ m.

Answer: 0.20 mm

(1)

(c)(ii)

Explain why the student measures the height of 40 layers instead of measuring one layer.

EXPLAINNotes

Show solution

one layer is much thinner than the smallest division of the ruler (1 mm), so it cannot be measured directly / accurately	✓ 1	Accept: the ruler's resolution is too large for one layer.
the reading error of the ruler is shared out over 40 layers, so the error in the thickness of one layer is (40 times) smaller	✓ 1	Accept: the percentage uncertainty is smaller. "More accurate" alone is not enough.

(2)

(c)(iii)

The student changes a setting so that each layer is now 0.10 mm thick.

Suggest how many layers should be measured with the same ruler to give the thickness of one layer with the same percentage uncertainty as before. Give a reason.

REASONNotes

Show solution

80 layers	✓ 1	Allow ECF from (c)(i): 8.0 mm ÷ new thickness.
the stack is then the same height (8.0 mm), so the same ruler reading error is the same fraction of the height measured and the percentage uncertainty in one layer is unchanged	✓ 1	Accept: the total height measured must stay the same.

Answer: 80 layers

(2)

(d)(i)

The student prints a small solid test piece and finds its mass on a balance. The mass is 15.6 g. The test piece is lowered on a thread into a measuring cylinder containing water. The water level rises from 50.0 cm³ to 63.0 cm³ and the test piece is completely under the water.

Show that the density of the plastic is about 1.2 g/cm³.

MULTINotes

Show solution

volume = 63.0 − 50.0 = 13.0 cm³	✓ 1	Using 63.0 cm³ as the volume scores 0 for this mark.
density = mass ÷ volume	✓ 1	Equation in words or symbols (ρ = m/V), or a correct substitution.
15.6 ÷ 13.0 = 1.20 g/cm³	✓ 1	The full substitution must be seen; 1.2 alone scores 0 for this mark.

Answer: 1.20 g/cm³

(3)

(d)(ii)

State the density of the plastic in kg/m³ and state whether the test piece floats or sinks in water. Give a reason. The density of water is 1000 kg/m³.

REASONNotes

Show solution

1200 kg/m³	✓ 1	Allow ECF from (d)(i).
sinks, because its density is greater than the density of water	✓ 1	Must compare densities. "Because it is heavy" scores 0. Allow ECF consistent with the student's density.

Answer: 1200 kg/m³; it sinks

(2)

(e)(i)

During a print, the printer feeds filament at a steady rate. The table shows the length L of filament fed into the printer after time t.

t / s	0	20	40	60	80	100
L / mm	0	35	67	103	136	170

On Figure 2, plot the points and draw the line of best fit.

Figure 2: grid for a graph of length of filament fed, L, against time, t

GRAPHNotes

Show solution

all six points plotted correctly to within half a small square	✓ 1	Five correct points: allow this mark.
single straight line of best fit through (or very close to) the origin, with points evenly scattered either side	✓ 1	Do not accept a dot-to-dot line or a curve.

(2)

(e)(ii)

Use your graph to determine the speed at which the filament is fed into the printer. Give your answer in mm/s.

GRAPHNotes

Show solution

gradient found from a large triangle on the line, e.g. 170 ÷ 100	✓ 1	Values read from the line, not just from the table, using a triangle covering at least half the line.
1.7 mm/s	✓ 1	Accept 1.6 mm/s to 1.8 mm/s. Unit required.

Answer: 1.7 mm/s

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
reading = 1.5 mm + 26 × 0.01 mm = 1.76 mm	✓ 1	Sleeve 1.5 mm and thimble 0.26 mm both needed.
diameter = 1.76 − 0.03 = 1.73 mm	✓ 1	Allow ECF from an incorrect reading. Unit required. 1.79 mm (zero error added) scores this mark 0.
Part (b)
yes: 1.73 mm lies within the range 1.72 mm to 1.78 mm	✓ 1	Allow ECF from (a). The range or the lower limit 1.72 mm must be seen.
Part (c)(i)
8.0 ÷ 40 = 0.20 mm	✓ 1	Accept 0.020 cm or 2.0 × 10⁻⁴ m.
Part (c)(ii)
one layer is much thinner than the smallest division of the ruler (1 mm), so it cannot be measured directly / accurately	✓ 1	Accept: the ruler's resolution is too large for one layer.
the reading error of the ruler is shared out over 40 layers, so the error in the thickness of one layer is (40 times) smaller	✓ 1	Accept: the percentage uncertainty is smaller. "More accurate" alone is not enough.
Part (c)(iii)
80 layers	✓ 1	Allow ECF from (c)(i): 8.0 mm ÷ new thickness.
the stack is then the same height (8.0 mm), so the same ruler reading error is the same fraction of the height measured and the percentage uncertainty in one layer is unchanged	✓ 1	Accept: the total height measured must stay the same.
Part (d)(i)
volume = 63.0 − 50.0 = 13.0 cm³	✓ 1	Using 63.0 cm³ as the volume scores 0 for this mark.
density = mass ÷ volume	✓ 1	Equation in words or symbols (ρ = m/V), or a correct substitution.
15.6 ÷ 13.0 = 1.20 g/cm³	✓ 1	The full substitution must be seen; 1.2 alone scores 0 for this mark.
Part (d)(ii)
1200 kg/m³	✓ 1	Allow ECF from (d)(i).
sinks, because its density is greater than the density of water	✓ 1	Must compare densities. "Because it is heavy" scores 0. Allow ECF consistent with the student's density.
Part (e)(i)
all six points plotted correctly to within half a small square	✓ 1	Five correct points: allow this mark.
single straight line of best fit through (or very close to) the origin, with points evenly scattered either side	✓ 1	Do not accept a dot-to-dot line or a curve.
Part (e)(ii)
gradient found from a large triangle on the line, e.g. 170 ÷ 100	✓ 1	Values read from the line, not just from the table, using a triangle covering at least half the line.
1.7 mm/s	✓ 1	Accept 1.6 mm/s to 1.8 mm/s. Unit required.

Answers: (a) 1.73 mm · (b) Yes, 1.73 mm is between 1.72 mm and 1.78 mm · (c)(i) 0.20 mm · (c)(iii) 80 layers · (d)(i) 1.20 g/cm³ · (d)(ii) 1200 kg/m³; it sinks · (e)(ii) 1.7 mm/s

Common error (a) Adding a positive zero error. If the gauge reads +0.03 mm with nothing in the jaws, every reading is 0.03 mm too large, so 0.03 mm is subtracted.

Exam tip (d)(i) In a "show that" question, give your answer to one more significant figure than the value quoted (here 1.20 g/cm³), so it is clear you have calculated it yourself.

More 1.1 questions in the question bank

+ Add to test✓ In your testCC-CIP-11-220 marks~20 min

A rowing crew on a river: scalars, vectors and timing

1.1 Physical quantities and measurement1.2 Speed, velocity and motion graphs1.4 Forces, friction and springsExtended

A rowing coach trains a crew on a straight, wide river. The river flows due east at a steady 1.5 m/s. The coach stands on the bank with a hand-operated stopwatch.

(a)

The coach's notebook contains the quantities below.

distance, force, mass, time, velocity, acceleration

State which of these quantities are vectors.

RECALLNotes

Show solution

any two of force, velocity and acceleration identified as vectors	✓ 1	Each scalar (distance, mass, time) wrongly listed as a vector cancels one mark, to a minimum of 0.
all three of force, velocity and acceleration, with no scalar listed	✓ 1

Answer: force, velocity, acceleration

(2)

(b)(i)

To find the stroke rate, the coach times 10 complete strokes. The stopwatch reads 25.0 s.

Calculate the time for one stroke and the frequency of the strokes.

CALCNotes

Show solution

time for one stroke = 25.0 ÷ 10 = 2.50 s	✓ 1	Accept 2.5 s.
frequency = 1 ÷ 2.50 = 0.40 Hz	✓ 1	Allow ECF from the time for one stroke. Unit required.

Answer: 2.50 s; 0.40 Hz

(2)

(b)(ii)

The coach's reaction time is about 0.2 s. Explain why timing 10 strokes gives a more reliable value for the time of one stroke than timing a single stroke.

EXPLAINNotes

Show solution

0.2 s is a large fraction (about 8%) of a single stroke of 2.5 s	✓ 1	Accept: the reaction-time error is large compared with the time for one stroke.
over 10 strokes the same 0.2 s error is shared over 25 s, so the error in one stroke is about 10 times smaller / about 0.8%	✓ 1	Accept: the percentage uncertainty is about 10 times smaller.

(2)

(c)

The crew rows a 500 m test piece upstream (towards the west), starting from rest beside a moored pontoon. The coach records a time of 3 min 24 s.

Calculate the average speed of the boat relative to the bank.

CALCNotes

Show solution

time = 3 × 60 + 24 = 204 s	✓ 1	Conversion to seconds must be seen or used.
average speed = 500 ÷ 204 = 2.45 m/s	✓ 1	Accept 2.5 m/s. Unit required.

Answer: 2.45 m/s

(2)

(d)(i)

Figure 1 shows how the speed of the boat relative to the bank changes during the first 20 s of the test piece.

Determine the acceleration of the boat during the first 8.0 s.

Figure 1: speed of the boat relative to the bank for the first 20 s of the upstream piece

Extended onlyGRAPHNotes

Show solution

acceleration = gradient = (2.5 − 0) ÷ 8.0	✓ 1	Values must be read from the graph.
0.31 m/s²	✓ 1	Accept 0.3125 m/s² or 0.3 m/s². Unit required.

Answer: 0.31 m/s²

(2)

(d)(ii)

After 8.0 s the boat moves at a constant speed. Explain, in terms of the forces on the boat, why the speed stops increasing.

EXPLAINNotes

Show solution

the drag (resistance) of the water on the boat increases as the speed increases	✓ 1	Accept water resistance / fluid resistance / friction from the water.
until the drag equals the forward force from the oars, so the resultant force is zero and there is no acceleration	✓ 1	Accept: forces are balanced so the speed is constant. "The forces stop" scores 0.

(2)

(d)(iii)

The velocity of the boat relative to the bank is the resultant of the crew's rowing velocity through the water and the velocity of the current.

Show that the crew's rowing speed through the water is 4.0 m/s while the boat moves at a constant speed upstream.

REASONNotes

Show solution

constant speed relative to the bank = 2.5 m/s (from Figure 1), towards the west	✓ 1	Must come from the graph.
rowing velocity and current are in opposite directions, so rowing speed = 2.5 + 1.5 = 4.0 m/s	✓ 1	The idea that the speeds subtract to give the resultant (so must be added back) must be clear.

Answer: 4.0 m/s

(2)

(e)(i)

The crew now points the boat due north, straight across the river, and rows at 4.0 m/s through the water, as shown in Figure 2. The current is still 1.5 m/s due east.

Show that the size of the resultant velocity of the boat is about 4.3 m/s.

Figure 2: plan view of the river as the crew crosses (not to scale)

Extended onlyCALCNotes

Show solution

the two velocities are at right angles, so R² = 4.0² + 1.5²	✓ 1	Accept a correct scale drawing with the resultant measured.
R = √18.25 = 4.27 m/s	✓ 1	The value 4.27 (or 4.272) must be seen. 5.5 m/s (adding the speeds) scores 0.

Answer: 4.27 m/s

(2)

(e)(ii)

Determine the direction of the resultant velocity, as an angle measured from due north.

Extended onlyCALCNotes

Show solution

tan θ = 1.5 ÷ 4.0	✓ 1	Accept sin θ = 1.5 ÷ 4.27 or cos θ = 4.0 ÷ 4.27 (ECF from (e)(i)), or a scale drawing.
θ = 21° east of north	✓ 1	Accept 20.6°. Direction (east of north, or a bearing of 021°) required.

Answer: 21° east of north

(2)

(f)

After heavy rain, the current increases to 2.5 m/s due east. The coach claims the crew can no longer row the 500 m upstream piece in under 5 minutes, even if the crew rows at 4.0 m/s through the water for the whole distance.

Determine whether the coach is correct.

PREDICTNotes

Show solution

speed relative to the bank = 4.0 − 2.5 = 1.5 m/s	✓ 1	Opposite directions, so subtract.
time = 500 ÷ 1.5 = 333 s (5 min 33 s), which is more than 5 minutes, so the coach is correct	✓ 1	A conclusion consistent with the calculated time is needed. Allow ECF from an incorrect speed.

Answer: 333 s; the coach is correct

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
any two of force, velocity and acceleration identified as vectors	✓ 1	Each scalar (distance, mass, time) wrongly listed as a vector cancels one mark, to a minimum of 0.
all three of force, velocity and acceleration, with no scalar listed	✓ 1
Part (b)(i)
time for one stroke = 25.0 ÷ 10 = 2.50 s	✓ 1	Accept 2.5 s.
frequency = 1 ÷ 2.50 = 0.40 Hz	✓ 1	Allow ECF from the time for one stroke. Unit required.
Part (b)(ii)
0.2 s is a large fraction (about 8%) of a single stroke of 2.5 s	✓ 1	Accept: the reaction-time error is large compared with the time for one stroke.
over 10 strokes the same 0.2 s error is shared over 25 s, so the error in one stroke is about 10 times smaller / about 0.8%	✓ 1	Accept: the percentage uncertainty is about 10 times smaller.
Part (c)
time = 3 × 60 + 24 = 204 s	✓ 1	Conversion to seconds must be seen or used.
average speed = 500 ÷ 204 = 2.45 m/s	✓ 1	Accept 2.5 m/s. Unit required.
Part (d)(i)
acceleration = gradient = (2.5 − 0) ÷ 8.0	✓ 1	Values must be read from the graph.
0.31 m/s²	✓ 1	Accept 0.3125 m/s² or 0.3 m/s². Unit required.
Part (d)(ii)
the drag (resistance) of the water on the boat increases as the speed increases	✓ 1	Accept water resistance / fluid resistance / friction from the water.
until the drag equals the forward force from the oars, so the resultant force is zero and there is no acceleration	✓ 1	Accept: forces are balanced so the speed is constant. "The forces stop" scores 0.
Part (d)(iii)
constant speed relative to the bank = 2.5 m/s (from Figure 1), towards the west	✓ 1	Must come from the graph.
rowing velocity and current are in opposite directions, so rowing speed = 2.5 + 1.5 = 4.0 m/s	✓ 1	The idea that the speeds subtract to give the resultant (so must be added back) must be clear.
Part (e)(i)
the two velocities are at right angles, so R² = 4.0² + 1.5²	✓ 1	Accept a correct scale drawing with the resultant measured.
R = √18.25 = 4.27 m/s	✓ 1	The value 4.27 (or 4.272) must be seen. 5.5 m/s (adding the speeds) scores 0.
Part (e)(ii)
tan θ = 1.5 ÷ 4.0	✓ 1	Accept sin θ = 1.5 ÷ 4.27 or cos θ = 4.0 ÷ 4.27 (ECF from (e)(i)), or a scale drawing.
θ = 21° east of north	✓ 1	Accept 20.6°. Direction (east of north, or a bearing of 021°) required.
Part (f)
speed relative to the bank = 4.0 − 2.5 = 1.5 m/s	✓ 1	Opposite directions, so subtract.
time = 500 ÷ 1.5 = 333 s (5 min 33 s), which is more than 5 minutes, so the coach is correct	✓ 1	A conclusion consistent with the calculated time is needed. Allow ECF from an incorrect speed.

Answers: (a) force, velocity, acceleration · (b)(i) 2.50 s; 0.40 Hz · (c) 2.45 m/s · (d)(i) 0.31 m/s² · (d)(iii) 4.0 m/s · (e)(i) 4.27 m/s · (e)(ii) 21° east of north · (f) 333 s; the coach is correct

Common error (e)(i) Adding the two speeds to get 5.5 m/s. Speeds add like this only when the two velocities lie along the same line; at right angles, use Pythagoras.

More 1.1 questions in the question bank

1.2 Speed, velocity and motion graphs

+ Add to test✓ In your testCC-CIP-12-119 marks~20 min

A skydiver jumping from a balloon: free fall, terminal speed and landing

1.2 Speed, velocity and motion graphs1.4 Forces, friction and springs1.7 Energy, work and powerCore

A skydiver of mass 85 kg steps out of a hot-air balloon that is not moving. Figure 1 shows how the speed of the skydiver changes during the first 60 s of the jump. The canopy (parachute) opens at time t = 30 s.

Take g = 9.8 N/kg.

Figure 1: speed–time graph for the first 60 s of the jump

(a)

State the acceleration of the skydiver at t = 0 and explain why it has this value.

RECALLNotes

Show solution

9.8 m/s²	✓ 1	Accept 9.8 N/kg or 10 m/s².
the speed is zero, so there is no air resistance (drag) and the only force is the weight	✓ 1	Accept: the skydiver is in free fall / falling only under gravity.

Answer: 9.8 m/s²

(2)

(b)

From t = 16 s until the canopy opens, the skydiver falls at a constant speed.

Use Figure 1 to calculate the distance fallen between t = 16 s and t = 30 s.

GRAPHNotes

Show solution

distance = area under graph = 50 × (30 − 16)	✓ 1	Speed of 50 m/s read from the graph and a time of 14 s.
700 m	✓ 1	Unit required.

Answer: 700 m

(2)

(c)

Explain, in terms of the forces on the skydiver, the shape of the graph between t = 0 and t = 16 s.

Extended onlyEXPLAINNotes

Show solution

as the speed increases, the air resistance (drag) increases, while the weight stays the same	✓ 1	Both ideas needed.
so the resultant force (weight − drag) decreases and the acceleration decreases: the gradient of the graph gets smaller	✓ 1	Accept: speed still increases but more slowly.
at about 16 s the drag equals the weight, so the resultant force is zero and the speed is constant (terminal speed)	✓ 1	"Gravity stops acting" scores 0 for this mark.

(3)

(d)(i)

State the size of the air resistance on the skydiver at t = 20 s.

CALCNotes

Show solution

833 N (equal to the weight, 85 × 9.8)

✓ 1

Accept 830 N or 850 N (with g = 10 N/kg).

Answer: 833 N

(1)

(d)(ii)

Show that the kinetic energy of the skydiver at t = 20 s is about 1.1 × 10⁵ J.

CALCNotes

Show solution

E_k = ½mv² = ½ × 85 × 50²	✓ 1	Speed of 50 m/s must come from the graph.
= 106 000 J (1.06 × 10⁵ J)	✓ 1	The unrounded value must be seen; 1.1 × 10⁵ J alone scores 0 for this mark.

Answer: 1.06 × 10⁵ J

(2)

(e)

Between t = 16 s and t = 30 s the skydiver loses gravitational potential energy.

Calculate the gravitational potential energy lost in this time, and state what happens to this energy.

MULTINotes

Show solution

ΔE_p = mgΔh = 85 × 9.8 × 700	✓ 1	Allow ECF from (b).
= 5.8 × 10⁵ J	✓ 1	Accept 583 000 J or 583 100 J.
the kinetic energy does not change (constant speed), so this energy is transferred to thermal energy of the air and the skydiver (by the drag)	✓ 1	Must state that the energy becomes thermal (internal) energy. "The energy is lost" scores 0.

Answer: 5.8 × 10⁵ J; transferred to thermal energy

(3)

(f)

Explain why the speed decreases quickly when the canopy opens, and why the skydiver then falls at a steady 6.0 m/s.

Extended onlyEXPLAINNotes

Show solution

the canopy greatly increases the air resistance, so the drag is now bigger than the weight and the resultant force is upwards: the skydiver decelerates	✓ 1	Upward resultant force or drag > weight needed.
as the speed falls the drag decreases until it equals the weight again, so the resultant force is zero and the speed is constant (a lower terminal speed)	✓ 1

(2)

(g)

By counting squares, estimate the distance fallen between t = 0 and t = 16 s.

Extended onlyGRAPHNotes

Show solution

area of one square = 2 s × 5 m/s = 10 m, and about 62 squares counted under the curve	✓ 1	Accept 58 to 67 squares; part squares estimated.
distance ≈ 620 m	✓ 1	Accept 580 m to 670 m. Unit required.

Answer: about 620 m

(2)

(h)

On a second jump, the skydiver falls head-down. In this position the air resistance is smaller at every speed. The canopy again opens at t = 30 s.

On Figure 1, sketch the graph for this jump from t = 0 to t = 30 s.

Extended onlySKETCHNotes

Show solution

curve starts at the origin with the same initial gradient (9.8 m/s²) as the original graph	✓ 1	The new curve must not start below the original.
curve lies above the original graph and levels off at a higher constant speed, reached later than 16 s	✓ 1	The curve may level off at any time after about 16 s, but it must be clearly flattening (gradient decreasing) and end above 50 m/s.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
9.8 m/s²	✓ 1	Accept 9.8 N/kg or 10 m/s².
the speed is zero, so there is no air resistance (drag) and the only force is the weight	✓ 1	Accept: the skydiver is in free fall / falling only under gravity.
Part (b)
distance = area under graph = 50 × (30 − 16)	✓ 1	Speed of 50 m/s read from the graph and a time of 14 s.
700 m	✓ 1	Unit required.
Part (c)
as the speed increases, the air resistance (drag) increases, while the weight stays the same	✓ 1	Both ideas needed.
so the resultant force (weight − drag) decreases and the acceleration decreases: the gradient of the graph gets smaller	✓ 1	Accept: speed still increases but more slowly.
at about 16 s the drag equals the weight, so the resultant force is zero and the speed is constant (terminal speed)	✓ 1	"Gravity stops acting" scores 0 for this mark.
Part (d)(i)
833 N (equal to the weight, 85 × 9.8)	✓ 1	Accept 830 N or 850 N (with g = 10 N/kg).
Part (d)(ii)
E_k = ½mv² = ½ × 85 × 50²	✓ 1	Speed of 50 m/s must come from the graph.
= 106 000 J (1.06 × 10⁵ J)	✓ 1	The unrounded value must be seen; 1.1 × 10⁵ J alone scores 0 for this mark.
Part (e)
ΔE_p = mgΔh = 85 × 9.8 × 700	✓ 1	Allow ECF from (b).
= 5.8 × 10⁵ J	✓ 1	Accept 583 000 J or 583 100 J.
the kinetic energy does not change (constant speed), so this energy is transferred to thermal energy of the air and the skydiver (by the drag)	✓ 1	Must state that the energy becomes thermal (internal) energy. "The energy is lost" scores 0.
Part (f)
the canopy greatly increases the air resistance, so the drag is now bigger than the weight and the resultant force is upwards: the skydiver decelerates	✓ 1	Upward resultant force or drag > weight needed.
as the speed falls the drag decreases until it equals the weight again, so the resultant force is zero and the speed is constant (a lower terminal speed)	✓ 1
Part (g)
area of one square = 2 s × 5 m/s = 10 m, and about 62 squares counted under the curve	✓ 1	Accept 58 to 67 squares; part squares estimated.
distance ≈ 620 m	✓ 1	Accept 580 m to 670 m. Unit required.
Part (h)
curve starts at the origin with the same initial gradient (9.8 m/s²) as the original graph	✓ 1	The new curve must not start below the original.
curve lies above the original graph and levels off at a higher constant speed, reached later than 16 s	✓ 1	The curve may level off at any time after about 16 s, but it must be clearly flattening (gradient decreasing) and end above 50 m/s.

Answers: (a) 9.8 m/s² · (b) 700 m · (d)(i) 833 N · (d)(ii) 1.06 × 10⁵ J · (e) 5.8 × 10⁵ J; transferred to thermal energy · (g) about 620 m

Exam tip (c) For terminal speed, write the steps in order: drag increases with speed, the resultant force decreases, drag becomes equal to weight, so the acceleration is zero and the speed is constant.

Common error (e) Saying that the lost potential energy becomes kinetic energy. At constant speed the kinetic energy stays the same, so the energy goes to the surroundings as thermal energy.

More 1.2 questions in the question bank

+ Add to test✓ In your testCC-CIP-12-217 marks~20 min

A driverless metro train between two stations

1.2 Speed, velocity and motion graphs1.1 Physical quantities and measurement1.4 Forces, friction and springsCore

A driverless metro train runs along a straight, level track between two stations. Its top speed between these stations is 54 km/h.

Leaving the first station, the train starts from rest and accelerates uniformly for 20 s until it reaches its top speed. It then travels at this constant speed for 64 s. Finally it decelerates uniformly and comes to rest at the second station 12 s later.

(a)

Show that 54 km/h is equal to 15 m/s.

CALCNotes

Show solution

54 × 1000 ÷ 3600 = 15 m/s

✓ 1

Accept 54 ÷ 3.6 = 15. The working must be seen.

Answer: 15 m/s

(1)

(b)

On Figure 1, draw the speed–time graph for the journey between the two stations.

Figure 1: grid for the speed–time graph of the train

GRAPHNotes

Show solution

straight line from (0, 0) to (20 s, 15 m/s), then horizontal at 15 m/s to 84 s	✓ 1	Correct times 20 s and 84 s needed.
straight line from (84 s, 15 m/s) down to (96 s, 0)	✓ 1	Allow ECF for a misplotted earlier point if the deceleration lasts 12 s.

(2)

(c)

Calculate the acceleration of the train during the first 20 s.

Extended onlyCALCNotes

Show solution

a = (v − u) ÷ t = (15 − 0) ÷ 20	✓ 1	Or the gradient of the graph.
0.75 m/s²	✓ 1	Unit required.

Answer: 0.75 m/s²

(2)

(d)

Use your graph to show that the two stations are 1.2 km apart.

MULTINotes

Show solution

distance = area under the speed–time graph	✓ 1	Stated or clearly used.
½ × 20 × 15 + 64 × 15 + ½ × 12 × 15 = 150 + 960 + 90	✓ 1	Accept the trapezium: ½ × (64 + 96) × 15.
= 1200 m = 1.2 km	✓ 1	Conversion to km (or 1.2 km to 1200 m) must be seen.

Answer: 1200 m

(3)

(e)

Calculate the average speed of the train between the two stations.

CALCNotes

Show solution

average speed = total distance ÷ total time = 1200 ÷ 96	✓ 1	Allow ECF from (d).
12.5 m/s	✓ 1	Accept 13 m/s. Unit required. 7.5 m/s (averaging 0 and 15) scores 0.

Answer: 12.5 m/s

(2)

(f)

The total mass of the train and passengers is 1.6 × 10⁵ kg.

Calculate the resultant force on the train while it decelerates.

Extended onlyCALCNotes

Show solution

deceleration = 15 ÷ 12 = 1.25 m/s²	✓ 1	Accept acceleration = −1.25 m/s².
F = ma = 1.6 × 10⁵ × 1.25 = 2.0 × 10⁵ N	✓ 1	Allow ECF from the deceleration. Unit required. Ignore a negative sign.

Answer: 2.0 × 10⁵ N

(2)

(g)

Describe the shape of the distance–time graph for the journey between the two stations, and explain each part of the shape.

EXPLAINNotes

Show solution

0 to 20 s: a curve getting steeper, because the speed is increasing	✓ 1	Gradient of a distance–time graph = speed, stated or implied.
20 s to 84 s: a straight sloping line, because the speed is constant	✓ 1
84 s to 96 s: a curve that becomes less steep and ends horizontal at 1200 m, because the train slows down and stops	✓ 1	Must mention the line becoming horizontal (at rest).

(3)

(h)

The stations on this line are all 1.2 km apart, and every journey between stations is the same as the one described. At each station the train stops for 24 s.

Calculate the average speed of the train over a long run along the line.

PREDICTNotes

Show solution

each station-to-station section takes 96 + 24 = 120 s for 1200 m	✓ 1	The time at rest must be included.
average speed = 1200 ÷ 120 = 10 m/s	✓ 1	Allow ECF from (d). Unit required.

Answer: 10 m/s

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
54 × 1000 ÷ 3600 = 15 m/s	✓ 1	Accept 54 ÷ 3.6 = 15. The working must be seen.
Part (b)
straight line from (0, 0) to (20 s, 15 m/s), then horizontal at 15 m/s to 84 s	✓ 1	Correct times 20 s and 84 s needed.
straight line from (84 s, 15 m/s) down to (96 s, 0)	✓ 1	Allow ECF for a misplotted earlier point if the deceleration lasts 12 s.
Part (c)
a = (v − u) ÷ t = (15 − 0) ÷ 20	✓ 1	Or the gradient of the graph.
0.75 m/s²	✓ 1	Unit required.
Part (d)
distance = area under the speed–time graph	✓ 1	Stated or clearly used.
½ × 20 × 15 + 64 × 15 + ½ × 12 × 15 = 150 + 960 + 90	✓ 1	Accept the trapezium: ½ × (64 + 96) × 15.
= 1200 m = 1.2 km	✓ 1	Conversion to km (or 1.2 km to 1200 m) must be seen.
Part (e)
average speed = total distance ÷ total time = 1200 ÷ 96	✓ 1	Allow ECF from (d).
12.5 m/s	✓ 1	Accept 13 m/s. Unit required. 7.5 m/s (averaging 0 and 15) scores 0.
Part (f)
deceleration = 15 ÷ 12 = 1.25 m/s²	✓ 1	Accept acceleration = −1.25 m/s².
F = ma = 1.6 × 10⁵ × 1.25 = 2.0 × 10⁵ N	✓ 1	Allow ECF from the deceleration. Unit required. Ignore a negative sign.
Part (g)
0 to 20 s: a curve getting steeper, because the speed is increasing	✓ 1	Gradient of a distance–time graph = speed, stated or implied.
20 s to 84 s: a straight sloping line, because the speed is constant	✓ 1
84 s to 96 s: a curve that becomes less steep and ends horizontal at 1200 m, because the train slows down and stops	✓ 1	Must mention the line becoming horizontal (at rest).
Part (h)
each station-to-station section takes 96 + 24 = 120 s for 1200 m	✓ 1	The time at rest must be included.
average speed = 1200 ÷ 120 = 10 m/s	✓ 1	Allow ECF from (d). Unit required.

Answers: (a) 15 m/s · (c) 0.75 m/s² · (d) 1200 m · (e) 12.5 m/s · (f) 2.0 × 10⁵ N · (h) 10 m/s

Common error (d) Multiplying the top speed by the total time (15 × 96 = 1440 m). The train is slower than 15 m/s while it speeds up and slows down, so split the area into triangles and a rectangle.

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1.3 Mass, weight and density

+ Add to test✓ In your testCC-CIP-13-119 marks~20 min

Is the gold bar real? Density in a jeweller's lab

1.3 Mass, weight and density1.1 Physical quantities and measurement1.8 Pressure in solids and liquidsCore

A jeweller is offered a small bar that is sold as pure gold. The bar is roughly a cuboid, but its edges and corners are rounded. She tests it in three ways.

An electronic balance gives the mass of the bar as 198 g.
Calipers give the outside dimensions of the bar as 50.0 mm × 24.0 mm × 15.0 mm.
The bar is lowered on a thin thread into a measuring cylinder of water. Figure 1 shows enlarged views of the scale before and after.

Take g = 9.8 N/kg on the Earth.

Figure 1: enlarged views of the measuring cylinder scale, A before and B after the bar is lowered in (readings in cm³)

(a)

Calculate the weight of the bar on the Earth.

CALCNotes

Show solution

m = 198 g = 0.198 kg	✓ 1	The conversion to kg may be seen in the substitution.
W = mg = 0.198 × 9.8 = 1.94 N	✓ 1	Accept 1.9 N. 198 × 9.8 = 1940 N (mass not converted to kg) scores 1 mark in total.

Answer: 1.94 N

(2)

(b)

Calculate the volume of the bar, in cm³, from the caliper measurements.

CALCNotes

Show solution

dimensions converted: 5.00 cm × 2.40 cm × 1.50 cm	✓ 1	Or 50.0 × 24.0 × 15.0 = 18 000 mm³ seen.
V = 18.0 cm³	✓ 1	Accept 18 cm³. 18 000 with no conversion to cm³ scores the first mark only.

Answer: 18.0 cm³

(2)

(c)

Use Figure 1 to find the volume of the bar measured by displacement.

DATANotes

Show solution

readings 24.0 cm³ (A) and 41.5 cm³ (B)	✓ 1	Both readings needed, taken at the bottom of the meniscus.
volume = 41.5 − 24.0 = 17.5 cm³	✓ 1	Allow ECF from misread values if the subtraction is correct. 41.5 cm³ alone scores 0.

Answer: 17.5 cm³

(2)

(d)

Show that the density of the bar, using the volume from (c), is about 11 g/cm³.

DERIVENotes

Show solution

ρ = m ÷ V = 198 ÷ 17.5	✓ 1	The full substitution must be seen.
= 11.3 g/cm³	✓ 1	At least 3 s.f. needed, since the answer is given. Accept 11 300 kg/m³.

Answer: 11.3 g/cm³

(2)

(e)(i)

Figure 2 shows how the mass depends on the volume for samples of three pure metals.

Plot a point on Figure 2 to represent the bar. Use your point to suggest what metal makes up most of the bar.

Figure 2: mass against volume for samples of gold, lead and copper

GRAPHNotes

Show solution

point plotted at (17.5 cm³, 198 g)	✓ 1	Within half a small square. Allow ECF from (c).
the point lies on the lead line, so the bar is mainly lead (with a thin coating of gold)	✓ 1	Accept lead alone if the point is on or very close to the lead line. Do not accept copper or gold.

Answer: lead (coated with gold)

(2)

(e)(ii)

Use Figure 2 to find the mass that a bar of pure gold with the same volume would have.

GRAPHNotes

Show solution

about 338 g

✓ 1

Accept 330–345 g read from the gold line at 17.5 cm³. Allow ECF from (c).

Answer: about 338 g

(1)

(f)

Using the caliper volume from (b) gives a density of 11.0 g/cm³ instead.

Explain why the displacement method gives the more reliable value of the volume for this bar.

EXPLAINNotes

Show solution

the rounded edges and corners mean the bar is not a true cuboid, so length × width × height includes space that the bar does not fill / the caliper volume is too large	✓ 1	Any clear statement that the calculated volume is too big because of the missing metal at the edges.
displacement measures the actual volume of water pushed aside, whatever the shape of the bar	✓ 1	Accept: displacement works for any irregular shape.

(2)

(g)

The bar rests on a flat face on the pan of the balance. The area of the bar in contact with the pan is 10.8 cm².

Calculate the pressure exerted by the bar on the pan.

MULTINotes

Show solution

A = 10.8 cm² = 10.8 × 10⁻⁴ m² (= 1.08 × 10⁻³ m²)	✓ 1
p = F ÷ A = 1.94 ÷ 1.08 × 10⁻³	✓ 1	Allow ECF from (a).
= 1800 Pa	✓ 1	Accept 1790–1800 Pa. 0.18 (N/cm²) with the area left in cm² scores the second mark only.

Answer: 1800 Pa

(3)

(h)

An identical bar is later tested in a laboratory on the Moon, where g = 1.6 N/kg. The bar rests on the same face.

State the mass of the bar on the Moon and calculate the pressure that it exerts on the pan there.

PREDICTNotes

Show solution

mass = 198 g (unchanged: the amount of matter is the same)	✓ 1
W = 0.198 × 1.6 = 0.317 N	✓ 1
p = 0.317 ÷ 1.08 × 10⁻³ = 293 Pa	✓ 1	Accept 290–295 Pa. Accept the ratio method: 1800 × 1.6 ÷ 9.8 = 294 Pa. Allow ECF from (g).

Answer: 198 g; 293 Pa

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
m = 198 g = 0.198 kg	✓ 1	The conversion to kg may be seen in the substitution.
W = mg = 0.198 × 9.8 = 1.94 N	✓ 1	Accept 1.9 N. 198 × 9.8 = 1940 N (mass not converted to kg) scores 1 mark in total.
Part (b)
dimensions converted: 5.00 cm × 2.40 cm × 1.50 cm	✓ 1	Or 50.0 × 24.0 × 15.0 = 18 000 mm³ seen.
V = 18.0 cm³	✓ 1	Accept 18 cm³. 18 000 with no conversion to cm³ scores the first mark only.
Part (c)
readings 24.0 cm³ (A) and 41.5 cm³ (B)	✓ 1	Both readings needed, taken at the bottom of the meniscus.
volume = 41.5 − 24.0 = 17.5 cm³	✓ 1	Allow ECF from misread values if the subtraction is correct. 41.5 cm³ alone scores 0.
Part (d)
ρ = m ÷ V = 198 ÷ 17.5	✓ 1	The full substitution must be seen.
= 11.3 g/cm³	✓ 1	At least 3 s.f. needed, since the answer is given. Accept 11 300 kg/m³.
Part (e)(i)
point plotted at (17.5 cm³, 198 g)	✓ 1	Within half a small square. Allow ECF from (c).
the point lies on the lead line, so the bar is mainly lead (with a thin coating of gold)	✓ 1	Accept lead alone if the point is on or very close to the lead line. Do not accept copper or gold.
Part (e)(ii)
about 338 g	✓ 1	Accept 330–345 g read from the gold line at 17.5 cm³. Allow ECF from (c).
Part (f)
the rounded edges and corners mean the bar is not a true cuboid, so length × width × height includes space that the bar does not fill / the caliper volume is too large	✓ 1	Any clear statement that the calculated volume is too big because of the missing metal at the edges.
displacement measures the actual volume of water pushed aside, whatever the shape of the bar	✓ 1	Accept: displacement works for any irregular shape.
Part (g)
A = 10.8 cm² = 10.8 × 10⁻⁴ m² (= 1.08 × 10⁻³ m²)	✓ 1
p = F ÷ A = 1.94 ÷ 1.08 × 10⁻³	✓ 1	Allow ECF from (a).
= 1800 Pa	✓ 1	Accept 1790–1800 Pa. 0.18 (N/cm²) with the area left in cm² scores the second mark only.
Part (h)
mass = 198 g (unchanged: the amount of matter is the same)	✓ 1
W = 0.198 × 1.6 = 0.317 N	✓ 1
p = 0.317 ÷ 1.08 × 10⁻³ = 293 Pa	✓ 1	Accept 290–295 Pa. Accept the ratio method: 1800 × 1.6 ÷ 9.8 = 294 Pa. Allow ECF from (g).

Answers: (a) 1.94 N · (b) 18.0 cm³ · (c) 17.5 cm³ · (d) 11.3 g/cm³ · (e)(i) lead (coated with gold) · (e)(ii) about 338 g · (g) 1800 Pa · (h) 198 g; 293 Pa

Common error (c) Giving the final reading, 41.5 cm³, as the volume of the bar. The bar only accounts for the rise in the water level, final reading minus starting reading.

Exam tip (g) Convert the area to m² before you substitute into p = F/A: 1 cm² = 0.0001 m². Only then is the answer in pascals.

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Why a hot-air balloon floats: density of hot and cold air

1.3 Mass, weight and density1.8 Pressure in solids and liquids2.1 Particle model and gas pressureCore

Figure 1 shows a hot-air balloon. The envelope has a volume of 2800 m³ and is open at the bottom. A gas burner heats the air inside the envelope.

The envelope has a mass of 90 kg. The basket, burner, fuel and passengers have a total mass of 360 kg.

On the morning of the flight the density of the outside air at ground level is 1.20 kg/m³. Take g = 9.8 N/kg.

Figure 1: a hot-air balloon (not to scale)

(a)

Before the envelope is attached, the loaded basket (basket, burner, fuel and passengers, total mass 360 kg) stands on soft grass. The base of the basket measures 1.50 m × 1.20 m.

Calculate the pressure exerted on the grass by the loaded basket.

CALCNotes

Show solution

F = W = 360 × 9.8 = 3530 N (3528 N)	✓ 1
p = F ÷ A = 3528 ÷ (1.50 × 1.20) = 1960 Pa	✓ 1	Accept 1.96 kPa. Using 360 N as the force scores 0 for this mark.

Answer: 1960 Pa

(2)

(b)

The envelope is first filled with air at the outside temperature.

Calculate the mass of air inside the envelope.

CALCNotes

Show solution

m = ρV = 1.20 × 2800	✓ 1
= 3360 kg	✓ 1	Unit required.

Answer: 3360 kg

(2)

(c)

The burner now heats the air inside the envelope. The pressure inside stays equal to the pressure outside.

Explain, in terms of particles, why the density of the air inside the envelope decreases.

EXPLAINNotes

Show solution

the air particles gain kinetic energy / move faster	✓ 1
the particles move further apart, so the air expands and some of it leaves through the opening	✓ 1	Accept: faster particles push harder on the envelope and on the air outside, so air is pushed out.
fewer particles (less mass of air) in each m³, so the density is smaller	✓ 1	Do not accept 'the particles get bigger' or 'the particles get lighter'.

(3)

(d)

When it is hot, the air inside the envelope has a density of 0.95 kg/m³.

Show that the weight of the air inside the envelope has decreased by about 7 kN.

DERIVENotes

Show solution

mass of hot air = 0.95 × 2800 = 2660 kg	✓ 1
decrease in mass = 3360 − 2660 = 700 kg	✓ 1	Or: weights 32 900 N and 26 100 N found, or (1.20 − 0.95) × 2800 = 700 kg. Allow ECF from (b).
decrease in weight = 700 × 9.8 = 6860 N (≈ 6.9 kN)	✓ 1	Final value to at least 2 s.f. needed, since the answer is given.

Answer: 6860 N

(3)

(e)

The loaded balloon is the envelope, the hot air, the basket, the burner, the fuel and the passengers. Take its volume to be 2800 m³.

Calculate the average density of the loaded balloon and state whether it rises. Give a reason.

REASONNotes

Show solution

total mass = 2660 + 90 + 360 = 3110 kg	✓ 1	Allow ECF from (d).
average density = 3110 ÷ 2800 = 1.11 kg/m³	✓ 1
1.11 kg/m³ is less than 1.20 kg/m³ (density of outside air), so the balloon rises	✓ 1	The comparison of densities is needed. Allow ECF from an incorrect average density.

Answer: 1.11 kg/m³; it rises

(3)

(f)(i)

The table shows how the density of the outside air changes with height on the morning of the flight.

height above the ground / m	0	500	1000	1500	2000	2500
density of outside air / kg/m³	1.20	1.14	1.08	1.03	0.97	0.92

On Figure 2, plot the points from the table and draw a smooth curve through them.

Figure 2: axes for the density of the outside air against height

GRAPHNotes

Show solution

at least 4 points plotted correctly	✓ 1	Within half a small square.
all 6 points plotted correctly	✓ 1
smooth curve (or line) through all the points, drawn with care	✓ 1	Not dot-to-dot with sharp corners.

(3)

(f)(ii)

In a simple model, the average density of the loaded balloon stays equal to your answer to (e) as it rises.

Use your graph to estimate the height at which the balloon stops rising. Explain your answer.

GRAPHNotes

Show solution

it stops rising where the density of the outside air equals the average density of the balloon (1.11 kg/m³)	✓ 1
about 750 m	✓ 1	Accept 700–800 m read from the candidate's own curve at 1.11 kg/m³. Allow ECF from (e).

Answer: about 750 m

(2)

(g)

On a cold winter morning the density of the outside air at ground level is 1.29 kg/m³. The air in the envelope is heated to the same density, 0.95 kg/m³, as before.

Calculate the greatest total mass of envelope, basket, burner, fuel and passengers that the balloon can just lift off the ground.

PREDICTNotes

Show solution

for lift-off, total mass ÷ 2800 must not exceed 1.29, so total mass ≤ 1.29 × 2800 = 3612 kg	✓ 1
greatest mass = 3612 − 2660 = 952 kg	✓ 1	Accept 950 kg. Allow ECF from (d).

Answer: 952 kg

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
F = W = 360 × 9.8 = 3530 N (3528 N)	✓ 1
p = F ÷ A = 3528 ÷ (1.50 × 1.20) = 1960 Pa	✓ 1	Accept 1.96 kPa. Using 360 N as the force scores 0 for this mark.
Part (b)
m = ρV = 1.20 × 2800	✓ 1
= 3360 kg	✓ 1	Unit required.
Part (c)
the air particles gain kinetic energy / move faster	✓ 1
the particles move further apart, so the air expands and some of it leaves through the opening	✓ 1	Accept: faster particles push harder on the envelope and on the air outside, so air is pushed out.
fewer particles (less mass of air) in each m³, so the density is smaller	✓ 1	Do not accept 'the particles get bigger' or 'the particles get lighter'.
Part (d)
mass of hot air = 0.95 × 2800 = 2660 kg	✓ 1
decrease in mass = 3360 − 2660 = 700 kg	✓ 1	Or: weights 32 900 N and 26 100 N found, or (1.20 − 0.95) × 2800 = 700 kg. Allow ECF from (b).
decrease in weight = 700 × 9.8 = 6860 N (≈ 6.9 kN)	✓ 1	Final value to at least 2 s.f. needed, since the answer is given.
Part (e)
total mass = 2660 + 90 + 360 = 3110 kg	✓ 1	Allow ECF from (d).
average density = 3110 ÷ 2800 = 1.11 kg/m³	✓ 1
1.11 kg/m³ is less than 1.20 kg/m³ (density of outside air), so the balloon rises	✓ 1	The comparison of densities is needed. Allow ECF from an incorrect average density.
Part (f)(i)
at least 4 points plotted correctly	✓ 1	Within half a small square.
all 6 points plotted correctly	✓ 1
smooth curve (or line) through all the points, drawn with care	✓ 1	Not dot-to-dot with sharp corners.
Part (f)(ii)
it stops rising where the density of the outside air equals the average density of the balloon (1.11 kg/m³)	✓ 1
about 750 m	✓ 1	Accept 700–800 m read from the candidate's own curve at 1.11 kg/m³. Allow ECF from (e).
Part (g)
for lift-off, total mass ÷ 2800 must not exceed 1.29, so total mass ≤ 1.29 × 2800 = 3612 kg	✓ 1
greatest mass = 3612 − 2660 = 952 kg	✓ 1	Accept 950 kg. Allow ECF from (d).

Answers: (a) 1960 Pa · (b) 3360 kg · (d) 6860 N · (e) 1.11 kg/m³; it rises · (f)(ii) about 750 m · (g) 952 kg

Common error (c) Writing that the particles themselves expand or become lighter when heated. They stay the same size and mass; they move faster and spread further apart, so fewer of them fit in each cubic metre.

Exam tip (e) In a float-or-sink answer, name both densities and compare them: 'the average density of the balloon is less than the density of the air, so it rises'.

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1.4 Forces, friction and springs

+ Add to test✓ In your testCC-CIP-14-120 marks~20 min

Testing a bungee cord: extension, limit of proportionality and safety

1.4 Forces, friction and springs1.2 Speed, velocity and motion graphs1.7 Energy, work and powerExtended

An adventure park tests a new bungee cord on a tall test tower. The unstretched cord is 10.0 m long. Loads are hung from its lower end and the extension of the cord is measured for each load.

load F / N	0	500	1000	1500	2000	2500	3000	3500
extension x / m	0	2.4	5.1	7.5	9.9	12.6	15.8	20.1

Take g = 9.8 N/kg. Ignore air resistance throughout.

(a)

On Figure 1, plot the results and draw the best-fit line.

Figure 1: axes for the force–extension graph of the bungee cord

GRAPHNotes

Show solution

all 8 points plotted correctly	✓ 1	Within half a small square.
straight line through the origin and the points up to 2500 N, curving away from it (towards the extension axis) for the last points	✓ 1	The straight line must not be forced through the points at 3000 N and 3500 N.

(2)

(b)

Use your graph to determine the spring constant k of the cord.

Extended onlyGRAPHNotes

Show solution

gradient of the straight part from a large triangle, e.g. 2500 ÷ 12.5	✓ 1	Values read from the line, not just one table pair, though a correct table pair on the line is allowed.
k = 200 N/m	✓ 1	Accept 195–205 N/m. Unit required.

Answer: 200 N/m

(2)

(c)

State the load at the limit of proportionality of the cord, and describe how the graph shows it.

Extended onlyGRAPHNotes

Show solution

about 2500 N	✓ 1	Accept 2500–2800 N from the candidate's graph.
beyond this load the graph is no longer a straight line / the extension increases faster than the load (no longer proportional)	✓ 1

Answer: about 2500 N

(2)

(d)

A jumper of mass 60 kg uses the cord.

Show that when the jumper hangs at rest at the end of the jump, the extension of the cord is about 2.9 m.

Extended onlyDERIVENotes

Show solution

W = mg = 60 × 9.8 = 588 N	✓ 1	Tension = weight when at rest must be clear from the working.
x = F ÷ k = 588 ÷ 200 = 2.94 m	✓ 1	Allow ECF from (b). Final value to at least 2 s.f.

Answer: 2.94 m

(2)

(e)

The jumper steps off the platform from rest and falls freely through 10.0 m before the cord starts to stretch.

Calculate the speed of the jumper at the moment the cord starts to stretch.

