Momentum: IGCSE Physics worked examples

Key ideas

Momentum = mass × velocity. Its unit is kg m/s.
Total momentum includes the signs of all velocities.
Kinetic energy can change during a collision even when total momentum is conserved.
Interacting objects exert equal and opposite forces for the same time. Their momentum changes are equal and opposite.

The method

Choose a positive direction and give each velocity its sign.
Calculate momentum using p = m v.
Find change in momentum as final momentum minus initial momentum.
For a collision with negligible external forces, set total momentum before equal to total momentum after.
Find average resultant force by dividing momentum change by time.

Worked examples

Rearrange momentum

A car has mass 1500 kg and momentum 24 000 kg m/s.
Velocity = momentum ÷ mass = 24 000 ÷ 1500 = 16 m/s.

Answer: 16 m/s

A rebound

A 0.50 kg ball approaches a wall at 12 m/s and rebounds at 8 m/s. Take towards the wall as positive.
Change = 0.50 × (−8 − 12) = −10 kg m/s.
Over 0.050 s, average force = −10 ÷ 0.050 = −200 N.

Answer: 10 kg m/s and 200 N away from the wall

Stopping a car

A 1000 kg car stops from 20 m/s. Take its initial motion as positive.
Momentum change = 1000 × (0 − 20) = −20 000 kg m/s.
Over 4.0 s: force = −20 000 ÷ 4.0 = −5000 N.
Over 0.10 s: force = −20 000 ÷ 0.10 = −200 000 N.
Force magnitude ratio = 200 000 ÷ 5000 = 40.

Answer: 5000 N and 200 000 N against the motion

A longer stopping time

A 60 kg dummy stops from 15 m/s. Momentum change magnitude = 60 × 15 = 900 kg m/s.
Over 0.050 s: force magnitude = 900 ÷ 0.050 = 18 000 N.
Over 0.30 s: force magnitude = 900 ÷ 0.30 = 3000 N.
18 000 ÷ 3000 = 6, so the second force is one sixth as large.

Answer: 18 000 N compared with 3000 N

Trolleys stick together

A 3.0 kg trolley at 4.0 m/s hits a stationary 1.0 kg trolley. Ignore external forces.
Before = 3.0 × 4.0 + 1.0 × 0 = 12 kg m/s.
After = (3.0 + 1.0)v = 4.0v.
Velocity = 12 ÷ 4.0 = 3.0 m/s forwards.

Answer: 3.0 m/s forwards

Find an unknown mass

A 0.20 kg trolley at 6.0 m/s joins a stationary block. They move at 2.0 m/s. Ignore external forces.
Before = 0.20 × 6.0 = 1.2 kg m/s.
(0.20 + m) × 2.0 = 1.2.
m = 1.2 ÷ 2.0 − 0.20 = 0.40 kg.

Answer: 0.40 kg

Objects bounce apart

A 2.0 kg trolley at 5.0 m/s hits a stationary 3.0 kg trolley. The first rebounds at 1.0 m/s. Ignore external forces.
Before = 2.0 × 5.0 = 10 kg m/s.
10 = 2.0 × (−1.0) + 3.0v.
v = (10 + 2.0) ÷ 3.0 = 4.0 m/s forwards.

Answer: 4.0 m/s forwards

Opposite momenta cancel

A 4.0 kg trolley moves right at 3.0 m/s and a 2.0 kg trolley moves left at 6.0 m/s. They stick. Ignore external forces.
Before = 4.0 × 3.0 + 2.0 × (−6.0) = 12 − 12 = 0.
Final velocity = 0 ÷ (4.0 + 2.0) = 0 m/s.

Answer: Both stop.

Two skaters push apart

Two skaters of masses 60 kg and 40 kg start at rest. The heavier skater moves left at 2.0 m/s. Ignore external forces.
0 = 60 × (−2.0) + 40v.
v = 120 ÷ 40 = 3.0 m/s to the right.

Answer: 3.0 m/s to the right

Recoil

A 500 kg launcher fires a 5.0 kg ball at 100 m/s from rest. Ignore external forces.
Ball momentum = 5.0 × 100 = 500 kg m/s forwards.
Launcher velocity = −500 ÷ 500 = −1.0 m/s.

Answer: 1.0 m/s backwards

Mistakes that lose marks

Using speed without direction in a rebound.
Using only one mass after two objects stick together.
Confusing momentum units, kg m/s, with force units, N.
Saying that safety features reduce the momentum change for the same initial and final motion.

EXAM-STYLE questions written by Apex tutors

Practise with 12 original questions and full solutions. The suggested marks are a guide from our tutors. M means method, A means accuracy and B means an independent point. Try each question, then open its solution.

