Momentum and changes in momentum (Higher) – GCSE Physics exam questions with mark scheme

1Grade 6–7HigherTriple only7 marks

Trolley A has a mass of 2.0 kg and moves at 3.0 m/s towards trolley B. Trolley B has a mass of 1.0 kg and is stationary. The trolleys collide and stick together.

(a) Calculate the momentum of trolley A before the collision. [2]

(b) Calculate the velocity of the trolleys immediately after the collision. [3]

(a) p = 2.0 × 3.0(1)
(a) 6.0 kg m/s(1)
(b) total momentum after = total momentum before = 6.0 kg m/s (allow ecf from (a))(1)
(b) 6.0 = (2.0 + 1.0) × v(1)
(b) v = 2.0 m/s (allow 2 m/s; direction not required)(1)
(c) KE before = ½ × 2.0 × 3.0² = 9.0 J(1)
(c) KE after = ½ × 3.0 × 2.0² = 6.0 J, which is less than 9.0 J (allow ecf from (b))(1)
Answer: (a) 6.0 kg m/s (b) 2.0 m/s (c) 9.0 J before, 6.0 J after
Examiner tip: After the trolleys stick together, use their COMBINED mass of 3.0 kg.

2Grade 6–7HigherTriple only5 marks

In a crash, a passenger of mass 70 kg moving at 15 m/s is brought to rest by an airbag. The airbag stops the passenger in 0.15 s.

(a) Calculate the average force exerted on the passenger. [3]

(b) Explain how the airbag reduces the force on the passenger. [2]

(a) change in momentum = 70 × 15 = 1050 kg m/s (allow F = 70 × 15 ÷ 0.15 for the first two marks; allow a = 15 ÷ 0.15 = 100 m/s² for this mark)(1)
(a) F = 1050 ÷ 0.15 (allow F = 70 × 100)(1)
(a) 7000 N (allow 7.0 × 10³ N; a correct answer with no working scores 3 marks)(1)
(b) increases the time taken for the passenger to stop (allow: increases the time of the impact/collision)(1)
(b) so the rate of change of momentum is smaller, so the force is smaller (allow: same change in momentum over a longer time; allow: so the deceleration is smaller, so the force is smaller (F = ma); ignore 'cushions' / 'softer')(1)
Answer: (a) 7000 N (b) longer stopping time → smaller rate of change of momentum → smaller force
Examiner tip: Say 'rate of change of momentum' — 'it cushions the passenger' scores nothing.

3Grade 4–5Higher5 marks

A cricket ball has a mass of 0.16 kg. A bowler throws the ball at a velocity of 25 m/s towards the batter.

(a) Write down the equation that links mass ($m$), momentum ($p$) and velocity ($v$). [1]

(b) Calculate the momentum of the ball. Give the unit. [3]

How does the momentum of the ball now compare with its momentum in part (b)? [1]

(a) momentum = mass × velocity (allow $p = m\,v$; allow any correct rearrangement)(1)
(b) $p = 0.16 \times 25$(1)
(b) 4.0 (allow 4)(1)
(b) kg m/s (allow kg m s$^{-1}$ or N s; the unit mark is independent of the calculation)(1)
(c) same size / 4.0 kg m/s and in the opposite direction (allow: −4.0 kg m/s; do not accept ‘the same’ alone)(1)
Answer: (a) $p = mv$ (b) 4.0 kg m/s (c) same size, opposite direction
Examiner tip: Momentum has a direction (it is a vector), so the ball travelling back the other way at the same speed has momentum of the same size but opposite direction – just writing ‘the same’ loses the mark.

4Grade 4–5Higher5 marks

A man is sitting in a small rowing boat next to a jetty. The boat and the man are both at rest. The man steps out of the boat, forwards onto the jetty. As he does so, the boat moves away from the jetty.

(a) State the principle of conservation of momentum. [1]

(b) Explain, in terms of momentum, why the boat moves away from the jetty. [3]

Suggest why. [1]

(a) in a closed system, the total momentum before an event is equal to the total momentum after the event (allow: total momentum is conserved / stays the same, if ‘closed system’ or ‘no external forces’ is stated; do not accept ‘momentum is conserved’ alone)(1)
(b) (total) momentum before is zero (because the man and the boat are both at rest)(1)
(b) the man gains momentum towards the jetty (allow: the man moves forwards so has momentum)(1)
(b) the boat gains an equal (amount of) momentum in the opposite direction, so the total momentum is still zero (allow: momentum of man and boat are equal and opposite; ignore references to Newton’s third law if the momentum argument is given)(1)
(c) the boat has a smaller mass than the man (so it needs a larger velocity to have the same size of momentum) (allow: the boat is lighter)(1)
Answer: (a) In a closed system total momentum before = total momentum after (b) momentum before = 0; man gains forward momentum; boat gains equal and opposite momentum (c) boat has smaller mass
Examiner tip: Always start a recoil explanation with ‘total momentum before is zero’ – that is the step most students skip, and it is the reason the boat must move in the opposite direction.