CALCNotes

Show solution

ΔE_p = mgΔh = 60 × 9.8 × 10.0 = 5880 J	✓ 1
½ × 60 × v² = 5880, so v² = 196	✓ 1	Or v² = 2gΔh = 196.
v = 14.0 m/s	✓ 1	Accept 14 m/s.

Answer: 14.0 m/s

(3)

(f)(i)

At the lowest point of the jump the jumper is momentarily at rest and the extension of the cord is 11.2 m.

Calculate the elastic (strain) energy stored in the cord at this point.

MULTINotes

Show solution

total drop = 10.0 + 11.2 = 21.2 m, all the gravitational potential energy lost becomes elastic (strain) energy	✓ 1
E = 60 × 9.8 × 21.2 = 12 500 J (1.25 × 10⁴ J)	✓ 1	Accept 12 466 J. Using 11.2 m only (6590 J) scores 0.

Answer: 12 500 J

(2)

(f)(ii)

Show that the cord still obeys Hooke's law at the lowest point of the jump.

Extended onlyREASONNotes

Show solution

F = kx = 200 × 11.2 = 2240 N	✓ 1	Or: the extension at the limit, 12.5 m, is greater than 11.2 m. Allow ECF from (b).
2240 N is less than the load at the limit of proportionality (2500 N), so the cord is on the straight part of the graph	✓ 1	Allow ECF from (c).

Answer: 2240 N < 2500 N

(2)

(g)

Describe and explain how the speed of the jumper changes from the moment the cord starts to stretch until the lowest point.

EXPLAINNotes

Show solution

at first the upward force from the cord is less than the weight, so the resultant force is downwards and the jumper keeps speeding up (with decreasing acceleration)	✓ 1
the force from the cord increases with the extension; when it equals the weight (extension 2.94 m) the resultant force is zero and the speed is at its maximum	✓ 1
beyond this the force from the cord is greater than the weight, so the resultant force is upwards and the jumper slows down, reaching zero speed at the lowest point	✓ 1

(3)

(h)

The park also offers a double-cord set-up: two identical cords hang side by side, both fixed to the same harness.

Calculate the extension of each cord when the 60 kg jumper hangs at rest from the pair.

PREDICTNotes

Show solution

each cord supports half the weight: 588 ÷ 2 = 294 N	✓ 1
x = 294 ÷ 200 = 1.47 m	✓ 1	Accept 1.5 m. Allow ECF from (b) and (d). Alternative: half the force on each cord gives half the extension, 2.94 ÷ 2 = 1.47 m (both marks).

Answer: 1.47 m

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
all 8 points plotted correctly	✓ 1	Within half a small square.
straight line through the origin and the points up to 2500 N, curving away from it (towards the extension axis) for the last points	✓ 1	The straight line must not be forced through the points at 3000 N and 3500 N.
Part (b)
gradient of the straight part from a large triangle, e.g. 2500 ÷ 12.5	✓ 1	Values read from the line, not just one table pair, though a correct table pair on the line is allowed.
k = 200 N/m	✓ 1	Accept 195–205 N/m. Unit required.
Part (c)
about 2500 N	✓ 1	Accept 2500–2800 N from the candidate's graph.
beyond this load the graph is no longer a straight line / the extension increases faster than the load (no longer proportional)	✓ 1
Part (d)
W = mg = 60 × 9.8 = 588 N	✓ 1	Tension = weight when at rest must be clear from the working.
x = F ÷ k = 588 ÷ 200 = 2.94 m	✓ 1	Allow ECF from (b). Final value to at least 2 s.f.
Part (e)
ΔE_p = mgΔh = 60 × 9.8 × 10.0 = 5880 J	✓ 1
½ × 60 × v² = 5880, so v² = 196	✓ 1	Or v² = 2gΔh = 196.
v = 14.0 m/s	✓ 1	Accept 14 m/s.
Part (f)(i)
total drop = 10.0 + 11.2 = 21.2 m, all the gravitational potential energy lost becomes elastic (strain) energy	✓ 1
E = 60 × 9.8 × 21.2 = 12 500 J (1.25 × 10⁴ J)	✓ 1	Accept 12 466 J. Using 11.2 m only (6590 J) scores 0.
Part (f)(ii)
F = kx = 200 × 11.2 = 2240 N	✓ 1	Or: the extension at the limit, 12.5 m, is greater than 11.2 m. Allow ECF from (b).
2240 N is less than the load at the limit of proportionality (2500 N), so the cord is on the straight part of the graph	✓ 1	Allow ECF from (c).
Part (g)
at first the upward force from the cord is less than the weight, so the resultant force is downwards and the jumper keeps speeding up (with decreasing acceleration)	✓ 1
the force from the cord increases with the extension; when it equals the weight (extension 2.94 m) the resultant force is zero and the speed is at its maximum	✓ 1
beyond this the force from the cord is greater than the weight, so the resultant force is upwards and the jumper slows down, reaching zero speed at the lowest point	✓ 1
Part (h)
each cord supports half the weight: 588 ÷ 2 = 294 N	✓ 1
x = 294 ÷ 200 = 1.47 m	✓ 1	Accept 1.5 m. Allow ECF from (b) and (d). Alternative: half the force on each cord gives half the extension, 2.94 ÷ 2 = 1.47 m (both marks).

Answers: (b) 200 N/m · (c) about 2500 N · (d) 2.94 m · (e) 14.0 m/s · (f)(i) 12 500 J · (f)(ii) 2240 N < 2500 N · (h) 1.47 m

Common error (f)(i) Using only the stretch, 11.2 m, as the height fallen. The jumper falls the slack 10.0 m as well, so the energy stored comes from the whole 21.2 m drop.

More 1.4 questions in the question bank

+ Add to test✓ In your testCC-CIP-14-220 marks~20 min

An electric racing car: thrust, drag and cornering

1.4 Forces, friction and springs1.2 Speed, velocity and motion graphs1.6 Momentum and impulseExtended

An electric racing car, including its driver, has a mass of 900 kg. It starts from rest on a straight, level track. Figure 1 shows the car moving along the straight.

Figure 1: the car on the straight (not to scale)

(a)

On Figure 1, draw and label arrows to show the four forces acting on the car while it is speeding up along the straight.

SKETCHNotes

Show solution

all four forces correct: weight (down), upward (contact) force from the track, driving force (forwards), resistive force / drag and friction (backwards)	✓ 1	Arrows must start on the car and be labelled.
driving force arrow drawn longer than the resistive force arrow, and the upward and weight arrows of equal length	✓ 1	Two or three forces correct and labelled: award 1 mark in total.

(2)

(b)

Just after the start, the driving force on the car is 5400 N and the total resistive force is 900 N.

Calculate the acceleration of the car at this moment.

Extended onlyCALCNotes

Show solution

resultant force = 5400 − 900 = 4500 N	✓ 1
a = F ÷ m = 4500 ÷ 900	✓ 1
= 5.0 m/s²	✓ 1	Using 5400 N gives 6.0 m/s²: award the second mark only.

Answer: 5.0 m/s²

(3)

(c)(i)

Figure 2 shows the speed of the car during the first 80 s. The driving force stays at 5400 N.

State the greatest speed that the car reaches.

Figure 2: speed–time graph for the car from the start

GRAPHNotes

Show solution

70 m/s

✓ 1

Accept 69–71 m/s.

Answer: 70 m/s

(1)

(c)(ii)

State the size of the total resistive force on the car when it moves at its greatest speed.

RECALLNotes

Show solution

5400 N

✓ 1

Equal to the driving force.

Answer: 5400 N

(1)

(c)(iii)

Describe how the acceleration of the car changes during the first 80 s, and explain this change in terms of the forces on the car.

Extended onlyEXPLAINNotes

Show solution

the gradient of the graph decreases, so the acceleration decreases (to zero)	✓ 1
as the speed increases the drag increases, so the resultant force decreases	✓ 1	Accept 'resistive force' for drag.
at about 70 m/s the resistive force equals the driving force, the resultant force is zero, so the acceleration is zero and the speed stays constant	✓ 1

(3)

(d)

Later in the race the car goes round a hairpin bend, a semicircle with centre C, at a constant speed of 20 m/s (Figure 3).

Explain why a resultant force must act on the car on the bend, and state what provides this force and its direction.

Figure 3: plan view of the hairpin bend; C is the centre of the semicircle (not to scale)

Extended onlyEXPLAINNotes

Show solution

the direction of the car's velocity is changing, so the car is accelerating and (by F = ma) needs a resultant force	✓ 1	Constant speed but changing velocity must be clear.
friction between the tyres and the track, directed towards the centre C	✓ 1	Both the source and the direction needed.

(2)

(e)

At point P the car drives onto a patch of oil, where the friction between the tyres and the track is almost zero.

On Figure 3, draw the path of the car just after P. Explain your answer.

Extended onlyPREDICTNotes

Show solution

straight line drawn from P along the tangent to the bend, in the direction of travel	✓ 1	A path curving outwards or drawn directly away from C scores 0.
there is no force towards the centre (no resultant force), so the car continues with constant velocity, in a straight line	✓ 1

(2)

(f)

In a later race the track is wet, and the greatest friction force that the tyres can provide is smaller.

Explain why the driver must take the hairpin bend at a lower speed.

Extended onlyREASONNotes

Show solution

the force needed towards the centre is larger at a higher speed (for the same radius and mass)	✓ 1
the smaller friction on the wet track cannot provide the force needed at 20 m/s, so the car would not follow the bend; a lower speed needs a smaller force	✓ 1

(2)

(g)(i)

In a safety test the car, moving at 20 m/s, is driven straight into a crash barrier and is brought to rest.

Show that the impulse on the car is 18 000 N s.

Extended onlyDERIVENotes

Show solution

change in momentum = mv − mu = 900 × 0 − 900 × 20	✓ 1
size = 18 000 kg m/s = 18 000 N s	✓ 1	Accept −18 000 N s, with the sign explained as opposite to the motion.

Answer: 18 000 N s

(2)

(g)(ii)

The car takes 0.25 s to come to rest.

Calculate the average force exerted by the barrier on the car.

Extended onlyCALCNotes

Show solution

F = impulse ÷ time = 18 000 ÷ 0.25	✓ 1
= 72 000 N	✓ 1	Accept 72 kN or 7.2 × 10⁴ N.

Answer: 72 000 N

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
all four forces correct: weight (down), upward (contact) force from the track, driving force (forwards), resistive force / drag and friction (backwards)	✓ 1	Arrows must start on the car and be labelled.
driving force arrow drawn longer than the resistive force arrow, and the upward and weight arrows of equal length	✓ 1	Two or three forces correct and labelled: award 1 mark in total.
Part (b)
resultant force = 5400 − 900 = 4500 N	✓ 1
a = F ÷ m = 4500 ÷ 900	✓ 1
= 5.0 m/s²	✓ 1	Using 5400 N gives 6.0 m/s²: award the second mark only.
Part (c)(i)
70 m/s	✓ 1	Accept 69–71 m/s.
Part (c)(ii)
5400 N	✓ 1	Equal to the driving force.
Part (c)(iii)
the gradient of the graph decreases, so the acceleration decreases (to zero)	✓ 1
as the speed increases the drag increases, so the resultant force decreases	✓ 1	Accept 'resistive force' for drag.
at about 70 m/s the resistive force equals the driving force, the resultant force is zero, so the acceleration is zero and the speed stays constant	✓ 1
Part (d)
the direction of the car's velocity is changing, so the car is accelerating and (by F = ma) needs a resultant force	✓ 1	Constant speed but changing velocity must be clear.
friction between the tyres and the track, directed towards the centre C	✓ 1	Both the source and the direction needed.
Part (e)
straight line drawn from P along the tangent to the bend, in the direction of travel	✓ 1	A path curving outwards or drawn directly away from C scores 0.
there is no force towards the centre (no resultant force), so the car continues with constant velocity, in a straight line	✓ 1
Part (f)
the force needed towards the centre is larger at a higher speed (for the same radius and mass)	✓ 1
the smaller friction on the wet track cannot provide the force needed at 20 m/s, so the car would not follow the bend; a lower speed needs a smaller force	✓ 1
Part (g)(i)
change in momentum = mv − mu = 900 × 0 − 900 × 20	✓ 1
size = 18 000 kg m/s = 18 000 N s	✓ 1	Accept −18 000 N s, with the sign explained as opposite to the motion.
Part (g)(ii)
F = impulse ÷ time = 18 000 ÷ 0.25	✓ 1
= 72 000 N	✓ 1	Accept 72 kN or 7.2 × 10⁴ N.

Answers: (b) 5.0 m/s² · (c)(i) 70 m/s · (c)(ii) 5400 N · (g)(i) 18 000 N s · (g)(ii) 72 000 N

Exam tip (b) Write down the resultant force first. F = ma uses the resultant, not the driving force alone.

Common error (e) Drawing the car flying straight outwards, away from the centre. With no inward force nothing pushes it outwards; it simply keeps the velocity it had at P, along the tangent.

More 1.4 questions in the question bank

1.5 Turning effect and centre of gravity

+ Add to test✓ In your testCC-CIP-15-119 marks~20 min

A tower crane on a building site: moments and counterweights

1.5 Turning effect and centre of gravity1.3 Mass, weight and density1.4 Forces, friction and springsCore

Fig. 1 shows a tower crane on a building site. The horizontal jib reaches 40 m to the right of the centre line of the tower. A trolley runs along the jib and carries the load on a vertical cable. The distance of the trolley from the centre line is d. On the left, a concrete counterweight of mass 15 000 kg acts 12 m from the centre line.

T is the top of the tower, on its centre line. The weight of the jib acts through the centre line, so it has no moment about T. Ignore the weights of the trolley and the cable. To be safe, the moment of the load about T must never be greater than the moment of the counterweight about T.

Take g = 9.8 N/kg.

Fig. 1: the tower crane (not to scale)

(a)

State what is meant by the moment of a force about a pivot.

RECALLNotes

Show solution

force × perpendicular distance from the pivot to the line of action of the force

✓ 1

Accept the turning effect of a force, given as force × perpendicular distance. 'Force × distance' without 'perpendicular' scores 0.

(1)

(b)

The counterweight is made from concrete blocks. The density of concrete is 2400 kg/m³.

Calculate the total volume of the concrete blocks.

CALCNotes

Show solution

V = m ÷ ρ = 15 000 ÷ 2400	✓ 1	Rearrangement may be implied by the substitution.
6.25 m³	✓ 1	Accept 6.3 m³. Unit required.

Answer: 6.25 m³ (6.3 m³)

(2)

(c)

Calculate the moment of the weight of the counterweight about T.

CALCNotes

Show solution

weight = mg = 15 000 × 9.8 = 147 000 N	✓ 1	Must be seen or used.
moment = 147 000 × 12 = 1 764 000 N m	✓ 1	Accept 1.76 × 10⁶ N m or 1.8 × 10⁶ N m. Unit required. 180 000 (mass used instead of weight) scores 1 out of 2.

Answer: 1.76 × 10⁶ N m

(2)

(d)

Show that the largest load that can be carried safely with the trolley at the end of the jib is 4500 kg.

DERIVENotes

Show solution

moment of load = moment of counterweight: W × 40 = 1 764 000, so W = 44 100 N	✓ 1	Allow ECF from (c) for the method. Accept the ratio method: 15 000 × 12 ÷ 40.
m = 44 100 ÷ 9.8 = 4500 kg	✓ 1	The working must lead to 4500 kg; the answer alone scores 0.

Answer: 4500 kg

(2)

(e)

The maximum safe load depends on the distance d of the trolley from the centre line.

On Fig. 2, sketch a graph to show how the maximum safe load varies with d for values of d from 10 m to 40 m. Your graph should pass through correct values at d = 20 m and d = 40 m.

Fig. 2: axes for the maximum safe load against trolley distance

SKETCHNotes

Show solution

a curve that falls as d increases and becomes less steep (not a straight line)	✓ 1	Maximum load × d = 15 000 × 12 = 180 000 kg m is constant.
curve passes through (40 m, 4500 kg)	✓ 1	Allow ± half a small square.
curve passes through (20 m, 9000 kg)	✓ 1	Also accept (10 m, 18 000 kg) or (30 m, 6000 kg) if the curve shows it clearly.

(3)

(f)(i)

The crane lifts a load of mass 6000 kg.

Calculate the distance d at which the trolley must be placed so that the moment of this load about T is exactly equal to the moment of the counterweight.

CALCNotes

Show solution

6000 × 9.8 × d = 1 764 000 (or 6000 × d = 15 000 × 12)	✓ 1	Allow ECF from (c).
d = 30 m	✓ 1	Unit required.

Answer: 30 m

(2)

(f)(ii)

With the trolley at this position, the 6000 kg load is raised at a constant speed.

State the resultant force on the load and determine the tension in the cable. Give a reason for your answer.

REASONNotes

Show solution

constant speed, so the resultant force on the load is zero / the forces on the load are balanced	✓ 1	Accept 'tension = weight because the forces are balanced'.
tension = weight = 6000 × 9.8 = 58 800 N	✓ 1	Accept 5.9 × 10⁴ N. Unit required.

Answer: resultant force 0; tension 58 800 N

(2)

(g)(i)

While the crane holds the 6000 kg load at this position, the centre of gravity of the whole crane, including its load and its concrete base, lies on the centre line of the tower. The total weight of the crane, load and base is 1.2 × 10⁶ N. The base is 8.0 m wide.

A storm starts. The wind pushes horizontally from left to right on the crane. Its total force can be treated as a single force acting 40 m above the ground. The crane would start to topple by turning about edge E of the base.

Calculate the largest wind force that the crane can withstand without toppling.

MULTINotes

Show solution

perpendicular distance from E to the line of action of the weight = 4.0 m	✓ 1	Half the base width.
wind force × 40 = 1.2 × 10⁶ × 4.0	✓ 1	Moments about E. Allow ECF for a wrong distance.
wind force = 1.2 × 10⁵ N	✓ 1	Accept 120 000 N. Unit required. Using 8.0 m gives 2.4 × 10⁵ N and scores 1 out of 3.

Answer: 1.2 × 10⁵ N

(3)

(g)(ii)

Explain, in terms of the centre of gravity, why a wider base would make the crane more stable.

EXPLAINNotes

Show solution

the crane can tilt further before the vertical line through the centre of gravity passes outside the base / beyond E	✓ 1	Accept: the weight's line of action stays inside the base for a bigger tilt.
the weight then has a larger (perpendicular) distance from the edge, so it gives a larger moment opposing toppling / a larger wind force is needed to topple it	✓ 1	Accept: so the weight turns the crane back rather than over.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
force × perpendicular distance from the pivot to the line of action of the force	✓ 1	Accept the turning effect of a force, given as force × perpendicular distance. 'Force × distance' without 'perpendicular' scores 0.
Part (b)
V = m ÷ ρ = 15 000 ÷ 2400	✓ 1	Rearrangement may be implied by the substitution.
6.25 m³	✓ 1	Accept 6.3 m³. Unit required.
Part (c)
weight = mg = 15 000 × 9.8 = 147 000 N	✓ 1	Must be seen or used.
moment = 147 000 × 12 = 1 764 000 N m	✓ 1	Accept 1.76 × 10⁶ N m or 1.8 × 10⁶ N m. Unit required. 180 000 (mass used instead of weight) scores 1 out of 2.
Part (d)
moment of load = moment of counterweight: W × 40 = 1 764 000, so W = 44 100 N	✓ 1	Allow ECF from (c) for the method. Accept the ratio method: 15 000 × 12 ÷ 40.
m = 44 100 ÷ 9.8 = 4500 kg	✓ 1	The working must lead to 4500 kg; the answer alone scores 0.
Part (e)
a curve that falls as d increases and becomes less steep (not a straight line)	✓ 1	Maximum load × d = 15 000 × 12 = 180 000 kg m is constant.
curve passes through (40 m, 4500 kg)	✓ 1	Allow ± half a small square.
curve passes through (20 m, 9000 kg)	✓ 1	Also accept (10 m, 18 000 kg) or (30 m, 6000 kg) if the curve shows it clearly.
Part (f)(i)
6000 × 9.8 × d = 1 764 000 (or 6000 × d = 15 000 × 12)	✓ 1	Allow ECF from (c).
d = 30 m	✓ 1	Unit required.
Part (f)(ii)
constant speed, so the resultant force on the load is zero / the forces on the load are balanced	✓ 1	Accept 'tension = weight because the forces are balanced'.
tension = weight = 6000 × 9.8 = 58 800 N	✓ 1	Accept 5.9 × 10⁴ N. Unit required.
Part (g)(i)
perpendicular distance from E to the line of action of the weight = 4.0 m	✓ 1	Half the base width.
wind force × 40 = 1.2 × 10⁶ × 4.0	✓ 1	Moments about E. Allow ECF for a wrong distance.
wind force = 1.2 × 10⁵ N	✓ 1	Accept 120 000 N. Unit required. Using 8.0 m gives 2.4 × 10⁵ N and scores 1 out of 3.
Part (g)(ii)
the crane can tilt further before the vertical line through the centre of gravity passes outside the base / beyond E	✓ 1	Accept: the weight's line of action stays inside the base for a bigger tilt.
the weight then has a larger (perpendicular) distance from the edge, so it gives a larger moment opposing toppling / a larger wind force is needed to topple it	✓ 1	Accept: so the weight turns the crane back rather than over.

Answers: (b) 6.25 m³ (6.3 m³) · (c) 1.76 × 10⁶ N m · (d) 4500 kg · (f)(i) 30 m · (f)(ii) resultant force 0; tension 58 800 N · (g)(i) 1.2 × 10⁵ N

Exam tip (d) In a 'show that' question, write the equation 'clockwise moment = anticlockwise moment' with the numbers in it, then finish with the given value. The working earns the marks, not the final number.

Common error (e) A common error is to draw a straight line. Halving d doubles the maximum load, so the graph is a curve that gets flatter as d increases.

More 1.5 questions in the question bank

+ Add to test✓ In your testCC-CIP-15-219 marks~20 min

Designing a playground seesaw for two children and a parent

1.5 Turning effect and centre of gravity1.3 Mass, weight and density1.8 Pressure in solids and liquidsCore

A designer is testing a playground seesaw. It is a uniform plank of length 3.6 m and mass 25 kg, resting on a pivot at its centre, as shown in Fig. 1.

A child of mass 25 kg sits at the left-hand end of the plank. A parent of mass 75 kg sits on the right-hand side, a distance x from the pivot.

Take g = 9.8 N/kg. Treat the weight of each person as acting at a single point on the plank.

Fig. 1: the seesaw (not to scale)

(a)

State the principle of moments.

RECALLNotes

Show solution

when an object is in equilibrium / balanced	✓ 1	Accept 'not turning'.
the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about the same point	✓ 1	'Moments are equal' alone scores 0 for this mark; clockwise and anticlockwise must both be mentioned.

(2)

(b)

The plank is horizontal and balanced. Calculate the distance x.

CALCNotes

Show solution

child's moment = 25 × 9.8 × 1.8 = 441 N m (anticlockwise)	✓ 1	1.8 m is half the plank length. Accept working in kg throughout: 25 × 1.8 = 75 × x.
735 × x = 441	✓ 1	The plank's weight acts at the pivot and has no moment; any moment for it scores 0 for this mark.
x = 0.60 m	✓ 1	Unit required.

Answer: 0.60 m

(3)

(c)

Show that the upward force exerted by the pivot on the plank is 1225 N.

DERIVENotes

Show solution

forces up = forces down, so pivot force = total weight of child, parent and plank	✓ 1	The plank's weight must be included.
(25 + 75 + 25) × 9.8 = 125 × 9.8 = 1225 N	✓ 1	Or 245 + 735 + 245 = 1225 N. The answer alone scores 0.

Answer: 1225 N

(2)

(d)(i)

The pivot stand has a mass of 25 kg. It rests on a flat rectangular base plate measuring 0.50 m × 0.30 m, which lies on soft ground. The mass of the base plate is included in the mass of the stand.

Calculate the pressure exerted on the ground by the base plate.

CALCNotes

Show solution

force on ground = 1225 + 25 × 9.8 = 1470 N	✓ 1	Allow ECF from (c). Using 1225 N only scores this mark 0 but allow the next two marks.
area = 0.50 × 0.30 = 0.15 m²	✓ 1
p = F ÷ A = 1470 ÷ 0.15 = 9800 Pa	✓ 1	Accept 9.8 kPa or 9800 N/m². 1225 ÷ 0.15 = 8170 Pa scores 2 out of 3.

Answer: 9800 Pa

(3)

(d)(ii)

After heavy rain, the base plate starts to sink into the ground. Suggest one change to the base plate that would reduce this, and give a reason.

REASONNotes

Show solution

use a base plate with a larger area, so the same force gives a smaller pressure

✓ 1

Both the change and the reason are needed for the mark.

(1)

(e)

To find the weight of a second child, B, the designer removes the parent and the first child and keeps the plank level by resting its left-hand end on a force sensor, 1.80 m from the pivot. Child B sits on the left-hand side at different distances d from the pivot. For each position the sensor reading F is recorded. Fig. 2 shows the results and the line of best fit.

Use the gradient of the line in Fig. 2 to determine the weight of child B. Show your working.

Fig. 2: sensor reading F against the distance d of child B from the pivot

GRAPHNotes

Show solution

gradient from a large triangle on the line, e.g. 294 ÷ 1.80 = 163 N/m	✓ 1	Accept 155–170 N/m. Points read from the line, not from the data points.
moments about the pivot: F × 1.80 = W × d, so W = gradient × 1.80	✓ 1	May be seen as a reading: when d = 1.80 m, F = W.
W = 294 N	✓ 1	Accept 280–305 N. Allow ECF from their gradient.

Answer: 294 N (about 290 N)

(3)

(f)

The first child (25 kg) and the parent (75 kg) return. The pivot is now moved 0.30 m towards the parent's side, so it is no longer at the centre of the plank. The child sits at the left-hand end again.

Calculate the new distance from the pivot at which the parent must sit for the plank to balance.

MULTINotes

Show solution

child is now 2.1 m from the pivot and the plank's weight acts 0.30 m from the pivot, on the child's side	✓ 1	Both distances needed.
anticlockwise moments: 245 × 2.1 + 245 × 0.30 = 588 N m	✓ 1	Allow ECF for one wrong distance.
735 × x = 588, so x = 0.80 m	✓ 1	Unit required. 0.70 m (plank's weight left out) scores 1 out of 3.

Answer: 0.80 m

(3)

(g)

A student says: 'The plank is uniform, so its weight can still be left out of the moments calculation after the pivot is moved.'

Explain why the student is wrong.

EXPLAINNotes

Show solution

the weight of a uniform plank acts at its centre (its centre of gravity), which is now 0.30 m from the pivot	✓ 1	Accept 'the centre of gravity is no longer above the pivot'.
so the plank's weight has a (non-zero) moment about the pivot, which must be included	✓ 1	Accept: it adds to the anticlockwise moment on the child's side.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
when an object is in equilibrium / balanced	✓ 1	Accept 'not turning'.
the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about the same point	✓ 1	'Moments are equal' alone scores 0 for this mark; clockwise and anticlockwise must both be mentioned.
Part (b)
child's moment = 25 × 9.8 × 1.8 = 441 N m (anticlockwise)	✓ 1	1.8 m is half the plank length. Accept working in kg throughout: 25 × 1.8 = 75 × x.
735 × x = 441	✓ 1	The plank's weight acts at the pivot and has no moment; any moment for it scores 0 for this mark.
x = 0.60 m	✓ 1	Unit required.
Part (c)
forces up = forces down, so pivot force = total weight of child, parent and plank	✓ 1	The plank's weight must be included.
(25 + 75 + 25) × 9.8 = 125 × 9.8 = 1225 N	✓ 1	Or 245 + 735 + 245 = 1225 N. The answer alone scores 0.
Part (d)(i)
force on ground = 1225 + 25 × 9.8 = 1470 N	✓ 1	Allow ECF from (c). Using 1225 N only scores this mark 0 but allow the next two marks.
area = 0.50 × 0.30 = 0.15 m²	✓ 1
p = F ÷ A = 1470 ÷ 0.15 = 9800 Pa	✓ 1	Accept 9.8 kPa or 9800 N/m². 1225 ÷ 0.15 = 8170 Pa scores 2 out of 3.
Part (d)(ii)
use a base plate with a larger area, so the same force gives a smaller pressure	✓ 1	Both the change and the reason are needed for the mark.
Part (e)
gradient from a large triangle on the line, e.g. 294 ÷ 1.80 = 163 N/m	✓ 1	Accept 155–170 N/m. Points read from the line, not from the data points.
moments about the pivot: F × 1.80 = W × d, so W = gradient × 1.80	✓ 1	May be seen as a reading: when d = 1.80 m, F = W.
W = 294 N	✓ 1	Accept 280–305 N. Allow ECF from their gradient.
Part (f)
child is now 2.1 m from the pivot and the plank's weight acts 0.30 m from the pivot, on the child's side	✓ 1	Both distances needed.
anticlockwise moments: 245 × 2.1 + 245 × 0.30 = 588 N m	✓ 1	Allow ECF for one wrong distance.
735 × x = 588, so x = 0.80 m	✓ 1	Unit required. 0.70 m (plank's weight left out) scores 1 out of 3.
Part (g)
the weight of a uniform plank acts at its centre (its centre of gravity), which is now 0.30 m from the pivot	✓ 1	Accept 'the centre of gravity is no longer above the pivot'.
so the plank's weight has a (non-zero) moment about the pivot, which must be included	✓ 1	Accept: it adds to the anticlockwise moment on the child's side.

Answers: (b) 0.60 m · (c) 1225 N · (d)(i) 9800 Pa · (e) 294 N (about 290 N) · (f) 0.80 m

Common error (c) A common error is to leave out the weight of the plank when finding the pivot force. The plank's weight has no moment about the pivot, but it still pushes down on the pivot.

More 1.5 questions in the question bank

1.6 Momentum and impulse

+ Add to test✓ In your testCC-CIP-16-120 marks~20 min

Curling stones: collisions on the ice

1.6 Momentum and impulse1.2 Speed, velocity and motion graphs1.7 Energy, work and powerExtended

In the sport of curling, heavy granite stones slide across a flat sheet of ice. Stone A and stone B each have a mass of 20.0 kg.

Stone A slides at 2.0 m/s and hits stone B head-on. Stone B is at rest before the collision. After the collision, stone A continues in the same direction at 0.40 m/s, as shown in Fig. 1.

During the very short collision, friction from the ice can be ignored.

Fig. 1: the stones before and after the collision (not to scale)

(a)

State what is meant by the momentum of an object.

Extended onlyRECALLNotes

Show solution

mass × velocity

✓ 1

Accept p = mv with symbols explained. 'Mass × speed' scores 0.

(1)

(b)

Calculate the velocity of stone B immediately after the collision.

Extended onlyCALCNotes

Show solution

momentum before = 20.0 × 2.0 = 40 kg m/s	✓ 1
40 = 20.0 × 0.40 + 20.0 × v (momentum is conserved)	✓ 1
v = 1.6 m/s in the original direction of A	✓ 1	Direction required for the mark: accept 'to the right' or 'same direction as A'.

Answer: 1.6 m/s in the direction A was moving

(3)

(c)

Show that 12.8 J of kinetic energy is transferred to other energy stores in the collision.

Extended onlyDERIVENotes

Show solution

kinetic energy before = ½ × 20.0 × 2.0² = 40 J	✓ 1
kinetic energy after = ½ × 20.0 × 0.40² + ½ × 20.0 × 1.6² = 1.6 + 25.6 = 27.2 J	✓ 1	Allow ECF from (b) for the method only.
40 − 27.2 = 12.8 J	✓ 1	The answer alone scores 0.

Answer: 12.8 J

(3)

(d)(i)

Fig. 2 shows how the force F exerted on stone B by stone A varies with time t during the collision.

Use Fig. 2 to determine the impulse on stone B.

Fig. 2: force on stone B during the collision

Extended onlyGRAPHNotes

Show solution

impulse = area under the graph = ½ × 4.0 × 10⁻³ × 16 000	✓ 1	Time must be converted from ms to s.
32 N s	✓ 1	Accept 32 kg m/s. Unit required. 32 000 (time left in ms) scores 1 out of 2.

Answer: 32 N s

(2)

(d)(ii)

State the impulse on stone A during the collision. Explain your answer.

Extended onlyREASONNotes

Show solution

32 N s in the opposite direction (backwards / −32 N s)	✓ 1	Size and direction both needed. Allow ECF from (d)(i).
the force of B on A is equal and opposite to the force of A on B and acts for the same time	✓ 1	Accept: A loses the momentum that B gains, 20.0 × (0.40 − 2.0) = −32 kg m/s.

Answer: 32 N s, opposite to the motion of A

(2)

(e)(i)

After the collision, stone B slides along the ice and slows down. Fig. 3 shows the speed–time graph for stone B from the end of the collision.

Determine the deceleration of stone B.

Fig. 3: speed–time graph for stone B after the collision

Extended onlyGRAPHNotes

Show solution

gradient = (0 − 1.6) ÷ 20	✓ 1
deceleration = 0.080 m/s²	✓ 1	Accept acceleration = −0.080 m/s². Unit required.

Answer: 0.080 m/s²

(2)

(e)(ii)

Calculate the distance that stone B travels before it stops.

CALCNotes

Show solution

distance = area under the graph = ½ × 20 × 1.6	✓ 1
16 m	✓ 1	Unit required. 32 m (rectangle used) scores 0.

Answer: 16 m

(2)

(f)

Players can sweep the ice in front of a moving stone. This reduces the friction force on the stone.

Explain how sweeping in front of stone B would change its motion after the collision. Refer to the graph in Fig. 3.

EXPLAINNotes

Show solution

smaller (resultant) force so smaller deceleration / the line on the speed–time graph is less steep	✓ 1
so the stone takes longer to stop and the area under the graph is larger: it travels further	✓ 1

(2)

(g)

In a test, the front of a stone C of mass 20.0 kg is covered with soft putty. Stone C slides at 2.0 m/s and hits stone B, which is at rest. The two stones stick together.

Calculate the velocity of the stones just after the collision and the kinetic energy transferred to other stores. Compare this energy with your answer to (c).

Extended onlyPREDICTNotes

Show solution

20.0 × 2.0 = (20.0 + 20.0) × v, so v = 1.0 m/s	✓ 1	Direction not needed here.
kinetic energy after = ½ × 40.0 × 1.0² = 20 J, so 40 − 20 = 20 J is transferred	✓ 1
more energy (20 J instead of 12.8 J) is transferred when the stones stick together	✓ 1	Accept: half the kinetic energy is transferred. Allow ECF from (c).

Answer: 1.0 m/s; 20 J (more than 12.8 J)

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
mass × velocity	✓ 1	Accept p = mv with symbols explained. 'Mass × speed' scores 0.
Part (b)
momentum before = 20.0 × 2.0 = 40 kg m/s	✓ 1
40 = 20.0 × 0.40 + 20.0 × v (momentum is conserved)	✓ 1
v = 1.6 m/s in the original direction of A	✓ 1	Direction required for the mark: accept 'to the right' or 'same direction as A'.
Part (c)
kinetic energy before = ½ × 20.0 × 2.0² = 40 J	✓ 1
kinetic energy after = ½ × 20.0 × 0.40² + ½ × 20.0 × 1.6² = 1.6 + 25.6 = 27.2 J	✓ 1	Allow ECF from (b) for the method only.
40 − 27.2 = 12.8 J	✓ 1	The answer alone scores 0.
Part (d)(i)
impulse = area under the graph = ½ × 4.0 × 10⁻³ × 16 000	✓ 1	Time must be converted from ms to s.
32 N s	✓ 1	Accept 32 kg m/s. Unit required. 32 000 (time left in ms) scores 1 out of 2.
Part (d)(ii)
32 N s in the opposite direction (backwards / −32 N s)	✓ 1	Size and direction both needed. Allow ECF from (d)(i).
the force of B on A is equal and opposite to the force of A on B and acts for the same time	✓ 1	Accept: A loses the momentum that B gains, 20.0 × (0.40 − 2.0) = −32 kg m/s.
Part (e)(i)
gradient = (0 − 1.6) ÷ 20	✓ 1
deceleration = 0.080 m/s²	✓ 1	Accept acceleration = −0.080 m/s². Unit required.
Part (e)(ii)
distance = area under the graph = ½ × 20 × 1.6	✓ 1
16 m	✓ 1	Unit required. 32 m (rectangle used) scores 0.
Part (f)
smaller (resultant) force so smaller deceleration / the line on the speed–time graph is less steep	✓ 1
so the stone takes longer to stop and the area under the graph is larger: it travels further	✓ 1
Part (g)
20.0 × 2.0 = (20.0 + 20.0) × v, so v = 1.0 m/s	✓ 1	Direction not needed here.
kinetic energy after = ½ × 40.0 × 1.0² = 20 J, so 40 − 20 = 20 J is transferred	✓ 1
more energy (20 J instead of 12.8 J) is transferred when the stones stick together	✓ 1	Accept: half the kinetic energy is transferred. Allow ECF from (c).

Answers: (b) 1.6 m/s in the direction A was moving · (c) 12.8 J · (d)(i) 32 N s · (d)(ii) 32 N s, opposite to the motion of A · (e)(i) 0.080 m/s² · (e)(ii) 16 m · (g) 1.0 m/s; 20 J (more than 12.8 J)

Exam tip (b) Write 'total momentum before = total momentum after' first, with every velocity signed, then substitute. A velocity needs a direction in the answer.

Common error (g) A common error is to use only one mass after a collision in which the objects stick together. After sticking they move as one object, so use the total mass, 40.0 kg.

More 1.6 questions in the question bank

+ Add to test✓ In your testCC-CIP-16-219 marks~20 min

Crash-test dummy: airbags, crumple zones and impulse

1.6 Momentum and impulse1.4 Forces, friction and springs1.2 Speed, velocity and motion graphsExtended

In a crash test, a car of mass 1200 kg travelling at 14 m/s is driven into a rigid concrete barrier. A crash-test dummy sits in the driver's seat, held by a seat belt. The head of the dummy has a mass of 5.0 kg.

Assume that the head is still moving forwards at 14 m/s when it first touches the airbag, and that the airbag brings it to rest.

(a)

The car is brought to rest by the barrier. Calculate the change in momentum of the car. State its direction.

Extended onlyCALCNotes

Show solution

Δp = 0 − 1200 × 14	✓ 1
16 800 kg m/s, opposite to the direction the car was moving	✓ 1	Accept −16 800 kg m/s with forwards taken as positive. Unit required.

Answer: 16 800 kg m/s backwards (−16 800 kg m/s)

(2)

(b)

With the airbag, the head comes to rest in 0.080 s. In a test without an airbag, the head hits the steering wheel and comes to rest in 0.010 s.

Show that the airbag reduces the average force on the head by a factor of 8.

Extended onlyDERIVENotes

Show solution

change in momentum of head = 5.0 × 14 = 70 kg m/s	✓ 1	The same in both tests.
F = Δp ÷ Δt: 70 ÷ 0.080 = 875 N and 70 ÷ 0.010 = 7000 N	✓ 1	Both forces needed.
7000 ÷ 875 = 8	✓ 1	Accept 0.080 ÷ 0.010 = 8 with a statement that Δp is the same, so the force is inversely proportional to the time.

Answer: 875 N and 7000 N; ratio 8

(3)

(c)(i)

Fig. 1 shows how the force F on the head varies with time t while the airbag stops it.

Treat this force as the resultant force on the head.

Use Fig. 1 to determine the impulse on the head.

Fig. 1: force on the dummy's head while the airbag stops it

Extended onlyGRAPHNotes

Show solution

impulse = area under graph = ½ × 0.010 × 1000 + 0.060 × 1000 + ½ × 0.010 × 1000	✓ 1	Accept the trapezium: ½ × (0.080 + 0.060) × 1000.
70 N s	✓ 1	Accept 70 kg m/s. Unit required.

Answer: 70 N s

(2)

(c)(ii)

The maximum force on the head in Fig. 1 is greater than the average force you calculated in (b). Use Fig. 1 to explain why.

Extended onlyREASONNotes

Show solution

the force is less than 1000 N while it rises and falls (in the first and last 0.010 s), so its average over 0.080 s is less than 1000 N

✓ 1

Accept: average force = impulse ÷ total time = 70 ÷ 0.080 = 875 N, which is below the 1000 N peak because of the sloping parts.

(1)

(d)

Calculate the deceleration of the head between t = 0.010 s and t = 0.070 s.

Extended onlyCALCNotes

Show solution

a = F ÷ m = 1000 ÷ 5.0	✓ 1	Force read from Fig. 1.
200 m/s²	✓ 1	Accept acceleration = −200 m/s². Unit required.

Answer: 200 m/s²

(2)

(e)

On Fig. 2, sketch the speed–time graph of the head from t = 0 until it comes to rest.

Fig. 2: axes for the speed–time graph of the head

Extended onlySKETCHNotes

Show solution

starts at 14 m/s at t = 0 and reaches zero at t = 0.080 s	✓ 1
straight line sloping down between t = 0.010 s and t = 0.070 s (constant deceleration)	✓ 1	Steepest part of the graph; it falls from about 13 m/s to about 1 m/s.
curved at the start and at the end, less steep there (gradient getting steeper from 0 to 0.010 s and flatter from 0.070 s to 0.080 s)	✓ 1	Because the force, and so the deceleration, is smaller in these intervals.

(3)

(f)

In a rebound test, a car of the same mass hits the barrier at 14 m/s and bounces back at 2.0 m/s. The car is in contact with the barrier for 0.12 s.

Calculate the average force exerted by the barrier on the car.

Extended onlyMULTINotes

Show solution

taking forwards as positive: Δp = 1200 × (−2.0) − 1200 × 14	✓ 1	Signs must show that the velocity reverses.
Δp = −19 200 kg m/s	✓ 1	Size 19 200 kg m/s. 14 400 kg m/s (directions ignored) scores 0 for this mark; allow ECF for the force.
F = 19 200 ÷ 0.12 = 160 000 N, backwards (away from the barrier)	✓ 1	Accept 1.6 × 10⁵ N. Unit required.

Answer: 1.6 × 10⁵ N, away from the barrier

(3)

(g)

Explain how a crumple zone at the front of a car reduces the force on the car in a collision.

Extended onlyEXPLAINNotes

Show solution

the change in momentum of the car is the same (with or without a crumple zone)	✓ 1
the crumple zone collapses gradually, so the time taken to stop is longer	✓ 1
force = change in momentum ÷ time (rate of change of momentum), so the average force is smaller	✓ 1	The link to force = change in momentum ÷ time is needed for this mark; a longer time alone does not score it.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
Δp = 0 − 1200 × 14	✓ 1
16 800 kg m/s, opposite to the direction the car was moving	✓ 1	Accept −16 800 kg m/s with forwards taken as positive. Unit required.
Part (b)
change in momentum of head = 5.0 × 14 = 70 kg m/s	✓ 1	The same in both tests.
F = Δp ÷ Δt: 70 ÷ 0.080 = 875 N and 70 ÷ 0.010 = 7000 N	✓ 1	Both forces needed.
7000 ÷ 875 = 8	✓ 1	Accept 0.080 ÷ 0.010 = 8 with a statement that Δp is the same, so the force is inversely proportional to the time.
Part (c)(i)
impulse = area under graph = ½ × 0.010 × 1000 + 0.060 × 1000 + ½ × 0.010 × 1000	✓ 1	Accept the trapezium: ½ × (0.080 + 0.060) × 1000.
70 N s	✓ 1	Accept 70 kg m/s. Unit required.
Part (c)(ii)
the force is less than 1000 N while it rises and falls (in the first and last 0.010 s), so its average over 0.080 s is less than 1000 N	✓ 1	Accept: average force = impulse ÷ total time = 70 ÷ 0.080 = 875 N, which is below the 1000 N peak because of the sloping parts.
Part (d)
a = F ÷ m = 1000 ÷ 5.0	✓ 1	Force read from Fig. 1.
200 m/s²	✓ 1	Accept acceleration = −200 m/s². Unit required.
Part (e)
starts at 14 m/s at t = 0 and reaches zero at t = 0.080 s	✓ 1
straight line sloping down between t = 0.010 s and t = 0.070 s (constant deceleration)	✓ 1	Steepest part of the graph; it falls from about 13 m/s to about 1 m/s.
curved at the start and at the end, less steep there (gradient getting steeper from 0 to 0.010 s and flatter from 0.070 s to 0.080 s)	✓ 1	Because the force, and so the deceleration, is smaller in these intervals.
Part (f)
taking forwards as positive: Δp = 1200 × (−2.0) − 1200 × 14	✓ 1	Signs must show that the velocity reverses.
Δp = −19 200 kg m/s	✓ 1	Size 19 200 kg m/s. 14 400 kg m/s (directions ignored) scores 0 for this mark; allow ECF for the force.
F = 19 200 ÷ 0.12 = 160 000 N, backwards (away from the barrier)	✓ 1	Accept 1.6 × 10⁵ N. Unit required.
Part (g)
the change in momentum of the car is the same (with or without a crumple zone)	✓ 1
the crumple zone collapses gradually, so the time taken to stop is longer	✓ 1
force = change in momentum ÷ time (rate of change of momentum), so the average force is smaller	✓ 1	The link to force = change in momentum ÷ time is needed for this mark; a longer time alone does not score it.

Answers: (a) 16 800 kg m/s backwards (−16 800 kg m/s) · (b) 875 N and 7000 N; ratio 8 · (c)(i) 70 N s · (d) 200 m/s² · (f) 1.6 × 10⁵ N, away from the barrier

Exam tip (e) The gradient of a speed–time graph is the acceleration. Where the force is largest, the speed–time line is steepest; where the force is small, the line is nearly flat.

Common error (f) A common error is to subtract the speeds and get 12 m/s. When the car bounces back, its velocity changes from +14 m/s to −2.0 m/s, a change of 16 m/s.

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1.7 Energy, work and power

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A pumped-storage hydroelectric plant: energy, power and efficiency

1.7 Energy, work and power1.4 Forces, friction and springs4.2 Charge, current, potential difference, resistance and powerCore

A pumped-storage power station uses two reservoirs. The water surface of the upper reservoir is 300 m vertically above the water surface of the lower reservoir (Figure 1).

At night, electric pumps in the power station take energy from the grid and pump water up the pipe into the upper reservoir. During the day the water flows back down the same pipe and turns turbines, which drive generators.

Assume that the vertical height between the two water surfaces stays 300 m. Take g = 9.8 N/kg.

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

(a)

During the day, water from the upper reservoir flows down the pipe and turns the turbines, which drive the generators.

State the useful energy transfers that take place, starting with the energy store of the water in the upper reservoir.

RECALLNotes

Show solution

gravitational potential energy (of the water) → kinetic energy (of the moving water / turbines)	✓ 1	Accept "gravitational energy". "Potential energy" alone is not enough.
kinetic energy → electrical energy (in the generators)	✓ 1	Both stores in the correct order needed for each mark. Ignore thermal energy or sound added as wasted energy.

(2)

(b)

At night, while the pumps are working, the water moves up the straight section of pipe at a constant speed.

State how the force exerted on the water by the pumps, up the pipe, compares with the total force on the water down the pipe (from its weight and the friction from the pipe). Explain your answer.

REASONNotes

Show solution

the force from the pumps up the pipe is equal to the total force down the pipe	✓ 1	Accept "the forces are balanced".
the speed is constant (no acceleration), so the resultant force on the water is zero	✓ 1	The reason must be linked to the constant speed. "The force from the pumps is bigger because the water moves up" scores 0 for both marks.

(2)

(c)

During the day, 1.2 × 10⁵ kg of water flows down through the turbines every second.

Show that the water loses gravitational potential energy at a rate of about 3.5 × 10⁸ W.

DERIVENotes

Show solution

ΔE_p = mgΔh = 1.2 × 10⁵ × 9.8 × 300 (= 3.528 × 10⁸ J) for the water flowing in 1.0 s	✓ 1	Full substitution must be seen.
P = E/t = 3.528 × 10⁸ ÷ 1.0 = 3.53 × 10⁸ W	✓ 1	The answer must be given to at least 3 s.f. (3.53 × 10⁸ W or 352 800 000 W). A bare 3.5 × 10⁸ W scores 0.

Answer: 3.53 × 10⁸ W

(2)

(d)(i)

At time t = 0 the generators are switched on. A meter records the total electrical energy E supplied to the grid since t = 0. The readings are shown in the table.

t / min	0	5	10	15	20	25	30
E / 10⁹ J	0	30	108	197	289	378	468

For the first few minutes the turbines are still speeding up. From t = 10 min they run at full speed.

On the grid in Figure 2, plot the points from the table and draw the best-fit straight line through the points from t = 10 min to t = 30 min.

Figure 2: grid for the graph of E against t

GRAPHNotes

Show solution

all seven points plotted correctly to within half a small square	✓ 1	Award this mark and the next for all seven points correct.
at least five points plotted correctly to within half a small square	✓ 1	So: 7 correct = 2 marks, 5 or 6 correct = 1 mark.
a single, thin, ruled best-fit straight line through the points from 10 min to 30 min	✓ 1	The line must not be forced through the origin or through the points at 0 and 5 min.

