IGCSE Physics 0625 · Topic 1.6

IGCSE Physics Momentum Questions and Worked Solutions

50 topic-wise practice questions on Momentum (Unit 1: Motion, Forces and Energy), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions0 Core40 Extended10 Challenge148 marksTextbook §1.6 · pp. 55–59

Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Momentum

Momentum combines mass and velocity. In a collision or explosion with no external forces, total momentum is conserved. A force acting for a time changes momentum — this is impulse. (This whole subtopic is Extended.)

Key formulas

p = m v	momentum p (kg m/s), mass m (kg), velocity v (m/s)	Extended
impulse = F Δt = Δ(mv)	force × time = change in momentum (N s)	Extended
F = Δp ÷ Δt	resultant force = rate of change of momentum	Extended
total p before = total p after	conservation of momentum	Extended

Prerequisite knowledge

Velocity as a vector (1.1, 1.2)
F = ma (1.5.1)
Kinetic energy (1.7.1) for comparing collisions

Exam tips & common mistakes

Use a negative sign for velocities in the opposite direction
For rebounds, Δv = final − initial with signs (e.g. −6 − 8 = −14 m/s)
Kinetic energy is usually NOT conserved in collisions, but momentum always is

Key facts from the syllabus (1.6)

Momentum is a vector — direction matters; choose one direction as positive
Momentum is conserved in all collisions and explosions if no external forces act
Objects that start at rest and push apart have equal and opposite momenta
Increasing the time of impact reduces the force (airbags, crumple zones, crash mats, catching with moving hands)
1 N s = 1 kg m/s

Textbook: §1.6 Momentum, pp. 55–59

Extended[1 mark]

Define momentum.

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Method: p = mv.

Mass × velocity

Extended[2 marks]

Calculate the momentum of a 1200 kg car moving at 15 m/s.

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Method: p = mv.

p = 1200 × 15 = 18000 kg m/s

Extended[2 marks]

A 0.045 kg golf ball has momentum 3.15 kg m/s. Calculate its velocity.

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Method: v = p ÷ m.

v = 3.15 ÷ 0.045 = 70.0 m/s

Extended[1 mark]

Define impulse.

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Method: Impulse = FΔt.

Force × time for which the force acts

Extended[2 marks]

A 50 N force acts on a ball for 0.12 s. Calculate the impulse.

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Method: Impulse = FΔt.

Impulse = 50 × 0.12 = 6.00 N s

Extended[3 marks]

A 0.40 kg ball travelling at 12 m/s is stopped in 0.080 s. Calculate the average force.

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Method: F = Δp ÷ Δt.

Δp = 0.40 × 12 = 4.80 kg m/s; F = 4.8 ÷ 0.080 = 60.0 N

Extended[1 mark]

State the principle of conservation of momentum.

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Method: Closed system.

Total momentum before an interaction equals total momentum after, if no external forces act

Extended[3 marks]

A 2.0 kg trolley at 3.0 m/s hits a stationary 1.0 kg trolley and they stick together. Calculate their velocity.

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Method: Momentum conserved.

v = (2.0 × 3.0) ÷ 3.0 = 2.00 m/s

Extended[4 marks]

Train A (0.45 kg, 0.34 m/s right) collides with train B (0.21 kg, 0.12 m/s left). They stick together. Calculate their velocity.

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Method: Take right as positive.

Total p = 0.45×0.34 − 0.21×0.12 = 0.128 kg m/s; v = 0.128 ÷ 0.66 = 0.194 m/s to the right

Extended[2 marks]

A train has momentum 0.26 kg m/s and stops in 2.1 s. Calculate the average force on it.

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Method: F = Δp ÷ Δt.

F = 0.26 ÷ 2.1 = 0.124 N

Extended[3 marks]

A 0.16 kg ball hits a wall at 8.0 m/s and rebounds at 6.0 m/s. Calculate the change in momentum.

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Method: Directions opposite: Δp = m(v − u) with u = +8, v = −6.

Δp = 0.16 × (−6.0 − 8.0) = -2.24 kg m/s (magnitude 2.24 kg m/s)

Extended[4 marks]

A 60 kg skater and a 40 kg skater, at rest, push apart. The 60 kg skater moves at 2.0 m/s. Calculate the speed of the 40 kg skater and state the direction.

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Method: Total momentum stays zero.

v = (60 × 2.0) ÷ 40 = 3.00 m/s in the opposite direction

Extended[3 marks]

A gun of mass 3.0 kg fires a 0.010 kg bullet at 400 m/s. Calculate the recoil speed of the gun.

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Method: Momentum conserved from rest.

v = (0.010 × 400) ÷ 3.0 = 1.33 m/s backwards

Extended[2 marks]

Define resultant force in terms of momentum.

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Method: F = Δp/Δt.

The rate of change of momentum

Extended[3 marks]

Explain how a car airbag reduces injury in a crash.

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Method: Same change in momentum over a longer time.