5Grade 6–7HigherTriple only4 marks

Two ice skaters, P and Q, stand face to face on an ice rink. They are at rest. They push against each other and move apart in opposite directions.

Skater P has a mass of 64 kg. Skater Q has a mass of 48 kg.

Immediately after they push apart, skater Q moves to the left at 2.0 m/s.

Calculate the velocity of skater P immediately after they push apart. Assume there is no friction between the skates and the ice. [4]

total momentum before = 0 (allow: momentum of P = momentum of Q in size, stated or used)(1)
momentum of Q = $48 \times 2.0 = 96$ (kg m/s)(1)
$64 \times v = 96$ (allow $v = \dfrac{96}{64}$)(1)
1.5 m/s to the right (allow: in the opposite direction to Q; allow −1.5 m/s if the left is clearly taken as positive; direction needed for this mark)(1)
Answer: 1.5 m/s to the right
Examiner tip: The question asks for a velocity, so the direction is part of the answer – ‘1.5 m/s’ on its own loses the last mark.

6Grade 6–7HigherTriple only5 marks

The diagram shows two bumper cars, A and B, at a funfair just before they collide head-on. Each mass includes the rider.

After the collision, car A moves to the left at 0.60 m/s.

Calculate the velocity of car B after the collision. Assume that no external forces act on the cars during the collision. [5]

momentum of A before $= 220 \times 3.0 = 660$ (kg m/s) and momentum of B before $= 180 \times 2.0 = 360$ (kg m/s)(1)
total momentum before $= 660 - 360 = 300$ (kg m/s) (to the right) (allow: correct use of signs for opposite directions)(1)
momentum of A after $= 220 \times 0.60 = 132$ (kg m/s) to the left / $= -132$ (kg m/s)(1)
$300 = -132 + 180\,v_B$ (allow: $180\,v_B = 432$; allow ecf from their total momentum)(1)
2.4 m/s to the right (allow: 2.4 m/s in the direction car A was originally moving; direction needed; an answer of 0.93 m/s from ignoring the sign of A’s momentum after scores max 3)(1)
Answer: 2.4 m/s to the right
Examiner tip: Choose one direction as positive at the start and give every velocity to the left a minus sign – most errors here come from adding 660 and 360 instead of subtracting.

7Grade 8–9HigherTriple only5 marks

A tennis ball of mass 0.058 kg is hit horizontally at a wall. The ball hits the wall at 20 m/s and rebounds along the same line at 15 m/s.

The ball is in contact with the wall for 5.0 milliseconds.

(a) Calculate the size of the change in momentum of the ball. [3]

(b) Calculate the mean force exerted by the wall on the ball. [2]

(a) change in velocity $= 15 - (-20) = 35$ m/s (allow: $15 + 20$)(1)
(a) $\Delta p = 0.058 \times 35$ (allow: $0.058 \times 15 - 0.058 \times (-20)$, i.e. $0.87 + 1.16$)(1)
(a) 2.03 kg m/s (allow 2.0 kg m/s; an answer of 0.29 kg m/s from using 20 − 15 scores 1 mark only if the method is shown)(1)
(b) $F = \dfrac{2.03}{0.0050}$ (allow ecf from (a); time must be converted to seconds)(1)
(b) 406 N (allow 410 N or 400 N; allow ecf: 0.29 kg m/s gives 58 N)(1)
Answer: (a) 2.03 kg m/s (b) 406 N (≈ 410 N)
Examiner tip: When an object bounces back, its velocity changes sign, so the change in velocity is 20 + 15 = 35 m/s, not 20 − 15 = 5 m/s; also convert 5.0 ms to 0.0050 s.

8Grade 6–7HigherTriple only6 marks

Many children’s playgrounds have a thick, soft rubber surface under the climbing equipment instead of concrete.

The rubber surface under the tallest climbing frame is thicker than the surface under the smaller equipment.

Explain how the rubber surface reduces the risk of injury to a child who falls, and why a thicker surface is used under the tallest climbing frame.

Your answer should refer to momentum. [6]

when the child lands, their momentum is reduced to zero (they have a change in momentum)(1)
the change in momentum is the same whether the child lands on concrete or rubber (for the same height of fall)(1)
the rubber squashes / deforms, so the child takes a longer time to stop (allow: stops over a longer distance)(1)
force = rate of change of momentum / $F = \dfrac{m\,\Delta v}{\Delta t}$(1)
so a longer time means a smaller rate of change of momentum and a smaller force on the child(1)
smaller force means less chance of injury (e.g. broken bones)(1)
a child falling from a greater height hits the ground at a greater speed(1)
so the child has more momentum / a larger change in momentum(1)
a thicker surface can squash more, increasing the stopping time further, so the force is kept low (allow: so the force is not too large)(1)
ignore references to energy unless linked to momentum or force; do not credit ‘the rubber absorbs the force’(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, logically linked explanation that uses momentum ideas to explain both why the rubber surface reduces the force on the child and why the surface must be thicker under the tallest frame. The link from a longer stopping time to a smaller rate of change of momentum and a smaller force is clear.
Level 2	3–4	An explanation of how the rubber surface reduces the force, linking a longer stopping time to a smaller force, with some reference to momentum. The thicker surface may be mentioned but is not fully explained, or the chain of reasoning has a gap.
Level 1	1–2	Simple relevant statements, e.g. the rubber squashes so the child takes longer to stop, or the force is smaller. Links between the points are not made clear.
0	0	No relevant content.