(3)

(d)(ii)

Use your graph to determine the electrical power output of the power station when the turbines are running at full speed. Give your answer in W.

GRAPHNotes

Show solution

power = gradient of the line, using a large triangle, e.g. (468 − 108) × 10⁹ J ÷ (30 − 10) min = 18 × 10⁹ J per minute	✓ 1	Readings must come from the line, over at least half of its length.
time converted to seconds: 20 min = 1200 s (or 18 × 10⁹ J ÷ 60 s)	✓ 1
P = 360 × 10⁹ ÷ 1200 = 3.0 × 10⁸ W	✓ 1	Accept 2.9 × 10⁸ W to 3.1 × 10⁸ W. 1.8 × 10¹⁰ W (time left in minutes) scores 1 mark only.

Answer: 3.0 × 10⁸ W

(3)

(e)

Calculate the efficiency with which the power station transfers the gravitational potential energy lost by the water into electrical energy.

Extended onlyCALCNotes

Show solution

efficiency = useful power output ÷ total power input × 100% = (3.0 × 10⁸ ÷ 3.53 × 10⁸) × 100%	✓ 1	Allow ECF from (c) and (d)(ii). Using 3.5 × 10⁸ W is allowed.
85%	✓ 1	Accept 84% to 86%. An answer above 100% scores 0.

Answer: 85%

(2)

(f)

The power station supplies its electrical output to the grid at a potential difference of 400 kV.

Calculate the current supplied to the grid.

CALCNotes

Show solution

I = P ÷ V = 3.0 × 10⁸ ÷ 4.0 × 10⁵	✓ 1	Allow ECF from (d)(ii). The p.d. must be converted to volts.
750 A	✓ 1	7.5 × 10⁵ A (kV not converted) scores 1 mark.

Answer: 750 A

(2)

(g)

During a long drought, less water is available and the vertical height between the two water surfaces falls to 260 m. The mass of water flowing through the turbines each second and the efficiency stay the same as before.

Calculate the new electrical power output.

PREDICTNotes

Show solution

the energy lost by each kilogram of water is proportional to the height, so P = 3.0 × 10⁸ × 260 ÷ 300	✓ 1	Or (Extended route) P = 0.85 × 1.2 × 10⁵ × 9.8 × 260. Allow ECF from (d)(ii) and (e).
2.6 × 10⁸ W	✓ 1	Accept 2.5 × 10⁸ W to 2.6 × 10⁸ W.

Answer: 2.6 × 10⁸ W

(2)

(h)

Hydroelectric power is a renewable energy resource. Give one advantage and one disadvantage of generating electricity in this power station compared with a coal-fired power station.

EXPLAINNotes

Show solution

advantage, e.g. no carbon dioxide released while it runs / the water is not used up (renewable) / output can be supplied when it is needed	✓ 1	One advantage only for this mark.
disadvantage, e.g. high cost to build / large area of land flooded for the reservoirs / needs a suitable hilly site / output falls in a drought, as in (g)	✓ 1	"It is bad for the environment" with no detail scores 0.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
gravitational potential energy (of the water) → kinetic energy (of the moving water / turbines)	✓ 1	Accept "gravitational energy". "Potential energy" alone is not enough.
kinetic energy → electrical energy (in the generators)	✓ 1	Both stores in the correct order needed for each mark. Ignore thermal energy or sound added as wasted energy.
Part (b)
the force from the pumps up the pipe is equal to the total force down the pipe	✓ 1	Accept "the forces are balanced".
the speed is constant (no acceleration), so the resultant force on the water is zero	✓ 1	The reason must be linked to the constant speed. "The force from the pumps is bigger because the water moves up" scores 0 for both marks.
Part (c)
ΔE_p = mgΔh = 1.2 × 10⁵ × 9.8 × 300 (= 3.528 × 10⁸ J) for the water flowing in 1.0 s	✓ 1	Full substitution must be seen.
P = E/t = 3.528 × 10⁸ ÷ 1.0 = 3.53 × 10⁸ W	✓ 1	The answer must be given to at least 3 s.f. (3.53 × 10⁸ W or 352 800 000 W). A bare 3.5 × 10⁸ W scores 0.
Part (d)(i)
all seven points plotted correctly to within half a small square	✓ 1	Award this mark and the next for all seven points correct.
at least five points plotted correctly to within half a small square	✓ 1	So: 7 correct = 2 marks, 5 or 6 correct = 1 mark.
a single, thin, ruled best-fit straight line through the points from 10 min to 30 min	✓ 1	The line must not be forced through the origin or through the points at 0 and 5 min.
Part (d)(ii)
power = gradient of the line, using a large triangle, e.g. (468 − 108) × 10⁹ J ÷ (30 − 10) min = 18 × 10⁹ J per minute	✓ 1	Readings must come from the line, over at least half of its length.
time converted to seconds: 20 min = 1200 s (or 18 × 10⁹ J ÷ 60 s)	✓ 1
P = 360 × 10⁹ ÷ 1200 = 3.0 × 10⁸ W	✓ 1	Accept 2.9 × 10⁸ W to 3.1 × 10⁸ W. 1.8 × 10¹⁰ W (time left in minutes) scores 1 mark only.
Part (e)
efficiency = useful power output ÷ total power input × 100% = (3.0 × 10⁸ ÷ 3.53 × 10⁸) × 100%	✓ 1	Allow ECF from (c) and (d)(ii). Using 3.5 × 10⁸ W is allowed.
85%	✓ 1	Accept 84% to 86%. An answer above 100% scores 0.
Part (f)
I = P ÷ V = 3.0 × 10⁸ ÷ 4.0 × 10⁵	✓ 1	Allow ECF from (d)(ii). The p.d. must be converted to volts.
750 A	✓ 1	7.5 × 10⁵ A (kV not converted) scores 1 mark.
Part (g)
the energy lost by each kilogram of water is proportional to the height, so P = 3.0 × 10⁸ × 260 ÷ 300	✓ 1	Or (Extended route) P = 0.85 × 1.2 × 10⁵ × 9.8 × 260. Allow ECF from (d)(ii) and (e).
2.6 × 10⁸ W	✓ 1	Accept 2.5 × 10⁸ W to 2.6 × 10⁸ W.
Part (h)
advantage, e.g. no carbon dioxide released while it runs / the water is not used up (renewable) / output can be supplied when it is needed	✓ 1	One advantage only for this mark.
disadvantage, e.g. high cost to build / large area of land flooded for the reservoirs / needs a suitable hilly site / output falls in a drought, as in (g)	✓ 1	"It is bad for the environment" with no detail scores 0.

Answers: (c) 3.53 × 10⁸ W · (d)(ii) 3.0 × 10⁸ W · (e) 85% · (f) 750 A · (g) 2.6 × 10⁸ W

Common error (b) Thinking that an object moving upwards must have a resultant force upwards. A resultant force changes the motion; at constant speed in a straight line the resultant force is zero.

Exam tip (d)(ii) A power is in watts only when the energy is in joules and the time in seconds. Convert minutes to seconds before you divide.

More 1.7 questions in the question bank

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An electric scooter on a hill: energy changes, power and braking

1.7 Energy, work and power1.2 Speed, velocity and motion graphs1.4 Forces, friction and springsCore

A rider on an electric scooter climbs a straight hill (Figure 1). The total mass of the rider and scooter is 95 kg.

The road along the slope is 600 m long and rises through a vertical height of 40 m. At the bottom of the hill the speed of the scooter is 6.0 m/s, and at the top it is 4.0 m/s. The climb takes 120 s.

Take g = 9.8 N/kg.

Figure 1: the hill climbed by the scooter (not to scale)

(a)

Calculate the decrease in the kinetic energy of the rider and scooter between the bottom and the top of the hill.

CALCNotes

Show solution

E_k at bottom = ½ × 95 × 6.0² = 1710 J and E_k at top = ½ × 95 × 4.0² = 760 J	✓ 1	Either value correct, with working, scores this mark.
decrease = 1710 − 760 = 950 J	✓ 1	½ × 95 × (6.0 − 4.0)² = 190 J scores 0.

Answer: 950 J

(2)

(b)

Calculate the gain in gravitational potential energy of the rider and scooter.

CALCNotes

Show solution

ΔE_p = mgΔh = 95 × 9.8 × 40	✓ 1	Using 600 m scores 0 for both marks.
37 240 J	✓ 1	Accept 37 000 J or 3.7 × 10⁴ J.

Answer: 37 240 J (3.72 × 10⁴ J)

(2)

(c)

Friction and air resistance together exert an average force of 35 N on the scooter, opposite to its motion, all the way up the slope.

Calculate the work done against this resistive force during the climb.

CALCNotes

Show solution

W = Fd = 35 × 600	✓ 1	The distance is the 600 m along the slope, in the direction of the force.
21 000 J	✓ 1	35 × 40 = 1400 J scores 0.

Answer: 21 000 J

(2)

(d)

Show that the useful power output of the motor during the climb is about 480 W.

DERIVENotes

Show solution

energy supplied by the motor = gain in E_p + work against resistive force − decrease in E_k	✓ 1	Can be implied by correct numbers.
= 37 240 + 21 000 − 950 = 57 290 J	✓ 1	Allow ECF from (a), (b) and (c).
P = W/t = 57 290 ÷ 120 = 477 W	✓ 1	Answer must be given to 3 s.f. (477 W). Adding the 950 J instead of subtracting (59 190 J, 493 W) scores max 2.

Answer: 477 W

(3)

(e)

The battery supplies energy to the motor at a rate of 600 W during the climb.

Calculate the efficiency of the motor.

Extended onlyCALCNotes

Show solution

efficiency = useful power output ÷ total power input × 100% = 477 ÷ 600 × 100%	✓ 1	Using 480 W is allowed.
80%	✓ 1	Accept 79% to 80%. Allow ECF from (d).

Answer: 80%

(2)

(f)(i)

The rider then rides down a hill on the other side. Figure 2 shows the speed–time graph for the descent.

Calculate the acceleration of the scooter during the first 10 s.

Figure 2: speed–time graph for the descent

Extended onlyGRAPHNotes

Show solution

a = Δv ÷ Δt = (10.0 − 4.0) ÷ 10	✓ 1	Gradient of the first section of the graph.
0.60 m/s²	✓ 1	10.0 ÷ 10 = 1.0 m/s² (initial speed ignored) scores 0.

Answer: 0.60 m/s²

(2)

(f)(ii)

From t = 10 s to t = 63 s the rider uses the brakes so that the scooter keeps a constant speed down the slope.

Explain, in terms of the forces on the scooter, why the speed is constant.

EXPLAINNotes

Show solution

the resultant force on the rider and scooter is zero / the forces are balanced	✓ 1	"There is no force" scores 0.
the part of the weight acting down the slope is balanced by the friction of the brakes plus air resistance (acting up the slope)	✓ 1	Accept "the forward force equals the backward forces" if the forces are named.

(2)

(f)(iii)

Some scooters have regenerative braking. When the rider brakes, the motor works as a generator and charges the battery.

Describe what happens to the gravitational potential energy lost by the rider and scooter between t = 10 s and t = 63 s with ordinary brakes, and suggest why regenerative braking lets the scooter travel further on one charge of the battery.

EXPLAINNotes

Show solution

the kinetic energy does not change, so the gravitational potential energy is transferred to thermal energy (in the brakes and the surroundings)	✓ 1	Accept "wasted as heat in the brakes". "The energy is used up" scores 0.
with regenerative braking some of this energy is stored as chemical energy in the battery, so more energy is available to drive the motor later	✓ 1	Must refer to energy being stored in the battery rather than dissipated.

(2)

(g)

The rider climbs the same hill again carrying a rucksack of mass 10 kg. The speeds at the bottom and top, the resistive force and the useful power output of the motor are the same as in (d).

Calculate the time taken for the climb.

PREDICTNotes

Show solution

new gain in E_p = 105 × 9.8 × 40 = 41 160 J (an extra 3920 J) and new decrease in E_k = ½ × 105 × (6.0² − 4.0²) = 1050 J	✓ 1	Using the decrease in kinetic energy from (a), 950 J, instead of 1050 J is allowed: it changes the answer by less than 1 s.
energy supplied by the motor = 41 160 + 21 000 − 1050 = 61 110 J	✓ 1	Allow ECF from (c) and (d).
t = E ÷ P = 61 110 ÷ 477 = 128 s	✓ 1	Accept 127 s to 129 s.

Answer: 128 s

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
E_k at bottom = ½ × 95 × 6.0² = 1710 J and E_k at top = ½ × 95 × 4.0² = 760 J	✓ 1	Either value correct, with working, scores this mark.
decrease = 1710 − 760 = 950 J	✓ 1	½ × 95 × (6.0 − 4.0)² = 190 J scores 0.
Part (b)
ΔE_p = mgΔh = 95 × 9.8 × 40	✓ 1	Using 600 m scores 0 for both marks.
37 240 J	✓ 1	Accept 37 000 J or 3.7 × 10⁴ J.
Part (c)
W = Fd = 35 × 600	✓ 1	The distance is the 600 m along the slope, in the direction of the force.
21 000 J	✓ 1	35 × 40 = 1400 J scores 0.
Part (d)
energy supplied by the motor = gain in E_p + work against resistive force − decrease in E_k	✓ 1	Can be implied by correct numbers.
= 37 240 + 21 000 − 950 = 57 290 J	✓ 1	Allow ECF from (a), (b) and (c).
P = W/t = 57 290 ÷ 120 = 477 W	✓ 1	Answer must be given to 3 s.f. (477 W). Adding the 950 J instead of subtracting (59 190 J, 493 W) scores max 2.
Part (e)
efficiency = useful power output ÷ total power input × 100% = 477 ÷ 600 × 100%	✓ 1	Using 480 W is allowed.
80%	✓ 1	Accept 79% to 80%. Allow ECF from (d).
Part (f)(i)
a = Δv ÷ Δt = (10.0 − 4.0) ÷ 10	✓ 1	Gradient of the first section of the graph.
0.60 m/s²	✓ 1	10.0 ÷ 10 = 1.0 m/s² (initial speed ignored) scores 0.
Part (f)(ii)
the resultant force on the rider and scooter is zero / the forces are balanced	✓ 1	"There is no force" scores 0.
the part of the weight acting down the slope is balanced by the friction of the brakes plus air resistance (acting up the slope)	✓ 1	Accept "the forward force equals the backward forces" if the forces are named.
Part (f)(iii)
the kinetic energy does not change, so the gravitational potential energy is transferred to thermal energy (in the brakes and the surroundings)	✓ 1	Accept "wasted as heat in the brakes". "The energy is used up" scores 0.
with regenerative braking some of this energy is stored as chemical energy in the battery, so more energy is available to drive the motor later	✓ 1	Must refer to energy being stored in the battery rather than dissipated.
Part (g)
new gain in E_p = 105 × 9.8 × 40 = 41 160 J (an extra 3920 J) and new decrease in E_k = ½ × 105 × (6.0² − 4.0²) = 1050 J	✓ 1	Using the decrease in kinetic energy from (a), 950 J, instead of 1050 J is allowed: it changes the answer by less than 1 s.
energy supplied by the motor = 41 160 + 21 000 − 1050 = 61 110 J	✓ 1	Allow ECF from (c) and (d).
t = E ÷ P = 61 110 ÷ 477 = 128 s	✓ 1	Accept 127 s to 129 s.

Answers: (a) 950 J · (b) 37 240 J (3.72 × 10⁴ J) · (c) 21 000 J · (d) 477 W · (e) 80% · (f)(i) 0.60 m/s² · (g) 128 s

Common error (a) Squaring the change in speed. Work out each kinetic energy separately, then subtract: (6.0 − 4.0)² is not the same as 6.0² − 4.0².

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1.8 Pressure in solids and liquids

+ Add to test✓ In your testCC-CIP-18-118 marks~20 min

A deep-sea submersible: pressure on its window and hatch

1.8 Pressure in solids and liquids1.3 Mass, weight and density1.5 Turning effect and centre of gravityExtended

A research submersible dives to a depth of 2500 m in the sea (Figure 1). The density of the sea water is 1030 kg/m³.

The air inside the crew cabin is kept at atmospheric pressure, 1.0 × 10⁵ Pa. This is also the pressure of the air above the sea surface.

Take g = 9.8 N/kg.

Figure 1: the submersible at its working depth (not to scale)

(a)

Explain why the pressure on the hull of the submersible increases as it goes deeper.

EXPLAINNotes

Show solution

there is a greater depth / more water above the submersible	✓ 1
so a greater weight of water presses on each square metre (unit area) of the hull	✓ 1	"More force" alone, with no reference to the water above or to area, scores 1 mark only.

(2)

(b)(i)

Calculate the pressure due to the sea water at a depth of 2500 m.

Extended onlyCALCNotes

Show solution

p = ρgh = 1030 × 9.8 × 2500	✓ 1
2.52 × 10⁷ Pa	✓ 1	Accept 25 235 000 Pa or 2.5 × 10⁷ Pa.

Answer: 2.52 × 10⁷ Pa

(2)

(b)(ii)

Calculate the total pressure on the outside of the hull at this depth.

Extended onlyCALCNotes

Show solution

total pressure = 2.52 × 10⁷ + 1.0 × 10⁵ = 2.53 × 10⁷ Pa

✓ 1

Allow ECF from (b)(i). Accept 25 335 000 Pa.

Answer: 2.53 × 10⁷ Pa

(1)

(c)

The viewing window is a flat disc of diameter 0.20 m.

Show that the resultant force on the window, due to the water outside and the air inside, is about 8 × 10⁵ N.

Extended onlyDERIVENotes

Show solution

area of window = π × 0.10² = 0.0314 m²	✓ 1	Using the diameter as the radius (0.126 m²) loses this mark.
atmospheric pressure acts on both sides (the air inside pushes out with 1.0 × 10⁵ Pa), so the pressure difference across the window is the pressure due to the water, 2.52 × 10⁷ Pa	✓ 1	Must be stated or clearly used.
F = pA = 2.52 × 10⁷ × 0.0314 = 7.9 × 10⁵ N	✓ 1	Answer to at least 2 s.f. (7.9 × 10⁵ N); a bare 8 × 10⁵ N does not score this mark.

Answer: 7.9 × 10⁵ N

(3)

(d)(i)

The circular hatch on top of the cabin has a diameter of 0.50 m. It is hinged at one edge and opens outwards (Figure 1). At 2500 m, the resultant force on the hatch, due to the water outside and the air inside, can be taken to act at the centre of the hatch.

Calculate the moment of this resultant force about the hinge.

Extended onlyMULTINotes

Show solution

area = π × 0.25² = 0.196 m²	✓ 1
F = pA = 2.52 × 10⁷ × 0.196 = 4.95 × 10⁶ N	✓ 1	Allow ECF from (b)(i). Use of the total pressure (4.97 × 10⁶ N) is allowed here.
moment = F × d = 4.95 × 10⁶ × 0.25 = 1.24 × 10⁶ N m	✓ 1	The perpendicular distance from the hinge to the centre is 0.25 m. Using 0.50 m (2.48 × 10⁶ N m) loses this mark.

Answer: 1.24 × 10⁶ N m

(3)

(d)(ii)

A crew member inside the cabin can push upwards on the hatch handle with a force of up to 800 N. The handle is at the edge of the hatch opposite the hinge, 0.50 m from the hinge.

Use moments to explain why the crew cannot open the hatch until the submersible has returned to the surface.

Extended onlyREASONNotes

Show solution

largest moment of the crew member's push = 800 × 0.50 = 400 N m	✓ 1
this is far smaller than the opposite moment of the water force (1.24 × 10⁶ N m), so the hatch cannot turn about the hinge; at the surface the water pressure on the hatch is (almost) zero	✓ 1	Allow ECF from (d)(i). The comparison of the two moments is needed for this mark.

(2)

(e)(i)

Before the deep dive, the pressure sensor of the submersible is tested in a freshwater lake. The sensor measures the pressure due to the water only. Figure 2 shows its readings plotted against depth, with the best-fit line.

Determine the gradient of the line. Give your answer in Pa/m.

Figure 2: readings of the pressure sensor in the lake, with the best-fit line

Extended onlyGRAPHNotes

Show solution

gradient from a large triangle on the line, e.g. 980 kPa ÷ 100 m	✓ 1	Readings must be taken from the line, over at least half its length.
9800 Pa/m	✓ 1	Accept 9700 Pa/m to 9900 Pa/m. 9.8 Pa/m (kPa not converted) scores 1 mark.

Answer: 9800 Pa/m

(2)

(e)(ii)

The gradient of the line is equal to ρg, where ρ is the density of the lake water.

Use your answer to (e)(i) to find the density of the lake water.

Extended onlyCALCNotes

Show solution

ρ = 9800 ÷ 9.8 = 1000 kg/m³

✓ 1

Allow ECF from (e)(i). Unit required.

Answer: 1000 kg/m³

(1)

(f)

The window is designed to withstand the pressure due to sea water at a depth of 2500 m. The same submersible is now used in fresh water of the density found in (e)(ii).

Calculate the depth in the fresh water at which the pressure due to the water is the same as the pressure due to the sea water at 2500 m.

Extended onlyPREDICTNotes

Show solution

h = p ÷ ρg = 2.52 × 10⁷ ÷ (1000 × 9.8)	✓ 1	Or 2500 × 1030 ÷ 1000. Allow ECF from (b)(i) and (e)(ii).
2580 m	✓ 1	Accept 2570 m to 2580 m. An answer less than 2500 m scores 0: the less dense water must be deeper to give the same pressure.

Answer: 2580 m

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
there is a greater depth / more water above the submersible	✓ 1
so a greater weight of water presses on each square metre (unit area) of the hull	✓ 1	"More force" alone, with no reference to the water above or to area, scores 1 mark only.
Part (b)(i)
p = ρgh = 1030 × 9.8 × 2500	✓ 1
2.52 × 10⁷ Pa	✓ 1	Accept 25 235 000 Pa or 2.5 × 10⁷ Pa.
Part (b)(ii)
total pressure = 2.52 × 10⁷ + 1.0 × 10⁵ = 2.53 × 10⁷ Pa	✓ 1	Allow ECF from (b)(i). Accept 25 335 000 Pa.
Part (c)
area of window = π × 0.10² = 0.0314 m²	✓ 1	Using the diameter as the radius (0.126 m²) loses this mark.
atmospheric pressure acts on both sides (the air inside pushes out with 1.0 × 10⁵ Pa), so the pressure difference across the window is the pressure due to the water, 2.52 × 10⁷ Pa	✓ 1	Must be stated or clearly used.
F = pA = 2.52 × 10⁷ × 0.0314 = 7.9 × 10⁵ N	✓ 1	Answer to at least 2 s.f. (7.9 × 10⁵ N); a bare 8 × 10⁵ N does not score this mark.
Part (d)(i)
area = π × 0.25² = 0.196 m²	✓ 1
F = pA = 2.52 × 10⁷ × 0.196 = 4.95 × 10⁶ N	✓ 1	Allow ECF from (b)(i). Use of the total pressure (4.97 × 10⁶ N) is allowed here.
moment = F × d = 4.95 × 10⁶ × 0.25 = 1.24 × 10⁶ N m	✓ 1	The perpendicular distance from the hinge to the centre is 0.25 m. Using 0.50 m (2.48 × 10⁶ N m) loses this mark.
Part (d)(ii)
largest moment of the crew member's push = 800 × 0.50 = 400 N m	✓ 1
this is far smaller than the opposite moment of the water force (1.24 × 10⁶ N m), so the hatch cannot turn about the hinge; at the surface the water pressure on the hatch is (almost) zero	✓ 1	Allow ECF from (d)(i). The comparison of the two moments is needed for this mark.
Part (e)(i)
gradient from a large triangle on the line, e.g. 980 kPa ÷ 100 m	✓ 1	Readings must be taken from the line, over at least half its length.
9800 Pa/m	✓ 1	Accept 9700 Pa/m to 9900 Pa/m. 9.8 Pa/m (kPa not converted) scores 1 mark.
Part (e)(ii)
ρ = 9800 ÷ 9.8 = 1000 kg/m³	✓ 1	Allow ECF from (e)(i). Unit required.
Part (f)
h = p ÷ ρg = 2.52 × 10⁷ ÷ (1000 × 9.8)	✓ 1	Or 2500 × 1030 ÷ 1000. Allow ECF from (b)(i) and (e)(ii).
2580 m	✓ 1	Accept 2570 m to 2580 m. An answer less than 2500 m scores 0: the less dense water must be deeper to give the same pressure.

Answers: (b)(i) 2.52 × 10⁷ Pa · (b)(ii) 2.53 × 10⁷ Pa · (c) 7.9 × 10⁵ N · (d)(i) 1.24 × 10⁶ N m · (e)(i) 9800 Pa/m · (e)(ii) 1000 kg/m³ · (f) 2580 m

Common error (c) Using the total pressure outside when there is air at atmospheric pressure on the inside. The atmosphere pushes on both faces of the window, so its effect cancels and only ρgh is left.

Exam tip (f) Check the direction of a change before you calculate it: fresh water is less dense than sea water, so the same pressure needs a greater depth.

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Snowboards, stilettos and tractors: pressure on soft ground

1.8 Pressure in solids and liquids1.3 Mass, weight and density1.4 Forces, friction and springsCore

Fresh powder snow gives way, so that a person sinks into it, when the pressure on it is greater than 2.0 kPa.

A snowboarder has a mass of 70 kg. The masses of the snowboard and boots can be ignored.

Take g = 9.8 N/kg.

Figure 1: the snowboard seen from above, treated as a rectangle (not to scale)

(a)

Calculate the weight of the snowboarder.

CALCNotes

Show solution

W = mg = 70 × 9.8 = 686 N

✓ 1

Unit required.

Answer: 686 N

(1)

(b)

Standing in snowboard boots without the board, the total area of the boot soles in contact with the snow is 560 cm².

Calculate the pressure on the snow.

CALCNotes

Show solution

area = 560 cm² = 560 × 0.0001 = 0.056 m²	✓ 1
p = F/A = 686 ÷ 0.056 = 12 250 Pa	✓ 1	Accept 12 000 Pa, 1.2 × 10⁴ Pa or 12 kPa. Allow ECF from (a). 1.225 (area left in cm², which gives N/cm²) scores 0 for this mark.

Answer: 12 250 Pa (12 kPa)

(2)

(c)

The snowboard can be treated as a rectangle 160 cm long and 25 cm wide (Figure 1).

Show that the snowboarder standing on the snowboard does not sink into the snow.

DERIVENotes

Show solution

area = 1.60 × 0.25 = 0.40 m²	✓ 1	Or 160 × 25 = 4000 cm² = 0.40 m².
p = 686 ÷ 0.40 = 1715 Pa	✓ 1	Allow ECF from (a).
1715 Pa (1.7 kPa) is less than 2.0 kPa (2000 Pa), so the snow does not give way	✓ 1	A clear comparison with 2.0 kPa is needed.

Answer: 1715 Pa, below 2.0 kPa

(3)

(d)(i)

Figure 2 shows how the pressure exerted by the snowboarder depends on the contact area A with the snow.

Use Figure 2 to find the smallest contact area on which the snowboarder can stand without sinking into this snow.

Figure 2: pressure p exerted by the snowboarder against contact area A

GRAPHNotes

Show solution

reading taken where the curve crosses p = 2.0 kPa	✓ 1	Can be shown by a line drawn on the graph.
0.34 m²	✓ 1	Accept 0.33 m² to 0.35 m².

Answer: 0.34 m²

(2)

(d)(ii)

Use two readings from Figure 2 to describe the relationship between p and A.

DATANotes

Show solution

two correct readings in which the area doubles and the pressure halves, e.g. 0.10 m² → 6.9 kPa and 0.20 m² → 3.4 kPa	✓ 1	Readings within half a small square.
the pressure is inversely proportional to the area / p × A stays the same (686 N)	✓ 1	"As the area increases the pressure decreases" alone scores 0 for this mark.

(2)

(e)

The snowboarder stands still on the board on level snow.

State the size and direction of the force that the snow exerts on the snowboard. Explain your answer.

REASONNotes

Show solution

686 N, upwards	✓ 1	Allow ECF from (a). Both size and direction needed.
the snowboarder is at rest, so the resultant force is zero: the upward force from the snow balances the weight	✓ 1

(2)

(f)

The snowboarder now carries a rucksack of mass 15 kg and stands on the same snowboard.

Determine whether the snowboarder now sinks into the snow.

PREDICTNotes

Show solution

new weight = 85 × 9.8 = 833 N	✓ 1	Or extra weight 15 × 9.8 = 147 N added to 686 N.
p = 833 ÷ 0.40 = 2080 Pa	✓ 1	Accept 2082.5 Pa or 2.1 kPa. Allow ECF from (c).
2080 Pa is greater than 2.0 kPa, so the snowboarder sinks (slightly) into the snow	✓ 1	The conclusion must follow from a calculated pressure.

Answer: 2080 Pa: sinks

(3)

(g)

A person of mass 60 kg walks across a wooden floor in shoes with stiletto heels. At one instant the whole weight of the person acts on one heel, of contact area 0.80 cm².

Calculate the pressure under the heel.

CALCNotes

Show solution

weight = 60 × 9.8 = 588 N and area = 0.80 × 0.0001 = 8.0 × 10⁻⁵ m²	✓ 1	Both needed.
p = 588 ÷ 8.0 × 10⁻⁵ = 7.4 × 10⁶ Pa	✓ 1	Accept 7 350 000 Pa. Allow ECF from an incorrect area conversion for this mark only if the method is correct.

Answer: 7.4 × 10⁶ Pa

(2)

(h)

Tractors that work on soft, muddy fields are fitted with very wide tyres.

Explain why.

EXPLAINNotes

Show solution

the wide tyres give a larger area in contact with the ground	✓ 1
the same weight (force) is spread over this larger area, so the pressure on the ground is smaller and the tractor does not sink in	✓ 1	"The tyres reduce the weight/force" scores 0 for this mark.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
W = mg = 70 × 9.8 = 686 N	✓ 1	Unit required.
Part (b)
area = 560 cm² = 560 × 0.0001 = 0.056 m²	✓ 1
p = F/A = 686 ÷ 0.056 = 12 250 Pa	✓ 1	Accept 12 000 Pa, 1.2 × 10⁴ Pa or 12 kPa. Allow ECF from (a). 1.225 (area left in cm², which gives N/cm²) scores 0 for this mark.
Part (c)
area = 1.60 × 0.25 = 0.40 m²	✓ 1	Or 160 × 25 = 4000 cm² = 0.40 m².
p = 686 ÷ 0.40 = 1715 Pa	✓ 1	Allow ECF from (a).
1715 Pa (1.7 kPa) is less than 2.0 kPa (2000 Pa), so the snow does not give way	✓ 1	A clear comparison with 2.0 kPa is needed.
Part (d)(i)
reading taken where the curve crosses p = 2.0 kPa	✓ 1	Can be shown by a line drawn on the graph.
0.34 m²	✓ 1	Accept 0.33 m² to 0.35 m².
Part (d)(ii)
two correct readings in which the area doubles and the pressure halves, e.g. 0.10 m² → 6.9 kPa and 0.20 m² → 3.4 kPa	✓ 1	Readings within half a small square.
the pressure is inversely proportional to the area / p × A stays the same (686 N)	✓ 1	"As the area increases the pressure decreases" alone scores 0 for this mark.
Part (e)
686 N, upwards	✓ 1	Allow ECF from (a). Both size and direction needed.
the snowboarder is at rest, so the resultant force is zero: the upward force from the snow balances the weight	✓ 1
Part (f)
new weight = 85 × 9.8 = 833 N	✓ 1	Or extra weight 15 × 9.8 = 147 N added to 686 N.
p = 833 ÷ 0.40 = 2080 Pa	✓ 1	Accept 2082.5 Pa or 2.1 kPa. Allow ECF from (c).
2080 Pa is greater than 2.0 kPa, so the snowboarder sinks (slightly) into the snow	✓ 1	The conclusion must follow from a calculated pressure.
Part (g)
weight = 60 × 9.8 = 588 N and area = 0.80 × 0.0001 = 8.0 × 10⁻⁵ m²	✓ 1	Both needed.
p = 588 ÷ 8.0 × 10⁻⁵ = 7.4 × 10⁶ Pa	✓ 1	Accept 7 350 000 Pa. Allow ECF from an incorrect area conversion for this mark only if the method is correct.
Part (h)
the wide tyres give a larger area in contact with the ground	✓ 1
the same weight (force) is spread over this larger area, so the pressure on the ground is smaller and the tractor does not sink in	✓ 1	"The tyres reduce the weight/force" scores 0 for this mark.

Answers: (a) 686 N · (b) 12 250 Pa (12 kPa) · (c) 1715 Pa, below 2.0 kPa · (d)(i) 0.34 m² · (f) 2080 Pa: sinks · (g) 7.4 × 10⁶ Pa

Common error (b) Dividing by 100 to change cm² into m². A square metre is 100 cm × 100 cm, so 1 m² = 10 000 cm².

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Topic 2 · Thermal physics

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

Setup	2.1	2.2	2.3	Other
A scuba diver: gas pressure, bubbles and depth	●	○		1.8
Pumping up a racing-bike tyre	●	○		1.1
Testing an electric kettle: heating and boiling water		●	○	4.2
Railway rails in a heatwave: expansion and stress	○	●		1.4
A passive house: insulation, windows and heat loss		○	●	1.7
A rooftop solar water heater		○	●	3.3

2.1 Particle model and gas pressure

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A scuba diver: gas pressure, bubbles and depth

2.1 Particle model and gas pressure1.8 Pressure in solids and liquids2.2 Thermal expansion, specific heat capacity and changes of stateExtended

A scuba diver swims at a depth of 25 m in the sea. She breathes out a small bubble of air of volume 5.0 cm³, which rises to the surface, as shown in Figure 1.

Density of sea water = 1030 kg/m³. Atmospheric pressure at the sea surface = 1.0 × 10⁵ Pa. g = 9.8 N/kg.

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

(a)

Describe the arrangement and the motion of the particles of the air inside the bubble.

RECALLNotes

Show solution

particles far apart / irregular arrangement (no pattern)	✓ 1	Accept: lots of space between the particles.
move quickly / at high speed and randomly / in all directions	✓ 1	'Vibrate about fixed positions' scores 0 for this mark.

(2)

(b)

Explain, in terms of particles, how the air inside the bubble exerts a pressure on the water around it.

EXPLAINNotes

Show solution

air particles collide with / hit the surface (wall) of the bubble and exert a force on it	✓ 1	Collisions between particles only: 0 for this mark.
pressure = force per unit area / many collisions every second add up to a steady force on each unit of area	✓ 1

(2)

(c)

Show that the total pressure on the bubble when it is released at a depth of 25 m is about 3.5 × 10⁵ Pa.

Extended onlyCALCNotes

Show solution

pressure due to water = ρgh = 1030 × 9.8 × 25 = 2.52 × 10⁵ Pa	✓ 1	Accept 252 350 Pa.
total = 1.0 × 10⁵ + 2.52 × 10⁵ = 3.52 × 10⁵ Pa	✓ 1	Atmospheric pressure must be added. Answer must be given to more than 2 s.f. or the addition shown.

Answer: 3.52 × 10⁵ Pa

(2)

(d)(i)

An underwater camera films the bubble as it rises. The table shows the total pressure p on the bubble and its volume V at different depths.

depth / m	25	20	15	10	5.0	0
p / 10⁵ Pa	3.5	3.0	2.5	2.0	1.5	1.0
V / cm³	5.0	5.8	7.0	8.8	11.7	17.6
1/V / cm⁻³	0.200		0.143		0.085	0.057

Complete the table by calculating 1/V at depths of 20 m and 10 m.

DATANotes

Show solution

0.172 (20 m) and 0.114 (10 m)

✓ 1

Both needed. Accept 0.17 and 0.11.

Answer: 0.172 cm⁻³ and 0.114 cm⁻³

(1)

(d)(ii)

Figure 2 shows four of the results plotted on a grid.

On Figure 2, plot the two remaining points and draw the best-fit line.

Figure 2: four of the results plotted as pressure p against 1/V

GRAPHNotes

Show solution

(0.172, 3.0) and (0.114, 2.0) plotted to within half a small square	✓ 1	Allow ECF from (d)(i).
single straight best-fit line passing through (or very close to) the origin	✓ 1	Do not accept dot-to-dot or a curve.

(2)

(d)(iii)

Explain how the graph shows that, for the air in the bubble, pV = constant.

Extended onlyREASONNotes

Show solution

the graph is a straight line through the origin	✓ 1	Both 'straight' and 'through the origin' needed.
so p is (directly) proportional to 1/V, so p × V is constant	✓ 1	Accept: the gradient (= pV) is constant, about 17.5 (× 10⁵ Pa cm³).

(2)

(e)

Later in the dive, the diver releases a bubble of volume 5.0 cm³ at a depth of 30 m.

Calculate the volume of this bubble when it reaches the surface. Assume that the temperature of the air in the bubble stays constant.

Extended onlyMULTINotes

Show solution

total pressure at 30 m = 1.0 × 10⁵ + 1030 × 9.8 × 30 = 4.03 × 10⁵ Pa	✓ 1
p₁V₁ = p₂V₂: V₂ = 4.03 × 10⁵ × 5.0 ÷ 1.0 × 10⁵	✓ 1	Allow ECF from an incorrect pressure.
20 cm³	✓ 1	Accept 20.1 cm³. Using 3.03 × 10⁵ Pa (atmosphere left out) gives 15 cm³: max 1 mark.

Answer: 20 cm³

(3)

(f)(i)

After the dive, a full scuba cylinder is left in a car parked in the sun. The temperature of the air inside the cylinder rises from 15 °C to 45 °C.

Convert both of these temperatures to kelvin.

Extended onlyRECALLNotes

Show solution

288 K and 318 K

✓ 1

Both needed.

Answer: 288 K and 318 K

(1)

(f)(ii)

The volume of the steel cylinder does not change. Explain, in terms of particles, why the pressure of the air in the cylinder increases.

EXPLAINNotes

Show solution

the particles move faster / have more kinetic energy (on average)	✓ 1
they hit the walls more often and with greater force, so the force per unit area (pressure) increases	✓ 1	Must refer to collisions with the walls. 'Particles collide with each other more' scores 0 for this mark.

(2)

(g)

At the dive centre, cylinders are filled with compressed air while standing in a tank of water to keep them cool. While one cylinder is filled, the 40 kg of water in the tank warms from 16 °C to 19 °C.

Calculate the thermal energy gained by the water. The specific heat capacity of water is 4200 J/(kg °C).

Extended onlyCALCNotes

Show solution

Q = mcΔT = 40 × 4200 × (19 − 16)	✓ 1	ΔT = 3 °C must be used.
5.0 × 10⁵ J	✓ 1	Accept 504 000 J or 504 kJ. Unit required.

Answer: 5.0 × 10⁵ J

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
particles far apart / irregular arrangement (no pattern)	✓ 1	Accept: lots of space between the particles.
move quickly / at high speed and randomly / in all directions	✓ 1	'Vibrate about fixed positions' scores 0 for this mark.
Part (b)
air particles collide with / hit the surface (wall) of the bubble and exert a force on it	✓ 1	Collisions between particles only: 0 for this mark.
pressure = force per unit area / many collisions every second add up to a steady force on each unit of area	✓ 1
Part (c)
pressure due to water = ρgh = 1030 × 9.8 × 25 = 2.52 × 10⁵ Pa	✓ 1	Accept 252 350 Pa.
total = 1.0 × 10⁵ + 2.52 × 10⁵ = 3.52 × 10⁵ Pa	✓ 1	Atmospheric pressure must be added. Answer must be given to more than 2 s.f. or the addition shown.
Part (d)(i)
0.172 (20 m) and 0.114 (10 m)	✓ 1	Both needed. Accept 0.17 and 0.11.
Part (d)(ii)
(0.172, 3.0) and (0.114, 2.0) plotted to within half a small square	✓ 1	Allow ECF from (d)(i).
single straight best-fit line passing through (or very close to) the origin	✓ 1	Do not accept dot-to-dot or a curve.
Part (d)(iii)
the graph is a straight line through the origin	✓ 1	Both 'straight' and 'through the origin' needed.
so p is (directly) proportional to 1/V, so p × V is constant	✓ 1	Accept: the gradient (= pV) is constant, about 17.5 (× 10⁵ Pa cm³).
Part (e)
total pressure at 30 m = 1.0 × 10⁵ + 1030 × 9.8 × 30 = 4.03 × 10⁵ Pa	✓ 1
p₁V₁ = p₂V₂: V₂ = 4.03 × 10⁵ × 5.0 ÷ 1.0 × 10⁵	✓ 1	Allow ECF from an incorrect pressure.
20 cm³	✓ 1	Accept 20.1 cm³. Using 3.03 × 10⁵ Pa (atmosphere left out) gives 15 cm³: max 1 mark.
Part (f)(i)
288 K and 318 K	✓ 1	Both needed.
Part (f)(ii)
the particles move faster / have more kinetic energy (on average)	✓ 1
they hit the walls more often and with greater force, so the force per unit area (pressure) increases	✓ 1	Must refer to collisions with the walls. 'Particles collide with each other more' scores 0 for this mark.
Part (g)
Q = mcΔT = 40 × 4200 × (19 − 16)	✓ 1	ΔT = 3 °C must be used.
5.0 × 10⁵ J	✓ 1	Accept 504 000 J or 504 kJ. Unit required.

Answers: (c) 3.52 × 10⁵ Pa · (d)(i) 0.172 cm⁻³ and 0.114 cm⁻³ · (e) 20 cm³ · (f)(i) 288 K and 318 K · (g) 5.0 × 10⁵ J

Common error (b) Gas pressure comes from particles hitting the walls of the container (here the surface of the bubble), not from particles colliding with each other.

Exam tip (e) In a Boyle's law calculation for a bubble, use the total pressure (atmospheric pressure plus ρgh). The pressure units on both sides just have to match.

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Pumping up a racing-bike tyre

2.1 Particle model and gas pressure1.1 Physical quantities and measurement2.2 Thermal expansion, specific heat capacity and changes of stateCore

A cyclist uses a track pump to inflate a racing-bike tyre. Figure 1 shows the pump. When the handle is pulled fully up, the barrel is filled with air at atmospheric pressure, 100 kPa.

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

(a)

The inside of the barrel has a length of 50.0 cm and a cross-sectional area of 6.0 cm².

Calculate the volume of air in the barrel when the handle is fully up. Give your answer in m³.

CALCNotes

Show solution

volume = 6.0 × 50.0 = 300 cm³	✓ 1
300 × 10⁻⁶ = 3.0 × 10⁻⁴ m³	✓ 1	Allow ECF from an incorrect volume in cm³. Dividing by 100 (3.0 m³) scores 0 for this mark.

Answer: 3.0 × 10⁻⁴ m³

(2)

(b)

The cyclist blocks the end of the hose and pushes the handle down slowly, so that the temperature of the air in the barrel does not change.

Explain, in terms of particles, why the pressure of the air in the barrel increases.

EXPLAINNotes

Show solution

the particles move with the same (average) speed / the temperature is constant	✓ 1
the volume is smaller, so the particles hit the walls (of the barrel) more often / more collisions per second	✓ 1	Collisions with each other: 0 for this mark.
so the force per unit area on the walls is greater (so the pressure increases)	✓ 1

(3)

(c)

The handle is pushed down until the volume of the trapped air is 120 cm³. The temperature is constant.

Show that the pressure of the trapped air is now 250 kPa.

Extended onlyCALCNotes

Show solution

p₁V₁ = p₂V₂ stated or used: 100 × 300 = p₂ × 120	✓ 1	Allow ECF from (a) for the volume in consistent units.
p₂ = 30 000 ÷ 120 = 250 kPa	✓ 1	The rearrangement or division must be seen for this mark.

Answer: 250 kPa

(2)

(d)

The cyclist repeats (c), but this time pushes the handle down very quickly. The air in the barrel warms up as it is compressed.

Explain why the pressure reached at a volume of 120 cm³ is now greater than 250 kPa.

Extended onlyREASONNotes

Show solution

the temperature is not constant, so p₁V₁ = p₂V₂ does not apply	✓ 1	Accept: Boyle's law needs constant temperature.
the warmer particles move faster, so they hit the walls harder and more often (than in (c) at the same volume)	✓ 1

(2)

(e)(i)

The cyclist now connects the pump to an empty tyre. Figure 2 shows how the pressure of the air in the tyre changes with the number of strokes of the pump.

The recommended pressure for the tyre is 700 kPa. Use Figure 2 to find the number of strokes needed to reach this pressure.

Figure 2: pressure of the air in the tyre against the number of strokes of the pump

GRAPHNotes

Show solution

24 or 25 strokes

✓ 1

Accept 24 to 25.

Answer: 24 to 25 strokes

(1)

(e)(ii)

Use Figure 2 to determine the increase in pressure for each stroke on the straight part of the graph.

GRAPHNotes

Show solution

gradient from two points on the straight section, e.g. (865 − 265) ÷ (30 − 10)	✓ 1	Points read from the graph, at least 10 strokes apart.
30 kPa per stroke	✓ 1	Accept 28 to 32 kPa.

Answer: 30 kPa per stroke

(2)

(e)(iii)

During the first few strokes, the pressure in the tyre hardly increases. Suggest why.

REASONNotes

Show solution

the (flat) tyre / inner tube is filling out, so its volume increases	✓ 1	Accept: the tyre expands to its full size.
the extra particles have more space, so the rate of collisions per unit area of wall hardly changes	✓ 1	Accept: the larger volume cancels the effect of adding more air.

(2)

(f)

The tyre is pumped to 700 kPa in a garage at 20 °C. The cyclist then rides on a hot summer afternoon and the air in the tyre warms up. The volume of the tyre stays almost the same.

State and explain, in terms of particles, what happens to the pressure in the tyre.

PREDICTNotes

Show solution

the pressure increases, because the particles move faster / have more kinetic energy	✓ 1
they hit the walls more often and harder, so there is a larger force per unit area	✓ 1	Must refer to the walls (of the tyre).

(2)

(g)

Before the ride, the air inside the tyre and the air in the garage were both at 20 °C. The cyclist says, 'The air in the tyre and the air in the garage have the same thermal energy because they are at the same temperature.'

Explain why the cyclist is wrong.

EXPLAINNotes

Show solution

temperature depends on the average kinetic energy of the particles, but thermal energy is the total energy of all the particles	✓ 1	Accept: thermal energy depends on mass (and material) as well as temperature.
the garage contains far more air (particles / mass), so the air in the garage has more thermal energy	✓ 1

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
volume = 6.0 × 50.0 = 300 cm³	✓ 1
300 × 10⁻⁶ = 3.0 × 10⁻⁴ m³	✓ 1	Allow ECF from an incorrect volume in cm³. Dividing by 100 (3.0 m³) scores 0 for this mark.
Part (b)
the particles move with the same (average) speed / the temperature is constant	✓ 1
the volume is smaller, so the particles hit the walls (of the barrel) more often / more collisions per second	✓ 1	Collisions with each other: 0 for this mark.
so the force per unit area on the walls is greater (so the pressure increases)	✓ 1
Part (c)
p₁V₁ = p₂V₂ stated or used: 100 × 300 = p₂ × 120	✓ 1	Allow ECF from (a) for the volume in consistent units.
p₂ = 30 000 ÷ 120 = 250 kPa	✓ 1	The rearrangement or division must be seen for this mark.
Part (d)
the temperature is not constant, so p₁V₁ = p₂V₂ does not apply	✓ 1	Accept: Boyle's law needs constant temperature.
the warmer particles move faster, so they hit the walls harder and more often (than in (c) at the same volume)	✓ 1
Part (e)(i)
24 or 25 strokes	✓ 1	Accept 24 to 25.
Part (e)(ii)
gradient from two points on the straight section, e.g. (865 − 265) ÷ (30 − 10)	✓ 1	Points read from the graph, at least 10 strokes apart.
30 kPa per stroke	✓ 1	Accept 28 to 32 kPa.
Part (e)(iii)
the (flat) tyre / inner tube is filling out, so its volume increases	✓ 1	Accept: the tyre expands to its full size.
the extra particles have more space, so the rate of collisions per unit area of wall hardly changes	✓ 1	Accept: the larger volume cancels the effect of adding more air.
Part (f)
the pressure increases, because the particles move faster / have more kinetic energy	✓ 1
they hit the walls more often and harder, so there is a larger force per unit area	✓ 1	Must refer to the walls (of the tyre).
Part (g)
temperature depends on the average kinetic energy of the particles, but thermal energy is the total energy of all the particles	✓ 1	Accept: thermal energy depends on mass (and material) as well as temperature.
the garage contains far more air (particles / mass), so the air in the garage has more thermal energy	✓ 1

Answers: (a) 3.0 × 10⁻⁴ m³ · (c) 250 kPa · (e)(i) 24 to 25 strokes · (e)(ii) 30 kPa per stroke

Common error (a) 1 cm³ is 10⁻⁶ m³, not 10⁻² m³: the factor of 10⁻² for one centimetre is cubed.