It increases the time to stop, so the rate of change of momentum (force) on the person is smaller

Extended[3 marks]

A 70 kg person moving at 14 m/s is stopped by a seat belt in 0.20 s. Calculate the average force.

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Method: F = mΔv ÷ Δt.

F = (70 × 14) ÷ 0.20 = 4900 N

Extended[4 marks]

A 840 kg spacecraft at 7800 m/s ejects a 120 kg nose cone forwards at 7850 m/s. Calculate the speed of the remaining 720 kg capsule.

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Method: Momentum conserved.

Total p = 6552000 kg m/s; v = (6 552 000 − 120 × 7850) ÷ 720 = 7792 m/s

Extended[3 marks]

A 0.50 kg ball moving at 4.0 m/s hits a 1.5 kg ball at rest. The 0.50 kg ball stops. Calculate the velocity of the 1.5 kg ball.

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Method: Momentum conserved.

v = (0.50 × 4.0) ÷ 1.5 = 1.33 m/s

Extended[2 marks]

An impulse of 18 N s acts on a 3.0 kg trolley at rest. Calculate its final speed.

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Method: Impulse = Δ(mv).

v = 18 ÷ 3.0 = 6.00 m/s

Extended[3 marks]

Explain why a cricketer moves their hands backwards when catching a fast ball.

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Method: Increase stopping time.

Longer time to stop the ball, so the rate of change of momentum (force on the hands) is reduced

Extended[2 marks]

State the unit of momentum.

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Method: p = mv.

kg m/s

Extended[2 marks]

State the unit of impulse and show it is equivalent to the unit of momentum.

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Method: N s.

N s; since 1 N = 1 kg m/s², 1 N s = 1 kg m/s

Extended[2 marks]

Calculate the momentum of a 75 kg sprinter running at 10 m/s.

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Method: p = mv.

p = 750 kg m/s

Extended[2 marks]

A bullet of mass 0.020 kg has momentum 8.0 kg m/s. Calculate its speed.

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Method: v = p ÷ m.

v = 400 m/s

Extended[3 marks]

A 1000 kg car increases speed from 10 m/s to 25 m/s. Calculate the change in momentum.

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Method: Δp = mΔv.

Δp = 1000 × 15 = 15000 kg m/s

Extended[3 marks]

A force of 300 N acts on a 60 kg trolley for 4.0 s from rest. Calculate the final speed.

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Method: Impulse = Δp.

Δp = 1200 kg m/s; v = 1200 ÷ 60 = 20.0 m/s

Extended[3 marks]

A 4.0 kg ball at 5.0 m/s hits a stationary 1.0 kg ball. The 4.0 kg ball continues at 3.0 m/s. Calculate the velocity of the 1.0 kg ball.

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Method: Momentum conserved.

20 = 12 + 1.0v; v = 8.00 m/s

Extended[3 marks]

A 0.20 kg ball moving at 3.0 m/s collides head-on with a 0.10 kg ball moving at 4.0 m/s towards it. They stick. Calculate the final velocity.

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Method: Take first direction positive.

p = 0.60 − 0.40 = 0.20; v = 0.20 ÷ 0.30 = 0.667 m/s in the original direction of the 0.20 kg ball

Extended[3 marks]

A 0.058 kg tennis ball arrives at 30 m/s and leaves the racket at 40 m/s in the opposite direction. Calculate the impulse on the ball.

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Method: Δp = m(v − u).

Impulse = 0.058 × 70 = 4.06 N s

Extended[3 marks]

The ball in the previous question is in contact with the racket for 0.0050 s. Calculate the average force.

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Method: F = Δp ÷ Δt.

F = 4.06 ÷ 0.0050 = 812 N

Extended[3 marks]

Explain how crumple zones in a car reduce injury.

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Method: Longer collision time.

They crumple, increasing the time for the car to stop; the same change in momentum over a longer time means a smaller force

Extended[2 marks]

Explain why gymnastics mats are thick and soft.

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Method: Increase stopping time.

They increase the time to stop, reducing the rate of change of momentum and so the force

Extended[3 marks]

A rocket ejects 50 kg of gas each second at 2000 m/s. Calculate the thrust force.

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Method: F = Δp/Δt.

F = 50 × 2000 = 100000 N

Extended[3 marks]

A 2.0 kg trolley at 3.0 m/s hits a 2.0 kg trolley at rest. The first stops. Calculate the second trolley's speed and state whether kinetic energy is conserved.

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Method: Momentum conserved; compare Ek.

v = 3.0 m/s; Ek before = 9.00 J, after = 9.00 J, so kinetic energy is conserved

Extended[3 marks]

For the stick-together collision of a 2.0 kg trolley at 3.0 m/s with a stationary 1.0 kg trolley (final 2.0 m/s), calculate the kinetic energy lost.

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Method: Compare Ek.

Before = 9.00 J; after = ½ × 3.0 × 2.0² = 6.00 J; lost = 3.00 J

Extended[2 marks]

Explain why momentum is conserved in an explosion even though objects fly apart.