Apex original core A 0.040 kg arrow travels at 60 m/s to the right. Find its momentum. [2 marks]
Show full solution
1. Momentum = mass × velocity. M1
2. 0.040 × 60 = 2.4 kg m/s to the right. A1
Answer: 2.4 kg m/s to the right
Apex original extended A 1500 kg car has momentum 24 000 kg m/s forwards. State the momentum equation and find its velocity. [3 marks]
Show full solution
1. p = m v. M1
2. v = 24 000 ÷ 1500 = 16 m/s forwards. A2
Answer: 16 m/s forwards
Apex original extended A 0.30 kg ball slows from 8.0 m/s to 3.0 m/s to the right. Find its change in momentum. [3 marks]
Show full solution
1. Take right as positive. M1
2. Change = 0.30 × (3.0 − 8.0) = 0.30 × (−5.0) = −1.5 kg m/s. M1
3. Its momentum to the right decreases by 1.5 kg m/s. A1
Answer: −1.5 kg m/s
Apex original extended A 0.25 kg ball falls at 6.0 m/s and rebounds upwards at 4.0 m/s. Contact lasts 0.020 s. Find the momentum change and average resultant force. [4 marks]
Show full solution
1. Take up as positive: initial velocity = −6.0 m/s. M1
2. Change = 0.25 × (4.0 − (−6.0)) = 2.5 kg m/s upwards. M1
3. Average resultant force = 2.5 ÷ 0.020 = 125 N upwards. A2
Answer: 2.5 kg m/s upwards; 125 N upwards
Apex original extended An 8000 kg wagon moving at 1.5 m/s hits a stationary 12 000 kg wagon. They couple. Ignore external forces. Find their final velocity. [4 marks]
Show full solution
1. Before = 8000 × 1.5 + 12 000 × 0 = 12 000 kg m/s. M1
2. Combined mass = 8000 + 12 000 = 20 000 kg. M1
3. Final velocity = 12 000 ÷ 20 000 = 0.60 m/s in the original direction. A2
Answer: 0.60 m/s in the original direction
Apex original extended A 1.0 kg trolley moves right at 5.0 m/s. A 4.0 kg trolley moves left at 2.0 m/s. They collide and stick. Ignore external forces. Find the final velocity. [4 marks]
Show full solution
1. Take right as positive. M1
2. Before = 1.0 × 5.0 + 4.0 × (−2.0) = 5.0 − 8.0 = −3.0 kg m/s. M1
3. Final velocity = −3.0 ÷ (1.0 + 4.0) = −0.60 m/s. A2
Answer: 0.60 m/s to the left
Apex original extended A 3.0 kg launcher initially at rest fires a 0.015 kg projectile forwards at 400 m/s. Ignore external forces. Find the launcher recoil velocity. [4 marks]
Show full solution
1. Initial total momentum = 0. M1
2. Projectile momentum = 0.015 × 400 = 6.0 kg m/s forwards. M1
3. Launcher velocity = −6.0 ÷ 3.0 = −2.0 m/s. A2
Answer: 2.0 m/s backwards
Apex original extended Explain how an airbag reduces injury using momentum, stopping time and force. [4 marks]
Show full solution
1. Compare the same initial motion and final rest, so the momentum change is the same. M1
2. The airbag increases the stopping time. M1
3. Average force = momentum change ÷ time. A longer time gives a smaller average force. M1
4. The smaller force reduces the risk of injury. A1
Answer: A longer stopping time reduces the average force.
Apex original extended A rocket pushes gas backwards in space. Explain why it accelerates forwards. [3 marks]
Show full solution
1. The rocket pushes the gas backwards. M1
2. The gas exerts an equal force forwards on the rocket. The forces act on different objects. M1
3. The forward force accelerates the rocket. With no external force, total momentum is conserved. A1
Answer: The gas exerts a forward force on the rocket.
Apex original extended Trolley A has mass 0.80 kg and moves right at 5.0 m/s. It hits stationary trolley B of mass 1.2 kg. A rebounds left at 1.0 m/s. Ignore external forces. Find B's velocity and the average force on B over 0.050 s. [5 marks]
Show full solution
1. Take right as positive. Before = 0.80 × 5.0 = 4.0 kg m/s. M1
2. After = 0.80 × (−1.0) + 1.2v = −0.80 + 1.2v. M1
3. 1.2v = 4.0 + 0.80 = 4.8, so v = 4.8 ÷ 1.2 = 4.0 m/s. M1
4. Force on B = (1.2 × 4.0 − 0) ÷ 0.050 = 96 N to the right. A2
Answer: 4.0 m/s right; 96 N right
Apex original extended A 0.010 kg projectile moving at 400 m/s embeds in a stationary 2.0 kg block on frictionless ice. Find their common speed to 3 significant figures. [4 marks]
Show full solution
1. Before = 0.010 × 400 = 4.0 kg m/s. M1
2. Combined mass = 2.0 + 0.010 = 2.010 kg. M1
3. Speed = 4.0 ÷ 2.010 = 1.990049… m/s = 1.99 m/s. A2
Answer: 1.99 m/s
Apex original extended A 0.16 kg ball travelling at 30 m/s is caught. Find the average force magnitude for stopping times of 0.020 s and 0.10 s. [4 marks]
Show full solution
1. Momentum change magnitude = 0.16 × 30 = 4.8 kg m/s. M1
2. First force magnitude = 4.8 ÷ 0.020 = 240 N. M1
3. Second force magnitude = 4.8 ÷ 0.10 = 48 N. Both act against the initial motion. A2
Answer: 240 N; 48 N

Where this is in the syllabus

Explore the matching courses: Cambridge IGCSE Physics 0625; Pearson Edexcel International GCSE Physics 4PH1.

For the official syllabus, use the board links: Cambridge IGCSE Physics 0625; Pearson Edexcel International GCSE Physics 4PH1.

Questions students ask

Is momentum a vector?

Yes. Its direction is the same as velocity.

When is momentum conserved?

Total momentum is conserved when there is no resultant external force on the system.

Does an airbag reduce momentum change?

For the same initial motion and final rest, the change is the same. The longer stopping time reduces average force.