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2.2 Thermal expansion, specific heat capacity and changes of state

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Testing an electric kettle: heating and boiling water

2.2 Thermal expansion, specific heat capacity and changes of state2.3 Conduction, convection and radiation4.2 Charge, current, potential difference, resistance and powerExtended

A consumer group tests a new electric kettle. The kettle is rated at 230 V, 2.2 kW. It is filled with 1.50 kg of water at 18 °C and switched on. A temperature sensor records the temperature of the water. The automatic switch is disabled, so the kettle stays on after the water starts to boil.

Figure 1 shows the results.

Specific heat capacity of water = 4200 J/(kg °C). Specific latent heat of vaporisation of water = 2.26 × 10⁶ J/kg.

Figure 1: temperature of the water in the kettle against time

(a)(i)

Calculate the current in the heating element when the kettle is connected to the 230 V supply.

CALCNotes

Show solution

I = P ÷ V = 2200 ÷ 230 = 9.6 A

✓ 1

Accept 9.57 A. Using 2.2 (kW) gives 0.0096: 0 marks.

Answer: 9.6 A

(1)

(a)(ii)

Use Figure 1 to find the time taken for the water to reach 100 °C. Hence calculate the electrical energy supplied to the kettle in this time.

GRAPHNotes

Show solution

time = 270 s read from the graph	✓ 1	Accept 265 s to 275 s.
E = Pt = 2200 × 270 = 5.9 × 10⁵ J	✓ 1	Accept 594 000 J. Allow ECF from the time read.

Answer: 270 s; 5.9 × 10⁵ J

(2)

(b)

Calculate the thermal energy needed to raise the temperature of the water from 18 °C to 100 °C.

Extended onlyCALCNotes

Show solution

ΔT = 100 − 18 = 82 °C	✓ 1
Q = mcΔT = 1.50 × 4200 × 82 = 5.2 × 10⁵ J	✓ 1	Accept 516 600 J. Using 100 °C in place of ΔT scores 0 for this mark.

Answer: 5.2 × 10⁵ J

(2)

(c)

Show that about 87% of the energy supplied by the kettle is transferred to the water while it is heated to 100 °C.

Extended onlyCALCNotes

Show solution

516 600 ÷ 594 000 (× 100%)	✓ 1	Allow ECF from (a)(ii) and (b).
= 0.870 or 87.0%	✓ 1	Answer to at least 3 s.f., or full working shown.

Answer: 87%

(2)

(d)

Some of the energy supplied is transferred through the plastic wall of the kettle to the room.

Describe how thermal energy passes from the hot water, through the wall, and into the air of the room.

EXPLAINNotes

Show solution

conduction through the plastic wall: particles vibrate more and pass energy on to their neighbours	✓ 1	Plastic has no free electrons, so conduction is slow; accept 'the wall is a poor conductor'.
air next to the outside of the wall is warmed, becomes less dense and rises (convection current) carrying energy away	✓ 1	'Heat rises' does not score.
the outer surface emits infrared radiation	✓ 1

(3)

(e)(i)

Explain, in terms of particles, why the temperature of the water stays at 100 °C after 270 s, although the kettle is still supplying energy.

EXPLAINNotes

Show solution

the energy is used to separate the particles / overcome the forces (bonds) between them as the water changes to steam	✓ 1	Accept: the energy is latent heat used to change state.
the (average) kinetic energy of the particles does not increase, so the temperature stays the same	✓ 1	'The energy is lost' scores 0.

(2)

(e)(ii)

The kettle stays on for a further 150 s after the water reaches 100 °C. Assume that the same fraction of the energy supplied is transferred to the water as in (c).

Calculate the mass of water that boils away in this time.

Extended onlyMULTINotes

Show solution

energy supplied = 2200 × 150 = 330 000 J	✓ 1
energy to water = 0.87 × 330 000 = 2.87 × 10⁵ J	✓ 1	Allow ECF from (c).
m = Q ÷ L = 2.87 × 10⁵ ÷ 2.26 × 10⁶ = 0.13 kg	✓ 1	Accept 0.127 kg. Omitting the 87% gives 0.146 kg: max 2 marks.

Answer: 0.13 kg

(3)

(f)(i)

The test is repeated with 0.75 kg of water, again starting at 18 °C. Predict the time taken for this water to reach 100 °C, assuming that 87% of the energy supplied is still transferred to the water.

Extended onlyPREDICTNotes

Show solution

Q = 0.75 × 4200 × 82 = 258 300 J, and energy supplied = 258 300 ÷ 0.87 = 2.97 × 10⁵ J	✓ 1	Accept the argument: half the mass needs half the energy, so half the time.
t = 2.97 × 10⁵ ÷ 2200 = 135 s	✓ 1	Accept 130 s to 140 s. Allow ECF from (c).

Answer: 135 s

(2)

(f)(ii)

The measured time is longer than the prediction in (f)(i). Suggest a reason.

REASONNotes

Show solution

the heating element and the kettle itself must still be warmed, so a smaller fraction of the energy goes to the (smaller mass of) water

✓ 1

Accept: energy losses to the room are a larger fraction of the energy supplied.

(1)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
I = P ÷ V = 2200 ÷ 230 = 9.6 A	✓ 1	Accept 9.57 A. Using 2.2 (kW) gives 0.0096: 0 marks.
Part (a)(ii)
time = 270 s read from the graph	✓ 1	Accept 265 s to 275 s.
E = Pt = 2200 × 270 = 5.9 × 10⁵ J	✓ 1	Accept 594 000 J. Allow ECF from the time read.
Part (b)
ΔT = 100 − 18 = 82 °C	✓ 1
Q = mcΔT = 1.50 × 4200 × 82 = 5.2 × 10⁵ J	✓ 1	Accept 516 600 J. Using 100 °C in place of ΔT scores 0 for this mark.
Part (c)
516 600 ÷ 594 000 (× 100%)	✓ 1	Allow ECF from (a)(ii) and (b).
= 0.870 or 87.0%	✓ 1	Answer to at least 3 s.f., or full working shown.
Part (d)
conduction through the plastic wall: particles vibrate more and pass energy on to their neighbours	✓ 1	Plastic has no free electrons, so conduction is slow; accept 'the wall is a poor conductor'.
air next to the outside of the wall is warmed, becomes less dense and rises (convection current) carrying energy away	✓ 1	'Heat rises' does not score.
the outer surface emits infrared radiation	✓ 1
Part (e)(i)
the energy is used to separate the particles / overcome the forces (bonds) between them as the water changes to steam	✓ 1	Accept: the energy is latent heat used to change state.
the (average) kinetic energy of the particles does not increase, so the temperature stays the same	✓ 1	'The energy is lost' scores 0.
Part (e)(ii)
energy supplied = 2200 × 150 = 330 000 J	✓ 1
energy to water = 0.87 × 330 000 = 2.87 × 10⁵ J	✓ 1	Allow ECF from (c).
m = Q ÷ L = 2.87 × 10⁵ ÷ 2.26 × 10⁶ = 0.13 kg	✓ 1	Accept 0.127 kg. Omitting the 87% gives 0.146 kg: max 2 marks.
Part (f)(i)
Q = 0.75 × 4200 × 82 = 258 300 J, and energy supplied = 258 300 ÷ 0.87 = 2.97 × 10⁵ J	✓ 1	Accept the argument: half the mass needs half the energy, so half the time.
t = 2.97 × 10⁵ ÷ 2200 = 135 s	✓ 1	Accept 130 s to 140 s. Allow ECF from (c).
Part (f)(ii)
the heating element and the kettle itself must still be warmed, so a smaller fraction of the energy goes to the (smaller mass of) water	✓ 1	Accept: energy losses to the room are a larger fraction of the energy supplied.

Answers: (a)(i) 9.6 A · (a)(ii) 270 s; 5.9 × 10⁵ J · (b) 5.2 × 10⁵ J · (c) 87% · (e)(ii) 0.13 kg · (f)(i) 135 s

Common error (b) ΔT is the final temperature minus the starting temperature (82 °C here), not the final temperature (100 °C).

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Railway rails in a heatwave: expansion and stress

2.2 Thermal expansion, specific heat capacity and changes of state2.1 Particle model and gas pressure1.4 Forces, friction and springsCore

An older railway line is made of steel rail sections, each 18 m long, laid end to end with small gaps between them, as shown in Figure 1. The rails were laid on a day when their temperature was 10 °C. In a heatwave the temperature of the rails can reach 50 °C.

Figure 1: rail sections on sleepers, side view (not to scale; the gap is drawn much larger than it is)

(a)

Explain, in terms of particles, why a steel rail becomes longer when its temperature rises.

EXPLAINNotes

Show solution

the particles (gain kinetic energy and) vibrate more strongly / with larger movements about their fixed positions	✓ 1	'Particles move faster' without 'vibrate' is acceptable only if linked to a solid.
so the average separation of the particles increases	✓ 1	'The particles expand / get bigger' scores 0 for the whole part.

(2)

(b)

State what happens to the mass and to the density of a rail as it expands.

RECALLNotes

Show solution

mass stays the same	✓ 1
density decreases (because the same mass occupies a larger volume)	✓ 1

Answer: mass unchanged; density decreases

(2)

(c)(i)

An engineer heats a steel test bar of length 1.000 m and measures how much longer it becomes. The table shows the results.

temperature / °C	20	30	40	50	60	70
increase in length / mm	0.00	0.13	0.24	0.35	0.49	0.60

On Figure 2, plot the results and draw the best-fit line.

Figure 2: grid for a graph of increase in length against temperature

GRAPHNotes

Show solution

all six points plotted correctly to within half a small square	✓ 1	Five correct: allow this mark.
single straight best-fit line through the points (starting at 20 °C)	✓ 1	The line does not pass through the origin; do not accept a line forced through (0, 0).

(2)

(c)(ii)

Use your graph to determine the increase in length of the 1.000 m bar for each 1 °C rise in temperature.

GRAPHNotes

Show solution

gradient from a large triangle on the line, e.g. 0.60 ÷ (70 − 20)	✓ 1	Values read from the line, covering at least half of it.
0.012 mm per °C	✓ 1	Accept 0.011 to 0.013 mm per °C.

Answer: 0.012 mm per °C

(2)

(d)

Each gap must be wide enough for the expansion of one full 18 m rail section between 10 °C and 50 °C.

Use your answer to (c)(ii) to show that the gap must be at least about 9 mm wide when the rails are laid.

CALCNotes

Show solution

temperature rise = 50 − 10 = 40 °C	✓ 1
expansion = 0.012 × 18 × 40 = 8.6 mm (about 9 mm)	✓ 1	Allow ECF from (c)(ii) in the range 0.011 to 0.013 (7.9 mm to 9.4 mm).

Answer: 8.6 mm

(2)

(e)(i)

Modern lines use continuous welded rail: lengths are welded into one long rail with no gaps between them. A 120 m welded length is clamped firmly to concrete sleepers.

Calculate the increase in length that this rail would have between 10 °C and 50 °C if it were free to expand.

CALCNotes

Show solution

0.012 × 120 × 40 = 58 mm

✓ 1

Accept 57.6 mm. Allow ECF from (c)(ii).

Answer: 58 mm

(1)

(e)(ii)

The clamps stop the welded rail from expanding. On a very hot day, the rail sometimes bends sideways and the track buckles.

Explain, in terms of forces, why this happens.

REASONNotes

Show solution

the rail pushes against the clamps / the next part of the rail, so very large (compressive) forces act along the rail	✓ 1	Accept: the expansion forces are very large.
forces can change the shape of an object, so when these forces become larger than the clamps can hold, the rail bends sideways	✓ 1

(2)

(f)

One metre of rail has a mass of 60 kg. The specific heat capacity of steel is 450 J/(kg °C).

Calculate the thermal energy needed to raise the temperature of one metre of rail from 10 °C to 50 °C.

Extended onlyCALCNotes

Show solution

Q = mcΔT = 60 × 450 × 40	✓ 1
1.1 × 10⁶ J	✓ 1	Accept 1 080 000 J. Unit required.

Answer: 1.1 × 10⁶ J

(2)

(g)

On a hot afternoon the rail and a short steel bolt lying beside the track are both at 50 °C. State, with a reason, which has more thermal energy.

REASONNotes

Show solution

the rail	✓ 1
same temperature and same material, but the rail has a much larger mass / far more particles	✓ 1	Accept: thermal energy depends on mass as well as temperature.

Answer: the rail

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the particles (gain kinetic energy and) vibrate more strongly / with larger movements about their fixed positions	✓ 1	'Particles move faster' without 'vibrate' is acceptable only if linked to a solid.
so the average separation of the particles increases	✓ 1	'The particles expand / get bigger' scores 0 for the whole part.
Part (b)
mass stays the same	✓ 1
density decreases (because the same mass occupies a larger volume)	✓ 1
Part (c)(i)
all six points plotted correctly to within half a small square	✓ 1	Five correct: allow this mark.
single straight best-fit line through the points (starting at 20 °C)	✓ 1	The line does not pass through the origin; do not accept a line forced through (0, 0).
Part (c)(ii)
gradient from a large triangle on the line, e.g. 0.60 ÷ (70 − 20)	✓ 1	Values read from the line, covering at least half of it.
0.012 mm per °C	✓ 1	Accept 0.011 to 0.013 mm per °C.
Part (d)
temperature rise = 50 − 10 = 40 °C	✓ 1
expansion = 0.012 × 18 × 40 = 8.6 mm (about 9 mm)	✓ 1	Allow ECF from (c)(ii) in the range 0.011 to 0.013 (7.9 mm to 9.4 mm).
Part (e)(i)
0.012 × 120 × 40 = 58 mm	✓ 1	Accept 57.6 mm. Allow ECF from (c)(ii).
Part (e)(ii)
the rail pushes against the clamps / the next part of the rail, so very large (compressive) forces act along the rail	✓ 1	Accept: the expansion forces are very large.
forces can change the shape of an object, so when these forces become larger than the clamps can hold, the rail bends sideways	✓ 1
Part (f)
Q = mcΔT = 60 × 450 × 40	✓ 1
1.1 × 10⁶ J	✓ 1	Accept 1 080 000 J. Unit required.
Part (g)
the rail	✓ 1
same temperature and same material, but the rail has a much larger mass / far more particles	✓ 1	Accept: thermal energy depends on mass as well as temperature.

Answers: (b) mass unchanged; density decreases · (c)(ii) 0.012 mm per °C · (d) 8.6 mm · (e)(i) 58 mm · (f) 1.1 × 10⁶ J · (g) the rail

Common error (a) The particles do not get bigger when a solid is heated. It is the spacing between them that increases.

More 2.2 questions in the question bank

2.3 Conduction, convection and radiation

+ Add to test✓ In your testCC-CIP-23-120 marks~20 min

A passive house: insulation, windows and heat loss

2.3 Conduction, convection and radiation2.2 Thermal expansion, specific heat capacity and changes of state1.7 Energy, work and powerCore

An architect is improving an old house so that it loses much less thermal energy in winter. The outside walls are cavity walls: an inner wall and an outer brick wall with a gap between them. Figure 1 shows the wall as it is now (wall A) and after the gap has been filled with plastic foam (wall B). The foam is full of tiny bubbles of trapped air.

The table shows the energy lost each hour through the walls and through the windows when the inside of the house is kept at 20 °C and the outside temperature is 5 °C.

Part of the house	Energy lost each hour now	Energy lost each hour after the improvement	Cost of the improvement
walls	9.0 MJ (empty cavity, wall A)	3.6 MJ (foam-filled cavity, wall B)	€2400
windows	2.7 MJ (double glazing)	0.9 MJ (triple glazing)	€7200

Figure 1: cross-sections of the cavity wall before (A) and after (B) the cavity is filled with foam (not to scale)

(a)

The foam is a good thermal insulator because most of it is trapped air.

State why air is a poor thermal conductor.

RECALLNotes

Show solution

the particles (molecules) of a gas are far apart, so they collide only occasionally / energy is passed on very slowly

✓ 1

Accept 'particles are widely spaced'. 'Air has no particles' or 'air is a vacuum' scores 0.

(1)

(b)

In wall A, thermal energy crosses the air gap mainly by convection.

Explain how a convection current in the air gap of wall A carries thermal energy from the inner wall to the outer wall, and why this does not happen in wall B.

EXPLAINNotes

Show solution

air next to the warm inner wall is heated, expands and becomes less dense, so it rises	✓ 1	Must link warming to lower density. 'Heat rises' scores 0.
air next to the cold outer wall cools, becomes denser and sinks, so a circulating current carries energy across the gap	✓ 1	The idea of a continuous loop that transfers energy is needed.
in wall B the air is trapped in small bubbles, so it cannot circulate (flow) across the gap	✓ 1	Accept 'the air cannot move from place to place'.

(3)

(c)(i)

Calculate the average rate at which thermal energy is lost through the walls of the house now (wall A). Give your answer in watts.

CALCNotes

Show solution

P = E ÷ t = 9.0 × 10⁶ ÷ 3600	✓ 1	Award for the equation used with substituted values, even if a conversion is wrong.
= 2500 W	✓ 1	Both conversions needed: 9.0 MJ = 9.0 × 10⁶ J and 1 hour = 3600 s. A conversion error (e.g. 150 000 W from t = 60 s, or 0.0025 W with E left in MJ) scores 1 mark only. Unit not required as 'watts' is asked.

Answer: 2500 W

(2)

(c)(ii)

Show that filling the cavity with foam reduces the energy lost through the walls by 60%.

DERIVENotes

Show solution

energy saved each hour = 9.0 − 3.6 = 5.4 MJ	✓ 1	Accept 3.6 ÷ 9.0 = 0.40 seen, i.e. 40% of the original loss remains.
(5.4 ÷ 9.0) × 100 = 60%	✓ 1	Full working must be seen for a 'show that'. 1 − 0.40 = 0.60 = 60% is fine.

Answer: 5.4 MJ ÷ 9.0 MJ × 100 = 60%

(2)

(d)

Both improvements are made. The heating system keeps the inside of the house at a steady 20 °C while the outside temperature stays at 5 °C.

Use conservation of energy to determine the power the heating system must supply to replace the energy lost through the walls and windows.

MULTINotes

Show solution

the temperature is constant, so the energy supplied each hour = energy lost each hour = 3.6 + 0.9 = 4.5 MJ	✓ 1	The idea that supply must equal loss, or the total 4.5 MJ, is needed.
P = 4.5 × 10⁶ ÷ 3600 = 1250 W	✓ 1	Accept 1300 W or 1.25 kW. Unit required. Allow ECF from a wrong total energy.

Answer: 1250 W

(2)

(e)(i)

Before the improvements, the architect switches the heating off at 22:00 one night and records the temperature inside the house. The outside temperature stays at 5 °C all night. Figure 2 shows the results.

Use Figure 2 to determine the fall in temperature during the first 3 hours and during the last 3 hours (from 9 h to 12 h).

Figure 2: temperature inside the house after the heating is switched off at 22:00

GRAPHNotes

Show solution

first 3 hours: 20 − 16.7 = 3.3 °C	✓ 1	Accept 3.0 °C to 3.6 °C.
last 3 hours: 12.1 − 10.5 = 1.6 °C	✓ 1	Accept 1.3 °C to 1.9 °C. Unit needed at least once.

Answer: about 3.3 °C (first 3 h); about 1.6 °C (last 3 h)

(2)

(e)(ii)

Explain why the temperature falls more slowly as the night goes on.

EXPLAINNotes

Show solution

the temperature difference between the inside of the house and the outside becomes smaller	✓ 1	Must refer to a difference in temperature, not just 'the house is colder'.
a smaller temperature difference means a smaller rate of energy transfer (less energy lost each second)	✓ 1	Accept 'energy flows out more slowly'.

(2)

(e)(iii)

The test is repeated on a night when the outside temperature is −5 °C. The house again starts at 20 °C.

State and explain how the fall in temperature during the first 3 hours compares with your first answer to (e)(i).

PREDICTNotes

Show solution

the fall is larger (the temperature drops faster)	✓ 1	Do not accept 'the same'.
the temperature difference between inside and outside is greater (25 °C instead of 15 °C), so energy is lost at a greater rate	✓ 1	Reason must refer to the larger temperature difference.

(2)

(f)

The architect also wants a thick concrete floor. On a winter evening the concrete floor and the air in the living room are both at 20 °C.

Explain why the floor stores much more thermal energy than the air in the room, even though they are at the same temperature.

REASONNotes

Show solution

thermal energy depends on the mass (number of particles) and the material, not only on the temperature	✓ 1	Temperature alone does not decide the thermal energy.
the concrete floor has a much greater mass than the air in the room, so its particles have far more energy in total	✓ 1	Accept 'many more particles'. 'Concrete is hotter' scores 0.

(2)

(g)

The owner can afford only one improvement this year. Use the table to suggest which improvement should be made first. Explain your answer.

REASONNotes

Show solution

foam saves 9.0 − 3.6 = 5.4 MJ each hour; triple glazing saves 2.7 − 0.9 = 1.8 MJ each hour	✓ 1	Both savings needed, or a clear comparison such as 'three times as much'.
foam first: it saves three times as much energy and costs only a third as much (€2400 against €7200)	✓ 1	Conclusion must use both the saving and the cost. A conclusion without numbers scores 0.

Answer: Fill the cavity with foam first

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the particles (molecules) of a gas are far apart, so they collide only occasionally / energy is passed on very slowly	✓ 1	Accept 'particles are widely spaced'. 'Air has no particles' or 'air is a vacuum' scores 0.
Part (b)
air next to the warm inner wall is heated, expands and becomes less dense, so it rises	✓ 1	Must link warming to lower density. 'Heat rises' scores 0.
air next to the cold outer wall cools, becomes denser and sinks, so a circulating current carries energy across the gap	✓ 1	The idea of a continuous loop that transfers energy is needed.
in wall B the air is trapped in small bubbles, so it cannot circulate (flow) across the gap	✓ 1	Accept 'the air cannot move from place to place'.
Part (c)(i)
P = E ÷ t = 9.0 × 10⁶ ÷ 3600	✓ 1	Award for the equation used with substituted values, even if a conversion is wrong.
= 2500 W	✓ 1	Both conversions needed: 9.0 MJ = 9.0 × 10⁶ J and 1 hour = 3600 s. A conversion error (e.g. 150 000 W from t = 60 s, or 0.0025 W with E left in MJ) scores 1 mark only. Unit not required as 'watts' is asked.
Part (c)(ii)
energy saved each hour = 9.0 − 3.6 = 5.4 MJ	✓ 1	Accept 3.6 ÷ 9.0 = 0.40 seen, i.e. 40% of the original loss remains.
(5.4 ÷ 9.0) × 100 = 60%	✓ 1	Full working must be seen for a 'show that'. 1 − 0.40 = 0.60 = 60% is fine.
Part (d)
the temperature is constant, so the energy supplied each hour = energy lost each hour = 3.6 + 0.9 = 4.5 MJ	✓ 1	The idea that supply must equal loss, or the total 4.5 MJ, is needed.
P = 4.5 × 10⁶ ÷ 3600 = 1250 W	✓ 1	Accept 1300 W or 1.25 kW. Unit required. Allow ECF from a wrong total energy.
Part (e)(i)
first 3 hours: 20 − 16.7 = 3.3 °C	✓ 1	Accept 3.0 °C to 3.6 °C.
last 3 hours: 12.1 − 10.5 = 1.6 °C	✓ 1	Accept 1.3 °C to 1.9 °C. Unit needed at least once.
Part (e)(ii)
the temperature difference between the inside of the house and the outside becomes smaller	✓ 1	Must refer to a difference in temperature, not just 'the house is colder'.
a smaller temperature difference means a smaller rate of energy transfer (less energy lost each second)	✓ 1	Accept 'energy flows out more slowly'.
Part (e)(iii)
the fall is larger (the temperature drops faster)	✓ 1	Do not accept 'the same'.
the temperature difference between inside and outside is greater (25 °C instead of 15 °C), so energy is lost at a greater rate	✓ 1	Reason must refer to the larger temperature difference.
Part (f)
thermal energy depends on the mass (number of particles) and the material, not only on the temperature	✓ 1	Temperature alone does not decide the thermal energy.
the concrete floor has a much greater mass than the air in the room, so its particles have far more energy in total	✓ 1	Accept 'many more particles'. 'Concrete is hotter' scores 0.
Part (g)
foam saves 9.0 − 3.6 = 5.4 MJ each hour; triple glazing saves 2.7 − 0.9 = 1.8 MJ each hour	✓ 1	Both savings needed, or a clear comparison such as 'three times as much'.
foam first: it saves three times as much energy and costs only a third as much (€2400 against €7200)	✓ 1	Conclusion must use both the saving and the cost. A conclusion without numbers scores 0.

Answers: (c)(i) 2500 W · (c)(ii) 5.4 MJ ÷ 9.0 MJ × 100 = 60% · (d) 1250 W · (e)(i) about 3.3 °C (first 3 h); about 1.6 °C (last 3 h) · (g) Fill the cavity with foam first

Exam tip (b) Write 'warm air is less dense and rises', never 'heat rises'. Then describe the cooler, denser air sinking so that the loop is complete.

More 2.3 questions in the question bank

+ Add to test✓ In your testCC-CIP-23-220 marks~20 min

A rooftop solar water heater

2.3 Conduction, convection and radiation3.3 The electromagnetic spectrum2.2 Thermal expansion, specific heat capacity and changes of stateCore

Figure 1 shows a solar water heater on the roof of a house. Radiation from the Sun passes through a glass cover and falls on a matt black metal plate. Water pipes are fixed to the back of the plate. The heated water rises from the top of the panel to a storage tank, and cooler water returns from the bottom of the tank to the bottom of the panel. There is no pump.

The panel has an area of 2.0 m². On a clear day, each square metre of the panel receives 900 J of energy from the Sun every second. The tank holds 80 kg of water.

Figure 1: a solar water heater on a sloping roof (not to scale)

(a)

Explain why the plate is painted matt black and not shiny silver.

EXPLAINNotes

Show solution

a matt (dull) black surface is the best absorber of (infrared) radiation	✓ 1	Accept 'good absorber'. 'Attracts heat' scores 0.
a shiny silver surface would reflect most of the radiation, so less energy would reach the water	✓ 1	Accept 'more energy is passed to the water' for a black surface.

(2)

(b)(i)

Much of the energy from the Sun arrives as infrared radiation, which is part of the electromagnetic spectrum.

Name the region of the electromagnetic spectrum that is next to infrared and has a higher frequency than infrared.

RECALLNotes

Show solution

visible light

✓ 1

Accept 'light' or 'red light'. Microwaves scores 0 (lower frequency).

Answer: visible light

(1)

(b)(ii)

Some of the infrared radiation from the Sun has a wavelength of 1.2 μm. All electromagnetic waves travel at 3.0 × 10⁸ m/s in a vacuum.

Calculate the frequency of this infrared radiation.

CALCNotes

Show solution

f = v ÷ λ = 3.0 × 10⁸ ÷ 1.2 × 10⁻⁶	✓ 1	Award for the rearranged equation with substituted values, even if μm is not converted.
= 2.5 × 10¹⁴ Hz	✓ 1	Conversion 1.2 μm = 1.2 × 10⁻⁶ m needed. Unit required. 2.5 × 10⁸ Hz (μm not converted) scores 1 mark only.

Answer: 2.5 × 10¹⁴ Hz

(2)

(c)

The storage tank is fixed higher than the top of the panel. Explain how the water circulates between the panel and the tank without a pump.

EXPLAINNotes

Show solution

water in the panel is heated, expands and becomes less dense	✓ 1	Lower density must be linked to the warming.
so the hot water rises up the pipe into the tank (which is above the panel)	✓ 1	Accept 'warm water rises because it is less dense'. 'Heat rises' scores 0.
cooler, denser water from the bottom of the tank sinks down to the bottom of the panel to replace it, forming a convection current	✓ 1	The idea of a continuous circulation is needed.

(3)

(d)

Show that the panel receives about 6.5 MJ of energy from the Sun in 1 hour.

DERIVENotes

Show solution

power received = 900 × 2.0 = 1800 W (1800 J each second)	✓ 1	Accept 1.8 kW.
energy = P × t = 1800 × 3600 = 6.48 × 10⁶ J (≈ 6.5 MJ)	✓ 1	6.48 × 10⁶ J (or 6 480 000 J) must be seen; the time must be in seconds.

Answer: 1800 W × 3600 s = 6.48 × 10⁶ J ≈ 6.5 MJ

(2)

(e)(i)

During the first hour, the temperature of the 80 kg of water rises from 20 °C to 30 °C. The specific heat capacity of water is 4200 J/(kg °C).

Calculate the thermal energy gained by the water.

Extended onlyCALCNotes

Show solution

Q = mcΔT = 80 × 4200 × (30 − 20)	✓ 1	Temperature change of 10 °C must be used.
= 3.36 × 10⁶ J	✓ 1	Accept 3.4 MJ or 3 360 000 J. Unit required. Using 30 °C gives 1.0 × 10⁷ J: 1 mark only.

Answer: 3.36 × 10⁶ J (3.4 MJ)

(2)

(e)(ii)

Only part of the energy received by the panel ends up in the water in the tank.

Suggest two ways in which energy is transferred to the surroundings instead, and for each one name the method of thermal energy transfer.

REASONNotes

Show solution

any one of: some radiation is reflected by the glass cover; the hot plate emits infrared radiation; warm air above / around the panel or tank moves away (convection)	✓ 1	Each way must name a method (radiation or convection) to score.
any second, different one: conduction through the back of the panel, the frame or the pipes; conduction / convection / radiation from the tank or the pipes to the air	✓ 1	Accept 'energy conducted through the pipe walls to the cold air'. Two ways with the same method can score both marks only if they are different routes.

(2)

(f)(i)

On another clear day, the temperature of the water in the tank is recorded every hour, starting at 10:00. No hot water is used during this time.

time / h	0	1	2	3	4	5	6
temperature / °C	20	30	37	42	46	48	50

Plot the results on the grid in Figure 2 and draw a smooth curve through the points.

Figure 2: grid for plotting the temperature of the water against time

GRAPHNotes

Show solution

all seven points plotted correctly to within half a small square	✓ 1	Six correct points scores this mark only if the curve mark is also gained.
a single smooth curve through (or very close to) the points, getting less steep as time goes on	✓ 1	Not dot-to-dot straight lines.

(2)

(f)(ii)

Explain why the temperature of the water rises more and more slowly and starts to level off, even though the sunshine stays the same.

EXPLAINNotes

Show solution

as the water (and panel) gets hotter, the temperature difference between them and the surroundings increases, so they lose energy faster	✓ 1	Accept 'a hotter surface emits more radiation each second'.
the energy absorbed each second stays the same, so the net gain each second falls; it levels off when the rate of loss equals the rate of absorption	✓ 1	Idea of loss approaching the input is needed.

(2)

(g)

The next day is cloudy, and each square metre of the panel receives only 450 J of energy each second. The water again starts at 20 °C at 10:00.

Describe how the curve for the cloudy day would differ from the curve you drew in (f)(i). Give a reason.

PREDICTNotes

Show solution

the curve rises less steeply (more slowly) and levels off at a lower temperature	✓ 1	Either feature is enough for this mark.
the panel receives energy at half the rate (900 W instead of 1800 W), so less energy reaches the water each second	✓ 1	Accept 'half as much energy each second'.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
a matt (dull) black surface is the best absorber of (infrared) radiation	✓ 1	Accept 'good absorber'. 'Attracts heat' scores 0.
a shiny silver surface would reflect most of the radiation, so less energy would reach the water	✓ 1	Accept 'more energy is passed to the water' for a black surface.
Part (b)(i)
visible light	✓ 1	Accept 'light' or 'red light'. Microwaves scores 0 (lower frequency).
Part (b)(ii)
f = v ÷ λ = 3.0 × 10⁸ ÷ 1.2 × 10⁻⁶	✓ 1	Award for the rearranged equation with substituted values, even if μm is not converted.
= 2.5 × 10¹⁴ Hz	✓ 1	Conversion 1.2 μm = 1.2 × 10⁻⁶ m needed. Unit required. 2.5 × 10⁸ Hz (μm not converted) scores 1 mark only.
Part (c)
water in the panel is heated, expands and becomes less dense	✓ 1	Lower density must be linked to the warming.
so the hot water rises up the pipe into the tank (which is above the panel)	✓ 1	Accept 'warm water rises because it is less dense'. 'Heat rises' scores 0.
cooler, denser water from the bottom of the tank sinks down to the bottom of the panel to replace it, forming a convection current	✓ 1	The idea of a continuous circulation is needed.
Part (d)
power received = 900 × 2.0 = 1800 W (1800 J each second)	✓ 1	Accept 1.8 kW.
energy = P × t = 1800 × 3600 = 6.48 × 10⁶ J (≈ 6.5 MJ)	✓ 1	6.48 × 10⁶ J (or 6 480 000 J) must be seen; the time must be in seconds.
Part (e)(i)
Q = mcΔT = 80 × 4200 × (30 − 20)	✓ 1	Temperature change of 10 °C must be used.
= 3.36 × 10⁶ J	✓ 1	Accept 3.4 MJ or 3 360 000 J. Unit required. Using 30 °C gives 1.0 × 10⁷ J: 1 mark only.
Part (e)(ii)
any one of: some radiation is reflected by the glass cover; the hot plate emits infrared radiation; warm air above / around the panel or tank moves away (convection)	✓ 1	Each way must name a method (radiation or convection) to score.
any second, different one: conduction through the back of the panel, the frame or the pipes; conduction / convection / radiation from the tank or the pipes to the air	✓ 1	Accept 'energy conducted through the pipe walls to the cold air'. Two ways with the same method can score both marks only if they are different routes.
Part (f)(i)
all seven points plotted correctly to within half a small square	✓ 1	Six correct points scores this mark only if the curve mark is also gained.
a single smooth curve through (or very close to) the points, getting less steep as time goes on	✓ 1	Not dot-to-dot straight lines.
Part (f)(ii)
as the water (and panel) gets hotter, the temperature difference between them and the surroundings increases, so they lose energy faster	✓ 1	Accept 'a hotter surface emits more radiation each second'.
the energy absorbed each second stays the same, so the net gain each second falls; it levels off when the rate of loss equals the rate of absorption	✓ 1	Idea of loss approaching the input is needed.
Part (g)
the curve rises less steeply (more slowly) and levels off at a lower temperature	✓ 1	Either feature is enough for this mark.
the panel receives energy at half the rate (900 W instead of 1800 W), so less energy reaches the water each second	✓ 1	Accept 'half as much energy each second'.

Answers: (b)(i) visible light · (b)(ii) 2.5 × 10¹⁴ Hz · (d) 1800 W × 3600 s = 6.48 × 10⁶ J ≈ 6.5 MJ · (e)(i) 3.36 × 10⁶ J (3.4 MJ)

Common error (e)(i) Using the final temperature, 30 °C, in place of the temperature change. ΔT is final minus starting temperature, here 10 °C.

More 2.3 questions in the question bank

Topic 3 · Waves

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

Setup	3.1	3.2	3.3	3.4	Other
Ocean swell entering a harbour	●			○	1.2
The stadium 'wave': transverse and longitudinal waves	●			○	1.1
An endoscope: optical fibres and total internal reflection	○	●	○		—
A phone camera and a magnifying glass: converging lenses	○	●			1.1
Satellite TV and a delayed live broadcast	○		●		6.1
A hospital imaging department: X-rays, UV and infrared			●		5.2 2.3
Bat echolocation: hunting a moth with ultrasound	○			●	1.2
An ultrasound scan of an unborn baby	○			●	1.1

3.1 Wave properties, reflection, refraction and diffraction

+ Add to test✓ In your testCC-CIP-31-117 marks~20 min

Ocean swell entering a harbour

3.1 Wave properties, reflection, refraction and diffraction1.2 Speed, velocity and motion graphs3.4 Sound and ultrasoundCore

A small fishing harbour is protected by two harbour walls with an entrance 30 m wide, as shown in Figure 1. A buoy out at sea, 2.4 km from the entrance, measures the waves that pass it. The buoy records that a wave crest passes it every 5.0 s.

Figure 2 shows the shape of the sea surface near the buoy at one instant.

Figure 1: plan view of the harbour and the approaching waves (not to scale)

Figure 2: displacement of the sea surface against distance near the buoy at one instant

(a)

Use Figure 2 to determine the amplitude and the wavelength of the waves.

GRAPHNotes

Show solution

amplitude = 1.2 m	✓ 1	Measured from the rest position to a crest. 2.4 m (crest to trough) scores 0.
wavelength = 40 m	✓ 1	Crest to crest, e.g. 50 − 10 = 40 m.

Answer: amplitude 1.2 m; wavelength 40 m

(2)

(b)

Show that the speed of the waves near the buoy is 8.0 m/s.

DERIVENotes

Show solution

f = 1 ÷ T = 1 ÷ 5.0 = 0.20 Hz	✓ 1	Accept v = λ ÷ T used directly for both marks.
v = fλ = 0.20 × 40 = 8.0 m/s	✓ 1	The calculation must be shown; stating 8.0 m/s alone scores 0. Allow ECF for λ from (a) if the working is shown.

Answer: v = 0.20 Hz × 40 m = 8.0 m/s

(2)

(c)(i)

A crest takes 6 minutes 40 seconds to travel from the buoy to the harbour entrance.

Calculate the average speed of the crest between the buoy and the entrance.

CALCNotes

Show solution

time = 6 × 60 + 40 = 400 s and distance = 2400 m	✓ 1	Both conversions needed for this mark.
average speed = 2400 ÷ 400 = 6.0 m/s	✓ 1	Unit required. Allow ECF from a wrong conversion.

Answer: 6.0 m/s

(2)

(c)(ii)

Suggest why this average speed is less than the speed near the buoy.

REASONNotes

Show solution

the water becomes shallower towards the coast and waves travel more slowly in shallow water

✓ 1

Accept 'the waves slow down as the water gets shallower'.

(1)

(d)

In the shallow water at the harbour entrance the waves travel at 5.0 m/s.

Determine the wavelength of the waves at the entrance. Explain your method.

CALCNotes

Show solution

the frequency stays the same, 0.20 Hz, because it is set by the source / the same number of waves arrive each second	✓ 1	The constant frequency must be stated.
λ = v ÷ f = 5.0 ÷ 0.20 = 25 m	✓ 1	Unit required. Allow ECF from the frequency in (b).

Answer: 25 m

(2)

(e)

Further along the coast, the same waves reach a region of shallower water at an angle to the boundary between the deep and the shallow water.

Explain why the direction in which the waves travel changes, and state which way they bend.

EXPLAINNotes

Show solution

one end of each wavefront enters the shallow water first and slows down while the rest is still moving faster (the speed changes at the boundary)	✓ 1	Accept 'refraction is caused by the change in speed'.
the waves bend towards the normal	✓ 1	Accept 'the wavelength gets shorter and the waves turn towards the normal'.

(2)

(f)

The harbour master checks the depth of water at the entrance with an echo sounder on a boat. A pulse of sound is sent straight down, and the echo from the sea bed returns 3.4 ms later. The speed of sound in sea water is 1500 m/s.

Calculate the depth of the water at the entrance.

CALCNotes

Show solution

distance there and back = 1500 × 3.4 × 10⁻³ = 5.1 m	✓ 1	Conversion 3.4 ms = 3.4 × 10⁻³ s needed.
depth = 5.1 ÷ 2 = 2.55 m	✓ 1	Accept 2.6 m. 5.1 m (not halved) scores 1 mark.

Answer: 2.55 m (2.6 m)

(2)

(g)

During a storm far out at sea, long swell waves of frequency 0.10 Hz reach the harbour. At the entrance they also travel at 5.0 m/s.

Calculate the wavelength of the storm swell at the entrance, and describe and explain how its diffraction through the entrance differs from that of the waves in (d).

PREDICTNotes

Show solution

λ = 5.0 ÷ 0.10 = 50 m	✓ 1	Unit required.
the storm waves spread out more inside the harbour / curve further round behind the harbour walls	✓ 1	Accept 'more diffraction' with a description of the spreading.
diffraction increases for a longer wavelength: 50 m is now larger than the 30 m gap, compared with 25 m before	✓ 1	The comparison of wavelength with the gap width is needed. Allow ECF from (d).

Answer: 50 m; the storm swell diffracts more

(3)

(h)

State one way in which the sound pulse from the echo sounder is different from the water waves, apart from its speed and frequency.

RECALLNotes

Show solution

sound is a longitudinal wave (vibrations parallel to the direction of travel) but water waves are transverse (vibrations at right angles)

✓ 1

Accept 'sound has compressions and rarefactions; water waves have crests and troughs'.

(1)

Show full solution

Marking point	Mark	Notes
Part (a)
amplitude = 1.2 m	✓ 1	Measured from the rest position to a crest. 2.4 m (crest to trough) scores 0.
wavelength = 40 m	✓ 1	Crest to crest, e.g. 50 − 10 = 40 m.
Part (b)
f = 1 ÷ T = 1 ÷ 5.0 = 0.20 Hz	✓ 1	Accept v = λ ÷ T used directly for both marks.
v = fλ = 0.20 × 40 = 8.0 m/s	✓ 1	The calculation must be shown; stating 8.0 m/s alone scores 0. Allow ECF for λ from (a) if the working is shown.
Part (c)(i)
time = 6 × 60 + 40 = 400 s and distance = 2400 m	✓ 1	Both conversions needed for this mark.
average speed = 2400 ÷ 400 = 6.0 m/s	✓ 1	Unit required. Allow ECF from a wrong conversion.
Part (c)(ii)
the water becomes shallower towards the coast and waves travel more slowly in shallow water	✓ 1	Accept 'the waves slow down as the water gets shallower'.
Part (d)
the frequency stays the same, 0.20 Hz, because it is set by the source / the same number of waves arrive each second	✓ 1	The constant frequency must be stated.
λ = v ÷ f = 5.0 ÷ 0.20 = 25 m	✓ 1	Unit required. Allow ECF from the frequency in (b).
Part (e)
one end of each wavefront enters the shallow water first and slows down while the rest is still moving faster (the speed changes at the boundary)	✓ 1	Accept 'refraction is caused by the change in speed'.
the waves bend towards the normal	✓ 1	Accept 'the wavelength gets shorter and the waves turn towards the normal'.
Part (f)
distance there and back = 1500 × 3.4 × 10⁻³ = 5.1 m	✓ 1	Conversion 3.4 ms = 3.4 × 10⁻³ s needed.
depth = 5.1 ÷ 2 = 2.55 m	✓ 1	Accept 2.6 m. 5.1 m (not halved) scores 1 mark.
Part (g)
λ = 5.0 ÷ 0.10 = 50 m	✓ 1	Unit required.
the storm waves spread out more inside the harbour / curve further round behind the harbour walls	✓ 1	Accept 'more diffraction' with a description of the spreading.
diffraction increases for a longer wavelength: 50 m is now larger than the 30 m gap, compared with 25 m before	✓ 1	The comparison of wavelength with the gap width is needed. Allow ECF from (d).
Part (h)
sound is a longitudinal wave (vibrations parallel to the direction of travel) but water waves are transverse (vibrations at right angles)	✓ 1	Accept 'sound has compressions and rarefactions; water waves have crests and troughs'.

Answers: (a) amplitude 1.2 m; wavelength 40 m · (b) v = 0.20 Hz × 40 m = 8.0 m/s · (c)(i) 6.0 m/s · (d) 25 m · (f) 2.55 m (2.6 m) · (g) 50 m; the storm swell diffracts more

Exam tip (a) On a displacement–distance graph, crest to crest gives the wavelength. On a displacement–time graph the same gap would be the period, so read the axis label first.

Common error (d) Changing the frequency when the waves enter shallow water. The frequency is fixed by the source; only the speed and the wavelength change.

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The stadium 'wave': transverse and longitudinal waves

3.1 Wave properties, reflection, refraction and diffraction3.4 Sound and ultrasound1.1 Physical quantities and measurementCore

At a football match, the fans make a crowd wave. Each fan stands up and raises their arms, then sits down again, just after the person next to them. The pattern travels round the stadium, but every fan stays in their own seat. Figure 1 shows a plan view of the stadium. The ring of seats used for the wave is 540 m long all the way round.

Figure 1: plan view of the stadium during a crowd wave (not to scale)

(a)

The crowd wave behaves like a transverse wave.

Explain why it is a transverse wave, and state what is transferred round the stadium by the wave.

EXPLAINNotes

Show solution

the fans move up and down, at right angles to the direction in which the wave travels (round the stadium)	✓ 1	Both directions, or 'perpendicular to the direction of travel', must be clear.
energy / the pattern is transferred, not the fans (matter)	✓ 1	Accept 'the disturbance moves but the people stay in their seats'.

(2)

(b)(i)

A student in the stand starts a stopwatch when the standing fans pass the flag. She stops it when they pass the flag for the third time after that. The stopwatch reads 135.0 s.

Explain why the student times three laps of the wave rather than one.

EXPTNotes

Show solution

the reaction-time error (about 0.2 s) is shared over a longer time, so the percentage uncertainty is smaller

✓ 1

Accept 'a more accurate / reliable time for one lap'. 'To be fair' scores 0.

(1)

(b)(ii)

Calculate the speed of the crowd wave.

CALCNotes

Show solution

time for one lap = 135.0 ÷ 3 = 45.0 s	✓ 1	Or total distance 3 × 540 = 1620 m.
speed = 540 ÷ 45.0 = 12 m/s	✓ 1	Accept 12.0 m/s. Unit required. 4.0 m/s (540 ÷ 135) scores 0.

Answer: 12 m/s

(2)

(c)(i)

The fans now make a continuous series of waves, so that each fan stands up and sits down over and over again at a steady rate. One fan counts that she stands up 15 times in 60 s.

Show that the frequency of the crowd wave is 0.25 Hz.

DERIVENotes

Show solution

frequency = number of complete waves ÷ time = 15 ÷ 60	✓ 1	Or period = 60 ÷ 15 = 4.0 s.
= 0.25 Hz	✓ 1	Or f = 1 ÷ 4.0 = 0.25 Hz. Working must be seen.

Answer: 15 ÷ 60 s = 0.25 Hz

(2)

(c)(ii)

The wave still travels at the speed you found in (b)(ii). Calculate the wavelength of the crowd wave.

CALCNotes

Show solution

λ = v ÷ f = 12 ÷ 0.25	✓ 1	Rearranged equation or substitution.
= 48 m	✓ 1	Unit required. Allow ECF from (b)(ii).

Answer: 48 m

(2)

(d)

Treat the motion of the arms of one fan as a smooth up-and-down vibration about a middle position.

Sketch, on the axes in Figure 2, a graph of the displacement of this fan's arms against time for the first 12 s.

Figure 2: axes for a displacement–time graph of one fan

SKETCHNotes

Show solution

a smooth, repeating wave shape with constant amplitude	✓ 1	A wave about the time axis, with crests of equal height.
period 4.0 s: exactly 3 complete cycles between 0 and 12 s	✓ 1	Crests (or zero crossings) 4.0 s apart, read from the time axis.

(2)

(e)

The fans are now asked to stand up twice as often. The wave still travels round the stadium at the same speed.

State the new frequency and the new wavelength of the crowd wave.

PREDICTNotes

Show solution

frequency = 0.50 Hz	✓ 1	Allow ECF from (c)(i).
wavelength = 12 ÷ 0.50 = 24 m (halved)	✓ 1	Accept 'halved' with the value. Allow ECF from (c)(ii).

Answer: 0.50 Hz; 24 m

(2)

(f)

Before the match, a teacher demonstrates a longitudinal wave by pushing and pulling one end of a long stretched spring (a slinky) along its length.

Describe how the coils of the spring move as the wave passes, and state one way in which a sound wave in air is like this wave.

EXPLAINNotes

Show solution

each coil vibrates backwards and forwards along the length of the spring (parallel to the direction of travel), forming compressions and rarefactions	✓ 1	Both the direction of vibration and the idea of compressions / rarefactions are needed.
sound is also longitudinal: air particles vibrate parallel to the direction of travel, forming compressions and rarefactions	✓ 1	Accept 'sound also transfers energy without moving the air along with it'.

(2)

(g)

A drummer in the stand beats a drum once. The sound travels straight across the stadium, reflects from the stand opposite, 170 m away, and returns to the drummer. The speed of sound in air is 340 m/s.

Calculate the time between the drum beat and its echo reaching the drummer.

CALCNotes

Show solution

distance there and back = 2 × 170 = 340 m	✓ 1	The doubling must be seen.
time = 340 ÷ 340 = 1.0 s	✓ 1	Unit required. 0.50 s (not doubled) scores 1 mark.