Answer: Rubber deforms, so the same change in momentum takes a longer time; smaller rate of change of momentum means smaller force. A higher fall gives a greater speed and greater momentum change, so a thicker surface is needed to make the stopping time longer still.
Examiner tip: The top level needs a full chain: same change in momentum → longer time → smaller rate of change of momentum → smaller force; saying the rubber is ‘soft’ or ‘absorbs the impact’ earns nothing.

9Grade 8–9HigherTriple only6 marks

A student investigated whether momentum is conserved when two trolleys collide on a level track.

Trolley A has a card of length 0.10 m fixed to it. Trolley A is pushed towards trolley B, which is at rest. The trolleys stick together and move off as one.

The student’s results are:

mass of trolley A = 0.60 kg; mass of trolley B = 0.50 kg
time for the card to pass through light gate 1 (before the collision) = 0.040 s
time for the card to pass through light gate 2 (after the collision) = 0.075 s

(a) Calculate the momentum of trolley A before the collision. [2]

(b) Calculate the momentum of the two trolleys after the collision. [2]

Suggest one reason why the total momentum after the collision was less than before. [1]

(d) Give one reason why light gates are used in this investigation instead of a stopwatch. [1]

(a) speed of A $= \dfrac{0.10}{0.040} = 2.5$ m/s(1)
(a) momentum $= 0.60 \times 2.5 = 1.5$ kg m/s (allow ecf from their speed)(1)
(b) speed after $= \dfrac{0.10}{0.075} = 1.33$ m/s (allow 1.3 m/s)(1)
(b) momentum $= 1.10 \times 1.33 = 1.47$ kg m/s (allow 1.46–1.47 kg m/s, or 1.43 kg m/s from 1.3 m/s; mass must be 1.10 kg – using 0.60 kg (0.80 kg m/s) scores 1 mark max)(1)
(c) friction (between the trolleys / wheels and the track) or air resistance acts on the trolleys (allow: there is an external force, so the system is not closed; allow: the track is not perfectly level / slopes upwards; ignore ‘human error’ or ‘energy lost as heat’ alone)(1)
(d) light gates remove (human) reaction time (allow: measure very short times more accurately / precisely; allow: reduce random error in timing; do not accept ‘more accurate’ alone)(1)
Answer: (a) 1.5 kg m/s (b) 1.47 kg m/s (c) friction / air resistance acts, so the system is not closed (d) no reaction time
Examiner tip: After the trolleys stick together, use the COMBINED mass (1.10 kg) for the momentum after – using only trolley A’s mass is the most common slip.

10Grade 6–7HigherTriple only6 marks

Cycle helmets are tested by dropping a model head onto a hard surface. The model head has a mass of 5.0 kg.

The graph shows how the velocity of the model head changes after it hits the surface, with and without a helmet. The graph has been simplified.

(a) Use the graph to determine the time taken for the model head to stop when it is wearing the helmet. Give your answer in seconds. [1]

(b) Calculate the mean force on the model head when it is wearing the helmet.

Use the Physics Equations Sheet. [3]

(c) Without doing a further calculation, compare the mean force on the head without the helmet with your answer to part (b). Explain your answer. [2]

(a) 0.012 s (do not accept 12 or 12 s: the answer must be in seconds)(1)
(b) change in velocity $= 6.0$ m/s (read from graph) (allow: change in momentum $= 5.0 \times 6.0 = 30$ kg m/s)(1)
(b) $F = \dfrac{5.0 \times 6.0}{0.012}$ (allow ecf from (a))(1)
(b) 2500 N (allow ecf from (a): if 12 s was given in (a), 2.5 N scores 3 marks; otherwise 2.5 N from using 12 instead of 0.012 scores 2 marks)(1)
(c) the force without the helmet is 3 times bigger (allow: 7500 N)(1)
(c) the change in momentum is the same and it happens in one third of the time (allow: 0.004 s instead of 0.012 s), so the rate of change of momentum is three times greater(1)
Answer: (a) 0.012 s (b) 2500 N (c) 3 times bigger, because the same change in momentum happens in a third of the time
Examiner tip: Read the time from the graph in milliseconds, then convert to seconds before using F = mΔv/Δt – forgetting the conversion gives a force 1000 times too small.

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