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Method: Equal and opposite momenta.

Total momentum before is zero; afterwards the momenta of the pieces are equal and opposite, adding to zero

Extended[3 marks]

A 50 kg girl jumps off a 25 kg stationary boat at 2.0 m/s. Calculate the boat's recoil speed.

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Method: Momentum conserved from rest.

v = 50 × 2.0 ÷ 25 = 4.00 m/s in the opposite direction

Extended[3 marks]

A 3.0 kg object's momentum changes from 12 kg m/s to 30 kg m/s in 6.0 s. Calculate the resultant force.

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Method: F = Δp/Δt.

F = 18 ÷ 6.0 = 3.00 N

Extended[2 marks]

State the relationship between the impulse on an object and its change in momentum.

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Method: Equal.

Impulse = change in momentum (FΔt = Δ(mv))

Extended[3 marks]

A 0.15 kg cricket ball moving at 20 m/s is caught and stopped. Calculate the impulse, and the force if stopped in 0.10 s and in 0.020 s.

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Method: F = Δp/Δt.

Impulse = 3.00 N s; F = 30.0 N and 150 N

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

ChallengeExtended[5 marks]

A 0.50 kg ball moving east at 6.0 m/s collides with a 1.5 kg ball moving west at 2.0 m/s. After the collision the 1.5 kg ball moves east at 1.5 m/s. Calculate the velocity of the 0.50 kg ball afterwards.

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Method: Take east positive.

0.50 × 6.0 − 1.5 × 2.0 = 0.50v + 1.5 × 1.5, so 0 = 0.50v + 2.25; v = −4.5 m/s, i.e. 4.5 m/s west

ChallengeExtended[4 marks]

A 60 kg astronaut floating at rest throws a 2.0 kg tool at 9.0 m/s. Calculate the astronaut's recoil speed and the time to drift 3.0 m.

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Method: Momentum conservation.

v = 18 ÷ 60 = 0.300 m/s; t = 3.0 ÷ 0.30 = 10.0 s

ChallengeExtended[5 marks]

A 1200 kg car at 20 m/s crashes into a wall and stops in 0.15 s. (a) Calculate the average force. (b) With crumple zones the stopping time is 0.60 s; calculate the new force and explain the benefit.

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Method: F = Δp/Δt.

(a) 160000 N (b) 40000 N — a quarter of the force for the same change in momentum

ChallengeExtended[4 marks]

A 0.045 kg golf ball leaves a club at 70 m/s. The contact time is 0.50 ms. Calculate the impulse and the average force.

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Method: Impulse = mΔv.

Impulse = 3.15 N s; F = 3.15 ÷ 0.0005 = 6300 N

ChallengeExtended[5 marks]

A 2.0 kg trolley at 4.0 m/s collides with a stationary 3.0 kg trolley and they stick. Calculate (a) the common velocity, (b) the kinetic energy lost, (c) where the energy goes.

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Method: Momentum conserved; compare Ek.

(a) 1.60 m/s (b) 16 − 6.4 = 9.60 J (c) To thermal energy and sound (and deformation)

ChallengeExtended[4 marks]

A rocket of total mass 500 kg at rest in space ejects 20 kg of gas at 400 m/s in one burst. Calculate the rocket's speed afterwards.

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Method: Momentum conserved from zero.

v = 20 × 400 ÷ 480 = 16.7 m/s

ChallengeExtended[4 marks]

A force–time graph for a kick is a triangle with peak 600 N and duration 0.020 s. Calculate the impulse and the speed of a 0.43 kg ball initially at rest.

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Method: Impulse = area under graph.

Impulse = ½ × 0.020 × 600 = 6.00 N s; v = 6 ÷ 0.43 = 14.0 m/s

ChallengeExtended[4 marks]

Hail stones of total mass 0.20 kg hit a roof each second at 15 m/s and stop. Calculate the average force on the roof. How would the force change if they bounced back at 5.0 m/s?

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Method: F = rate of change of momentum.

F = 0.20 × 15 = 3.00 N; bouncing: 0.20 × 20 = 4.00 N

ChallengeExtended[4 marks]

A 5.0 kg gun fires a 0.050 kg bullet horizontally at 300 m/s. The gun is stopped by the shoulder in 0.10 s. Calculate the gun's recoil velocity and the average force on the shoulder.

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Method: Recoil from momentum conservation.

v = 15 ÷ 5.0 = 3.00 m/s; F = 15 ÷ 0.10 = 150 N

ChallengeExtended[5 marks]

Explain, using momentum, why a long-jumper lands in a sand pit, and estimate the force on a 70 kg jumper landing vertically at 6.0 m/s if they stop in 0.25 s compared with 0.025 s on hard ground.

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Method: F = Δp/Δt.

Sand: 1680 N; hard ground: 16800 N — sand increases stopping time, reducing the force ten times

All IGCSE Physics topics ← 1.5.3 Centre of Gravity 1.7.1 Energy →

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