Answer: 1.0 s

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the fans move up and down, at right angles to the direction in which the wave travels (round the stadium)	✓ 1	Both directions, or 'perpendicular to the direction of travel', must be clear.
energy / the pattern is transferred, not the fans (matter)	✓ 1	Accept 'the disturbance moves but the people stay in their seats'.
Part (b)(i)
the reaction-time error (about 0.2 s) is shared over a longer time, so the percentage uncertainty is smaller	✓ 1	Accept 'a more accurate / reliable time for one lap'. 'To be fair' scores 0.
Part (b)(ii)
time for one lap = 135.0 ÷ 3 = 45.0 s	✓ 1	Or total distance 3 × 540 = 1620 m.
speed = 540 ÷ 45.0 = 12 m/s	✓ 1	Accept 12.0 m/s. Unit required. 4.0 m/s (540 ÷ 135) scores 0.
Part (c)(i)
frequency = number of complete waves ÷ time = 15 ÷ 60	✓ 1	Or period = 60 ÷ 15 = 4.0 s.
= 0.25 Hz	✓ 1	Or f = 1 ÷ 4.0 = 0.25 Hz. Working must be seen.
Part (c)(ii)
λ = v ÷ f = 12 ÷ 0.25	✓ 1	Rearranged equation or substitution.
= 48 m	✓ 1	Unit required. Allow ECF from (b)(ii).
Part (d)
a smooth, repeating wave shape with constant amplitude	✓ 1	A wave about the time axis, with crests of equal height.
period 4.0 s: exactly 3 complete cycles between 0 and 12 s	✓ 1	Crests (or zero crossings) 4.0 s apart, read from the time axis.
Part (e)
frequency = 0.50 Hz	✓ 1	Allow ECF from (c)(i).
wavelength = 12 ÷ 0.50 = 24 m (halved)	✓ 1	Accept 'halved' with the value. Allow ECF from (c)(ii).
Part (f)
each coil vibrates backwards and forwards along the length of the spring (parallel to the direction of travel), forming compressions and rarefactions	✓ 1	Both the direction of vibration and the idea of compressions / rarefactions are needed.
sound is also longitudinal: air particles vibrate parallel to the direction of travel, forming compressions and rarefactions	✓ 1	Accept 'sound also transfers energy without moving the air along with it'.
Part (g)
distance there and back = 2 × 170 = 340 m	✓ 1	The doubling must be seen.
time = 340 ÷ 340 = 1.0 s	✓ 1	Unit required. 0.50 s (not doubled) scores 1 mark.

Answers: (b)(ii) 12 m/s · (c)(i) 15 ÷ 60 s = 0.25 Hz · (c)(ii) 48 m · (e) 0.50 Hz; 24 m · (g) 1.0 s

Exam tip (d) On a displacement–time graph the gap from crest to crest is the period, not the wavelength. Use the frequency to find it before you draw.

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3.2 Light, refraction, lenses and dispersion

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An endoscope: optical fibres and total internal reflection

3.2 Light, refraction, lenses and dispersion3.1 Wave properties, reflection, refraction and diffraction3.3 The electromagnetic spectrumExtended

A doctor uses an endoscope to look inside a patient's stomach. Light travels down a bundle of thin glass fibres to light up the stomach wall, and a second bundle carries light back to a camera.

The glass of each fibre has a refractive index of 1.52. In this question, treat each fibre as glass surrounded by air. (Real fibres also have a thin outer layer of glass, the cladding; ignore it here.)

Figure 1 shows a ray of light in air meeting the flat end face of one fibre with an angle of incidence of 30°. The ray refracts into the fibre and travels to point P on the side wall.

Figure 1: a ray entering the end of a glass fibre (not to scale)

(a)

Calculate the angle of refraction of the ray at the end face.

Extended onlyCALCNotes

Show solution

sin r = sin i ÷ n = sin 30° ÷ 1.52 (= 0.329)	✓ 1	Rearrangement and substitution seen.
r = 19°	✓ 1	Accept 19.2°. Answers that are larger than 30° score 0 for this mark.

Answer: 19° (19.2°)

(2)

(b)

Show that the critical angle for the glass of the fibre is about 41°.

Extended onlyDERIVENotes

Show solution

sin c = 1 ÷ n = 1 ÷ 1.52 = 0.658	✓ 1	Equation and substitution must be seen.
c = sin⁻¹(0.658) = 41.1°	✓ 1	Value to at least 3 s.f. needed, since 41° is given.

Answer: 41.1°

(2)

(c)(i)

The end face is at 90° to the side wall. Use your answer to (a) to determine the angle of incidence of the ray at P.

Extended onlyMULTINotes

Show solution

90° − 19.2° = 70.8°

✓ 1

Allow ECF from (a): 90° − their (a). Accept 71°.

Answer: 70.8°

(1)

(c)(ii)

Explain what happens to the ray at P.

EXPLAINNotes

Show solution

the ray is in the glass (the denser material) and its angle of incidence (70.8°) is greater than the critical angle (41°)	✓ 1	Both conditions needed: travelling in glass towards air, and i > c. Allow ECF from (b) and (c)(i).
so all the light is reflected back into the fibre / total internal reflection, at 70.8° to the normal	✓ 1	"It reflects" alone is not enough: total (internal) reflection, or no light escaping, must be stated.

(2)

(d)

Figure 2 shows how the angle of refraction r in air depends on the angle of incidence i inside the glass, for light leaving the fibre glass through a flat face into air.

Use Figure 2 to determine:

the critical angle of the glass
the angle of refraction in air when the angle of incidence in the glass is 30°.

Figure 2: angle of refraction in air against angle of incidence in the glass, for light leaving the fibre glass through a flat face

GRAPHNotes

Show solution

critical angle = value of i where r reaches 90° = 41°	✓ 1	Accept 40°–42°.
r = 49° when i = 30°	✓ 1	Accept 48°–51°.

Answer: critical angle 41°; r = 49°

(2)

(e)

The light source of the endoscope sends green light of wavelength 5.4 × 10⁻⁷ m in air into the fibres. The speed of light is 3.0 × 10⁸ m/s in air and 2.0 × 10⁸ m/s in the glass.

Calculate the wavelength of the green light inside the glass.

MULTINotes

Show solution

f = v ÷ λ = 3.0 × 10⁸ ÷ 5.4 × 10⁻⁷ = 5.6 × 10¹⁴ Hz	✓ 1	Accept 5.56 × 10¹⁴ Hz.
the frequency does not change when the light enters the glass	✓ 1	May be shown by using the same f in the next step.
λ = 2.0 × 10⁸ ÷ 5.56 × 10¹⁴ = 3.6 × 10⁻⁷ m	✓ 1	Allow the ratio method: 5.4 × 10⁻⁷ × (2.0 ÷ 3.0) = 3.6 × 10⁻⁷ m for all 3 marks.

Answer: 3.6 × 10⁻⁷ m

(3)

(f)

Optical fibres that carry internet signals use pulses of infrared radiation instead of visible light.

State one property that infrared and visible light share, and one way in which infrared is different from visible light.

RECALLNotes

Show solution

shared: any one of: both are transverse waves / both travel at 3.0 × 10⁸ m/s in a vacuum / both can travel through a vacuum / both are electromagnetic waves / both transfer energy	✓ 1
different: infrared has a longer wavelength / a lower frequency	✓ 1	Do not accept "infrared is slower" or "infrared is invisible" alone for this mark.

(2)

(g)(i)

During a procedure the endoscope is bent sharply. At one bend, a ray inside a fibre meets the side wall with an angle of incidence of 38°.

Describe what happens to this ray, and suggest the effect on the image seen by the doctor.

PREDICTNotes

Show solution

38° is less than the critical angle (41°), so the ray is refracted out of the fibre / escapes into the air (with only a weak reflection)	✓ 1	Allow ECF from (b).
less light reaches the far end, so the image is dimmer	✓ 1	Accept "darker" or "less bright".

(2)

(g)(ii)

A manufacturer makes a new endoscope with fibres made of glass of refractive index 1.65.

Calculate the critical angle for this glass and deduce whether the ray in (g)(i) now stays inside the fibre.

Extended onlyMULTINotes

Show solution

sin c = 1 ÷ 1.65, so c = 37.3°	✓ 1	Accept 37°.
38° is greater than 37.3°, so the ray is totally internally reflected and stays in the fibre	✓ 1	Allow ECF from their c if the comparison is correct.

Answer: 37.3°; the ray stays inside the fibre

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
sin r = sin i ÷ n = sin 30° ÷ 1.52 (= 0.329)	✓ 1	Rearrangement and substitution seen.
r = 19°	✓ 1	Accept 19.2°. Answers that are larger than 30° score 0 for this mark.
Part (b)
sin c = 1 ÷ n = 1 ÷ 1.52 = 0.658	✓ 1	Equation and substitution must be seen.
c = sin⁻¹(0.658) = 41.1°	✓ 1	Value to at least 3 s.f. needed, since 41° is given.
Part (c)(i)
90° − 19.2° = 70.8°	✓ 1	Allow ECF from (a): 90° − their (a). Accept 71°.
Part (c)(ii)
the ray is in the glass (the denser material) and its angle of incidence (70.8°) is greater than the critical angle (41°)	✓ 1	Both conditions needed: travelling in glass towards air, and i > c. Allow ECF from (b) and (c)(i).
so all the light is reflected back into the fibre / total internal reflection, at 70.8° to the normal	✓ 1	"It reflects" alone is not enough: total (internal) reflection, or no light escaping, must be stated.
Part (d)
critical angle = value of i where r reaches 90° = 41°	✓ 1	Accept 40°–42°.
r = 49° when i = 30°	✓ 1	Accept 48°–51°.
Part (e)
f = v ÷ λ = 3.0 × 10⁸ ÷ 5.4 × 10⁻⁷ = 5.6 × 10¹⁴ Hz	✓ 1	Accept 5.56 × 10¹⁴ Hz.
the frequency does not change when the light enters the glass	✓ 1	May be shown by using the same f in the next step.
λ = 2.0 × 10⁸ ÷ 5.56 × 10¹⁴ = 3.6 × 10⁻⁷ m	✓ 1	Allow the ratio method: 5.4 × 10⁻⁷ × (2.0 ÷ 3.0) = 3.6 × 10⁻⁷ m for all 3 marks.
Part (f)
shared: any one of: both are transverse waves / both travel at 3.0 × 10⁸ m/s in a vacuum / both can travel through a vacuum / both are electromagnetic waves / both transfer energy	✓ 1
different: infrared has a longer wavelength / a lower frequency	✓ 1	Do not accept "infrared is slower" or "infrared is invisible" alone for this mark.
Part (g)(i)
38° is less than the critical angle (41°), so the ray is refracted out of the fibre / escapes into the air (with only a weak reflection)	✓ 1	Allow ECF from (b).
less light reaches the far end, so the image is dimmer	✓ 1	Accept "darker" or "less bright".
Part (g)(ii)
sin c = 1 ÷ 1.65, so c = 37.3°	✓ 1	Accept 37°.
38° is greater than 37.3°, so the ray is totally internally reflected and stays in the fibre	✓ 1	Allow ECF from their c if the comparison is correct.

Answers: (a) 19° (19.2°) · (b) 41.1° · (c)(i) 70.8° · (d) critical angle 41°; r = 49° · (e) 3.6 × 10⁻⁷ m · (g)(ii) 37.3°; the ray stays inside the fibre

Exam tip (a) Light going from air into glass bends towards the normal, so the answer must be smaller than 30°. If it is not, check the rearrangement and that the calculator is in degree mode.

Common error (c)(ii) Comparing the angle of incidence with the angle of 30° at the end face. Total internal reflection depends on the angle at the wall where the light meets the glass–air boundary, compared with the critical angle.

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A phone camera and a magnifying glass: converging lenses

3.2 Light, refraction, lenses and dispersion3.1 Wave properties, reflection, refraction and diffraction1.1 Physical quantities and measurementCore

A phone camera uses a converging lens to form an image on a light sensor. A student models the camera in the laboratory with a converging lens of focal length 5.0 cm, an illuminated arrow 2.0 cm tall as the object, and a white screen in place of the sensor.

Figure 1: grid with 1 square = 1 cm. Object O is 2.0 cm tall and 15.0 cm from the lens; F is 5.0 cm from the lens

(a)

State what is meant by

the principal focus of a converging lens
the focal length of the lens.

RECALLNotes

Show solution

principal focus: the point (on the principal axis) where rays parallel to the axis meet after passing through the lens	✓ 1	"Where the light focuses" alone is not enough: parallel rays must be mentioned.
focal length: the distance from the centre of the lens to the principal focus	✓ 1

(2)

(b)(i)

In Figure 1 the object O stands on the principal axis, 15.0 cm from the lens. On Figure 1, draw two rays from the top of O to locate the image.

SKETCHNotes

Show solution

any one correct construction ray from the top of O: parallel to the axis, then through F on the right / straight through the centre of the lens / through F on the left, then parallel to the axis	✓ 1	Rays drawn with a ruler; refraction shown at the lens line.
a second correct ray, crossing the first at 7.5 cm to the right of the lens, 1.0 cm below the axis	✓ 1	Image arrow drawn from the axis to the crossing point.

(2)

(b)(ii)

Use your diagram to determine the distance of the image from the lens and the height of the image. Describe the image.

DATANotes

Show solution

image distance = 7.5 cm	✓ 1	Accept 7.2–7.8 cm.
image height = 1.0 cm	✓ 1	Accept 0.9–1.1 cm.
real, inverted and diminished (smaller than the object)	✓ 1	All three needed for the mark.

Answer: 7.5 cm; 1.0 cm; real, inverted, diminished

(3)

(c)

Magnification = image height ÷ object height.

Show that the magnification of this image is 0.50, and check that it agrees with the ratio image distance ÷ object distance.

DERIVENotes

Show solution

magnification = 1.0 cm ÷ 2.0 cm = 0.50	✓ 1	Allow ECF from (b)(ii) if their values give 0.50 within the tolerance.
7.5 cm ÷ 15.0 cm = 0.50, the same value	✓ 1

Answer: 0.50

(2)

(d)(i)

The student moves the object to different distances u from the lens and finds the image distance v each time by moving the screen until the image is sharp. The results are:

u / cm	6.0	7.5	10.0	15.0	20.0	25.0	30.0
v / cm	29.6	15.2	10.1	7.4	6.7	6.2	6.0

Figure 2 shows the results with a best-fit curve.

Use Figure 2 to find the image distance when the object is 12.0 cm from the lens.

Figure 2: image distance v against object distance u for the lens, with a best-fit curve

GRAPHNotes

Show solution

v = 8.6 cm

✓ 1

Accept 8.3–8.9 cm.

Answer: 8.6 cm

(1)

(d)(ii)

When the object is at 2F, the image is also at 2F, so u = v.

Use Figure 2 to find the value of u at which u = v, and hence show that the results agree with the focal length of 5.0 cm.

GRAPHNotes

Show solution

u = v = 10 cm, read where the curve crosses the line v = u	✓ 1	Accept 9.6–10.4 cm.
focal length = 10 ÷ 2 = 5.0 cm, which agrees with the value given	✓ 1	Allow ECF: half their value of u, with a comparison with 5.0 cm.

Answer: u = 10 cm; 10 ÷ 2 = 5.0 cm, as given

(2)

(e)

The student measures u and v with a metre rule.

State the resolution of a metre rule, and suggest how the student can make her value of v more reliable.

EXPTNotes

Show solution

resolution = 1 mm (0.1 cm)	✓ 1
repeat the measurement of v (moving the screen in from both sides) and take the average	✓ 1	Accept: find the range of screen positions where the image looks sharp and use the middle of the range. Accept: look straight down at the scale to avoid parallax.

Answer: 1 mm

(2)

(f)

The lens is now used as a magnifying glass. The object is placed 3.0 cm from the lens. A scale ray diagram shows that the image is 7.5 cm from the lens, on the same side as the object.

Describe the image, and calculate its height.

Extended onlyMULTINotes

Show solution

virtual, upright and larger (magnified)	✓ 1	All three needed.
height = (7.5 ÷ 3.0) × 2.0 cm = 5.0 cm	✓ 1	Magnification 2.5 must be applied to the 2.0 cm object.

Answer: virtual, upright, larger; 5.0 cm

(2)

(g)

Light from the object passes from the air into the glass of the lens. State what happens to the speed, the frequency and the wavelength of the light.

RECALLNotes

Show solution

speed decreases and wavelength decreases	✓ 1	Both needed.
frequency stays the same	✓ 1

(2)

(h)

When the student uses white light, the edges of the image on the screen have coloured fringes.

Explain why the lens produces coloured edges.

EXPLAINNotes

Show solution

white light is a mixture of colours, and each colour is refracted by a different amount in glass / violet is refracted more than red (dispersion)	✓ 1	Accept: different colours travel at different speeds in glass.
so the colours are brought together at slightly different distances from the lens (violet closer than red), and they cannot all be sharp on the screen at once	✓ 1	Accept: the colours do not all meet at the same point.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
principal focus: the point (on the principal axis) where rays parallel to the axis meet after passing through the lens	✓ 1	"Where the light focuses" alone is not enough: parallel rays must be mentioned.
focal length: the distance from the centre of the lens to the principal focus	✓ 1
Part (b)(i)
any one correct construction ray from the top of O: parallel to the axis, then through F on the right / straight through the centre of the lens / through F on the left, then parallel to the axis	✓ 1	Rays drawn with a ruler; refraction shown at the lens line.
a second correct ray, crossing the first at 7.5 cm to the right of the lens, 1.0 cm below the axis	✓ 1	Image arrow drawn from the axis to the crossing point.
Part (b)(ii)
image distance = 7.5 cm	✓ 1	Accept 7.2–7.8 cm.
image height = 1.0 cm	✓ 1	Accept 0.9–1.1 cm.
real, inverted and diminished (smaller than the object)	✓ 1	All three needed for the mark.
Part (c)
magnification = 1.0 cm ÷ 2.0 cm = 0.50	✓ 1	Allow ECF from (b)(ii) if their values give 0.50 within the tolerance.
7.5 cm ÷ 15.0 cm = 0.50, the same value	✓ 1
Part (d)(i)
v = 8.6 cm	✓ 1	Accept 8.3–8.9 cm.
Part (d)(ii)
u = v = 10 cm, read where the curve crosses the line v = u	✓ 1	Accept 9.6–10.4 cm.
focal length = 10 ÷ 2 = 5.0 cm, which agrees with the value given	✓ 1	Allow ECF: half their value of u, with a comparison with 5.0 cm.
Part (e)
resolution = 1 mm (0.1 cm)	✓ 1
repeat the measurement of v (moving the screen in from both sides) and take the average	✓ 1	Accept: find the range of screen positions where the image looks sharp and use the middle of the range. Accept: look straight down at the scale to avoid parallax.
Part (f)
virtual, upright and larger (magnified)	✓ 1	All three needed.
height = (7.5 ÷ 3.0) × 2.0 cm = 5.0 cm	✓ 1	Magnification 2.5 must be applied to the 2.0 cm object.
Part (g)
speed decreases and wavelength decreases	✓ 1	Both needed.
frequency stays the same	✓ 1
Part (h)
white light is a mixture of colours, and each colour is refracted by a different amount in glass / violet is refracted more than red (dispersion)	✓ 1	Accept: different colours travel at different speeds in glass.
so the colours are brought together at slightly different distances from the lens (violet closer than red), and they cannot all be sharp on the screen at once	✓ 1	Accept: the colours do not all meet at the same point.

Answers: (b)(ii) 7.5 cm; 1.0 cm; real, inverted, diminished · (c) 0.50 · (d)(i) 8.6 cm · (d)(ii) u = 10 cm; 10 ÷ 2 = 5.0 cm, as given · (e) 1 mm · (f) virtual, upright, larger; 5.0 cm

Exam tip (b)(ii) Describe every image with three words: real or virtual, upright or inverted, larger or smaller. Two out of three does not earn the mark.

More 3.2 questions in the question bank

3.3 The electromagnetic spectrum

+ Add to test✓ In your testCC-CIP-33-118 marks~20 min

Satellite TV and a delayed live broadcast

3.3 The electromagnetic spectrum3.1 Wave properties, reflection, refraction and diffraction6.1 Earth, Moon and the Solar SystemCore

A football match is shown live on satellite TV. A TV station sends the pictures as microwaves up to a satellite that orbits 3.6 × 10⁷ m above the equator. The satellite sends the signal back down to dishes on viewers' homes, using microwaves of frequency 12 GHz.

Figure 1 shows the paths of the signal to and from the satellite. The speed of electromagnetic waves is 3.0 × 10⁸ m/s.

Figure 1: the signal path through the satellite (not to scale)

(a)(i)

Microwaves are one region of the electromagnetic spectrum. Name the two regions on either side of microwaves.

RECALLNotes

Show solution

radio waves and infrared

✓ 1

Both needed.

Answer: radio waves and infrared

(1)

(a)(ii)

State two properties shared by all electromagnetic waves.

RECALLNotes

Show solution

any two of: transverse / travel at the same speed (3.0 × 10⁸ m/s) in a vacuum / can travel through a vacuum / transfer energy / can be reflected, refracted and diffracted	✓ 1	One mark for each property, up to 2.
a second, different property from the list	✓ 1	Do not accept "same frequency" or "same wavelength".

(2)

(b)

Calculate the wavelength of the 12 GHz microwaves.

CALCNotes

Show solution

12 GHz = 1.2 × 10¹⁰ Hz and λ = v ÷ f	✓ 1
λ = 3.0 × 10⁸ ÷ 1.2 × 10¹⁰ = 0.025 m	✓ 1	Accept 2.5 cm. 2.5 × 10⁷ m (GHz not converted to Hz) scores 1 mark at most.

Answer: 0.025 m (2.5 cm)

(2)

(c)

Use Figure 1 to show that the signal takes about 0.26 s to travel from the TV station to the viewer's dish.

DERIVENotes

Show solution

total distance = 3.80 × 10⁷ + 3.90 × 10⁷ = 7.70 × 10⁷ m	✓ 1
t = d ÷ v = 7.70 × 10⁷ ÷ 3.0 × 10⁸ = 0.257 s	✓ 1	Value to at least 3 s.f. needed, since 0.26 s is given. Using only one of the two distances scores 0.

Answer: 0.257 s

(2)

(d)(i)

Satellite links use a range of microwave frequencies. Figure 2 shows how the wavelength depends on the frequency.

Use Figure 2 to find the wavelength of microwaves of frequency 8.0 GHz.

Figure 2: wavelength against frequency for microwaves used by satellite links

GRAPHNotes

Show solution

λ = 3.75 cm

✓ 1

Accept 3.6–3.9 cm.

Answer: 3.75 cm

(1)

(d)(ii)

Describe the relationship shown in Figure 2, and explain it using the wave equation.

EXPLAINNotes

Show solution

as the frequency increases the wavelength decreases / doubling the frequency halves the wavelength	✓ 1	Accept "inversely proportional". "Negative correlation" alone is not enough.
all the waves travel at the same speed, so f × λ (= v) is constant	✓ 1	Accept a check from the graph, e.g. 6.0 GHz × 5.0 cm = 12 GHz × 2.5 cm.

(2)

(e)

The TV pictures are sent as a digital signal. The signal becomes weaker and picks up noise on its long journey, so it is amplified on the way.

Explain why a digital signal arrives with better quality than an analogue signal would.

EXPLAINNotes

Show solution

an analogue signal varies continuously, so the noise cannot be told apart from the signal and is amplified with it	✓ 1
a digital signal has only two values (1 and 0), so the receiver can decide which each pulse is and regenerate clean pulses / remove the noise	✓ 1

(2)

(f)

The satellite orbits the Earth once every 24 hours, so it stays above the same point on the equator. The radius of the Earth is 6.4 × 10⁶ m.

Calculate the orbital speed of the satellite.

MULTINotes

Show solution

orbit radius r = 6.4 × 10⁶ + 3.6 × 10⁷ = 4.24 × 10⁷ m	✓ 1
T = 24 × 3600 = 86 400 s	✓ 1
v = 2πr ÷ T = 2π × 4.24 × 10⁷ ÷ 86 400 = 3.1 × 10³ m/s	✓ 1	Accept 3080 m/s. Using r = 3.6 × 10⁷ m gives 2.6 × 10³ m/s: 2 marks at most.

Answer: 3.1 × 10³ m/s

(3)

(g)

A video-call service uses a different satellite in a low orbit, 1.2 × 10⁶ m above the ground.

Calculate the time for a signal to travel up to this satellite and back down when the satellite is directly overhead. Suggest why the low orbit is an advantage for live video calls.

PREDICTNotes

Show solution

distance = 2 × 1.2 × 10⁶ = 2.4 × 10⁶ m	✓ 1
t = 2.4 × 10⁶ ÷ 3.0 × 10⁸ = 8.0 × 10⁻³ s (0.0080 s)	✓ 1
the delay is much shorter (about 30 times less than via the high satellite), so there is no noticeable pause when people talk to each other	✓ 1	Accept any sensible comparison with the answer to (c).

Answer: 8.0 × 10⁻³ s

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
radio waves and infrared	✓ 1	Both needed.
Part (a)(ii)
any two of: transverse / travel at the same speed (3.0 × 10⁸ m/s) in a vacuum / can travel through a vacuum / transfer energy / can be reflected, refracted and diffracted	✓ 1	One mark for each property, up to 2.
a second, different property from the list	✓ 1	Do not accept "same frequency" or "same wavelength".
Part (b)
12 GHz = 1.2 × 10¹⁰ Hz and λ = v ÷ f	✓ 1
λ = 3.0 × 10⁸ ÷ 1.2 × 10¹⁰ = 0.025 m	✓ 1	Accept 2.5 cm. 2.5 × 10⁷ m (GHz not converted to Hz) scores 1 mark at most.
Part (c)
total distance = 3.80 × 10⁷ + 3.90 × 10⁷ = 7.70 × 10⁷ m	✓ 1
t = d ÷ v = 7.70 × 10⁷ ÷ 3.0 × 10⁸ = 0.257 s	✓ 1	Value to at least 3 s.f. needed, since 0.26 s is given. Using only one of the two distances scores 0.
Part (d)(i)
λ = 3.75 cm	✓ 1	Accept 3.6–3.9 cm.
Part (d)(ii)
as the frequency increases the wavelength decreases / doubling the frequency halves the wavelength	✓ 1	Accept "inversely proportional". "Negative correlation" alone is not enough.
all the waves travel at the same speed, so f × λ (= v) is constant	✓ 1	Accept a check from the graph, e.g. 6.0 GHz × 5.0 cm = 12 GHz × 2.5 cm.
Part (e)
an analogue signal varies continuously, so the noise cannot be told apart from the signal and is amplified with it	✓ 1
a digital signal has only two values (1 and 0), so the receiver can decide which each pulse is and regenerate clean pulses / remove the noise	✓ 1
Part (f)
orbit radius r = 6.4 × 10⁶ + 3.6 × 10⁷ = 4.24 × 10⁷ m	✓ 1
T = 24 × 3600 = 86 400 s	✓ 1
v = 2πr ÷ T = 2π × 4.24 × 10⁷ ÷ 86 400 = 3.1 × 10³ m/s	✓ 1	Accept 3080 m/s. Using r = 3.6 × 10⁷ m gives 2.6 × 10³ m/s: 2 marks at most.
Part (g)
distance = 2 × 1.2 × 10⁶ = 2.4 × 10⁶ m	✓ 1
t = 2.4 × 10⁶ ÷ 3.0 × 10⁸ = 8.0 × 10⁻³ s (0.0080 s)	✓ 1
the delay is much shorter (about 30 times less than via the high satellite), so there is no noticeable pause when people talk to each other	✓ 1	Accept any sensible comparison with the answer to (c).

Answers: (a)(i) radio waves and infrared · (b) 0.025 m (2.5 cm) · (c) 0.257 s · (d)(i) 3.75 cm · (f) 3.1 × 10³ m/s · (g) 8.0 × 10⁻³ s

Exam tip (b) Convert GHz to Hz before using v = fλ: 1 GHz = 10⁹ Hz.

Common error (f) Using the height above the ground as the radius. The radius of an orbit is measured from the centre of the Earth, so add the Earth's radius.

More 3.3 questions in the question bank

+ Add to test✓ In your testCC-CIP-33-218 marks~20 min

A hospital imaging department: X-rays, UV and infrared

3.3 The electromagnetic spectrum5.2 Radioactivity, half-life and safety2.3 Conduction, convection and radiationCore

A hospital uses three types of electromagnetic radiation:

X-rays of wavelength 2.0 × 10⁻¹¹ m to take images of bones
ultraviolet lamps of wavelength 254 nm to sterilise surfaces in empty operating theatres
infrared thermometers to measure patients' temperatures.

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

(a)(i)

Write the three types of radiation in order of increasing frequency.

RECALLNotes

Show solution

infrared, ultraviolet, X-rays

✓ 1

Correct order needed.

Answer: infrared, ultraviolet, X-rays

(1)

(a)(ii)

Explain how X-rays produce an image of the bones in a patient's arm.

EXPLAINNotes

Show solution

X-rays pass through soft tissue (skin and muscle) but are absorbed by bone	✓ 1
so a shadow of the bone forms on the detector (film) behind the arm	✓ 1

(2)

(b)

Calculate the frequency of the X-rays.

CALCNotes

Show solution

f = v ÷ λ = 3.0 × 10⁸ ÷ 2.0 × 10⁻¹¹	✓ 1
f = 1.5 × 10¹⁹ Hz	✓ 1

Answer: 1.5 × 10¹⁹ Hz

(2)

(c)

Show that the frequency of the ultraviolet radiation from the lamps is about 1.2 × 10¹⁵ Hz.

DERIVENotes

Show solution

254 nm = 254 × 10⁻⁹ m = 2.54 × 10⁻⁷ m	✓ 1
f = 3.0 × 10⁸ ÷ 2.54 × 10⁻⁷ = 1.18 × 10¹⁵ Hz	✓ 1	Value to at least 3 s.f. needed, since 1.2 × 10¹⁵ Hz is given.

Answer: 1.18 × 10¹⁵ Hz

(2)

(d)

The ultraviolet lamps are switched on only when nobody is in the operating theatre.

State one harmful effect of ultraviolet radiation on the human body, and give one other way to protect a person who has to enter the room while a lamp is on.

RECALLNotes

Show solution

any one of: skin burns / skin cancer / damage to the eyes	✓ 1	"It is dangerous" alone scores 0.
any one of: protective goggles or glasses / clothing covering the skin / limit the time spent in the room	✓ 1

(2)

(e)(i)

Lead screens protect the staff from X-rays. Figure 1 shows the percentage of the X-rays that pass through lead sheets of different thickness.

Use Figure 1 to find the thickness of lead that lets through 25% of the X-rays.

Figure 1: percentage of the X-rays that pass through a sheet of lead, against the thickness of the lead

GRAPHNotes

Show solution

0.50 mm

✓ 1

Accept 0.48–0.52 mm.

Answer: 0.50 mm

(1)

(e)(ii)

A radiographer says: "Every extra 0.25 mm of lead halves the X-rays that get through."

Use Figure 1 to test this statement.

DATANotes

Show solution

at least two readings, e.g. 100% at 0 mm, 50% at 0.25 mm, 25% at 0.50 mm, 12.5% at 0.75 mm	✓ 1	Readings within half a small square.
each 0.25 mm step halves the percentage, so the statement is correct	✓ 1	Conclusion must follow from the readings.

Answer: the statement is correct

(2)

(f)

The radiotherapy unit uses gamma rays to treat cancer. To halve the gamma rays that get through, the lead needs to be about 12 mm thick.

Compare this with the X-rays in Figure 1, and suggest why the gamma rays need much thicker lead.

REASONNotes

Show solution

gamma rays need about 50 times more lead to be halved, so they are much more penetrating	✓ 1	Accept "much thicker lead" with a correct comparison (12 mm against 0.25 mm).
gamma rays have a higher frequency (shorter wavelength) than these X-rays, so each wave carries more energy and passes through more lead	✓ 1	Do not accept "gamma rays are uncharged": the X-rays are uncharged too, so this does not explain the difference.

(2)

(g)

During a scan, the radiographer stands behind a lead screen. She now moves so that she is twice as far from the X-ray machine, still behind the screen.

Suggest how the amount of radiation she receives changes, and explain your answer.

PREDICTNotes

Show solution

the radiation she receives decreases	✓ 1	Accept "falls to about a quarter".
because the radiation spreads out and weakens with distance from the source	✓ 1	Do not accept "X-rays can only travel a short distance".

(2)

(h)

An infrared thermometer is pointed at a patient's forehead. It contains a sensor that absorbs infrared radiation.

Explain why the thermometer gives a higher reading for a patient with a fever.

EXPLAINNotes

Show solution

the skin (like every object) emits infrared radiation	✓ 1	Do not accept "the skin reflects infrared".
a hotter surface emits more infrared each second, so the sensor absorbs more energy and shows a higher temperature	✓ 1

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
infrared, ultraviolet, X-rays	✓ 1	Correct order needed.
Part (a)(ii)
X-rays pass through soft tissue (skin and muscle) but are absorbed by bone	✓ 1
so a shadow of the bone forms on the detector (film) behind the arm	✓ 1
Part (b)
f = v ÷ λ = 3.0 × 10⁸ ÷ 2.0 × 10⁻¹¹	✓ 1
f = 1.5 × 10¹⁹ Hz	✓ 1
Part (c)
254 nm = 254 × 10⁻⁹ m = 2.54 × 10⁻⁷ m	✓ 1
f = 3.0 × 10⁸ ÷ 2.54 × 10⁻⁷ = 1.18 × 10¹⁵ Hz	✓ 1	Value to at least 3 s.f. needed, since 1.2 × 10¹⁵ Hz is given.
Part (d)
any one of: skin burns / skin cancer / damage to the eyes	✓ 1	"It is dangerous" alone scores 0.
any one of: protective goggles or glasses / clothing covering the skin / limit the time spent in the room	✓ 1
Part (e)(i)
0.50 mm	✓ 1	Accept 0.48–0.52 mm.
Part (e)(ii)
at least two readings, e.g. 100% at 0 mm, 50% at 0.25 mm, 25% at 0.50 mm, 12.5% at 0.75 mm	✓ 1	Readings within half a small square.
each 0.25 mm step halves the percentage, so the statement is correct	✓ 1	Conclusion must follow from the readings.
Part (f)
gamma rays need about 50 times more lead to be halved, so they are much more penetrating	✓ 1	Accept "much thicker lead" with a correct comparison (12 mm against 0.25 mm).
gamma rays have a higher frequency (shorter wavelength) than these X-rays, so each wave carries more energy and passes through more lead	✓ 1	Do not accept "gamma rays are uncharged": the X-rays are uncharged too, so this does not explain the difference.
Part (g)
the radiation she receives decreases	✓ 1	Accept "falls to about a quarter".
because the radiation spreads out and weakens with distance from the source	✓ 1	Do not accept "X-rays can only travel a short distance".
Part (h)
the skin (like every object) emits infrared radiation	✓ 1	Do not accept "the skin reflects infrared".
a hotter surface emits more infrared each second, so the sensor absorbs more energy and shows a higher temperature	✓ 1

Answers: (a)(i) infrared, ultraviolet, X-rays · (b) 1.5 × 10¹⁹ Hz · (c) 1.18 × 10¹⁵ Hz · (e)(i) 0.50 mm · (e)(ii) the statement is correct

Exam tip (c) 1 nm = 10⁻⁹ m. Convert to metres before you divide.

More 3.3 questions in the question bank

3.4 Sound and ultrasound

+ Add to test✓ In your testCC-CIP-34-117 marks~20 min

Bat echolocation: hunting a moth with ultrasound

3.4 Sound and ultrasound3.1 Wave properties, reflection, refraction and diffraction1.2 Speed, velocity and motion graphsCore

A bat finds insects in the dark by echolocation. It sends out short pulses of sound called calls, and listens for the echoes that bounce back from a moth.

When it starts hunting, each call has a frequency of 50 kHz. The speed of sound in air is 340 m/s.

(a)

The calls are ultrasound.

(i) State the approximate range of frequencies that a healthy young person can hear.

(ii) Use your answer to (i) to explain why the bat's calls are described as ultrasound.

RECALLNotes

Show solution

about 20 Hz to 20 000 Hz	✓ 1	Accept 20 Hz to 20 kHz.
50 kHz (= 50 000 Hz) is above 20 000 Hz / above the upper limit of human hearing	✓ 1	Must compare 50 kHz with the upper limit. 'Too high to hear' alone is not enough without the comparison.

Answer: (i) 20 Hz to 20 000 Hz; (ii) 50 000 Hz is above 20 000 Hz

(2)

(b)

Calculate the wavelength of a 50 kHz call in air.

CALCNotes

Show solution

λ = v ÷ f or 340 ÷ 50 000	✓ 1	Award for the equation in any form or for the correct substitution with f in Hz.
0.0068 m	✓ 1	Accept 6.8 × 10⁻³ m or 6.8 mm. 6.8 m (kHz not converted to Hz) scores the first mark only if the equation is written correctly.

Answer: 0.0068 m (6.8 mm)

(2)

(c)

Describe how a call travels through the air from the bat to the moth. Your answer should refer to the air particles.

EXPLAINNotes

Show solution

sound is a longitudinal wave	✓ 1	Accept a description as a series of compressions and rarefactions.
the air particles vibrate (backwards and forwards) parallel to the direction the call travels, passing the vibration on; the particles themselves do not travel to the moth	✓ 1	Accept 'energy is transferred, not the air'. Do not accept particles vibrating at right angles.

(2)

(d)

At the start of the hunt the moth is 6.8 m from the bat.

Show that the echo reaches the bat 0.040 s after the call is sent out.

DERIVENotes

Show solution

distance there and back = 2 × 6.8 = 13.6 m	✓ 1	The doubling must be seen.
t = 13.6 ÷ 340 = 0.040 s	✓ 1	Full substitution needed; 0.040 s with no working scores 0. Accept t = 2d ÷ v with numbers.

Answer: 0.040 s

(2)

(e)(i)

The bat flies towards the moth. The bat uses the echo times to keep track of how far away the moth is. Figure 1 shows how the distance between the bat and the moth changes with time.

Use Figure 1 to calculate the speed at which the distance between the bat and the moth decreases between t = 0 and t = 0.80 s.

Figure 1: distance between the bat and the moth against time

GRAPHNotes

Show solution

gradient: (6.8 − 2.0) ÷ 0.80 or 4.8 ÷ 0.80	✓ 1	Values read from the graph must be seen.
6.0 m/s	✓ 1	Ignore a minus sign. Unit required.

Answer: 6.0 m/s

(2)

(e)(ii)

State what Figure 1 shows about the motion between t = 0.80 s and t = 1.20 s, compared with the first 0.80 s.

REASONNotes

Show solution

the line is less steep, so the distance decreases more slowly / at a smaller constant speed (3.0 m/s)

✓ 1

Must link the gradient to the speed. Accept 'the bat closes in at a lower steady speed'.

(1)

(f)(i)

As it closes in on the moth, the bat changes its calls. Figure 2 shows a microphone trace of an early call A and of a later call B. The horizontal scale is in microseconds (1 µs = 10⁻⁶ s).

Use Figure 2 to determine the frequency of call B.

Figure 2: microphone traces of call A and call B, drawn on the same scales (one small vertical square is the same displacement for both)

DATANotes

Show solution

8 complete cycles in 100 µs, so period T = 12.5 µs = 1.25 × 10⁻⁵ s	✓ 1	Accept a period read from any whole number of cycles.
f = 1 ÷ T = 80 000 Hz	✓ 1	Accept 80 kHz. Allow ECF from an incorrect period read from the trace.

Answer: 80 000 Hz (80 kHz)

(2)

(f)(ii)

Compare call B with call A in terms of loudness and pitch. Give a reason for each answer using Figure 2.

REASONNotes

Show solution

B is quieter (less loud), because its amplitude is smaller (half that of A)	✓ 1	Must link loudness to amplitude.
B has a higher pitch, because its frequency is higher / there are more waves in the same time	✓ 1	Must link pitch to frequency. Accept 'would sound higher if it could be heard'.

(2)

(f)(iii)

The speed of sound in the air does not change. Explain why call B helps the bat to detect small details of the moth better than call A.

EXPLAINNotes

Show solution

v = fλ with v constant, so a higher frequency gives a shorter wavelength	✓ 1	Accept a calculation: λ = 340 ÷ 80 000 = 0.00425 m (4.25 mm) compared with 6.8 mm.
a shorter wavelength allows smaller details / smaller objects to be detected	✓ 1	Must refer to wavelength for this mark.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
about 20 Hz to 20 000 Hz	✓ 1	Accept 20 Hz to 20 kHz.
50 kHz (= 50 000 Hz) is above 20 000 Hz / above the upper limit of human hearing	✓ 1	Must compare 50 kHz with the upper limit. 'Too high to hear' alone is not enough without the comparison.
Part (b)
λ = v ÷ f or 340 ÷ 50 000	✓ 1	Award for the equation in any form or for the correct substitution with f in Hz.
0.0068 m	✓ 1	Accept 6.8 × 10⁻³ m or 6.8 mm. 6.8 m (kHz not converted to Hz) scores the first mark only if the equation is written correctly.
Part (c)
sound is a longitudinal wave	✓ 1	Accept a description as a series of compressions and rarefactions.
the air particles vibrate (backwards and forwards) parallel to the direction the call travels, passing the vibration on; the particles themselves do not travel to the moth	✓ 1	Accept 'energy is transferred, not the air'. Do not accept particles vibrating at right angles.
Part (d)
distance there and back = 2 × 6.8 = 13.6 m	✓ 1	The doubling must be seen.
t = 13.6 ÷ 340 = 0.040 s	✓ 1	Full substitution needed; 0.040 s with no working scores 0. Accept t = 2d ÷ v with numbers.
Part (e)(i)
gradient: (6.8 − 2.0) ÷ 0.80 or 4.8 ÷ 0.80	✓ 1	Values read from the graph must be seen.
6.0 m/s	✓ 1	Ignore a minus sign. Unit required.
Part (e)(ii)
the line is less steep, so the distance decreases more slowly / at a smaller constant speed (3.0 m/s)	✓ 1	Must link the gradient to the speed. Accept 'the bat closes in at a lower steady speed'.
Part (f)(i)
8 complete cycles in 100 µs, so period T = 12.5 µs = 1.25 × 10⁻⁵ s	✓ 1	Accept a period read from any whole number of cycles.
f = 1 ÷ T = 80 000 Hz	✓ 1	Accept 80 kHz. Allow ECF from an incorrect period read from the trace.
Part (f)(ii)
B is quieter (less loud), because its amplitude is smaller (half that of A)	✓ 1	Must link loudness to amplitude.
B has a higher pitch, because its frequency is higher / there are more waves in the same time	✓ 1	Must link pitch to frequency. Accept 'would sound higher if it could be heard'.
Part (f)(iii)
v = fλ with v constant, so a higher frequency gives a shorter wavelength	✓ 1	Accept a calculation: λ = 340 ÷ 80 000 = 0.00425 m (4.25 mm) compared with 6.8 mm.
a shorter wavelength allows smaller details / smaller objects to be detected	✓ 1	Must refer to wavelength for this mark.

Answers: (a) (i) 20 Hz to 20 000 Hz; (ii) 50 000 Hz is above 20 000 Hz · (b) 0.0068 m (6.8 mm) · (d) 0.040 s · (e)(i) 6.0 m/s · (f)(i) 80 000 Hz (80 kHz)

Common error (b) Substituting 50 for the frequency. Convert 50 kHz to 50 000 Hz before using v = fλ.

Exam tip (f)(iii) When the medium stays the same, the speed stays the same. Then v = fλ tells you that frequency and wavelength change in opposite directions.

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An ultrasound scan of an unborn baby

3.4 Sound and ultrasound3.1 Wave properties, reflection, refraction and diffraction1.1 Physical quantities and measurementCore

A sonographer checks the growth of an unborn baby. A probe held on the mother's skin sends short pulses of ultrasound of frequency 3.5 MHz into her body. Part of each pulse is reflected at every boundary between different tissues, and the probe detects the echoes.

Assume that the speed of ultrasound is 1500 m/s in all the tissues and fluids that the pulse passes through.

(a)

X-rays can also be used to make images of the inside of the body. Explain why ultrasound, and not X-rays, is used to scan an unborn baby.

Extended onlyEXPLAINNotes

Show solution

ultrasound is not ionising / does not damage cells	✓ 1
X-rays are ionising and can damage the cells of the (developing) baby / may cause cancer	✓ 1	Accept 'X-rays are harmful to the baby' only if the damage to cells is mentioned in either mark.

(2)

(b)

Calculate the wavelength of the ultrasound in the body.

CALCNotes

Show solution

λ = v ÷ f = 1500 ÷ (3.5 × 10⁶)	✓ 1	Frequency must be in Hz.
4.3 × 10⁻⁴ m	✓ 1	Accept 0.00043 m or 0.43 mm; accept 4.29 × 10⁻⁴ m. Unit required.

Answer: 4.3 × 10⁻⁴ m (0.43 mm)

(2)

(c)

The speed of sound in air is about 340 m/s, much less than the speed of ultrasound in the body, which is mostly water. Explain why sound travels faster in a liquid than in a gas.

Extended onlyEXPLAINNotes

Show solution

in a liquid the particles are much closer together than in a gas	✓ 1
so each particle passes the vibration on to its neighbour more quickly / the vibration does not have to cross large gaps	✓ 1	Accept the reverse argument for a gas.

(2)

(d)(i)

Figure 1 shows the display for one pulse. The time-base is set to 20 µs per division. Times are measured from the peak of the pulse that is sent to the peak of each echo.

Use Figure 1 to find the time between the pulse being sent and echo E1. Give your answer in seconds, in standard form.

Figure 1: the scanner display. E1 is the echo from the near side of the baby's head and E2 the echo from the far side.

GRAPHNotes

Show solution

4.0 divisions × 20 µs = 80 µs = 8.0 × 10⁻⁵ s

✓ 1

Answer must be in seconds in standard form for the mark.

Answer: 8.0 × 10⁻⁵ s

(1)

(d)(ii)

Calculate the depth below the probe of the near side of the baby's head.

CALCNotes

Show solution

distance there and back = 1500 × 8.0 × 10⁻⁵ = 0.12 m	✓ 1	Allow ECF from (d)(i).
depth = 0.12 ÷ 2 = 0.060 m	✓ 1	Accept 6.0 cm. 0.12 m (not halved) scores the first mark only.

Answer: 0.060 m (6.0 cm)

(2)

(d)(iii)

Show that the width of the baby's head along the direction of the pulse is about 4.8 cm.

DERIVENotes

Show solution

time between E1 and E2 = (8.2 − 5.0) = 3.2 divisions × 20 µs = 64 µs	✓ 1	Or: E2 is 7.2 divisions after the pulse sent = 144 µs, so depth of far side = 1500 × 144 × 10⁻⁶ ÷ 2 = 0.108 m.
width = 1500 × 64 × 10⁻⁶ ÷ 2 = 0.048 m = 4.8 cm	✓ 1	Or 0.108 − 0.060 = 0.048 m. Full working needed for a 'show that'.

Answer: 0.048 m (4.8 cm)

(2)

(e)

The scanner could also be set to a time-base of 50 µs per division. Suggest why the sonographer uses 20 µs per division to measure the width of the head.

REASONNotes

Show solution

the echoes are more spread out across the screen / the time between E1 and E2 covers more divisions	✓ 1	Accept '3.2 divisions instead of 1.28 divisions'.
so the times can be read more precisely / the reading error is a smaller fraction (percentage) of the time measured	✓ 1

(2)

(f)

The sonographer changes to a probe of frequency 7.0 MHz to look at a small part of the baby more closely.

(i) State the new wavelength in the body.

(ii) Suggest why this probe gives a more detailed image.

PREDICTNotes

Show solution

2.1 × 10⁻⁴ m (half the value in (b), because the speed is unchanged)	✓ 1	Allow ECF: half the answer to (b).
shorter wavelength, so smaller details can be seen	✓ 1	Must refer to the wavelength.

Answer: (i) 2.1 × 10⁻⁴ m; (ii) shorter wavelength shows smaller details

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
ultrasound is not ionising / does not damage cells	✓ 1
X-rays are ionising and can damage the cells of the (developing) baby / may cause cancer	✓ 1	Accept 'X-rays are harmful to the baby' only if the damage to cells is mentioned in either mark.
Part (b)
λ = v ÷ f = 1500 ÷ (3.5 × 10⁶)	✓ 1	Frequency must be in Hz.
4.3 × 10⁻⁴ m	✓ 1	Accept 0.00043 m or 0.43 mm; accept 4.29 × 10⁻⁴ m. Unit required.
Part (c)
in a liquid the particles are much closer together than in a gas	✓ 1
so each particle passes the vibration on to its neighbour more quickly / the vibration does not have to cross large gaps	✓ 1	Accept the reverse argument for a gas.
Part (d)(i)
4.0 divisions × 20 µs = 80 µs = 8.0 × 10⁻⁵ s	✓ 1	Answer must be in seconds in standard form for the mark.
Part (d)(ii)
distance there and back = 1500 × 8.0 × 10⁻⁵ = 0.12 m	✓ 1	Allow ECF from (d)(i).
depth = 0.12 ÷ 2 = 0.060 m	✓ 1	Accept 6.0 cm. 0.12 m (not halved) scores the first mark only.
Part (d)(iii)
time between E1 and E2 = (8.2 − 5.0) = 3.2 divisions × 20 µs = 64 µs	✓ 1	Or: E2 is 7.2 divisions after the pulse sent = 144 µs, so depth of far side = 1500 × 144 × 10⁻⁶ ÷ 2 = 0.108 m.
width = 1500 × 64 × 10⁻⁶ ÷ 2 = 0.048 m = 4.8 cm	✓ 1	Or 0.108 − 0.060 = 0.048 m. Full working needed for a 'show that'.
Part (e)
the echoes are more spread out across the screen / the time between E1 and E2 covers more divisions	✓ 1	Accept '3.2 divisions instead of 1.28 divisions'.
so the times can be read more precisely / the reading error is a smaller fraction (percentage) of the time measured	✓ 1
Part (f)
2.1 × 10⁻⁴ m (half the value in (b), because the speed is unchanged)	✓ 1	Allow ECF: half the answer to (b).
shorter wavelength, so smaller details can be seen	✓ 1	Must refer to the wavelength.

Answers: (b) 4.3 × 10⁻⁴ m (0.43 mm) · (d)(i) 8.0 × 10⁻⁵ s · (d)(ii) 0.060 m (6.0 cm) · (d)(iii) 0.048 m (4.8 cm) · (f) (i) 2.1 × 10⁻⁴ m; (ii) shorter wavelength shows smaller details

Common error (d)(ii) Forgetting to halve. The echo time covers the journey to the boundary and back again.

More 3.4 questions in the question bank

Topic 4 · Electricity and magnetism

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

Setup	4.1	4.2	4.3	4.4	4.5	Other
A scrapyard electromagnet crane	●	○			○	—
Hiking compass errors near a steel fence and a buried cable	●	○			○	—
Charging an electric car at home		●	○	○		—
Inside a hair dryer: heating coil, fan and a thermal cut-out		●	○			2.2
A greenhouse frost alarm with a thermistor		○	●			1.1
Stage lighting for a music festival		○	●	○		—
Fuses and plugs in a student kitchen		○		●		1.7
A hedge trimmer, a wet garden and an RCD		○	○	●		—
An offshore wind farm and its transmission line		○			●	1.7
A bicycle dynamo and a hub motor	○	○			●	—

4.1 Magnets and magnetic fields

+ Add to test✓ In your testCC-CIP-41-120 marks~20 min

A scrapyard electromagnet crane

4.1 Magnets and magnetic fields4.5 Electromagnetic induction, motors and transformers4.2 Charge, current, potential difference, resistance and powerCore

A crane in a scrapyard uses a large electromagnet to pick up scrap metal and drop it into a container. The electromagnet is a coil of insulated copper wire wound around a core. The coil is connected to a 240 V d.c. supply and has a resistance of 16 Ω.

(a)

The scrap contains steel car parts, aluminium drinks cans and copper pipes. Explain why the crane can be used to separate the steel from the other metals.

EXPLAINNotes

Show solution

steel is a magnetic material, so it is attracted to the electromagnet (and lifted)	✓ 1
aluminium and copper are non-magnetic (not attracted), so they are left behind	✓ 1	Both metals, or 'the other metals', must be covered. 'Not all metals are magnetic' alone is not enough.

(2)

(b)

The engineers consider three materials for the core: iron, steel and copper. State which material they should choose. Explain why each of the other two would not be suitable.

REASONNotes

Show solution

iron	✓ 1	Mark independently of the explanations.
steel is a hard magnetic material: it would stay magnetised when the current is switched off, so the scrap would not be released	✓ 1	Must state the consequence for the crane.
copper is non-magnetic, so it would not be magnetised and would not make the electromagnet stronger	✓ 1	Accept 'copper cannot be magnetised'.

Answer: iron

(3)

(c)

Before the core is fitted, the coil is tested on its own. Figure 1 shows the coil. When the current flows, the bottom end of the coil is a N pole.

Sketch the magnetic field pattern inside and outside the coil. Show the direction of the field.

Figure 1: the coil of the electromagnet tested without its core (not to scale)

SKETCHNotes

Show solution

inside the coil: straight, parallel, evenly spaced lines along the axis	✓ 1	At least three lines inside.
outside: lines loop round from one end of the coil to the other, like the field of a bar magnet, and do not cross	✓ 1
arrows show the field leaving the bottom (N) end and entering the top end outside the coil, and pointing downwards inside the coil	✓ 1	Arrows must be consistent on every line.

(3)

(d)

Show that the current in the coil is 15 A.

CALCNotes

Show solution

I = V ÷ R = 240 ÷ 16	✓ 1	Equation may be in any form.
= 15 A	✓ 1	Substitution must be seen; 15 A alone scores 0.

Answer: 15 A

(2)

(e)

During each lift, the electromagnet is switched on for 45 s. Calculate the electrical energy transferred to the coil during one lift.

MULTINotes

Show solution

P = IV = 15 × 240 = 3600 W	✓ 1	Accept P = V² ÷ R if correctly used.
E = Pt = 3600 × 45	✓ 1	Allow ECF from (d) or from an incorrect power. Or E = IVt = 15 × 240 × 45.
1.6 × 10⁵ J	✓ 1	Accept 162 000 J or 160 kJ. Unit required.

Answer: 1.6 × 10⁵ J (162 000 J)

(3)

(f)

The iron core is fitted. An engineer measures the greatest force the electromagnet can lift for different currents. The results are shown in the table.

current I / A	0	3.0	6.0	9.0	12.0	15.0
lifting force F / kN	0	5.8	11.4	15.8	18.6	20.0

On the grid in Figure 2, plot the results and draw the best-fit curve.

Figure 2: grid for plotting the lifting force against the current

GRAPHNotes

Show solution

at least four points plotted correctly to within half a small square	✓ 1
all six points plotted correctly to within half a small square	✓ 1
smooth best-fit curve through the origin that rises steeply at first and becomes less steep	✓ 1	Do not accept a straight line through all points or dot-to-dot ruled lines.

(3)

(g)

Describe how the lifting force depends on the current. Use your graph.

EXPLAINNotes

Show solution

the force increases as the current increases	✓ 1
the force increases by smaller amounts for each extra ampere at larger currents / the force is not proportional to the current / roughly proportional only up to about 6 A	✓ 1	Must describe the change in gradient; quote data from the graph or table for support, e.g. from 0 to 3.0 A the force rises by 5.8 kN but from 12.0 to 15.0 A by only 1.4 kN.

(2)

(h)

A technician reconnects the coil the other way round, so the current in it is reversed. Explain whether the crane can still lift the steel scrap.

PREDICTNotes

Show solution

the magnetic field reverses / the poles of the electromagnet swap (the bottom becomes a S pole)	✓ 1
yes: the steel is magnetised by induction with the opposite pole facing the electromagnet, so it is still attracted	✓ 1	The answer 'yes' must be supported by the induced-pole idea. Accept 'a magnetic material is attracted to either pole'.

Answer: Yes; the poles swap, but steel is still attracted by induced magnetism

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
steel is a magnetic material, so it is attracted to the electromagnet (and lifted)	✓ 1
aluminium and copper are non-magnetic (not attracted), so they are left behind	✓ 1	Both metals, or 'the other metals', must be covered. 'Not all metals are magnetic' alone is not enough.
Part (b)
iron	✓ 1	Mark independently of the explanations.
steel is a hard magnetic material: it would stay magnetised when the current is switched off, so the scrap would not be released	✓ 1	Must state the consequence for the crane.
copper is non-magnetic, so it would not be magnetised and would not make the electromagnet stronger	✓ 1	Accept 'copper cannot be magnetised'.
Part (c)
inside the coil: straight, parallel, evenly spaced lines along the axis	✓ 1	At least three lines inside.
outside: lines loop round from one end of the coil to the other, like the field of a bar magnet, and do not cross	✓ 1
arrows show the field leaving the bottom (N) end and entering the top end outside the coil, and pointing downwards inside the coil	✓ 1	Arrows must be consistent on every line.
Part (d)
I = V ÷ R = 240 ÷ 16	✓ 1	Equation may be in any form.
= 15 A	✓ 1	Substitution must be seen; 15 A alone scores 0.
Part (e)
P = IV = 15 × 240 = 3600 W	✓ 1	Accept P = V² ÷ R if correctly used.
E = Pt = 3600 × 45	✓ 1	Allow ECF from (d) or from an incorrect power. Or E = IVt = 15 × 240 × 45.
1.6 × 10⁵ J	✓ 1	Accept 162 000 J or 160 kJ. Unit required.
Part (f)
at least four points plotted correctly to within half a small square	✓ 1
all six points plotted correctly to within half a small square	✓ 1
smooth best-fit curve through the origin that rises steeply at first and becomes less steep	✓ 1	Do not accept a straight line through all points or dot-to-dot ruled lines.
Part (g)
the force increases as the current increases	✓ 1
the force increases by smaller amounts for each extra ampere at larger currents / the force is not proportional to the current / roughly proportional only up to about 6 A	✓ 1	Must describe the change in gradient; quote data from the graph or table for support, e.g. from 0 to 3.0 A the force rises by 5.8 kN but from 12.0 to 15.0 A by only 1.4 kN.
Part (h)
the magnetic field reverses / the poles of the electromagnet swap (the bottom becomes a S pole)	✓ 1
yes: the steel is magnetised by induction with the opposite pole facing the electromagnet, so it is still attracted	✓ 1	The answer 'yes' must be supported by the induced-pole idea. Accept 'a magnetic material is attracted to either pole'.

Answers: (b) iron · (d) 15 A · (e) 1.6 × 10⁵ J (162 000 J) · (h) Yes; the poles swap, but steel is still attracted by induced magnetism

Exam tip (b) Link each property to the job: an electromagnet must lose its magnetism as soon as the current stops, so it needs a soft magnetic material.

Common error (h) Thinking that a reversed electromagnet will push the steel away. Unmagnetised steel always has the opposite pole induced in it, so the force is always attractive.

More 4.1 questions in the question bank

+ Add to test✓ In your testCC-CIP-41-217 marks~20 min

Hiking compass errors near a steel fence and a buried cable

4.1 Magnets and magnetic fields4.5 Electromagnetic induction, motors and transformers4.2 Charge, current, potential difference, resistance and powerCore

A hiker uses a magnetic compass to find her way. On one stretch of the path the compass gives wrong directions. The path runs beside a fence with steel posts and then crosses a buried cable that carries a large direct current (d.c.) from a solar farm to the electricity grid.

(a)

Away from the fence and the cable, the N pole of the compass needle points towards the geographic north. Explain why.

EXPLAINNotes

Show solution

the needle is a (small, freely turning) magnet and the Earth has its own magnetic field (like a bar magnet), which the needle lines up with	✓ 1
there is a magnetic S pole near the geographic north (pole), and unlike poles attract, so the N pole of the needle is pulled northwards	✓ 1	Accept 'the Earth's magnetic pole in the north is a south pole'.

(2)

(b)(i)

The steel fence posts are not magnets. When the hiker walks close to a post, the N pole of the needle turns towards it. Explain why.

EXPLAINNotes

Show solution

steel is a magnetic material, so the needle (a magnet) induces magnetism in the post	✓ 1
the end of the post nearest the N pole becomes a S pole; unlike poles attract, so the N pole is pulled towards the post	✓ 1	Accept the same argument for the S pole of the needle. Must state the induced pole is opposite.

(2)

(b)(ii)

The hiker's walking pole is made of aluminium. State and explain whether it affects the compass.

RECALLNotes

Show solution

no, because aluminium is not a magnetic material (so no magnetism is induced and there is no force)

✓ 1

The reason is needed for the mark.

(1)

(c)

The cable delivers a power of 2.0 MW at a potential difference of 8.0 kV.

Show that the current in the cable is about 250 A.

CALCNotes

Show solution

I = P ÷ V = 2.0 × 10⁶ ÷ (8.0 × 10³)	✓ 1	Both prefixes must be converted.
= 250 A	✓ 1	Full substitution needed for a 'show that'.

Answer: 250 A

(2)

(d)

To model the effect of the cable, a student passes a steady current of 5.0 A up through a long vertical wire. The wire passes through the centre of a horizontal card, shown from above in Figure 1.

On Figure 1, sketch the magnetic field pattern produced by the current in the wire. Show the direction of the field.

Figure 1: the card and the vertical wire seen from above. The current in the wire flows upwards, out of the card (shown by the dot).

SKETCHNotes

Show solution

at least three complete circles centred on the wire	✓ 1	Lines must not touch or cross.
spacing between circles increasing with distance from the wire	✓ 1	The field is strongest close to the wire.
arrows anticlockwise on every circle (as seen from above)	✓ 1

(3)

(e)

Describe how the student could use a small plotting compass to plot one of these field lines on the card.

EXPTNotes

Show solution

place the compass near the wire and mark (with a dot) the position of the N end of the needle	✓ 1	Accept dots at both ends of the needle.
move the compass so its other end (tail) is on that dot, mark the new position of the N end, repeat, and join the dots with a smooth line	✓ 1	Must describe the repeated stepping. Arrow from the direction the N end points is not required.

(2)

(f)(i)

The student then places the compass at different distances d due north of the wire and measures how far the needle turns away from north. Figure 2 shows the results.

Describe how the deflection changes as d increases. Explain this trend using the magnetic field of the wire.

Figure 2: deflection of the compass needle from north against distance from the wire, for a current of 5.0 A

GRAPHNotes

Show solution

the deflection decreases as d increases, quickly close to the wire and then more slowly	✓ 1	Data from Figure 2 may support the answer, e.g. 45° at 5 cm but 14° at 20 cm.
the field of the wire is weaker further from the wire (the field lines are further apart), so it has less effect on the needle compared with the Earth's field	✓ 1

(2)

(f)(ii)

Use Figure 2 to find the distance from the wire at which the deflection is 30°.

GRAPHNotes

Show solution

8.7 cm

✓ 1

Accept 8.2 cm to 9.2 cm.

Answer: 8.7 cm

(1)

(g)

The direction of the current in the wire is now reversed, keeping its size the same. State and explain the effect on the compass at a distance of 10 cm from the wire.

PREDICTNotes

Show solution

the needle is deflected by the same angle (about 27°) but to the other side of north	✓ 1	Accept 'the deflection is reversed'. Do not accept 'no deflection'.
because reversing the current reverses the direction of the magnetic field of the wire (its size is unchanged)	✓ 1

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the needle is a (small, freely turning) magnet and the Earth has its own magnetic field (like a bar magnet), which the needle lines up with	✓ 1
there is a magnetic S pole near the geographic north (pole), and unlike poles attract, so the N pole of the needle is pulled northwards	✓ 1	Accept 'the Earth's magnetic pole in the north is a south pole'.
Part (b)(i)
steel is a magnetic material, so the needle (a magnet) induces magnetism in the post	✓ 1
the end of the post nearest the N pole becomes a S pole; unlike poles attract, so the N pole is pulled towards the post	✓ 1	Accept the same argument for the S pole of the needle. Must state the induced pole is opposite.
Part (b)(ii)
no, because aluminium is not a magnetic material (so no magnetism is induced and there is no force)	✓ 1	The reason is needed for the mark.
Part (c)
I = P ÷ V = 2.0 × 10⁶ ÷ (8.0 × 10³)	✓ 1	Both prefixes must be converted.
= 250 A	✓ 1	Full substitution needed for a 'show that'.
Part (d)
at least three complete circles centred on the wire	✓ 1	Lines must not touch or cross.
spacing between circles increasing with distance from the wire	✓ 1	The field is strongest close to the wire.
arrows anticlockwise on every circle (as seen from above)	✓ 1
Part (e)
place the compass near the wire and mark (with a dot) the position of the N end of the needle	✓ 1	Accept dots at both ends of the needle.
move the compass so its other end (tail) is on that dot, mark the new position of the N end, repeat, and join the dots with a smooth line	✓ 1	Must describe the repeated stepping. Arrow from the direction the N end points is not required.
Part (f)(i)
the deflection decreases as d increases, quickly close to the wire and then more slowly	✓ 1	Data from Figure 2 may support the answer, e.g. 45° at 5 cm but 14° at 20 cm.
the field of the wire is weaker further from the wire (the field lines are further apart), so it has less effect on the needle compared with the Earth's field	✓ 1
Part (f)(ii)
8.7 cm	✓ 1	Accept 8.2 cm to 9.2 cm.
Part (g)
the needle is deflected by the same angle (about 27°) but to the other side of north	✓ 1	Accept 'the deflection is reversed'. Do not accept 'no deflection'.
because reversing the current reverses the direction of the magnetic field of the wire (its size is unchanged)	✓ 1

Answers: (c) 250 A · (f)(ii) 8.7 cm

Exam tip (g) When asked what happens if the current reverses, give both the direction (reversed) and the size (unchanged) of the effect.

More 4.1 questions in the question bank

4.2 Charge, current, potential difference, resistance and power

+ Add to test✓ In your testCC-CIP-42-120 marks~20 min

Charging an electric car at home

4.2 Charge, current, potential difference, resistance and power4.3 Series and parallel circuits4.4 Electrical safetyCore

A family fits a wall-mounted charger in their garage to charge their electric car. The charger is connected to the 230 V mains supply by its own cable. While the car is charging, the power input to the charger is 7.4 kW.

The garage also has a workbench light, which uses two identical filament lamps P and Q connected to a 12 V battery as shown in Figure 1.

Figure 1: the workbench light circuit

(a)

State what is meant by a potential difference of 230 V across the charger.

RECALLNotes

Show solution

p.d. is the energy transferred per unit charge (work done per coulomb)	✓ 1	Accept “joules per coulomb”. Not “the push of the current” on its own.
230 J of energy is transferred for each coulomb of charge that passes through the charger	✓ 1	Both the number 230 J and “per coulomb” needed.

(2)

(b)

Show that the current in the charger cable is about 32 A.

DERIVENotes

Show solution

P = IV with P = 7400 W, so I = P / V = 7400 / 230	✓ 1	Conversion of 7.4 kW to 7400 W must be seen or implied.
I = 32.2 A	✓ 1	Answer to at least 3 s.f., or full substitution leading to 32, needed for a show that.

Answer: 32.2 A

(2)

(c)(i)

One evening the car is charged for 4.5 hours.

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

CALCNotes

Show solution

t = 4.5 × 3600 = 16 200 s and Q = It = 32.2 × 16 200	✓ 1	Time must be in seconds.
Q = 5.2 × 10⁵ C	✓ 1	Using 32 A gives 5.18 × 10⁵ C: allow. Unit required. Allow ECF from (b).

Answer: 5.2 × 10⁵ C

(2)

(c)(ii)

Calculate the energy transferred to the charger in this time. Give your answer in MJ.

CALCNotes

Show solution

E = Pt = 7400 × 16 200 = 1.2 × 10⁸ J	✓ 1	Or E = QV = 5.2 × 10⁵ × 230. Allow ECF from (c)(i).
E = 120 MJ	✓ 1	Correct conversion to MJ needed for this mark. Allow 119.9 MJ.

Answer: 120 MJ

(2)

(d)(i)

The electrician compares two copper cables of the same length for the connection to the charger. Cable X has a cross-sectional area of 2.5 mm² and a resistance of 0.12 Ω. Cable Y has a cross-sectional area of 6.0 mm².

Calculate the resistance of cable Y.

Extended onlyCALCNotes

Show solution

R ∝ 1/A for the same material and length, so R = 0.12 × 2.5 / 6.0	✓ 1	Ratio may be shown in any correct form.
R = 0.050 Ω	✓ 1	Allow 0.05 Ω. 0.29 Ω (ratio inverted) scores 0.

Answer: 0.050 Ω

(2)

(d)(ii)

Explain why the electrician chooses cable Y and not cable X for the charger.

Extended onlyEXPLAINNotes

Show solution

cable Y has the smaller resistance, so less energy is transferred to thermal energy in it each second (it heats up less) with the large current of about 32 A	✓ 1	Accept “smaller p.d. across the cable” as the link to less heating.
so the cable does not overheat, melt its insulation or start a fire	✓ 1	Need a safety consequence, not just “it is better”.

(2)

(e)(i)

Figure 2 shows how the current in one of the lamps P and Q varies with the p.d. across it.

Use Figure 2 to determine the reading on the ammeter in Figure 1.

Figure 2: current–p.d. graph for one of the lamps P and Q

GRAPHNotes

Show solution

each lamp has the full 12 V across it, so from the graph the current in one lamp is 1.5 A	✓ 1	Accept 1.48–1.52 A.
ammeter reading = 1.5 + 1.5 = 3.0 A	✓ 1	Branch currents add in parallel. Allow ECF from a misread current.

Answer: 3.0 A

(2)

(e)(ii)

The filament of lamp P breaks. State and explain what happens to the brightness of lamp Q and to the reading on the ammeter.

REASONNotes

Show solution

lamp Q stays at the same brightness because it is on its own branch and still has the full 12 V across it	✓ 1	Reason needed for the mark.
the ammeter reading falls to 1.5 A, because there is no longer any current in the branch with lamp P	✓ 1	Allow “halves”. Allow ECF from (e)(i).

(2)

(f)(i)

On a visit to relatives, the car is charged from an ordinary 230 V socket using a portable charger with a power of 2.3 kW. The plug of the portable charger contains a fuse. Fuses rated at 3 A, 5 A and 13 A are available.

Calculate the current in the portable charger and state which fuse should be fitted.

CALCNotes

Show solution

I = P / V = 2300 / 230 = 10 A	✓ 1	Unit required.
13 A, the rating just above the working current	✓ 1	Must be consistent with the calculated current. Allow ECF.

Answer: 10 A; 13 A fuse

(2)

(f)(ii)

Calculate the time, in hours, that the portable charger takes to transfer the same energy as in (c)(ii).

MULTINotes

Show solution

t = E / P = 1.2 × 10⁸ / 2300 = 5.2 × 10⁴ s	✓ 1	Allow ECF from (c)(ii). Or ratio 4.5 × 7400 / 2300.
t = 14 h	✓ 1	Accept 14.5 h. Allow 14.4–14.5 h.

Answer: 14 h

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
p.d. is the energy transferred per unit charge (work done per coulomb)	✓ 1	Accept “joules per coulomb”. Not “the push of the current” on its own.
230 J of energy is transferred for each coulomb of charge that passes through the charger	✓ 1	Both the number 230 J and “per coulomb” needed.
Part (b)
P = IV with P = 7400 W, so I = P / V = 7400 / 230	✓ 1	Conversion of 7.4 kW to 7400 W must be seen or implied.
I = 32.2 A	✓ 1	Answer to at least 3 s.f., or full substitution leading to 32, needed for a show that.
Part (c)(i)
t = 4.5 × 3600 = 16 200 s and Q = It = 32.2 × 16 200	✓ 1	Time must be in seconds.
Q = 5.2 × 10⁵ C	✓ 1	Using 32 A gives 5.18 × 10⁵ C: allow. Unit required. Allow ECF from (b).
Part (c)(ii)
E = Pt = 7400 × 16 200 = 1.2 × 10⁸ J	✓ 1	Or E = QV = 5.2 × 10⁵ × 230. Allow ECF from (c)(i).
E = 120 MJ	✓ 1	Correct conversion to MJ needed for this mark. Allow 119.9 MJ.
Part (d)(i)
R ∝ 1/A for the same material and length, so R = 0.12 × 2.5 / 6.0	✓ 1	Ratio may be shown in any correct form.
R = 0.050 Ω	✓ 1	Allow 0.05 Ω. 0.29 Ω (ratio inverted) scores 0.
Part (d)(ii)
cable Y has the smaller resistance, so less energy is transferred to thermal energy in it each second (it heats up less) with the large current of about 32 A	✓ 1	Accept “smaller p.d. across the cable” as the link to less heating.
so the cable does not overheat, melt its insulation or start a fire	✓ 1	Need a safety consequence, not just “it is better”.
Part (e)(i)
each lamp has the full 12 V across it, so from the graph the current in one lamp is 1.5 A	✓ 1	Accept 1.48–1.52 A.
ammeter reading = 1.5 + 1.5 = 3.0 A	✓ 1	Branch currents add in parallel. Allow ECF from a misread current.
Part (e)(ii)
lamp Q stays at the same brightness because it is on its own branch and still has the full 12 V across it	✓ 1	Reason needed for the mark.
the ammeter reading falls to 1.5 A, because there is no longer any current in the branch with lamp P	✓ 1	Allow “halves”. Allow ECF from (e)(i).
Part (f)(i)
I = P / V = 2300 / 230 = 10 A	✓ 1	Unit required.
13 A, the rating just above the working current	✓ 1	Must be consistent with the calculated current. Allow ECF.
Part (f)(ii)
t = E / P = 1.2 × 10⁸ / 2300 = 5.2 × 10⁴ s	✓ 1	Allow ECF from (c)(ii). Or ratio 4.5 × 7400 / 2300.
t = 14 h	✓ 1	Accept 14.5 h. Allow 14.4–14.5 h.

Answers: (b) 32.2 A · (c)(i) 5.2 × 10⁵ C · (c)(ii) 120 MJ · (d)(i) 0.050 Ω · (e)(i) 3.0 A · (f)(i) 10 A; 13 A fuse · (f)(ii) 14 h

Common error (c)(i) Substituting the time in hours. Every time in Q = It and E = Pt must be in seconds, so 4.5 h is 16 200 s.

Exam tip (f)(ii) When the energy is the same, the time is inversely proportional to the power: 7.4 kW / 2.3 kW ≈ 3.2, so the time is about 3.2 × 4.5 h. This is a quick check on your answer.

More 4.2 questions in the question bank

+ Add to test✓ In your testCC-CIP-42-219 marks~20 min

Inside a hair dryer: heating coil, fan and a thermal cut-out

4.2 Charge, current, potential difference, resistance and power4.3 Series and parallel circuits2.2 Thermal expansion, specific heat capacity and changes of stateCore

Figure 1 shows a simplified circuit for a hair dryer that is used on the 230 V mains supply. The heating coil is a long coil of thin metal wire. A fan motor blows air over the coil.

At its normal working temperature the current in the heating coil is 8.0 A. The current in the fan motor is 0.30 A.

Figure 1: simplified circuit of the hair dryer

(a)

State the p.d. across the fan motor and calculate the current in the supply cable when the hair dryer is switched on.

REASONNotes

Show solution

230 V, because the motor is in parallel with the heating coil across the supply	✓ 1	The value alone scores the mark only with “parallel” or “same as the supply”.
current = 8.0 + 0.30 = 8.3 A	✓ 1	Branch currents add.

Answer: 230 V; 8.3 A

(2)

(b)

Calculate the resistance of the heating coil at its normal working temperature.

CALCNotes

Show solution

R = V / I = 230 / 8.0	✓ 1
R = 29 Ω	✓ 1	Accept 28.75 Ω. Unit required.

Answer: 29 Ω

(2)

(c)

Show that the total power of the hair dryer is about 1.9 kW.

DERIVENotes

Show solution

P = IV = 8.3 × 230	✓ 1	Allow ECF from the total current in (a).
P = 1909 W (= 1.9 kW)	✓ 1	Unrounded value needed. Using 8.0 A only (1840 W) scores 1 mark at most.

Answer: 1909 W

(2)

(d)

The hair dryer is used for 4.0 minutes. Calculate the total energy it transfers.

CALCNotes

Show solution

t = 4.0 × 60 = 240 s and E = Pt = 1909 × 240	✓ 1	Using 1900 W is fine.
E = 4.6 × 10⁵ J	✓ 1	Accept 458 000 J or 456 000 J. Unit required.

Answer: 4.6 × 10⁵ J

(2)

(e)(i)

The owner takes the hair dryer abroad and uses it with a 120 V supply. Figure 2 shows how the current in the heating coil varies with the p.d. across it.

Use Figure 2 to find the current in the heating coil on the 120 V supply, and calculate its power.

Figure 2: current–p.d. graph for the heating coil

GRAPHNotes

Show solution

I = 4.9 A read from the graph at 120 V	✓ 1	Accept 4.8–5.0 A.
P = IV = 4.9 × 120 = 590 W	✓ 1	Allow ECF from the reading. Accept 576–600 W.

Answer: 4.9 A; 590 W

(2)

(e)(ii)

A student predicts that the current on the 120 V supply should be 120 / 230 × 8.0 A = 4.2 A. Explain why the current read from Figure 2 is larger than this.

EXPLAINNotes

Show solution

at the lower p.d. the current is smaller, so the coil is cooler (its ions vibrate less and there are fewer collisions with the free electrons)	✓ 1
so the resistance of the coil is smaller than at 230 V, and I = V / R is larger than predicted (the current is not proportional to the p.d.)	✓ 1	“Non-ohmic” on its own scores 0.

(2)

(f)

Back on the 230 V supply, a manufacturer makes a gentler model. Its heating coil is made from the same wire but is 25% longer. Assume the coil works at the same temperature as before.

Show that the power of the new heating coil is about 1.5 kW.

Extended onlyMULTINotes

Show solution

R ∝ l, so new R = 1.25 × 28.75 = 35.9 Ω	✓ 1	Allow ECF from (b): 1.25 × 29 = 36 Ω.
I = V / R = 230 / 35.9 = 6.4 A	✓ 1
P = IV = 6.4 × 230 = 1470 W (≈ 1.5 kW)	✓ 1	Unrounded answer needed. A longer wire giving more power scores 0 for this row.

Answer: 1470 W

(3)

(g)

A thermistor is fixed in the air outlet. If the outlet gets too hot, a control circuit switches the heating coil off. Describe how the resistance of the thermistor changes as the outlet gets hotter, and suggest why this change can be used to switch off the heating coil.

EXPLAINNotes

Show solution

the resistance of the (NTC) thermistor decreases as its temperature increases	✓ 1
so the current in the thermistor's circuit increases (or a p.d. in the control circuit changes); at a set temperature this change is large enough to switch the coil off	✓ 1	Any sensible link between the falling resistance and a changing current or p.d. used as a signal.

(2)

(h)

On the 230 V supply, 0.040 kg of air passes through the hair dryer every second. Assume all the energy transferred by the heating coil (current 8.0 A) goes into this air. The specific heat capacity of air is 1000 J/(kg °C).

Calculate the rise in the temperature of the air.

Extended onlyMULTINotes

Show solution

energy per second = IV = 8.0 × 230 = 1840 J and ΔT = Q / (mc) = 1840 / (0.040 × 1000)	✓ 1	Using 1909 W (fan included) gives 48 °C: this row only (the stem asks for the heating coil alone).
ΔT = 46 °C	✓ 1	Unit required.

Answer: 46 °C

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
230 V, because the motor is in parallel with the heating coil across the supply	✓ 1	The value alone scores the mark only with “parallel” or “same as the supply”.
current = 8.0 + 0.30 = 8.3 A	✓ 1	Branch currents add.
Part (b)
R = V / I = 230 / 8.0	✓ 1
R = 29 Ω	✓ 1	Accept 28.75 Ω. Unit required.
Part (c)
P = IV = 8.3 × 230	✓ 1	Allow ECF from the total current in (a).
P = 1909 W (= 1.9 kW)	✓ 1	Unrounded value needed. Using 8.0 A only (1840 W) scores 1 mark at most.
Part (d)
t = 4.0 × 60 = 240 s and E = Pt = 1909 × 240	✓ 1	Using 1900 W is fine.
E = 4.6 × 10⁵ J	✓ 1	Accept 458 000 J or 456 000 J. Unit required.
Part (e)(i)
I = 4.9 A read from the graph at 120 V	✓ 1	Accept 4.8–5.0 A.
P = IV = 4.9 × 120 = 590 W	✓ 1	Allow ECF from the reading. Accept 576–600 W.
Part (e)(ii)
at the lower p.d. the current is smaller, so the coil is cooler (its ions vibrate less and there are fewer collisions with the free electrons)	✓ 1
so the resistance of the coil is smaller than at 230 V, and I = V / R is larger than predicted (the current is not proportional to the p.d.)	✓ 1	“Non-ohmic” on its own scores 0.
Part (f)
R ∝ l, so new R = 1.25 × 28.75 = 35.9 Ω	✓ 1	Allow ECF from (b): 1.25 × 29 = 36 Ω.
I = V / R = 230 / 35.9 = 6.4 A	✓ 1
P = IV = 6.4 × 230 = 1470 W (≈ 1.5 kW)	✓ 1	Unrounded answer needed. A longer wire giving more power scores 0 for this row.
Part (g)
the resistance of the (NTC) thermistor decreases as its temperature increases	✓ 1
so the current in the thermistor's circuit increases (or a p.d. in the control circuit changes); at a set temperature this change is large enough to switch the coil off	✓ 1	Any sensible link between the falling resistance and a changing current or p.d. used as a signal.
Part (h)
energy per second = IV = 8.0 × 230 = 1840 J and ΔT = Q / (mc) = 1840 / (0.040 × 1000)	✓ 1	Using 1909 W (fan included) gives 48 °C: this row only (the stem asks for the heating coil alone).
ΔT = 46 °C	✓ 1	Unit required.

Common error (e)(ii) Treating the coil as a fixed resistor. Its resistance depends on its temperature, so you must read the current from the graph, not scale it in proportion to the p.d.

More 4.2 questions in the question bank

4.3 Series and parallel circuits

+ Add to test✓ In your testCC-CIP-43-118 marks~20 min

A greenhouse frost alarm with a thermistor

4.3 Series and parallel circuits4.2 Charge, current, potential difference, resistance and power1.1 Physical quantities and measurementExtended

A gardener builds a frost alarm for a greenhouse. A thermistor and a 3.0 kΩ fixed resistor are connected in series to a 9.0 V battery. A switching circuit is connected across the thermistor. It sounds a buzzer when the p.d. across the thermistor reaches 6.0 V. The switching circuit takes a negligible current.

(a)

Sketch the circuit diagram of the battery, thermistor and fixed resistor, with a voltmeter connected to measure the p.d. across the thermistor. Use standard circuit symbols.

SKETCHNotes

Show solution

battery, thermistor and fixed resistor in a single series loop, with correct symbols for all three	✓ 1	Thermistor: resistor box with a diagonal line through it (a short horizontal tail on the lower end of the line is also accepted). Battery: two or more cells.
voltmeter connected in parallel across the thermistor only	✓ 1	Voltmeter in series with the components scores 0 for this row.

(2)

(b)

Figure 1 shows how the resistance of the thermistor varies with temperature.

Use Figure 1 to find the resistance of the thermistor at 2 °C.

Figure 1: resistance of the thermistor against temperature

GRAPHNotes

Show solution

6.0 kΩ

✓ 1

Accept 5.9–6.1 kΩ. Unit required.

Answer: 6.0 kΩ

(1)

(c)

Calculate the current in the circuit at 2 °C. Give your answer in mA.

CALCNotes

Show solution

total resistance = 6.0 + 3.0 = 9.0 kΩ = 9000 Ω and I = V / R = 9.0 / 9000	✓ 1	Allow ECF from (b).
I = 1.0 × 10⁻³ A = 1.0 mA	✓ 1	Answer must be in mA for this mark.

Answer: 1.0 mA

(2)

(d)

Show that the p.d. across the thermistor at 2 °C is 6.0 V.

DERIVENotes

Show solution

V = IR = 1.0 × 10⁻³ × 6000	✓ 1	Or V_out = 6.0 / (3.0 + 6.0) × 9.0.
V = 6.0 V	✓ 1	Working must be complete for a show that.

Answer: 6.0 V

(2)

(e)

During the day the temperature in the greenhouse is 20 °C. Use Figure 1 to calculate the p.d. across the thermistor at this temperature.

MULTINotes

Show solution

resistance at 20 °C = 2.0 kΩ from the graph	✓ 1	Accept 1.95–2.1 kΩ.
V = 2.0 / (2.0 + 3.0) × 9.0	✓ 1	Or current = 9.0 / 5000 = 1.8 mA, then V = IR. Allow ECF from the reading.
V = 3.6 V	✓ 1	Accept 3.5–3.7 V.

Answer: 3.6 V

(3)

(f)

Explain why the p.d. across the thermistor increases as the temperature in the greenhouse falls.

EXPLAINNotes

Show solution

as the temperature falls, the resistance of the thermistor increases	✓ 1
the supply p.d. is shared in proportion to the resistances (or: the total resistance increases, so the current decreases and the p.d. across the fixed resistor decreases)	✓ 1
so the thermistor takes a larger share of the 9.0 V	✓ 1	Must refer to the share of the fixed 9.0 V, or the p.d.s adding to 9.0 V.

(3)

(g)

One frosty night the gardener measures the p.d. across the 3.0 kΩ fixed resistor as 2.97 V. She uses this to work out the current in the circuit, and her calculator shows 0.00099 A.

Write down this current in mA, to a suitable number of significant figures, and give a reason for your choice.

DATANotes

Show solution

0.99 mA	✓ 1	Accept 9.9 × 10⁻¹ mA. 0.990 mA scores 0.
2 significant figures, because the least precise value used (3.0 kΩ) has 2 significant figures	✓ 1	Reason must refer to the precision of the data.

Answer: 0.99 mA

(2)

(h)(i)

In spring the gardener uses the same circuit as an automatic light for the greenhouse path. The thermistor is replaced by an LDR, and the switching circuit now turns on a lamp when the p.d. across the LDR reaches 6.0 V.

State what happens to the p.d. across the LDR as it gets darker.

PREDICTNotes

Show solution

it increases (because the resistance of the LDR increases in the dark)

✓ 1

Reason not required.

(1)

(h)(ii)

Determine the resistance of the LDR at the moment the lamp switches on.

CALCNotes

Show solution

p.d. across LDR = 6.0 V and across the 3.0 kΩ resistor = 3.0 V, so the resistances are in the ratio 2 : 1	✓ 1	Or current = 3.0 / 3000 = 1.0 mA, then R = 6.0 / 1.0 × 10⁻³.
R = 6.0 kΩ	✓ 1	Accept 6000 Ω. Unit required.

Answer: 6.0 kΩ

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
battery, thermistor and fixed resistor in a single series loop, with correct symbols for all three	✓ 1	Thermistor: resistor box with a diagonal line through it (a short horizontal tail on the lower end of the line is also accepted). Battery: two or more cells.
voltmeter connected in parallel across the thermistor only	✓ 1	Voltmeter in series with the components scores 0 for this row.
Part (b)
6.0 kΩ	✓ 1	Accept 5.9–6.1 kΩ. Unit required.
Part (c)
total resistance = 6.0 + 3.0 = 9.0 kΩ = 9000 Ω and I = V / R = 9.0 / 9000	✓ 1	Allow ECF from (b).
I = 1.0 × 10⁻³ A = 1.0 mA	✓ 1	Answer must be in mA for this mark.
Part (d)
V = IR = 1.0 × 10⁻³ × 6000	✓ 1	Or V_out = 6.0 / (3.0 + 6.0) × 9.0.
V = 6.0 V	✓ 1	Working must be complete for a show that.
Part (e)
resistance at 20 °C = 2.0 kΩ from the graph	✓ 1	Accept 1.95–2.1 kΩ.
V = 2.0 / (2.0 + 3.0) × 9.0	✓ 1	Or current = 9.0 / 5000 = 1.8 mA, then V = IR. Allow ECF from the reading.
V = 3.6 V	✓ 1	Accept 3.5–3.7 V.
Part (f)
as the temperature falls, the resistance of the thermistor increases	✓ 1
the supply p.d. is shared in proportion to the resistances (or: the total resistance increases, so the current decreases and the p.d. across the fixed resistor decreases)	✓ 1
so the thermistor takes a larger share of the 9.0 V	✓ 1	Must refer to the share of the fixed 9.0 V, or the p.d.s adding to 9.0 V.
Part (g)
0.99 mA	✓ 1	Accept 9.9 × 10⁻¹ mA. 0.990 mA scores 0.
2 significant figures, because the least precise value used (3.0 kΩ) has 2 significant figures	✓ 1	Reason must refer to the precision of the data.
Part (h)(i)
it increases (because the resistance of the LDR increases in the dark)	✓ 1	Reason not required.
Part (h)(ii)
p.d. across LDR = 6.0 V and across the 3.0 kΩ resistor = 3.0 V, so the resistances are in the ratio 2 : 1	✓ 1	Or current = 3.0 / 3000 = 1.0 mA, then R = 6.0 / 1.0 × 10⁻³.
R = 6.0 kΩ	✓ 1	Accept 6000 Ω. Unit required.

Answers: (b) 6.0 kΩ · (c) 1.0 mA · (d) 6.0 V · (e) 3.6 V · (g) 0.99 mA · (h)(ii) 6.0 kΩ

Common error (f) Saying the p.d. across the thermistor rises “because the current rises”. The current actually falls; the p.d. across the thermistor rises because it takes a larger share of the fixed supply p.d.

More 4.3 questions in the question bank

+ Add to test✓ In your testCC-CIP-43-220 marks~20 min

Stage lighting for a music festival

4.3 Series and parallel circuits4.2 Charge, current, potential difference, resistance and power4.4 Electrical safetyCore

A lighting technician sets up the lights for a small festival stage using a 24 V d.c. supply.

(a)

The technician first tests a decorative string of 12 identical filament lamps connected in series to the 24 V supply.

State the p.d. across each lamp, and compare the current in the first lamp of the string with the current in the last lamp.

REASONNotes

Show solution

p.d. across each lamp = 24 / 12 = 2.0 V	✓ 1	The supply p.d. is shared equally between identical lamps.
the current is the same in the first and last lamps (the same at every point in a series circuit)	✓ 1	“The current is shared” scores 0.

Answer: 2.0 V; the currents are equal

(2)

(b)

Figure 1 shows the spotlight rig. Spotlights S1 and S2 are identical. When lit, each has a resistance of 16 Ω.

Calculate the current in S1 and the total current in S1 and S2 together.

Figure 1: the spotlight rig

CALCNotes

Show solution

each spotlight has 24 V across it, so I = V / R = 24 / 16 = 1.5 A	✓ 1
total = 1.5 + 1.5 = 3.0 A	✓ 1	Allow ECF.

Answer: 1.5 A; 3.0 A

(2)

(c)

Calculate the combined resistance of S1 and S2.

Extended onlyCALCNotes

Show solution

1/R = 1/16 + 1/16 = 2/16	✓ 1	Or R = 24 / 3.0 using (b).
R = 8.0 Ω	✓ 1	16 Ω or 32 Ω scores 0. Unit required.

Answer: 8.0 Ω

(2)

(d)

Show that the total power of S1 and S2 is 72 W.

DERIVENotes

Show solution

P = IV = 3.0 × 24	✓ 1	Or 2 × (1.5 × 24). Allow ECF from (b) for the method only.
P = 72 W	✓ 1	Working must be complete for a show that.

Answer: 72 W

(2)

(e)

A third spotlight S3, with a resistance of 24 Ω when lit, is now connected in parallel with S1 and S2.

Calculate the new total current in the three spotlights, and state what happens to the brightness of S1. Give a reason.

PREDICTNotes

Show solution

current in S3 = 24 / 24 = 1.0 A	✓ 1
new current = 3.0 + 1.0 = 4.0 A	✓ 1	Allow ECF from (b).
S1 stays the same brightness, because it still has the full 24 V across it	✓ 1	“Gets dimmer” scores 0. Reason required.

Answer: 4.0 A; S1 unchanged

(3)

(f)

One lamp in the decorative string fails and so does spotlight S2. Explain why all the lamps in the string go out but S1 and S3 stay lit.

EXPLAINNotes

Show solution

the string is a single loop, so a broken lamp breaks the only path and there is no current in any lamp	✓ 1
S1 and S3 are on separate branches, each still a complete path across the 24 V supply, so they still have current	✓ 1

(2)

(g)(i)

The LED in Figure 1 shows that the rig is switched on. It is in series with a resistor R. Figure 2 shows how the current in the LED varies with the p.d. across it.

The current in the LED is 20 mA. Use Figure 2 to find the p.d. across the LED, and calculate the resistance of R.

Figure 2: current–p.d. graph for the LED

GRAPHNotes

Show solution

p.d. across LED = 2.0 V from the graph	✓ 1	Accept 1.95–2.05 V.
p.d. across R = 24 − 2.0 = 22 V	✓ 1	Series p.d.s add to the supply p.d. Allow ECF from the reading.
R = 22 / 0.020 = 1100 Ω	✓ 1	Accept 1.1 kΩ. Using 20 instead of 0.020 A (1.1 Ω) scores 0 for this row.

Answer: 2.0 V; 1100 Ω

(3)

(g)(ii)

At the next stage, the supply leads are connected the wrong way round. Suggest what happens to the LED and explain your answer using Figure 2.

REASONNotes

Show solution

the LED does not light	✓ 1
the p.d. across it is now reversed, and Figure 2 shows zero current for a reverse p.d. (a diode conducts in one direction only / has a very large resistance in reverse)	✓ 1	Must use the graph or the one-way property.

(2)

(h)

The fuse in Figure 1 protects the rig with S1, S2, S3 and the LED all working. Fuses rated at 3 A, 5 A, 10 A and 13 A are available.

State which fuse should be fitted, and explain why the 13 A fuse would not be suitable.

REASONNotes

Show solution

5 A, because the total current is 4.0 A (4.02 A with the LED), and 5 A is the rating just above it	✓ 1	Allow ECF from (e).
a 13 A fuse would not melt until the current was much larger than normal, so the cables could overheat before it blew	✓ 1

Answer: 5 A

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
p.d. across each lamp = 24 / 12 = 2.0 V	✓ 1	The supply p.d. is shared equally between identical lamps.
the current is the same in the first and last lamps (the same at every point in a series circuit)	✓ 1	“The current is shared” scores 0.
Part (b)
each spotlight has 24 V across it, so I = V / R = 24 / 16 = 1.5 A	✓ 1
total = 1.5 + 1.5 = 3.0 A	✓ 1	Allow ECF.
Part (c)
1/R = 1/16 + 1/16 = 2/16	✓ 1	Or R = 24 / 3.0 using (b).
R = 8.0 Ω	✓ 1	16 Ω or 32 Ω scores 0. Unit required.
Part (d)
P = IV = 3.0 × 24	✓ 1	Or 2 × (1.5 × 24). Allow ECF from (b) for the method only.
P = 72 W	✓ 1	Working must be complete for a show that.
Part (e)
current in S3 = 24 / 24 = 1.0 A	✓ 1
new current = 3.0 + 1.0 = 4.0 A	✓ 1	Allow ECF from (b).
S1 stays the same brightness, because it still has the full 24 V across it	✓ 1	“Gets dimmer” scores 0. Reason required.
Part (f)
the string is a single loop, so a broken lamp breaks the only path and there is no current in any lamp	✓ 1
S1 and S3 are on separate branches, each still a complete path across the 24 V supply, so they still have current	✓ 1
Part (g)(i)
p.d. across LED = 2.0 V from the graph	✓ 1	Accept 1.95–2.05 V.
p.d. across R = 24 − 2.0 = 22 V	✓ 1	Series p.d.s add to the supply p.d. Allow ECF from the reading.
R = 22 / 0.020 = 1100 Ω	✓ 1	Accept 1.1 kΩ. Using 20 instead of 0.020 A (1.1 Ω) scores 0 for this row.
Part (g)(ii)
the LED does not light	✓ 1
the p.d. across it is now reversed, and Figure 2 shows zero current for a reverse p.d. (a diode conducts in one direction only / has a very large resistance in reverse)	✓ 1	Must use the graph or the one-way property.
Part (h)
5 A, because the total current is 4.0 A (4.02 A with the LED), and 5 A is the rating just above it	✓ 1	Allow ECF from (e).
a 13 A fuse would not melt until the current was much larger than normal, so the cables could overheat before it blew	✓ 1

Answers: (a) 2.0 V; the currents are equal · (b) 1.5 A; 3.0 A · (c) 8.0 Ω · (d) 72 W · (e) 4.0 A; S1 unchanged · (g)(i) 2.0 V; 1100 Ω · (h) 5 A

Exam tip (e) Adding a branch in parallel does not change the p.d. across the other branches. It only adds its own current to the current in the supply wire.

Common error (g)(i) Forgetting to convert 20 mA to 0.020 A before using R = V / I.

More 4.3 questions in the question bank

4.4 Electrical safety

+ Add to test✓ In your testCC-CIP-44-119 marks~20 min

Fuses and plugs in a student kitchen

4.4 Electrical safety4.2 Charge, current, potential difference, resistance and power1.7 Energy, work and powerCore

Four students share a kitchen. The mains supply is 230 V. The table shows the power of three appliances that they use.

appliance	power
kettle	2.3 kW
microwave oven	1.2 kW
desk lamp	60 W

Fuses with ratings of 3 A, 5 A and 13 A are available.

Figure 1: inside the plug of the microwave oven, cover removed (not to scale)

(a)(i)

Figure 1 shows the inside of the plug of the microwave oven. Wire X is connected to the fuse and wire Z is connected to the terminal at the top.

State the name of each of the wires X, Y and Z.

RECALLNotes

Show solution

X: live and Y: neutral	✓ 1	Both needed for this mark.
Z: earth	✓ 1

Answer: X live, Y neutral, Z earth

(2)

(a)(ii)

Explain why the fuse is connected in wire X and not in wire Y.

EXPLAINNotes

Show solution

wire X is the live wire, which carries the (230 V) supply p.d. to the appliance	✓ 1	Accept “X is at a high p.d.”.
when the fuse melts, the appliance is disconnected from the live supply, so it is safe to touch	✓ 1	Accept: a fuse in the neutral wire would leave the appliance connected to the live supply after it melts.

(2)

(b)(i)

Show that the normal working current of the kettle is 10 A, and explain why the kettle must be fitted with the 13 A fuse.

DERIVENotes

Show solution

2.3 kW = 2300 W and I = P / V = 2300 / 230 (= 10 A)	✓ 1	The conversion and the substitution must both be seen. Using 2.3 in place of 2300 scores 0.
10 A is above 5 A, so a 5 A fuse would melt in normal use; 13 A is the next rating above 10 A	✓ 1	Both ideas needed: why not 5 A, and 13 A as the next value up.

Answer: 10 A; 13 A fuse

(2)

(b)(ii)

Calculate the working current of the microwave oven and of the desk lamp. Hence choose a fuse for each appliance.

CALCNotes

Show solution

microwave oven: 1200 / 230 = 5.2 A and desk lamp: 60 / 230 = 0.26 A	✓ 1	Both currents needed. Accept 5.22 A and 0.261 A.
microwave oven: 13 A fuse and desk lamp: 3 A fuse	✓ 1	Allow ECF from the currents calculated. A 5 A fuse for the microwave oven scores 0: 5.2 A is above 5 A.

Answer: 5.2 A, 13 A fuse; 0.26 A, 3 A fuse

(2)

(c)(i)

Figure 2 shows the working current of the kettle as a bar.

On Figure 2, draw the bars for the microwave oven and the desk lamp.

Figure 2: working currents of the appliances on the 230 V mains

GRAPHNotes

Show solution

microwave oven bar at 5.2 A and desk lamp bar at 0.26 A (just above the axis), each correct to the nearest small square (0.2 A)

✓ 1

Allow ECF from (b)(ii). Both bars needed; the lamp bar is a very short bar between 0.2 A and 0.4 A.

Answer: bars at 5.2 A and 0.26 A

(1)

(c)(ii)

One student says that it would be safer to fit a 13 A fuse in the plug of every appliance.

Use Figure 2 to explain why a 13 A fuse is not suitable for the desk lamp.

REASONNotes

Show solution

the working current of the lamp (0.26 A) is very much smaller than 13 A (about 50 times smaller)	✓ 1	Allow ECF from (b)(ii). A comparison with 13 A must be made.
so in a fault a current many times the normal current could flow without the fuse melting, and the lamp's cable could overheat or catch fire	✓ 1	Accept “the fuse would not protect the lamp or its cable”.

(2)

(d)

The microwave oven has a metal case. A fault occurs inside the oven, and the live wire touches the metal case.

Explain how the earth wire and the fuse together protect a person who then touches the case.

Extended onlyEXPLAINNotes

Show solution

the earth wire connects the metal case to the ground through a very low resistance	✓ 1
a very large current flows from the live wire, through the case and the earth wire, to the ground	✓ 1	Not enough: “the current goes to earth” with no idea of a large current.
this large current melts the fuse, so the oven is disconnected from the live supply and the case is no longer live	✓ 1	The steps must be in a logical order.

(3)

(e)

The students have one extension lead. The plug of the extension lead is fitted with a 13 A fuse. The kettle is plugged into the extension lead and switched on.

Determine which other appliance in the table, if any, can be used from the same extension lead at the same time as the kettle without the fuse melting. Show your working.

MULTINotes

Show solution

kettle and microwave oven: 10 + 5.2 = 15.2 A, and kettle and lamp: 10 + 0.26 = 10.3 A	✓ 1	Accept “13 − 10 = 3 A spare” compared with each current, or total powers 3500 W and 2360 W compared with 13 × 230 = 2990 W.
only the desk lamp, because 10.3 A is below 13 A (15.2 A is above 13 A)	✓ 1	Allow ECF from (b). The conclusion must follow from the working.

Answer: only the desk lamp

(2)

(f)

Each day the kettle is switched on for a total of 10 minutes and the desk lamp is switched on for 8.0 hours.

A student says: “The kettle transfers much more energy than the lamp each week, because its power is much larger.”

Calculate the energy transferred by each appliance in one week (7 days) and comment on the student's statement.

REASONNotes

Show solution

kettle: E = Pt = 2300 × (10 × 60 × 7) = 9.7 × 10⁶ J	✓ 1	Accept 9 660 000 J. Time must be in seconds (or consistent units). Accept 2.7 kW h.
lamp: E = 60 × (8.0 × 3600 × 7) = 1.2 × 10⁷ J	✓ 1	Accept 12 096 000 J. Accept 3.4 kW h.
the statement is wrong: the lamp transfers more energy, because it is on for a much longer time (48 times longer)	✓ 1	The comment must use the two energies. Allow ECF from wrong energies if the comment matches them.

Answer: kettle 9.7 × 10⁶ J, lamp 1.2 × 10⁷ J; the lamp transfers more

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
X: live and Y: neutral	✓ 1	Both needed for this mark.
Z: earth	✓ 1
Part (a)(ii)
wire X is the live wire, which carries the (230 V) supply p.d. to the appliance	✓ 1	Accept “X is at a high p.d.”.
when the fuse melts, the appliance is disconnected from the live supply, so it is safe to touch	✓ 1	Accept: a fuse in the neutral wire would leave the appliance connected to the live supply after it melts.
Part (b)(i)
2.3 kW = 2300 W and I = P / V = 2300 / 230 (= 10 A)	✓ 1	The conversion and the substitution must both be seen. Using 2.3 in place of 2300 scores 0.
10 A is above 5 A, so a 5 A fuse would melt in normal use; 13 A is the next rating above 10 A	✓ 1	Both ideas needed: why not 5 A, and 13 A as the next value up.
Part (b)(ii)
microwave oven: 1200 / 230 = 5.2 A and desk lamp: 60 / 230 = 0.26 A	✓ 1	Both currents needed. Accept 5.22 A and 0.261 A.
microwave oven: 13 A fuse and desk lamp: 3 A fuse	✓ 1	Allow ECF from the currents calculated. A 5 A fuse for the microwave oven scores 0: 5.2 A is above 5 A.
Part (c)(i)
microwave oven bar at 5.2 A and desk lamp bar at 0.26 A (just above the axis), each correct to the nearest small square (0.2 A)	✓ 1	Allow ECF from (b)(ii). Both bars needed; the lamp bar is a very short bar between 0.2 A and 0.4 A.
Part (c)(ii)
the working current of the lamp (0.26 A) is very much smaller than 13 A (about 50 times smaller)	✓ 1	Allow ECF from (b)(ii). A comparison with 13 A must be made.
so in a fault a current many times the normal current could flow without the fuse melting, and the lamp's cable could overheat or catch fire	✓ 1	Accept “the fuse would not protect the lamp or its cable”.
Part (d)
the earth wire connects the metal case to the ground through a very low resistance	✓ 1
a very large current flows from the live wire, through the case and the earth wire, to the ground	✓ 1	Not enough: “the current goes to earth” with no idea of a large current.
this large current melts the fuse, so the oven is disconnected from the live supply and the case is no longer live	✓ 1	The steps must be in a logical order.
Part (e)
kettle and microwave oven: 10 + 5.2 = 15.2 A, and kettle and lamp: 10 + 0.26 = 10.3 A	✓ 1	Accept “13 − 10 = 3 A spare” compared with each current, or total powers 3500 W and 2360 W compared with 13 × 230 = 2990 W.
only the desk lamp, because 10.3 A is below 13 A (15.2 A is above 13 A)	✓ 1	Allow ECF from (b). The conclusion must follow from the working.
Part (f)
kettle: E = Pt = 2300 × (10 × 60 × 7) = 9.7 × 10⁶ J	✓ 1	Accept 9 660 000 J. Time must be in seconds (or consistent units). Accept 2.7 kW h.
lamp: E = 60 × (8.0 × 3600 × 7) = 1.2 × 10⁷ J	✓ 1	Accept 12 096 000 J. Accept 3.4 kW h.
the statement is wrong: the lamp transfers more energy, because it is on for a much longer time (48 times longer)	✓ 1	The comment must use the two energies. Allow ECF from wrong energies if the comment matches them.

Answers: (a)(i) X live, Y neutral, Z earth · (b)(i) 10 A; 13 A fuse · (b)(ii) 5.2 A, 13 A fuse; 0.26 A, 3 A fuse · (c)(i) bars at 5.2 A and 0.26 A · (e) only the desk lamp · (f) kettle 9.7 × 10⁶ J, lamp 1.2 × 10⁷ J; the lamp transfers more

Common error (b)(ii) Choosing the fuse rating nearest to the working current. The rating must be just ABOVE it: the microwave oven takes 5.2 A, so a 5 A fuse would melt every time it is used.

Exam tip (e) Appliances on one extension lead are in parallel, so their currents add. Add the currents (or the powers) first, then compare the total with the fuse rating.

More 4.4 questions in the question bank

+ Add to test✓ In your testCC-CIP-44-217 marks~20 min

A hedge trimmer, a wet garden and an RCD

4.4 Electrical safety4.3 Series and parallel circuits4.2 Charge, current, potential difference, resistance and powerCore

A gardener uses an electric hedge trimmer with a long cable. The trimmer is connected to the 230 V mains through a residual current device (RCD) plugged into an outdoor socket.

The power of the trimmer is 460 W. The trimmer has a plastic case and is marked with the double-square symbol.

Figure 1: two circuit diagrams drawn for the trimmer

(a)

State two hazards that could make using the trimmer in the garden dangerous.

RECALLNotes

Show solution

damp or wet conditions (wet grass, rain or wet hands), because water conducts and lowers the resistance of the skin	✓ 1	Any two of the three points, one mark each.
damaged insulation on the cable (for example worn, or cut by the blades), so a bare wire can be touched	✓ 1	Also accept: overheating of the cable because the current is too large (overloading). Not enough: “electric shock” on its own.

(2)

(b)

The double-square symbol shows that the trimmer is double insulated.

Explain why the trimmer does not need an earth wire.

EXPLAINNotes

Show solution

the live parts are enclosed in a second layer of insulation and the case is plastic, an insulator	✓ 1	Accept “two layers of insulation”.
so the case cannot become live and the user cannot touch a live part, even if one layer of insulation fails	✓ 1

(2)

(c)(i)

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

CALCNotes

Show solution

I = P / V = 460 / 230	✓ 1
= 2.0 A	✓ 1	Unit required.

Answer: 2.0 A

(2)

(c)(ii)

The plug of the trimmer can be fitted with a 3 A, 5 A or 13 A fuse.

State which fuse should be fitted. Give a reason for your answer.

REASONNotes

Show solution

3 A, because it is the rating just above the working current of 2.0 A

✓ 1

Allow ECF from (c)(i). The reason is needed for the mark.

Answer: 3 A

(1)

(d)

Figure 1 shows two circuit diagrams that a student draws for the trimmer. In both circuits the fuse is in the live wire.

State which circuit, P or Q, is wired safely. Explain your answer.

EXPLAINNotes

Show solution

circuit P	✓ 1	No mark for P without an explanation.
in P the switch is in the live wire, so when it is open the motor is disconnected from the live supply; in Q the motor stays connected to the live wire (at 230 V) even when switched off, so touching it could give a shock	✓ 1	Accept a clear explanation of either circuit.

Answer: circuit P

(2)

(e)(i)

While cutting a wet hedge, the gardener accidentally cuts into the cable with the blades. The insulation of the live wire is cut through, but the wires are not broken, so the trimmer keeps running normally. The gardener's wet hand touches the bare live wire.

The resistance of the path from the live wire through the gardener's body to the ground is 2300 Ω.

Show that the current through the gardener is about 0.1 A.

DERIVENotes

Show solution

I = V / R = 230 / 2300	✓ 1
= 0.10 A	✓ 1	The answer must be given to at least 2 s.f., or the full substitution seen, for a show that.

Answer: 0.10 A

(2)

(e)(ii)

Suppose the trimmer were used without the RCD. The current through the gardener would then flow in the live wire as well as the normal current through the trimmer.

Explain why the 3 A fuse in the plug would not melt.

REASONNotes

Show solution

current in the live wire = 2.0 + 0.10 = 2.1 A	✓ 1	Allow ECF from (c)(i) and (e)(i).
this is less than 3 A, so the fuse wire does not get hot enough to melt (and the current through the gardener continues)	✓ 1	The comparison with 3 A is needed.

Answer: 2.1 A, which is below 3 A

(2)

(e)(iii)

Explain how the RCD detects this fault.

Extended onlyEXPLAINNotes

Show solution

the RCD compares the current in the live wire with the current in the neutral wire	✓ 1
the current through the gardener does not return along the neutral wire, so the live current (2.1 A) is larger than the neutral current (2.0 A), and the RCD switches off the supply	✓ 1	Accept “the two currents are different” for the second mark, provided the reason (current leaking to earth through the gardener) is given.

(2)

(f)

Figure 2 shows how the size of the current through the gardener changes with time when the RCD is used.

A current of 0.10 A through the body can be dangerous to the heart if it lasts longer than about 0.2 s.

Use Figure 2 to determine for how long the current flowed through the gardener. State whether the RCD protected the gardener.

Figure 2: size of the current through the gardener against time

GRAPHNotes

Show solution

40 − 15 = 25 ms (0.025 s)	✓ 1	Accept 24 ms to 26 ms.
yes: 25 ms is much shorter than 0.2 s (about one eighth of it), so the gardener is protected	✓ 1	Allow ECF from the time read. Units must be consistent in the comparison.

Answer: 25 ms; yes, the RCD protected the gardener

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
damp or wet conditions (wet grass, rain or wet hands), because water conducts and lowers the resistance of the skin	✓ 1	Any two of the three points, one mark each.
damaged insulation on the cable (for example worn, or cut by the blades), so a bare wire can be touched	✓ 1	Also accept: overheating of the cable because the current is too large (overloading). Not enough: “electric shock” on its own.
Part (b)
the live parts are enclosed in a second layer of insulation and the case is plastic, an insulator	✓ 1	Accept “two layers of insulation”.
so the case cannot become live and the user cannot touch a live part, even if one layer of insulation fails	✓ 1
Part (c)(i)
I = P / V = 460 / 230	✓ 1
= 2.0 A	✓ 1	Unit required.
Part (c)(ii)
3 A, because it is the rating just above the working current of 2.0 A	✓ 1	Allow ECF from (c)(i). The reason is needed for the mark.
Part (d)
circuit P	✓ 1	No mark for P without an explanation.
in P the switch is in the live wire, so when it is open the motor is disconnected from the live supply; in Q the motor stays connected to the live wire (at 230 V) even when switched off, so touching it could give a shock	✓ 1	Accept a clear explanation of either circuit.
Part (e)(i)
I = V / R = 230 / 2300	✓ 1
= 0.10 A	✓ 1	The answer must be given to at least 2 s.f., or the full substitution seen, for a show that.
Part (e)(ii)
current in the live wire = 2.0 + 0.10 = 2.1 A	✓ 1	Allow ECF from (c)(i) and (e)(i).
this is less than 3 A, so the fuse wire does not get hot enough to melt (and the current through the gardener continues)	✓ 1	The comparison with 3 A is needed.
Part (e)(iii)
the RCD compares the current in the live wire with the current in the neutral wire	✓ 1
the current through the gardener does not return along the neutral wire, so the live current (2.1 A) is larger than the neutral current (2.0 A), and the RCD switches off the supply	✓ 1	Accept “the two currents are different” for the second mark, provided the reason (current leaking to earth through the gardener) is given.
Part (f)
40 − 15 = 25 ms (0.025 s)	✓ 1	Accept 24 ms to 26 ms.
yes: 25 ms is much shorter than 0.2 s (about one eighth of it), so the gardener is protected	✓ 1	Allow ECF from the time read. Units must be consistent in the comparison.

Answers: (c)(i) 2.0 A · (c)(ii) 3 A · (d) circuit P · (e)(i) 0.10 A · (e)(ii) 2.1 A, which is below 3 A · (f) 25 ms; yes, the RCD protected the gardener

Common error (e)(ii) Thinking that any fault blows the fuse. A fuse melts only when the current in the live wire goes above its rating; a current through a person of a tenth of an ampere is far too small to do this, but it is still dangerous.

More 4.4 questions in the question bank

4.5 Electromagnetic induction, motors and transformers

+ Add to test✓ In your testCC-CIP-45-119 marks~20 min

An offshore wind farm and its transmission line

4.5 Electromagnetic induction, motors and transformers4.2 Charge, current, potential difference, resistance and power1.7 Energy, work and powerExtended

A small offshore wind farm has five identical turbines. Each turbine turns an a.c. generator, in which a coil rotates in a magnetic field. The output p.d. of each generator is 3.3 kV.

A transformer in the tower of each turbine steps this p.d. up to 33 kV. The five transformers are connected to one export cable, which carries the electrical power to an onshore substation (Figure 1). Assume that the transformers are 100% efficient.

Figure 1: the wind farm and its export cable (not to scale)

(a)

Explain how an alternating e.m.f. is produced in the rotating coil of a generator.

EXPLAINNotes

Show solution

as the coil rotates, its sides cut the magnetic field lines	✓ 1	Accept “the magnetic field through the coil changes”.
so an e.m.f. is induced in the coil (electromagnetic induction)	✓ 1
each side moves one way through the field and then the other way every half turn, so the e.m.f. reverses direction	✓ 1	Accept “the e.m.f. changes direction every half turn” with a reason.

(3)

(b)

Figure 2 shows how the e.m.f. of a small laboratory model of the generator varies with time.

Explain why the e.m.f. is a maximum at point P and zero at point Q.

Figure 2: e.m.f. of the laboratory model generator against time

GRAPHNotes

Show solution

at P the sides of the coil are moving at right angles to the field lines, so they cut the lines fastest	✓ 1
at Q the sides are moving along (parallel to) the field lines, so they cut no lines	✓ 1	Accept “the coil is perpendicular to the field at Q” with the idea of no cutting.

(2)

(c)

The primary coil of the transformer in each tower has 400 turns.

Calculate the number of turns on the secondary coil.

Extended onlyCALCNotes

Show solution

V_p / V_s = N_p / N_s, so N_s = 400 × 33 000 / 3300	✓ 1	Any correct rearrangement with the numbers substituted.
= 4000 turns	✓ 1	40 turns (ratio inverted) scores 0 for this mark: a step-up transformer needs more turns on the secondary.

Answer: 4000

(2)

(d)(i)

When the wind speed is 12 m/s, the output power of each generator is 7.2 MW.

Show that the current in the export cable is about 1100 A.

Extended onlyDERIVENotes

Show solution

total power = 5 × 7.2 MW = 36 MW (3.6 × 10⁷ W)	✓ 1	Or, for one turbine: I_p = 7.2 × 10⁶ / 3300 = 2180 A and I_s = I_pV_p / V_s = 218 A.
I = P / V = 3.6 × 10⁷ / 33 000 = 1090 A	✓ 1	Or 5 × 218 A = 1090 A. A final value to at least 3 s.f. (or the full substitution) is needed for a show that.

Answer: 1090 A

(2)

(d)(ii)

The export cable has a total resistance of 0.80 Ω.

Calculate the power wasted as heat in the export cable.

Extended onlyCALCNotes

Show solution

P = I²R = 1090² × 0.80	✓ 1
= 9.5 × 10⁵ W	✓ 1	Accept 950 kW. Using 1100 A gives 9.7 × 10⁵ W: accept. Unit required.

Answer: 9.5 × 10⁵ W (950 kW)

(2)

(e)

The engineers consider using transformers with twice the turns ratio, so that the export cable works at 66 kV. The output power of the turbines and the cable are unchanged.

Determine the power that would be wasted as heat in the export cable.

Extended onlyPREDICTNotes

Show solution

for the same power, doubling the p.d. halves the current: I = 3.6 × 10⁷ / 66 000 = 545 A	✓ 1	Accept “current halves” with 545 A or 550 A.
P = I²R = 545² × 0.80, or the loss is (1/2)² = 1/4 of the value in (d)(ii)	✓ 1
= 2.4 × 10⁵ W	✓ 1	Allow ECF from (d)(ii) for one quarter of that value. A loss halved to 4.8 × 10⁵ W scores the first mark only.

Answer: 2.4 × 10⁵ W

(3)

(f)(i)

At a wind speed of 12 m/s, a mass of 2.5 × 10⁵ kg of air passes through the blades of each turbine every second.

Show that the kinetic energy of this mass of air is 18 MJ.

DERIVENotes

Show solution

E_k = ½mv² = ½ × 2.5 × 10⁵ × 12²	✓ 1
= 1.8 × 10⁷ J (= 18 MJ)	✓ 1	The full substitution, or 18 000 000 J, must be seen.

Answer: 1.8 × 10⁷ J

(2)

(f)(ii)

Assume that the energy input to each turbine every second is the kinetic energy of the air that passes through its blades.

Calculate the overall efficiency of the wind farm in transferring the kinetic energy of the air to electrical energy delivered to the onshore substation.

Extended onlyMULTINotes

Show solution

total input power = 5 × 18 MJ per second = 90 MW	✓ 1
useful output power = 36 MW − 0.95 MW = 35 MW	✓ 1	Allow ECF from (d)(ii).
efficiency = 35 / 90 × 100% = 39%	✓ 1	Allow ECF. Using 36 MW (cable loss ignored) gives 40%: scores the first and third marks only.

Answer: 39%

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
as the coil rotates, its sides cut the magnetic field lines	✓ 1	Accept “the magnetic field through the coil changes”.
so an e.m.f. is induced in the coil (electromagnetic induction)	✓ 1
each side moves one way through the field and then the other way every half turn, so the e.m.f. reverses direction	✓ 1	Accept “the e.m.f. changes direction every half turn” with a reason.
Part (b)
at P the sides of the coil are moving at right angles to the field lines, so they cut the lines fastest	✓ 1
at Q the sides are moving along (parallel to) the field lines, so they cut no lines	✓ 1	Accept “the coil is perpendicular to the field at Q” with the idea of no cutting.
Part (c)
V_p / V_s = N_p / N_s, so N_s = 400 × 33 000 / 3300	✓ 1	Any correct rearrangement with the numbers substituted.
= 4000 turns	✓ 1	40 turns (ratio inverted) scores 0 for this mark: a step-up transformer needs more turns on the secondary.
Part (d)(i)
total power = 5 × 7.2 MW = 36 MW (3.6 × 10⁷ W)	✓ 1	Or, for one turbine: I_p = 7.2 × 10⁶ / 3300 = 2180 A and I_s = I_pV_p / V_s = 218 A.
I = P / V = 3.6 × 10⁷ / 33 000 = 1090 A	✓ 1	Or 5 × 218 A = 1090 A. A final value to at least 3 s.f. (or the full substitution) is needed for a show that.
Part (d)(ii)
P = I²R = 1090² × 0.80	✓ 1
= 9.5 × 10⁵ W	✓ 1	Accept 950 kW. Using 1100 A gives 9.7 × 10⁵ W: accept. Unit required.
Part (e)
for the same power, doubling the p.d. halves the current: I = 3.6 × 10⁷ / 66 000 = 545 A	✓ 1	Accept “current halves” with 545 A or 550 A.
P = I²R = 545² × 0.80, or the loss is (1/2)² = 1/4 of the value in (d)(ii)	✓ 1
= 2.4 × 10⁵ W	✓ 1	Allow ECF from (d)(ii) for one quarter of that value. A loss halved to 4.8 × 10⁵ W scores the first mark only.
Part (f)(i)
E_k = ½mv² = ½ × 2.5 × 10⁵ × 12²	✓ 1
= 1.8 × 10⁷ J (= 18 MJ)	✓ 1	The full substitution, or 18 000 000 J, must be seen.
Part (f)(ii)
total input power = 5 × 18 MJ per second = 90 MW	✓ 1
useful output power = 36 MW − 0.95 MW = 35 MW	✓ 1	Allow ECF from (d)(ii).
efficiency = 35 / 90 × 100% = 39%	✓ 1	Allow ECF. Using 36 MW (cable loss ignored) gives 40%: scores the first and third marks only.

Answers: (c) 4000 · (d)(i) 1090 A · (d)(ii) 9.5 × 10⁵ W (950 kW) · (e) 2.4 × 10⁵ W · (f)(i) 1.8 × 10⁷ J · (f)(ii) 39%

Common error (e) Halving the power loss when the p.d. is doubled. The loss is I²R, so halving the current divides the loss by four.

More 4.5 questions in the question bank

+ Add to test✓ In your testCC-CIP-45-216 marks~20 min

A bicycle dynamo and a hub motor

4.5 Electromagnetic induction, motors and transformers4.1 Magnets and magnetic fields4.2 Charge, current, potential difference, resistance and powerCore

A commuter's bicycle has an electrical machine in each wheel.

In the front hub there is a dynamo. A ring of permanent magnets turns with the wheel, close to a fixed coil of insulated copper wire wound on a soft-iron core. The dynamo is connected to the headlamp.

In the rear hub there is a d.c. electric motor, powered by a battery, which helps the cyclist.

(a)

Explain why an e.m.f. is induced in the coil of the dynamo when the wheel turns, but not when the bicycle is standing still.

EXPLAINNotes

Show solution

when the wheel turns, the magnets move relative to the coil, so the magnetic field through the coil changes (field lines are cut) and an e.m.f. is induced	✓ 1	The idea of relative movement or a changing field is needed.
when the bicycle is still, the field through the coil does not change, so no e.m.f. is induced	✓ 1	Not enough: “the magnets are not moving” without reference to the field through the coil not changing.

(2)

(b)

State two changes to the design of the dynamo that would give a larger e.m.f. at the same speed.

RECALLNotes

Show solution

use stronger magnets	✓ 1
use more turns on the coil	✓ 1	Accept “put the magnets closer to the coil” as one of the two changes. Not accepted: “turn the wheel faster”, because the speed is the same, or “add an iron core”, because the coil already has one.

(2)

(c)

Figure 1 shows how the e.m.f. of the dynamo varies with time when the bicycle is wheeled slowly at walking pace.

The bicycle is now wheeled at twice this speed.

On Figure 1, sketch a graph to show how the e.m.f. now varies with time.

Figure 1: e.m.f. of the dynamo at walking pace

SKETCHNotes

Show solution

an alternating wave with peaks of 6.0 V (twice the original peak of 3.0 V)	✓ 1
a period of 30 ms, half the original, so twice as many cycles are shown (4 cycles in 120 ms)	✓ 1	Both features must be clear for the two marks; one feature correct scores 1.

Answer: peak 6.0 V, period 30 ms

(2)

(d)(i)

When the bicycle is ridden at normal speed, the p.d. across the headlamp is 6.0 V. The resistance of the headlamp is 15 Ω.

Calculate the current in the headlamp.

CALCNotes

Show solution

I = V / R = 6.0 / 15	✓ 1
= 0.40 A	✓ 1	Unit required.

Answer: 0.40 A

(2)

(d)(ii)

Show that the energy transferred by the headlamp during a 25-minute ride at normal speed is 3600 J.

DERIVENotes

Show solution

P = IV = 0.40 × 6.0 = 2.4 W	✓ 1	Allow ECF from (d)(i) for this mark only.
E = Pt = 2.4 × 25 × 60 = 3600 J	✓ 1	Or E = IVt = 0.40 × 6.0 × 1500. The time must be converted to 1500 s.

Answer: 3600 J

(2)

(e)(i)

Figure 2 shows an end view of the coil of the hub motor between two magnetic poles. The coil is free to turn about its centre.

Use Fleming's left-hand rule to state the direction of the force on side A of the coil. Hence state whether the coil turns clockwise or anticlockwise, as seen in Figure 2.

Figure 2: end view of the motor coil between the poles

Extended onlyPREDICTNotes

Show solution

force on side A is downwards (towards the bottom of the page)	✓ 1	First finger (field) to the left, second finger (current) out of the page, thumb down.
anticlockwise, because the force on side B is upwards	✓ 1	Allow ECF: a consistent turning direction from an upward force on A (clockwise).

Answer: downwards; anticlockwise

(2)

(e)(ii)

Explain why the motor needs a split-ring commutator to keep turning in the same direction.

EXPLAINNotes

Show solution

without it, after half a turn the forces on the sides would push the coil back, so it would only swing to and fro	✓ 1	Accept “the coil would stop at the vertical position or rock back”.
the commutator reverses the current in the coil every half turn, so the force on each side keeps turning the coil the same way	✓ 1

(2)

(f)

The magnets in the dynamo are made of a hard magnetic material, but the core of the dynamo coil is made of iron, a soft magnetic material.

Explain why each material is chosen.

EXPLAINNotes

Show solution

hard magnetic material keeps its magnetism (is difficult to demagnetise), so the magnets stay magnetised for years	✓ 1
soft iron is easily magnetised and demagnetised, so the magnetism of the core can change quickly as the magnets pass the coil	✓ 1	Accept “iron loses its magnetism easily, so it follows the changing field”.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
when the wheel turns, the magnets move relative to the coil, so the magnetic field through the coil changes (field lines are cut) and an e.m.f. is induced	✓ 1	The idea of relative movement or a changing field is needed.
when the bicycle is still, the field through the coil does not change, so no e.m.f. is induced	✓ 1	Not enough: “the magnets are not moving” without reference to the field through the coil not changing.
Part (b)
use stronger magnets	✓ 1
use more turns on the coil	✓ 1	Accept “put the magnets closer to the coil” as one of the two changes. Not accepted: “turn the wheel faster”, because the speed is the same, or “add an iron core”, because the coil already has one.
Part (c)
an alternating wave with peaks of 6.0 V (twice the original peak of 3.0 V)	✓ 1
a period of 30 ms, half the original, so twice as many cycles are shown (4 cycles in 120 ms)	✓ 1	Both features must be clear for the two marks; one feature correct scores 1.
Part (d)(i)
I = V / R = 6.0 / 15	✓ 1
= 0.40 A	✓ 1	Unit required.
Part (d)(ii)
P = IV = 0.40 × 6.0 = 2.4 W	✓ 1	Allow ECF from (d)(i) for this mark only.
E = Pt = 2.4 × 25 × 60 = 3600 J	✓ 1	Or E = IVt = 0.40 × 6.0 × 1500. The time must be converted to 1500 s.
Part (e)(i)
force on side A is downwards (towards the bottom of the page)	✓ 1	First finger (field) to the left, second finger (current) out of the page, thumb down.
anticlockwise, because the force on side B is upwards	✓ 1	Allow ECF: a consistent turning direction from an upward force on A (clockwise).
Part (e)(ii)
without it, after half a turn the forces on the sides would push the coil back, so it would only swing to and fro	✓ 1	Accept “the coil would stop at the vertical position or rock back”.
the commutator reverses the current in the coil every half turn, so the force on each side keeps turning the coil the same way	✓ 1
Part (f)
hard magnetic material keeps its magnetism (is difficult to demagnetise), so the magnets stay magnetised for years	✓ 1
soft iron is easily magnetised and demagnetised, so the magnetism of the core can change quickly as the magnets pass the coil	✓ 1	Accept “iron loses its magnetism easily, so it follows the changing field”.

Answers: (c) peak 6.0 V, period 30 ms · (d)(i) 0.40 A · (d)(ii) 3600 J · (e)(i) downwards; anticlockwise

Exam tip (a) Always link the e.m.f. to movement or change: “the magnet is moving, so the field through the coil is changing”. A magnet that is simply near a coil induces nothing.

More 4.5 questions in the question bank

Topic 5 · Nuclear physics

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

Setup	5.1	5.2	Other
Inside a nuclear power station: fission, control rods and spent fuel	●	○	1.7
Fusion: an experimental reactor and the stars	●	○	6.2 1.7
A medical tracer: technetium-99m in a bone scan	○	●	1.1
Radioactive sources in a paper mill: a beta thickness gauge and alpha smoke detectors	○	●	3.3

5.1 Atomic structure, isotopes, fission and fusion

+ Add to test✓ In your testCC-CIP-51-119 marks~20 min

Inside a nuclear power station: fission, control rods and spent fuel

5.1 Atomic structure, isotopes, fission and fusion5.2 Radioactivity, half-life and safety1.7 Energy, work and powerCore

The reactor of a nuclear power station contains fuel rods of uranium. Most of the uranium is uranium-238, but the energy comes mainly from the fission of uranium-235, ²³⁵₉₂U.

Control rods can be raised out of the reactor core or lowered into it. Figure 1 shows how the thermal power produced in the reactor changes when the operators partly withdraw (raise) the control rods.

Figure 1: thermal power output of the reactor; the control rods are partly withdrawn at time 10 min

(a)(i)

State the number of protons and the number of neutrons in a nucleus of uranium-235, and the number of electrons in a neutral atom of uranium-235.

RECALLNotes

Show solution

protons = 92 and electrons = 92	✓ 1	Both needed for this mark.
neutrons = 235 − 92 = 143	✓ 1

Answer: 92 protons, 143 neutrons, 92 electrons

(2)

(a)(ii)

Uranium-235 and uranium-238 are isotopes of uranium. Explain what this means, giving the numbers of particles in the nuclei.

EXPLAINNotes

Show solution

both nuclei have the same number of protons (92) / same proton number, so both are uranium	✓ 1
different numbers of neutrons: 143 in uranium-235 and 146 in uranium-238	✓ 1	Allow ECF for the uranium-235 value from (a)(i). Accept “uranium-238 has 3 more neutrons”.

(2)

(b)

Describe what happens when a nucleus of uranium-235 undergoes fission.

RECALLNotes

Show solution

a (slow-moving) neutron is absorbed by the uranium-235 nucleus	✓ 1
the nucleus splits into two smaller nuclei / two fission fragments	✓ 1
two or three neutrons and energy are released	✓ 1	Accept “energy as kinetic energy of the fragments”.

(3)

(c)

One possible fission of uranium-235 produces a xenon nucleus, a strontium nucleus and two neutrons:

²³⁵₉₂U + ¹₀n → ¹⁴⁰₅₄Xe + ^?_?Sr + 2 ¹₀n

Determine the nucleon number and the proton number of the strontium nucleus.

Extended onlyCALCNotes

Show solution

nucleon number = 235 + 1 − 140 − 2 = 94	✓ 1
proton number = 92 − 54 = 38	✓ 1

Answer: ⁹⁴₃₈Sr

(2)

(d)

Each kilogram of uranium-235 that undergoes fission releases 8.0 × 10¹³ J of thermal energy.

Use Figure 1 to show that, once the power is steady after t = 20 min, about 3 kg of uranium-235 undergoes fission each day.

MULTINotes

Show solution

energy per day E = Pt = 3000 × 10⁶ × 24 × 3600 = 2.6 × 10¹⁴ J	✓ 1	Power must be read as 3000 MW and converted to W; time in s.
mass = 2.6 × 10¹⁴ ÷ 8.0 × 10¹³ = 3.2 kg (≈ 3 kg)	✓ 1	Final value to at least 2 s.f. needed, as the answer is given.

Answer: 3.2 kg

(2)

(e)(i)

Use Figure 1 to determine the increase in the thermal power after the control rods are partly withdrawn.

GRAPHNotes

Show solution

3000 − 2400 = 600 MW

✓ 1

Answer: 600 MW

(1)

(e)(ii)

Explain why partly withdrawing the control rods causes the increase in power between t = 10 min and t = 20 min.

EXPLAINNotes

Show solution

control rods absorb neutrons, so with the rods raised fewer neutrons are absorbed	✓ 1
more neutrons go on to cause further fissions, so the rate of fission (energy released per second) increases	✓ 1	Accept “more fissions per second”. Not “the reaction is faster” alone.

(2)

(e)(iii)

After t = 20 min the operators adjust the control rods so that the power stays constant. State what is then true, on average, about the neutrons released in each fission.

REASONNotes

Show solution

(on average) only one neutron from each fission goes on to cause another fission

✓ 1

Accept “the number of fissions per second stays the same because one neutron per fission causes a new fission”.

(1)

(f)

When the thermal power is 3000 MW, the generators of the power station deliver 1020 MW of electrical power.

Calculate the efficiency of the power station.

Extended onlyCALCNotes

Show solution

efficiency = useful power output ÷ total power input (× 100%) = 1020 ÷ 3000	✓ 1
= 0.34 or 34%	✓ 1	An answer above 100% scores 0.

Answer: 34%

(2)

(g)

The fission fragments in used (spent) fuel rods are radioactive, and some have half-lives of thousands of years.

An engineer suggests heating the spent fuel to make it decay faster, so that it can be stored for a shorter time. Explain why this would not work, and why the spent fuel must be stored behind thick shielding.

EXPLAINNotes

Show solution

radioactive decay is spontaneous / is not affected by temperature, so heating does not change the half-life or the rate of decay	✓ 1
the fuel emits ionising radiation (such as gamma) that penetrates and can damage cells, so it must be shielded (and stored for a very long time)	✓ 1	Accept “gamma passes through most materials, so thick lead/concrete/water is needed”.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
protons = 92 and electrons = 92	✓ 1	Both needed for this mark.
neutrons = 235 − 92 = 143	✓ 1
Part (a)(ii)
both nuclei have the same number of protons (92) / same proton number, so both are uranium	✓ 1
different numbers of neutrons: 143 in uranium-235 and 146 in uranium-238	✓ 1	Allow ECF for the uranium-235 value from (a)(i). Accept “uranium-238 has 3 more neutrons”.
Part (b)
a (slow-moving) neutron is absorbed by the uranium-235 nucleus	✓ 1
the nucleus splits into two smaller nuclei / two fission fragments	✓ 1
two or three neutrons and energy are released	✓ 1	Accept “energy as kinetic energy of the fragments”.
Part (c)
nucleon number = 235 + 1 − 140 − 2 = 94	✓ 1
proton number = 92 − 54 = 38	✓ 1
Part (d)
energy per day E = Pt = 3000 × 10⁶ × 24 × 3600 = 2.6 × 10¹⁴ J	✓ 1	Power must be read as 3000 MW and converted to W; time in s.
mass = 2.6 × 10¹⁴ ÷ 8.0 × 10¹³ = 3.2 kg (≈ 3 kg)	✓ 1	Final value to at least 2 s.f. needed, as the answer is given.
Part (e)(i)
3000 − 2400 = 600 MW	✓ 1
Part (e)(ii)
control rods absorb neutrons, so with the rods raised fewer neutrons are absorbed	✓ 1
more neutrons go on to cause further fissions, so the rate of fission (energy released per second) increases	✓ 1	Accept “more fissions per second”. Not “the reaction is faster” alone.
Part (e)(iii)
(on average) only one neutron from each fission goes on to cause another fission	✓ 1	Accept “the number of fissions per second stays the same because one neutron per fission causes a new fission”.
Part (f)
efficiency = useful power output ÷ total power input (× 100%) = 1020 ÷ 3000	✓ 1
= 0.34 or 34%	✓ 1	An answer above 100% scores 0.
Part (g)
radioactive decay is spontaneous / is not affected by temperature, so heating does not change the half-life or the rate of decay	✓ 1
the fuel emits ionising radiation (such as gamma) that penetrates and can damage cells, so it must be shielded (and stored for a very long time)	✓ 1	Accept “gamma passes through most materials, so thick lead/concrete/water is needed”.

Answers: (a)(i) 92 protons, 143 neutrons, 92 electrons · (c) ⁹⁴₃₈Sr · (d) 3.2 kg · (e)(i) 600 MW · (f) 34%

Common error (a)(i) The nucleon number 235 is not the number of neutrons: it counts the protons as well, so subtract the proton number.

Exam tip (c) Count the neutrons on both sides: one neutron is absorbed on the left and two are released on the right, so the nucleon numbers are 236 on each side.

More 5.1 questions in the question bank

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Fusion: an experimental reactor and the stars

5.1 Atomic structure, isotopes, fission and fusion6.2 Stars, galaxies and the expanding Universe5.2 Radioactivity, half-life and safety1.7 Energy, work and powerCore

An experimental fusion reactor heats a gas of deuterium (hydrogen-2) and tritium (hydrogen-3) to about 150 million °C. The nuclei fuse in the reaction

²₁H + ³₁H → ⁴₂He + ¹₀n + energy

Each reaction releases 2.8 × 10⁻¹² J of energy.

(a)(i)

State the number of protons and the number of neutrons in a nucleus of deuterium and in a nucleus of tritium.

RECALLNotes

Show solution

deuterium: 1 proton and 1 neutron	✓ 1
tritium: 1 proton and 2 neutrons	✓ 1

Answer: deuterium 1 p, 1 n; tritium 1 p, 2 n

(2)

(a)(ii)

Deuterium and tritium are isotopes of hydrogen. Explain why they react chemically in the same way, even though their atoms have different masses.

EXPLAINNotes

Show solution

same number of protons, so a neutral atom of each has the same number of electrons (one)	✓ 1
chemical behaviour depends on the electrons; the different mass is due to the different number of neutrons	✓ 1	Both ideas needed for this mark.

(2)

(b)

Explain why fusion only happens at an extremely high temperature.

EXPLAINNotes

Show solution

nuclei are (all) positively charged, so they repel each other	✓ 1
to get close enough to fuse they must move very fast, and a very high temperature means very fast-moving nuclei	✓ 1

(2)

(c)

In one test, 1.8 × 10²⁰ fusion reactions take place each second.

Show that the power released by the fusion reactions is about 500 MW.

CALCNotes

Show solution

energy per second = 1.8 × 10²⁰ × 2.8 × 10⁻¹²	✓ 1
= 5.0 × 10⁸ J/s = 5.0 × 10⁸ W (≈ 500 MW)	✓ 1	Accept 5.04 × 10⁸ W. The answer must be seen in W or MW, to at least 2 s.f.

Answer: 5.0 × 10⁸ W

(2)

(d)

During the test the temperature of the gas in the reactor drops. Predict what happens to the power released by fusion, and explain your answer.

PREDICTNotes

Show solution

the power decreases	✓ 1
the nuclei move more slowly, so fewer get close enough to fuse / fewer fusion reactions each second	✓ 1	The mark needs the link to the number of reactions per second.

(2)

(e)(i)

Tritium is radioactive and decays by emitting beta particles.

State what a beta particle is, and name a material that stops beta particles.

RECALLNotes

Show solution

a fast-moving electron (emitted from the nucleus)	✓ 1
a few mm of aluminium	✓ 1	Do not accept paper. Accept a thicker sheet of another metal, e.g. lead.

(2)

(e)(ii)

Tritium decays to an isotope of helium. Write the nuclide equation for the decay of tritium.

Extended onlyCALCNotes

Show solution

³₁H → ³₂He + …	✓ 1	Helium nucleus with nucleon number 3 and proton number 2.
… + ⁰₋₁e	✓ 1	Beta particle with nucleon number 0 and proton number −1. Accept β for e if the numbers are shown.

Answer: ³₁H → ³₂He + ⁰₋₁e

(2)

(e)(iii)

The laboratory stores its tritium in a sealed container. Figure 2 shows how the mass of tritium in the container decreases with time.

Use Figure 2 to determine the half-life of tritium. Show on the graph how you obtained your answer.

Figure 2: mass of tritium remaining in a stored sample against time

GRAPHNotes

Show solution

reads a value and the time at which it has halved, e.g. 8.0 g at 0 and 4.0 g at 12 years (or 4.0 g at 12 and 2.0 g at 24 years)	✓ 1
half-life = 12 years	✓ 1	Accept 11–13 years.

Answer: 12 years

(2)

(f)

The Sun releases its energy by fusing hydrogen nuclei, without any heating by people. Describe how a cloud of gas in space becomes a star in which hydrogen fusion takes place.

EXPLAINNotes

Show solution

a nebula (cloud of gas and dust) is pulled together by gravity	✓ 1
it collapses into a protostar, which becomes denser and hotter as gravitational (potential) energy is transferred to thermal energy	✓ 1
when the core is hot and dense enough, hydrogen fusion starts and it becomes a (main-sequence) star	✓ 1	Accept “high enough temperature and pressure for nuclei to fuse”.

(3)

Show full solution

Marking point	Mark	Notes
Part (a)(i)
deuterium: 1 proton and 1 neutron	✓ 1
tritium: 1 proton and 2 neutrons	✓ 1
Part (a)(ii)
same number of protons, so a neutral atom of each has the same number of electrons (one)	✓ 1
chemical behaviour depends on the electrons; the different mass is due to the different number of neutrons	✓ 1	Both ideas needed for this mark.
Part (b)
nuclei are (all) positively charged, so they repel each other	✓ 1
to get close enough to fuse they must move very fast, and a very high temperature means very fast-moving nuclei	✓ 1
Part (c)
energy per second = 1.8 × 10²⁰ × 2.8 × 10⁻¹²	✓ 1
= 5.0 × 10⁸ J/s = 5.0 × 10⁸ W (≈ 500 MW)	✓ 1	Accept 5.04 × 10⁸ W. The answer must be seen in W or MW, to at least 2 s.f.
Part (d)
the power decreases	✓ 1
the nuclei move more slowly, so fewer get close enough to fuse / fewer fusion reactions each second	✓ 1	The mark needs the link to the number of reactions per second.
Part (e)(i)
a fast-moving electron (emitted from the nucleus)	✓ 1
a few mm of aluminium	✓ 1	Do not accept paper. Accept a thicker sheet of another metal, e.g. lead.
Part (e)(ii)
³₁H → ³₂He + …	✓ 1	Helium nucleus with nucleon number 3 and proton number 2.
… + ⁰₋₁e	✓ 1	Beta particle with nucleon number 0 and proton number −1. Accept β for e if the numbers are shown.
Part (e)(iii)
reads a value and the time at which it has halved, e.g. 8.0 g at 0 and 4.0 g at 12 years (or 4.0 g at 12 and 2.0 g at 24 years)	✓ 1
half-life = 12 years	✓ 1	Accept 11–13 years.
Part (f)
a nebula (cloud of gas and dust) is pulled together by gravity	✓ 1
it collapses into a protostar, which becomes denser and hotter as gravitational (potential) energy is transferred to thermal energy	✓ 1
when the core is hot and dense enough, hydrogen fusion starts and it becomes a (main-sequence) star	✓ 1	Accept “high enough temperature and pressure for nuclei to fuse”.

Answers: (a)(i) deuterium 1 p, 1 n; tritium 1 p, 2 n · (c) 5.0 × 10⁸ W · (e)(ii) ³₁H → ³₂He + ⁰₋₁e · (e)(iii) 12 years

More 5.1 questions in the question bank

5.2 Radioactivity, half-life and safety

+ Add to test✓ In your testCC-CIP-52-118 marks~20 min

A medical tracer: technetium-99m in a bone scan

5.2 Radioactivity, half-life and safety5.1 Atomic structure, isotopes, fission and fusion1.1 Physical quantities and measurementCore

A hospital uses technetium-99m as a tracer for bone scans. A small amount is injected into a patient, it collects in the bones, and a gamma camera outside the body detects the radiation. Technetium-99m emits gamma radiation only, and its half-life is 6.0 hours.

Before a delivery of tracer is used, a technician places a small sample of it in front of a GM tube and records the count rate over the next 30 hours. Figure 1 shows the measured count rate. Earlier, with no sample present, the technician measured the background count rate as 40 counts per minute.

Figure 1: measured count rate from the sample of tracer against time (background not subtracted)

(a)

Explain why a source that emits gamma radiation and has a half-life of a few hours is suitable as a tracer inside the body.

EXPLAINNotes

Show solution

gamma passes out through the body to reach the detector/camera	✓ 1	Accept “alpha or beta would be absorbed in the body and not detected”.
the half-life is short, so the activity falls quickly and the patient is exposed to radiation for only a short time	✓ 1
but the half-life is long enough for the source to be detected while the scan is carried out	✓ 1

(3)

(b)

Write the half-life of technetium-99m in seconds, in standard form, to an appropriate number of significant figures.

CALCNotes

Show solution

6.0 × 3600 = 21 600 s	✓ 1
2.2 × 10⁴ s	✓ 1	Must be standard form and 2 s.f. (to match 6.0). 2.16 × 10⁴ s scores the first mark only.

Answer: 2.2 × 10⁴ s

(2)

(c)

Use Figure 1 to determine the half-life of the sample, and so confirm the value of 6.0 hours. Show your working clearly.

GRAPHNotes

Show solution

subtracts background: corrected count rate at t = 0 is 1320 − 40 = 1280 counts per minute	✓ 1
half of this is 640, so looks for a measured count rate of 640 + 40 = 680 counts per minute	✓ 1
reads time = 6.0 hours	✓ 1	Accept 5.8–6.2 hours. Halving 1320 without subtracting the background (660 counts per minute, about 6.3 hours) scores 0 for this part. Any other pair of readings is equally valid if the background is subtracted from both, e.g. 680 counts per minute at 6.0 h and 360 counts per minute at 12 h (corrected 640 → 320).

Answer: 6.0 hours

(3)

(d)

Show that the corrected count rate from the sample falls to 80 counts per minute after 24 hours, and state the fraction of the technetium-99m nuclei that remain at this time.

CALCNotes

Show solution

number of half-lives = 24 ÷ 6.0 = 4; 1280 → 640 → 320 → 160 → 80 counts per minute	✓ 1	Halving steps or 1280 × (1/2)⁴ must be seen.
fraction remaining = 1/16	✓ 1

Answer: 80 counts per minute; 1/16

(2)

(e)

One week, the delivery of tracer from the supplier is delayed by 12 hours. The patient must still receive the same amount of technetium-99m at the time of injection.

Determine by what factor the amount sent by the supplier must be increased, and explain your answer.

REASONNotes

Show solution

12 hours is 2 half-lives, so the amount falls to 1/4 during the delay	✓ 1
so the amount sent must be 4 times larger (× 4)	✓ 1	Answer of × 2 (halving once) scores 0.

Answer: 4 times

(2)

(f)

Describe two precautions that the technician should take to reduce the radiation received while preparing the injections.

RECALLNotes

Show solution

shielding: keep the tracer in a lead-lined container / use a lead syringe shield / work behind a lead or concrete screen	✓ 1	Any two different points from shielding, distance and time, one mark each.
distance: handle with tongs or keep sources pointing away; or time: spend as little time as possible near the source	✓ 1	Accept wearing a film badge to monitor the dose as one point.

(2)

(g)

Technetium has proton number 43. State the number of protons and the number of neutrons in a nucleus of technetium-99, and write its nuclide notation.

RECALLNotes

Show solution

43 protons and 99 − 43 = 56 neutrons	✓ 1
⁹⁹₄₃Tc	✓ 1	Nucleon number on top, proton number below.

Answer: 43 protons, 56 neutrons; ⁹⁹₄₃Tc

(2)

(h)

When technetium-99m emits a gamma ray it becomes technetium-99. State what happens to the nucleon number and to the proton number of the nucleus in gamma emission, and explain why the element does not change.

Extended onlyEXPLAINNotes

Show solution

nucleon number and proton number both stay the same	✓ 1
gamma emission only carries away surplus energy from the nucleus; the proton number decides the element, so it is still technetium	✓ 1	Both ideas needed.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
gamma passes out through the body to reach the detector/camera	✓ 1	Accept “alpha or beta would be absorbed in the body and not detected”.
the half-life is short, so the activity falls quickly and the patient is exposed to radiation for only a short time	✓ 1
but the half-life is long enough for the source to be detected while the scan is carried out	✓ 1
Part (b)
6.0 × 3600 = 21 600 s	✓ 1
2.2 × 10⁴ s	✓ 1	Must be standard form and 2 s.f. (to match 6.0). 2.16 × 10⁴ s scores the first mark only.
Part (c)
subtracts background: corrected count rate at t = 0 is 1320 − 40 = 1280 counts per minute	✓ 1
half of this is 640, so looks for a measured count rate of 640 + 40 = 680 counts per minute	✓ 1
reads time = 6.0 hours	✓ 1	Accept 5.8–6.2 hours. Halving 1320 without subtracting the background (660 counts per minute, about 6.3 hours) scores 0 for this part. Any other pair of readings is equally valid if the background is subtracted from both, e.g. 680 counts per minute at 6.0 h and 360 counts per minute at 12 h (corrected 640 → 320).
Part (d)
number of half-lives = 24 ÷ 6.0 = 4; 1280 → 640 → 320 → 160 → 80 counts per minute	✓ 1	Halving steps or 1280 × (1/2)⁴ must be seen.
fraction remaining = 1/16	✓ 1
Part (e)
12 hours is 2 half-lives, so the amount falls to 1/4 during the delay	✓ 1
so the amount sent must be 4 times larger (× 4)	✓ 1	Answer of × 2 (halving once) scores 0.
Part (f)
shielding: keep the tracer in a lead-lined container / use a lead syringe shield / work behind a lead or concrete screen	✓ 1	Any two different points from shielding, distance and time, one mark each.
distance: handle with tongs or keep sources pointing away; or time: spend as little time as possible near the source	✓ 1	Accept wearing a film badge to monitor the dose as one point.
Part (g)
43 protons and 99 − 43 = 56 neutrons	✓ 1
⁹⁹₄₃Tc	✓ 1	Nucleon number on top, proton number below.
Part (h)
nucleon number and proton number both stay the same	✓ 1
gamma emission only carries away surplus energy from the nucleus; the proton number decides the element, so it is still technetium	✓ 1	Both ideas needed.

Answers: (b) 2.2 × 10⁴ s · (c) 6.0 hours · (d) 80 counts per minute; 1/16 · (e) 4 times · (g) 43 protons, 56 neutrons; ⁹⁹₄₃Tc

Common error (c) Halving the measured count rate (1320 → 660) gives the wrong half-life, because the GM tube counts the background as well. Subtract the background first, halve, then add the background back before reading the graph.

More 5.2 questions in the question bank

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Radioactive sources in a paper mill: a beta thickness gauge and alpha smoke detectors

5.2 Radioactivity, half-life and safety5.1 Atomic structure, isotopes, fission and fusion3.3 The electromagnetic spectrumCore

A paper mill makes card that should be 0.80 mm thick. The card passes between rollers and then between a beta source (strontium-90, half-life 29 years) and a GM tube, as shown in Figure 1. The counter and control unit uses the count rate to adjust the gap between the rollers.

The mill also has smoke detectors that each contain a small alpha source.

Figure 1: the thickness gauge on the card production line (not to scale)

(a)

Complete the table.

radiation	what it is	ionising effect	stopped by
alpha	helium nucleus	…	a sheet of paper
beta	…	medium	…
gamma	…	weakest	reduced by thick lead

RECALLNotes

Show solution

alpha: strongest (most) ionising	✓ 1
beta: fast-moving electron (from the nucleus) and stopped by a few mm of aluminium	✓ 1	Both needed.
gamma: electromagnetic radiation/wave (of very high frequency)	✓ 1

(3)

(b)(i)

Explain why a gamma source would not be suitable for the gauge in Figure 1.

REASONNotes

Show solution

gamma passes through the card almost completely, so the count rate would hardly change when the thickness changes

✓ 1

(1)

(b)(ii)

Gamma rays are at the high-frequency end of the electromagnetic spectrum. A gamma ray has a frequency of 3.0 × 10²⁰ Hz. The speed of electromagnetic waves is 3.0 × 10⁸ m/s.

Calculate the wavelength of the gamma ray, and name the region of the spectrum that lies next to gamma rays.

CALCNotes

Show solution

λ = v ÷ f = 3.0 × 10⁸ ÷ 3.0 × 10²⁰ = 1.0 × 10⁻¹² m	✓ 1
X-rays	✓ 1

Answer: 1.0 × 10⁻¹² m; X-rays

(2)

(c)

Figure 2 shows how the corrected count rate depends on the thickness of the card between the source and the GM tube.

During production the corrected count rate rises to 360 counts per second. Use Figure 2 to determine the thickness of the card now passing the gauge, and state how the control unit should change the gap between the rollers.

Figure 2: corrected count rate at the GM tube against the thickness of card between the source and the tube

GRAPHNotes

Show solution

reads thickness at 360 counts per second from the graph	✓ 1	Line or mark on the graph, or value stated.
thickness = 0.66 mm	✓ 1	Accept 0.62–0.70 mm.
card is too thin, so the gap between the rollers must be increased (rollers moved apart)	✓ 1	Allow ECF only for a thickness less than 0.80 mm.

Answer: 0.66 mm; increase the gap

(3)

(d)

The mill changes to making cardboard 2.4 mm thick, using the same source and GM tube. Use Figure 2 to explain why the gauge cannot control the thickness of the cardboard as precisely as it controlled the 0.80 mm card.

EXPLAINNotes

Show solution

the graph is less steep at 2.4 mm: the same change in thickness gives a much smaller change in count rate	✓ 1	Accept a quoted comparison, e.g. 0.10 mm changes the count rate by about 28 counts per second at 0.80 mm but only about 7 at 2.4 mm.
decay is random, so the count rate fluctuates; at the low count rate (about 80 counts per second) small changes in thickness are hidden by these fluctuations	✓ 1

(2)

(e)

When the gauge was new, the corrected count rate with no card in place was 640 counts per second.

Show that, after 58 years, this count rate falls to 160 counts per second, and explain why a source with a long half-life is chosen for the gauge.

CALCNotes

Show solution

58 ÷ 29 = 2 half-lives; 640 → 320 → 160 counts per second	✓ 1	Working must be seen.
long half-life: the count rate hardly changes over months, so changes are caused by thickness only / the source does not need replacing for many years	✓ 1

Answer: 160 counts per second

(2)

(f)(i)

Strontium-90 is radioactive, but strontium-88 is stable. Strontium has proton number 38.

State one similarity and one difference between a nucleus of strontium-90 and a nucleus of strontium-88.

RECALLNotes

Show solution

similarity: both have 38 protons	✓ 1
difference: strontium-90 has 52 neutrons and strontium-88 has 50 neutrons	✓ 1	Accept “strontium-90 has 2 more neutrons” or “different nucleon numbers”.

(2)

(f)(ii)

Strontium-90 decays by beta emission to yttrium (Y). Write the nuclide equation for this decay.

Extended onlyCALCNotes

Show solution

⁹⁰₃₈Sr → ⁹⁰₃₉Y + …	✓ 1	Nucleon number unchanged; proton number up by 1.
… + ⁰₋₁e	✓ 1	Accept β with 0 and −1 shown.

Answer: ⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁e

(2)

(g)

Workers are concerned about the alpha sources in the smoke detectors on the ceiling. Explain why these sources are not a hazard to people in the room, and state when an alpha source would be dangerous.

EXPLAINNotes

Show solution

alpha particles travel only a few cm in air / are stopped by the case of the detector or by the skin, so they do not reach people	✓ 1
dangerous if the source is swallowed or breathed in, because it is then in contact with living cells (and is strongly ionising)	✓ 1

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
alpha: strongest (most) ionising	✓ 1
beta: fast-moving electron (from the nucleus) and stopped by a few mm of aluminium	✓ 1	Both needed.
gamma: electromagnetic radiation/wave (of very high frequency)	✓ 1
Part (b)(i)
gamma passes through the card almost completely, so the count rate would hardly change when the thickness changes	✓ 1
Part (b)(ii)
λ = v ÷ f = 3.0 × 10⁸ ÷ 3.0 × 10²⁰ = 1.0 × 10⁻¹² m	✓ 1
X-rays	✓ 1
Part (c)
reads thickness at 360 counts per second from the graph	✓ 1	Line or mark on the graph, or value stated.
thickness = 0.66 mm	✓ 1	Accept 0.62–0.70 mm.
card is too thin, so the gap between the rollers must be increased (rollers moved apart)	✓ 1	Allow ECF only for a thickness less than 0.80 mm.
Part (d)
the graph is less steep at 2.4 mm: the same change in thickness gives a much smaller change in count rate	✓ 1	Accept a quoted comparison, e.g. 0.10 mm changes the count rate by about 28 counts per second at 0.80 mm but only about 7 at 2.4 mm.
decay is random, so the count rate fluctuates; at the low count rate (about 80 counts per second) small changes in thickness are hidden by these fluctuations	✓ 1
Part (e)
58 ÷ 29 = 2 half-lives; 640 → 320 → 160 counts per second	✓ 1	Working must be seen.
long half-life: the count rate hardly changes over months, so changes are caused by thickness only / the source does not need replacing for many years	✓ 1
Part (f)(i)
similarity: both have 38 protons	✓ 1
difference: strontium-90 has 52 neutrons and strontium-88 has 50 neutrons	✓ 1	Accept “strontium-90 has 2 more neutrons” or “different nucleon numbers”.
Part (f)(ii)
⁹⁰₃₈Sr → ⁹⁰₃₉Y + …	✓ 1	Nucleon number unchanged; proton number up by 1.
… + ⁰₋₁e	✓ 1	Accept β with 0 and −1 shown.
Part (g)
alpha particles travel only a few cm in air / are stopped by the case of the detector or by the skin, so they do not reach people	✓ 1
dangerous if the source is swallowed or breathed in, because it is then in contact with living cells (and is strongly ionising)	✓ 1

Answers: (b)(ii) 1.0 × 10⁻¹² m; X-rays · (c) 0.66 mm; increase the gap · (e) 160 counts per second · (f)(ii) ⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁e

More 5.2 questions in the question bank

Topic 6 · Space physics

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

Setup	6.1	6.2	Other
The International Space Station: orbit, sunrises and radio links	●		1.4 3.3
Planning a Mars rover mission: orbits, signals, weight and landing	●		1.3 1.2
A galaxy redshift survey: Hubble’s law and the age of the Universe	○	●	3.3
From nebula to supernova: the lives of stars	○	●	5.1

6.1 Earth, Moon and the Solar System

+ Add to test✓ In your testCC-CIP-61-118 marks~20 min

The International Space Station: orbit, sunrises and radio links

6.1 Earth, Moon and the Solar System1.4 Forces, friction and springs3.3 The electromagnetic spectrumCore

The International Space Station (ISS) moves around the Earth in a circular orbit at a height of 400 km above the Earth’s surface. The radius of the Earth is 6370 km. The ISS takes 92 minutes to complete one orbit.

Figure 1 shows the ISS in its orbit, with sunlight arriving from the Sun on the left.

Figure 1: the ISS in orbit around the Earth (not to scale)

(a)

State the name of the force that keeps the ISS in its orbit, and state the direction of this force.

RECALLNotes

Show solution

gravitational force / gravity (of the Earth) / weight	✓ 1	Do not accept “gravity of the Sun”.
towards the centre of the Earth	✓ 1	Accept: towards the Earth / inwards to the centre of the orbit. Do not accept “downwards” alone.

Answer: gravity of the Earth, towards the centre of the Earth

(2)

(b)

The ISS moves at a constant speed. Explain why its velocity is not constant, and why a force must act on it.

REASONNotes

Show solution

velocity includes direction, and the direction of motion of the ISS is changing all the time	✓ 1	Accept: it moves in a circle, so its direction changes.
a (resultant) force is needed to change the direction of motion; with no force the ISS would move in a straight line	✓ 1	Accept: zero resultant force would mean constant velocity.

(2)

(c)

Calculate the orbital speed of the ISS in m/s.

CALCNotes

Show solution

r = 6370 + 400 = 6770 km = 6.77 × 10⁶ m	✓ 1	The radius is measured from the centre of the Earth.
T = 92 × 60 = 5520 s and v = 2πr ÷ T = 2 × π × 6.77 × 10⁶ ÷ 5520	✓ 1	Allow ECF from a wrong radius (e.g. 6.37 × 10⁶ m gives 7.3 × 10³ m/s: max 2; 4.0 × 10⁵ m gives 455 m/s: max 1).
v = 7.7 × 10³ m/s	✓ 1	Accept 7700 m/s or 7710 m/s. 7.7 m/s (radius left in km) or 128 m/s (period in minutes) loses this mark.

Answer: 7.7 × 10³ m/s

(3)

(d)

Show that the ISS completes about 16 orbits in one day.

CALCNotes

Show solution

one day = 24 × 60 = 1440 minutes (or 86 400 s)	✓ 1
number of orbits = 1440 ÷ 92 = 15.7	✓ 1	The unrounded value (15.6 or 15.7) must be seen; “16” alone scores 0 for this mark.

Answer: 15.7 orbits

(2)

(e)

On the Earth, a person sees the Sun rise once every 24 hours. An astronaut on the ISS sees about 16 sunrises every 24 hours.

Use Figure 1 to explain this difference.

EXPLAINNotes

Show solution

on the Earth, a sunrise happens once per rotation of the Earth on its axis, which takes 24 hours	✓ 1	Accept: the Earth spins once in a day, carrying the person from the night side into the day side.
the ISS goes from the night side (the side facing away from the Sun, in the Earth’s shadow) back into sunlight once every orbit, and it completes about 16 orbits in 24 hours	✓ 1	Link to one sunrise per orbit needed. Allow ECF from (d).

(2)

(f)(i)

Figure 2 shows how the orbital period of a satellite depends on the radius of its orbit.

Describe how the orbital period changes as the orbit radius increases.

Figure 2: orbital period against orbit radius for satellites orbiting the Earth (radius measured from the centre of the Earth)

GRAPHNotes

Show solution

the period increases as the radius increases, and it increases more and more steeply / not in proportion

✓ 1

Both ideas needed: increase AND not proportional (curve gets steeper).

(1)

(f)(ii)

A weather satellite is to take exactly 24 hours to complete one orbit.

Use Figure 2 to find the radius of its orbit.

GRAPHNotes

Show solution

42 000 km

✓ 1

Accept 41 000 km to 43 000 km. Accept 42 thousand km. Unit required.

Answer: 42 000 km

(1)

(g)

Thin air at the height of the ISS slowly lowers its orbit. Every few months the engines of the ISS are fired for a short time in a reboost, which moves the ISS into a circular orbit of slightly larger radius.

State and explain the effect of the reboost on the orbital period of the ISS.

PREDICTNotes

Show solution

the orbital period increases / each orbit takes longer	✓ 1
the ISS has further to travel in one orbit (larger circumference) and moves more slowly in a larger orbit	✓ 1	Accept either reason: further to travel / slower speed / weaker gravitational pull further from the Earth. Accept a reference to Figure 2 (period increases with radius) as the reason.

Answer: the period increases

(2)

(h)

The ISS sends a radio signal to a ground station. At one moment, the straight-line distance between the ISS and the ground station is 2100 km.

State the speed of radio waves in space and calculate the time taken for the signal to reach the ground station.

CALCNotes

Show solution

speed = 3.0 × 10⁸ m/s	✓ 1	Accept 300 000 km/s.
t = d ÷ v = 2.1 × 10⁶ ÷ 3.0 × 10⁸	✓ 1	Distance must be in metres (or speed in km/s).
t = 7.0 × 10⁻³ s	✓ 1	Accept 0.007 s or 7 ms. Allow ECF for a wrong speed if stated in the first mark.

Answer: 7.0 × 10⁻³ s

(3)

Show full solution

Marking point	Mark	Notes
Part (a)
gravitational force / gravity (of the Earth) / weight	✓ 1	Do not accept “gravity of the Sun”.
towards the centre of the Earth	✓ 1	Accept: towards the Earth / inwards to the centre of the orbit. Do not accept “downwards” alone.
Part (b)
velocity includes direction, and the direction of motion of the ISS is changing all the time	✓ 1	Accept: it moves in a circle, so its direction changes.
a (resultant) force is needed to change the direction of motion; with no force the ISS would move in a straight line	✓ 1	Accept: zero resultant force would mean constant velocity.
Part (c)
r = 6370 + 400 = 6770 km = 6.77 × 10⁶ m	✓ 1	The radius is measured from the centre of the Earth.
T = 92 × 60 = 5520 s and v = 2πr ÷ T = 2 × π × 6.77 × 10⁶ ÷ 5520	✓ 1	Allow ECF from a wrong radius (e.g. 6.37 × 10⁶ m gives 7.3 × 10³ m/s: max 2; 4.0 × 10⁵ m gives 455 m/s: max 1).
v = 7.7 × 10³ m/s	✓ 1	Accept 7700 m/s or 7710 m/s. 7.7 m/s (radius left in km) or 128 m/s (period in minutes) loses this mark.
Part (d)
one day = 24 × 60 = 1440 minutes (or 86 400 s)	✓ 1
number of orbits = 1440 ÷ 92 = 15.7	✓ 1	The unrounded value (15.6 or 15.7) must be seen; “16” alone scores 0 for this mark.
Part (e)
on the Earth, a sunrise happens once per rotation of the Earth on its axis, which takes 24 hours	✓ 1	Accept: the Earth spins once in a day, carrying the person from the night side into the day side.
the ISS goes from the night side (the side facing away from the Sun, in the Earth’s shadow) back into sunlight once every orbit, and it completes about 16 orbits in 24 hours	✓ 1	Link to one sunrise per orbit needed. Allow ECF from (d).
Part (f)(i)
the period increases as the radius increases, and it increases more and more steeply / not in proportion	✓ 1	Both ideas needed: increase AND not proportional (curve gets steeper).
Part (f)(ii)
42 000 km	✓ 1	Accept 41 000 km to 43 000 km. Accept 42 thousand km. Unit required.
Part (g)
the orbital period increases / each orbit takes longer	✓ 1
the ISS has further to travel in one orbit (larger circumference) and moves more slowly in a larger orbit	✓ 1	Accept either reason: further to travel / slower speed / weaker gravitational pull further from the Earth. Accept a reference to Figure 2 (period increases with radius) as the reason.
Part (h)
speed = 3.0 × 10⁸ m/s	✓ 1	Accept 300 000 km/s.
t = d ÷ v = 2.1 × 10⁶ ÷ 3.0 × 10⁸	✓ 1	Distance must be in metres (or speed in km/s).
t = 7.0 × 10⁻³ s	✓ 1	Accept 0.007 s or 7 ms. Allow ECF for a wrong speed if stated in the first mark.

Answers: (a) gravity of the Earth, towards the centre of the Earth · (c) 7.7 × 10³ m/s · (d) 15.7 orbits · (f)(ii) 42 000 km · (g) the period increases · (h) 7.0 × 10⁻³ s

Common error (c) Using the height above the surface as the orbit radius. The radius r in v = 2πr/T is measured from the centre of the Earth, so add the Earth’s radius; then write r in metres and T in seconds before substituting.

More 6.1 questions in the question bank

+ Add to test✓ In your testCC-CIP-61-220 marks~20 min

Planning a Mars rover mission: orbits, signals, weight and landing

6.1 Earth, Moon and the Solar System1.3 Mass, weight and density1.2 Speed, velocity and motion graphsCore

A space agency is planning to land a rover on Mars. The planning team compares data for the Earth and Mars.

	Earth	Mars
average radius of orbit around the Sun / m	1.50 × 10¹¹	2.28 × 10¹¹
average orbital speed / km/s	29.8	24.1
tilt of axis	23.5°	25.2°
gravitational field strength at the surface / N/kg	9.8	3.7

(a)

The Earth’s axis is tilted at 23.5°. In June the northern hemisphere is tilted towards the Sun.

Explain why the northern hemisphere is warmer in June than in December.

EXPLAINNotes

Show solution

the Sun is higher in the sky, so the sunlight is more concentrated / falls on a smaller area and each square metre receives more energy	✓ 1	Accept: sunlight strikes the ground more directly.
the days are longer (nights shorter), so there are more hours of sunlight to heat the ground	✓ 1	Do not award any mark for “the Earth is closer to the Sun in June”.

(2)

(b)(i)

Signals from the rover are passed on to the Earth by a relay satellite. The satellite orbits Mars in a circle of radius 3.79 × 10⁶ m, measured from the centre of Mars, at a speed of 3.36 km/s.

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

CALCNotes

Show solution

T = 2πr ÷ v = 2 × π × 3.79 × 10⁶ ÷ 3360	✓ 1	Speed must be converted to 3360 m/s (or radius to km).
T = 7090 s	✓ 1	Accept 7087 s.
÷ 60 = 118 minutes	✓ 1	Allow ECF from a wrong value of T in seconds.

Answer: 118 minutes

(3)

(b)(ii)

Mars has seasons, like the Earth. Use the data in the table to suggest why each season on Mars lasts almost twice as long as the same season on the Earth.

REASONNotes

Show solution

Mars has a larger orbit (further to travel) and a lower orbital speed, so it takes longer to orbit the Sun / a year on Mars is longer	✓ 1	Both the larger radius and the lower speed from the table needed.
one full cycle of seasons takes one orbit, so each season lasts longer	✓ 1	Accept: Mars has a similar tilt, so it has the same pattern of seasons, spread over a longer year.

(2)

(c)(i)

At its closest, Mars is 5.6 × 10¹⁰ m from the Earth. Commands are sent to the rover by radio waves.

Show that a command takes about 3 minutes to reach Mars at this distance.

CALCNotes

Show solution

t = d ÷ v = 5.6 × 10¹⁰ ÷ 3.0 × 10⁸ = 187 s	✓ 1
187 ÷ 60 = 3.1 minutes	✓ 1	The value 3.1 min (or 187 s with the division by 60 seen) must be shown.

Answer: 187 s = 3.1 min

(2)

(c)(ii)

When the rover lands, Mars is on the far side of the Sun, 4.0 × 10¹¹ m from the Earth. The descent through the Martian atmosphere takes about 7 minutes.

Calculate the time for a radio signal to reach Mars at this distance, and suggest why the landing must be controlled by the rover’s own computer.

PREDICTNotes

Show solution

t = 4.0 × 10¹¹ ÷ 3.0 × 10⁸ = 1300 s (22 minutes)	✓ 1	Accept 1330 s, 1333 s or 22.2 min.
the signal takes longer than the whole descent (7 minutes), so a command from the Earth would arrive after the rover had landed / could not respond to what is happening	✓ 1	Must compare the delay with the descent time.

Answer: 1.3 × 10³ s (22 min)

(2)

(d)

The rover has a mass of 1025 kg.

Calculate the weight of the rover on Mars.

CALCNotes

Show solution

W = mg = 1025 × 3.7	✓ 1	Use of 9.8 N/kg scores 0.
W = 3800 N	✓ 1	Accept 3790 N or 3792.5 N. Unit required.

Answer: 3.8 × 10³ N

(2)

(e)(i)

Figure 1 shows the gravitational field strength g at the surface of four planets and the Moon, plotted against their mass.

Describe the general trend shown by Figure 1.

Figure 1: gravitational field strength at the surface against mass for four planets and the Moon

GRAPHNotes

Show solution

the greater the mass, the greater the gravitational field strength at the surface

✓ 1

Accept a correct comparison of two named bodies only if a general statement is made.

(1)

(e)(ii)

Mars has about twice the mass of Mercury, but the gravitational field strength at the surface is the same, 3.7 N/kg. The radius of Mercury is 2440 km and the radius of Mars is 3390 km.

Suggest why the gravitational field strength at the surface of Mars is not greater than at the surface of Mercury.

Extended onlyREASONNotes

Show solution

Mars has a larger radius, so its surface is further from its centre	✓ 1
gravitational field strength decreases as the distance from the centre of the planet increases, which cancels the effect of its larger mass	✓ 1	Accept: gravity gets weaker with distance.

(2)

(f)(i)

Figure 2 shows the speed of the descending rover during the final stage of its landing. From t = 10 s, rocket engines slow the rover down.

Calculate the deceleration of the rover between t = 10 s and t = 36 s.

Figure 2: speed-time graph for the final stage of the descent

GRAPHNotes

Show solution

deceleration = change in speed ÷ time = (80 − 2.0) ÷ (36 − 10)	✓ 1	Readings from the graph: 80 m/s, 2.0 m/s, 26 s.
= 3.0 m/s²	✓ 1	Accept −3.0 m/s² if called an acceleration. Unit required.

Answer: 3.0 m/s²

(2)

(f)(ii)

Use Figure 2 to calculate the distance travelled by the rover between t = 10 s and t = 46 s.

GRAPHNotes

Show solution

area from 10 s to 36 s = ½ × (80 + 2.0) × 26 = 1066 m, and area from 36 s to 46 s = 2.0 × 10 = 20 m	✓ 1	Accept the trapezium split into a triangle (1014 m) and a rectangle (52 m).
total = 1090 m	✓ 1	Accept 1086 m. Allow ECF for one arithmetic slip in one area.

Answer: 1.09 × 10³ m

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
the Sun is higher in the sky, so the sunlight is more concentrated / falls on a smaller area and each square metre receives more energy	✓ 1	Accept: sunlight strikes the ground more directly.
the days are longer (nights shorter), so there are more hours of sunlight to heat the ground	✓ 1	Do not award any mark for “the Earth is closer to the Sun in June”.
Part (b)(i)
T = 2πr ÷ v = 2 × π × 3.79 × 10⁶ ÷ 3360	✓ 1	Speed must be converted to 3360 m/s (or radius to km).
T = 7090 s	✓ 1	Accept 7087 s.
÷ 60 = 118 minutes	✓ 1	Allow ECF from a wrong value of T in seconds.
Part (b)(ii)
Mars has a larger orbit (further to travel) and a lower orbital speed, so it takes longer to orbit the Sun / a year on Mars is longer	✓ 1	Both the larger radius and the lower speed from the table needed.
one full cycle of seasons takes one orbit, so each season lasts longer	✓ 1	Accept: Mars has a similar tilt, so it has the same pattern of seasons, spread over a longer year.
Part (c)(i)
t = d ÷ v = 5.6 × 10¹⁰ ÷ 3.0 × 10⁸ = 187 s	✓ 1
187 ÷ 60 = 3.1 minutes	✓ 1	The value 3.1 min (or 187 s with the division by 60 seen) must be shown.
Part (c)(ii)
t = 4.0 × 10¹¹ ÷ 3.0 × 10⁸ = 1300 s (22 minutes)	✓ 1	Accept 1330 s, 1333 s or 22.2 min.
the signal takes longer than the whole descent (7 minutes), so a command from the Earth would arrive after the rover had landed / could not respond to what is happening	✓ 1	Must compare the delay with the descent time.
Part (d)
W = mg = 1025 × 3.7	✓ 1	Use of 9.8 N/kg scores 0.
W = 3800 N	✓ 1	Accept 3790 N or 3792.5 N. Unit required.
Part (e)(i)
the greater the mass, the greater the gravitational field strength at the surface	✓ 1	Accept a correct comparison of two named bodies only if a general statement is made.
Part (e)(ii)
Mars has a larger radius, so its surface is further from its centre	✓ 1
gravitational field strength decreases as the distance from the centre of the planet increases, which cancels the effect of its larger mass	✓ 1	Accept: gravity gets weaker with distance.
Part (f)(i)
deceleration = change in speed ÷ time = (80 − 2.0) ÷ (36 − 10)	✓ 1	Readings from the graph: 80 m/s, 2.0 m/s, 26 s.
= 3.0 m/s²	✓ 1	Accept −3.0 m/s² if called an acceleration. Unit required.
Part (f)(ii)
area from 10 s to 36 s = ½ × (80 + 2.0) × 26 = 1066 m, and area from 36 s to 46 s = 2.0 × 10 = 20 m	✓ 1	Accept the trapezium split into a triangle (1014 m) and a rectangle (52 m).
total = 1090 m	✓ 1	Accept 1086 m. Allow ECF for one arithmetic slip in one area.

Answers: (b)(i) 118 minutes · (c)(i) 187 s = 3.1 min · (c)(ii) 1.3 × 10³ s (22 min) · (d) 3.8 × 10³ N · (f)(i) 3.0 m/s² · (f)(ii) 1.09 × 10³ m

Exam tip (a) For the seasons, link the tilt to its two effects: a higher Sun (more concentrated sunlight) and longer days. Never use the Earth–Sun distance as the reason.

More 6.1 questions in the question bank

6.2 Stars, galaxies and the expanding Universe

+ Add to test✓ In your testCC-CIP-62-120 marks~20 min

A galaxy redshift survey: Hubble’s law and the age of the Universe

6.2 Stars, galaxies and the expanding Universe3.3 The electromagnetic spectrum6.1 Earth, Moon and the Solar SystemExtended

An astronomy team studies six distant galaxies, A to F. For each galaxy they measure its distance from the Earth, and they use the redshift of its light to find its speed of recession. The table shows their results. The speed of light is 3.0 × 10⁸ m/s.

galaxy	distance from the Earth / 10²⁴ m	speed of recession / 10⁶ m/s
A	0.40	0.95
B	1.00	2.3
C	1.60	3.6
D	2.00	3.2
E	2.40	see (c)
F	3.20	7.4

(a)

State what is meant by a galaxy.

RECALLNotes

Show solution

a very large group of (billions of) stars held together by gravity

✓ 1

Both “large group of stars” and “gravity” needed.

(1)

(b)

Explain what is meant by redshift, and state what it shows about the motion of the galaxies.

EXPLAINNotes

Show solution

the light received from the galaxy has a longer wavelength than the light emitted by the same element in the laboratory / the spectral lines are shifted towards the red end	✓ 1	Accept: increase in observed wavelength (or decrease in frequency). Do not accept “the galaxy looks red”.
the galaxies are moving away from the Earth (receding)	✓ 1	Accept: the Universe is expanding.

(2)

(c)

A spectral line of calcium has a wavelength of 393.4 nm when measured in the laboratory. In the light from galaxy E, the same line is observed at 400.6 nm.

Calculate the speed of recession of galaxy E.

Extended onlyCALCNotes

Show solution

Δλ = 400.6 − 393.4 = 7.2 nm	✓ 1
v = c × Δλ ÷ λ = 3.0 × 10⁸ × 7.2 ÷ 393.4	✓ 1	Using 400.6 nm for λ gives 5.39 × 10⁶ m/s: allow this mark.
v = 5.5 × 10⁶ m/s	✓ 1	Accept 5.49 × 10⁶ m/s. Using 400.6 nm in place of Δλ scores 0 for the first and last marks.

Answer: 5.5 × 10⁶ m/s

(3)

(d)

Figure 1 shows the results for galaxies A, B, C, D and F.

On Figure 1, plot the point for galaxy E and draw the line of best fit. The line should pass through the origin.

Figure 1: speed of recession against distance for galaxies A, B, C, D and F

Extended onlyGRAPHNotes

Show solution

point E plotted at (2.40, 5.5) to within half a small square	✓ 1	Allow ECF from (c).
straight line through the origin passing close to A, B, C, E and F, ignoring D	✓ 1	Line should pass through (3.0, 6.9) ± 0.2. A line drawn through D scores 0.

(2)

(e)

One galaxy does not fit the pattern of the other results.

Identify this galaxy and explain how you treated its result when drawing the line of best fit.

Extended onlyREASONNotes

Show solution

galaxy D	✓ 1
it lies well below (off) the line followed by the other points, so it is treated as anomalous and ignored when drawing the line	✓ 1	Accept: its result should be checked / measured again. “It is wrong” alone is not enough.

Answer: galaxy D

(2)

(f)

Hubble’s law states that v = H₀d.

Use your line of best fit to determine a value for the Hubble constant H₀.

Extended onlyGRAPHNotes

Show solution

gradient = Δv ÷ Δd using a large triangle on the line, with the powers of ten included, e.g. 6.9 × 10⁶ ÷ 3.0 × 10²⁴	✓ 1	Triangle should cover at least half of the line drawn. Using a single data point (not on the line) scores max 1.
H₀ = 2.3 × 10⁻¹⁸ s⁻¹	✓ 1	Accept 2.1 × 10⁻¹⁸ to 2.5 × 10⁻¹⁸. Unit: s⁻¹ or /s. Allow ECF from their line.

Answer: 2.3 × 10⁻¹⁸ s⁻¹

(2)

(g)

Assume that galaxy F started from the same place as our galaxy at the Big Bang and has moved away at a constant speed ever since. One year is 3.15 × 10⁷ s.

Show that the data for galaxy F give an age for the Universe of about 1.4 × 10¹⁰ years.

Extended onlyCALCNotes

Show solution

t = d ÷ v = 3.20 × 10²⁴ ÷ 7.4 × 10⁶ = 4.3 × 10¹⁷ s	✓ 1	Accept 4.32 × 10¹⁷ s. Accept 1 ÷ H₀ using their gradient from (f) as an alternative route.
4.32 × 10¹⁷ ÷ 3.15 × 10⁷ = 1.37 × 10¹⁰ years	✓ 1	The unrounded value must be seen; 1.4 × 10¹⁰ alone scores 0 for this mark.

Answer: 1.37 × 10¹⁰ years

(2)

(h)(i)

The team then measures a new galaxy, G, which is 4.5 × 10²⁴ m from the Earth.

Use H₀ = 2.3 × 10⁻¹⁸ s⁻¹ to calculate the speed at which galaxy G is moving away.

Extended onlyPREDICTNotes

Show solution

v = H₀d = 2.3 × 10⁻¹⁸ × 4.5 × 10²⁴	✓ 1
v = 1.0 × 10⁷ m/s	✓ 1	Accept 1.04 × 10⁷ m/s. Unit required.

Answer: 1.0 × 10⁷ m/s

(2)

(h)(ii)

One light-year is 9.5 × 10¹⁵ m.

Calculate the distance of galaxy G in light-years, and state what this tells you about the light from galaxy G that reaches the Earth today.

CALCNotes

Show solution

4.5 × 10²⁴ ÷ 9.5 × 10¹⁵ = 4.7 × 10⁸ light-years	✓ 1	Accept 470 million light-years.
the light left galaxy G about 4.7 × 10⁸ years ago, so we see the galaxy as it was then	✓ 1	Allow ECF from the first mark. Do not accept “the galaxy is 470 million years old”.

Answer: 4.7 × 10⁸ light-years

(2)

(i)

Redshift is one piece of evidence for the Big Bang model. Name one other piece of evidence, state the region of the electromagnetic spectrum in which it is detected, and explain why it is detected in this region today.

EXPLAINNotes

Show solution

cosmic microwave background radiation (CMBR), detected as microwaves	✓ 1	Both the name and the region needed.
radiation from the very hot early Universe has had its wavelength stretched as the Universe expanded, so its wavelength is now long, in the microwave region	✓ 1	Accept: the radiation was originally of short wavelength / high energy.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
a very large group of (billions of) stars held together by gravity	✓ 1	Both “large group of stars” and “gravity” needed.
Part (b)
the light received from the galaxy has a longer wavelength than the light emitted by the same element in the laboratory / the spectral lines are shifted towards the red end	✓ 1	Accept: increase in observed wavelength (or decrease in frequency). Do not accept “the galaxy looks red”.
the galaxies are moving away from the Earth (receding)	✓ 1	Accept: the Universe is expanding.
Part (c)
Δλ = 400.6 − 393.4 = 7.2 nm	✓ 1
v = c × Δλ ÷ λ = 3.0 × 10⁸ × 7.2 ÷ 393.4	✓ 1	Using 400.6 nm for λ gives 5.39 × 10⁶ m/s: allow this mark.
v = 5.5 × 10⁶ m/s	✓ 1	Accept 5.49 × 10⁶ m/s. Using 400.6 nm in place of Δλ scores 0 for the first and last marks.
Part (d)
point E plotted at (2.40, 5.5) to within half a small square	✓ 1	Allow ECF from (c).
straight line through the origin passing close to A, B, C, E and F, ignoring D	✓ 1	Line should pass through (3.0, 6.9) ± 0.2. A line drawn through D scores 0.
Part (e)
galaxy D	✓ 1
it lies well below (off) the line followed by the other points, so it is treated as anomalous and ignored when drawing the line	✓ 1	Accept: its result should be checked / measured again. “It is wrong” alone is not enough.
Part (f)
gradient = Δv ÷ Δd using a large triangle on the line, with the powers of ten included, e.g. 6.9 × 10⁶ ÷ 3.0 × 10²⁴	✓ 1	Triangle should cover at least half of the line drawn. Using a single data point (not on the line) scores max 1.
H₀ = 2.3 × 10⁻¹⁸ s⁻¹	✓ 1	Accept 2.1 × 10⁻¹⁸ to 2.5 × 10⁻¹⁸. Unit: s⁻¹ or /s. Allow ECF from their line.
Part (g)
t = d ÷ v = 3.20 × 10²⁴ ÷ 7.4 × 10⁶ = 4.3 × 10¹⁷ s	✓ 1	Accept 4.32 × 10¹⁷ s. Accept 1 ÷ H₀ using their gradient from (f) as an alternative route.
4.32 × 10¹⁷ ÷ 3.15 × 10⁷ = 1.37 × 10¹⁰ years	✓ 1	The unrounded value must be seen; 1.4 × 10¹⁰ alone scores 0 for this mark.
Part (h)(i)
v = H₀d = 2.3 × 10⁻¹⁸ × 4.5 × 10²⁴	✓ 1
v = 1.0 × 10⁷ m/s	✓ 1	Accept 1.04 × 10⁷ m/s. Unit required.
Part (h)(ii)
4.5 × 10²⁴ ÷ 9.5 × 10¹⁵ = 4.7 × 10⁸ light-years	✓ 1	Accept 470 million light-years.
the light left galaxy G about 4.7 × 10⁸ years ago, so we see the galaxy as it was then	✓ 1	Allow ECF from the first mark. Do not accept “the galaxy is 470 million years old”.
Part (i)
cosmic microwave background radiation (CMBR), detected as microwaves	✓ 1	Both the name and the region needed.
radiation from the very hot early Universe has had its wavelength stretched as the Universe expanded, so its wavelength is now long, in the microwave region	✓ 1	Accept: the radiation was originally of short wavelength / high energy.

Answers: (c) 5.5 × 10⁶ m/s · (e) galaxy D · (f) 2.3 × 10⁻¹⁸ s⁻¹ · (g) 1.37 × 10¹⁰ years · (h)(i) 1.0 × 10⁷ m/s · (h)(ii) 4.7 × 10⁸ light-years

Common error (c) Δλ is the change in wavelength (observed minus laboratory), not the observed wavelength. Keep both wavelengths in nm so the units cancel.

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From nebula to supernova: the lives of stars

6.2 Stars, galaxies and the expanding Universe5.1 Atomic structure, isotopes, fission and fusion6.1 Earth, Moon and the Solar SystemCore

A planetarium show follows the lives of two stars that form in the same nebula. One star has the same mass as the Sun. The other has ten times the mass of the Sun.

(a)

Describe how a star forms from a nebula.

RECALLNotes

Show solution

a nebula is a (huge) cloud of gas (mostly hydrogen) and dust	✓ 1
gravity pulls parts of the cloud together; the clump collapses to form a protostar	✓ 1	Accept: the gas is pulled together by gravitational attraction.
the protostar becomes hotter (and denser) until the core is hot enough for (hydrogen) fusion to start	✓ 1	Accept: gravitational energy is transferred to thermal energy.

(3)

(b)

In the core of a main-sequence star, hydrogen nuclei join to form helium nuclei.

Name this process, and explain why it can only happen at a very high temperature.

EXPLAINNotes

Show solution

(nuclear) fusion	✓ 1	Do not accept fission or burning.
nuclei are positively charged, so they repel each other	✓ 1
only at a very high temperature do the nuclei move fast enough to get close enough (to touch) and fuse	✓ 1	Link between temperature and speed of the nuclei needed.

Answer: nuclear fusion

(3)

(c)

The presenter says that a main-sequence star is ‘in a state of balance’.

State the two effects that are balanced, and suggest what would happen to the size of the star if the outward effect became greater than the inward effect.

REASONNotes

Show solution

inward: the pull of gravity on the gas of the star	✓ 1
outward: the pressure of the hot gas and radiation produced by fusion in the core	✓ 1	Accept: outward pressure from fusion energy.
the star would expand / swell (get larger)	✓ 1	Accept: and its surface would cool.

(3)

(d)(i)

Figure 1 shows how long stars of different masses spend on the main sequence.

Describe the pattern shown by Figure 1.

Figure 1: the time that stars of different masses spend on the main sequence

GRAPHNotes

Show solution

the more massive the star, the shorter the time it spends on the main sequence

✓ 1

Accept: lifetime decreases (rapidly) as mass increases.

(1)

(d)(ii)

The Sun is about 4.6 billion years old. Use Figure 1 to estimate how much longer the Sun will stay on the main sequence.

GRAPHNotes

Show solution

main-sequence lifetime of a star with the mass of the Sun read as 10 billion years	✓ 1
10 − 4.6 = 5.4 billion years	✓ 1	Allow ECF from their reading. Accept about 5 billion years if the working is shown.

Answer: about 5.4 billion years

(2)

(e)

When the hydrogen in its core is used up, the star of one solar mass becomes a red giant and then a white dwarf.

Describe the stages that follow the main sequence for the star with ten times the mass of the Sun, and state how its main-sequence lifetime compares with the stars in Figure 1.

PREDICTNotes

Show solution

it becomes a red supergiant	✓ 1	Do not accept red giant.
it explodes as a supernova, leaving a neutron star or a black hole	✓ 1	Both the supernova and one correct remnant needed.
its main-sequence lifetime is shorter than that of any star in Figure 1 / shorter than 0.18 billion years	✓ 1	Accept: much shorter than for the 5.0 solar-mass star.

(3)

(f)

The show then turns to Sirius, the brightest star in the night sky, which is 8.1 × 10¹⁶ m from the Earth. The speed of light is 3.0 × 10⁸ m/s and one year is 3.15 × 10⁷ s.

Show that light from Sirius takes about 8.6 years to reach the Earth.

CALCNotes

Show solution

t = d ÷ v = 8.1 × 10¹⁶ ÷ 3.0 × 10⁸ = 2.7 × 10⁸ s	✓ 1
2.7 × 10⁸ ÷ 3.15 × 10⁷ = 8.57 years	✓ 1	The unrounded value (8.57) must be seen.

Answer: 8.57 years

(2)

(g)

The Earth contains elements such as carbon, oxygen and iron, which are much heavier than hydrogen and helium.

Explain what this suggests about stars that existed before the Solar System formed.

EXPLAINNotes

Show solution

heavy elements were made by (nuclear) fusion inside earlier stars	✓ 1	Accept: only hydrogen and helium were made in the early Universe.
these stars exploded as supernovae, scattering the elements into the nebula from which the Sun and the Earth later formed	✓ 1	Accept: the Sun is a second-generation star formed from the remains of earlier stars.

(2)

Show full solution

Marking point	Mark	Notes
Part (a)
a nebula is a (huge) cloud of gas (mostly hydrogen) and dust	✓ 1
gravity pulls parts of the cloud together; the clump collapses to form a protostar	✓ 1	Accept: the gas is pulled together by gravitational attraction.
the protostar becomes hotter (and denser) until the core is hot enough for (hydrogen) fusion to start	✓ 1	Accept: gravitational energy is transferred to thermal energy.
Part (b)
(nuclear) fusion	✓ 1	Do not accept fission or burning.
nuclei are positively charged, so they repel each other	✓ 1
only at a very high temperature do the nuclei move fast enough to get close enough (to touch) and fuse	✓ 1	Link between temperature and speed of the nuclei needed.
Part (c)
inward: the pull of gravity on the gas of the star	✓ 1
outward: the pressure of the hot gas and radiation produced by fusion in the core	✓ 1	Accept: outward pressure from fusion energy.
the star would expand / swell (get larger)	✓ 1	Accept: and its surface would cool.
Part (d)(i)
the more massive the star, the shorter the time it spends on the main sequence	✓ 1	Accept: lifetime decreases (rapidly) as mass increases.
Part (d)(ii)
main-sequence lifetime of a star with the mass of the Sun read as 10 billion years	✓ 1
10 − 4.6 = 5.4 billion years	✓ 1	Allow ECF from their reading. Accept about 5 billion years if the working is shown.
Part (e)
it becomes a red supergiant	✓ 1	Do not accept red giant.
it explodes as a supernova, leaving a neutron star or a black hole	✓ 1	Both the supernova and one correct remnant needed.
its main-sequence lifetime is shorter than that of any star in Figure 1 / shorter than 0.18 billion years	✓ 1	Accept: much shorter than for the 5.0 solar-mass star.
Part (f)
t = d ÷ v = 8.1 × 10¹⁶ ÷ 3.0 × 10⁸ = 2.7 × 10⁸ s	✓ 1
2.7 × 10⁸ ÷ 3.15 × 10⁷ = 8.57 years	✓ 1	The unrounded value (8.57) must be seen.
Part (g)
heavy elements were made by (nuclear) fusion inside earlier stars	✓ 1	Accept: only hydrogen and helium were made in the early Universe.
these stars exploded as supernovae, scattering the elements into the nebula from which the Sun and the Earth later formed	✓ 1	Accept: the Sun is a second-generation star formed from the remains of earlier stars.

Answers: (b) nuclear fusion · (d)(ii) about 5.4 billion years · (f) 8.57 years

Exam tip (f) In a “show that” calculation, give your answer to one more significant figure than the value in the question, so that it is clear you have not just copied it.

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