OxfordAQA International AS and A-level Physics: Momentum, impulse and collisions (9630) – Revision Notes

These notes condense section 3.2.6 (Momentum) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. This is shared AS and A-level content. For full explanations and longer worked examples, use the study guide. When you are ready to test yourself, use the practice questions, the course hub, the printable checklist and the free diagnostics.

Key definitions and equations

Quantity	Statement	Unit
Momentum	p = mv, a vector in the direction of velocity	kg m s⁻¹
Conservation of linear momentum	Total momentum of a closed system is constant if no resultant external force acts	kg m s⁻¹
Force	F = Δ(mv)/Δt, the rate of change of momentum	N
Impulse	Change in momentum; FΔt = Δ(mv) for a constant force	N s
Area under a force-time graph	Impulse, so the change in momentum	N s
Average force	F = impulse / contact time	N

N s and kg m s⁻¹ are the same unit. When mass is constant, F = Δ(mv)/Δt becomes F = ma.

Conservation of momentum: method in steps

1. Sketch the situation before and after, marking the positive direction.
2. Add up every mv before the event, keeping signs.
3. Add up every mv after it, including the unknown.
4. Equate and solve.
5. State the answer with direction in words, and check the sign.

Why it holds: the third law gives equal and opposite forces for equal times, so the two impulses cancel.

Reminder 1: joining. A 0.90 kg trolley at 3.5 m s⁻¹ catches a 0.60 kg trolley moving the same way at 1.0 m s⁻¹, and they stick. Before: 0.90 × 3.5 + 0.60 × 1.0 = 3.75 kg m s⁻¹. After: 1.5v = 3.75, so v = 2.5 m s⁻¹. KE before = 5.81 J, after = 4.69 J, lost = 1.1 J.

Elastic, inelastic and explosions

Type	Momentum	Kinetic energy	Notes
Elastic	Conserved	Conserved	Bodies separate at the speed they approached (1D)
Inelastic	Conserved	Some becomes thermal, sound or deformation energy	Most real collisions
Perfectly inelastic	Conserved	Largest loss	Bodies move together afterwards
Explosion	Conserved (often total zero)	Increases	Stored chemical or elastic energy released

Test for elastic: compare total KE before and after. Never assume it.

Reminder 2: explosion. A spring pushes a 0.50 kg cart and a 2.0 kg cart apart from rest. The light cart moves at 1.2 m s⁻¹. 0 = 0.50 × 1.2 + 2.0v, so v = −0.30 m s⁻¹ (opposite direction, 0.30 m s⁻¹). Energy released = ½ × 0.50 × 1.2² + ½ × 2.0 × 0.30² = 0.36 + 0.09 = 0.45 J.

Useful fact: for equal momentum, the lighter body gets more kinetic energy, because KE = p²/2m.

Force and impulse

• Constant force: F = Δ(mv)/Δt, so FΔt = mv − mu.
• Changing mass: F = (Δm/Δt) × v for a stream that is stopped. A hose delivering 0.80 kg s⁻¹ of water at 6.0 m s⁻¹ onto a wall, the water stopping, gives 4.8 N.
• Rebound: impulse = m(v − u) with signs, so it is larger than for simply stopping.
• Vertical impact: contact force = (Δp/Δt) + weight when the body decelerates while falling or landing.

Reminder 3: bouncing ball. A 0.060 kg ball arrives at 8.0 m s⁻¹ downwards and leaves at 5.0 m s⁻¹ upwards. Contact is 15 ms. Take up as positive: Δp = 0.060 × (5.0 − (−8.0)) = 0.78 N s. Resultant force = 0.78 / 0.015 = 52 N upwards. Contact force = 52 + 0.060 × 9.81 = 53 N.

Force-time graphs

• Area = impulse = Δ(mv).
• Divide into rectangles, triangles and trapeziums, or use trapezium-rule strips for a table of data.
• Curved graph: count squares and multiply by the value of one square.
• Average force = area / total time. It is lower than the peak.
• If the body starts from rest, v = area / m.

Reminder 4: trapezium graph. A force rises uniformly from 0 to 300 N in 5.0 ms, stays at 300 N until 15 ms, then falls uniformly to zero at 20 ms. Area = ½ × 0.0050 × 300 + 0.010 × 300 + ½ × 0.0050 × 300 = 0.75 + 3.0 + 0.75 = 4.5 N s. For a 0.30 kg ball starting from rest, v = 4.5 / 0.30 = 15 m s⁻¹. Average force = 4.5 / 0.020 = 225 N, which is below the 300 N plateau.

Impact forces and contact times

For a given Δp, F is inversely proportional to Δt.

Situation	Contact time	Effect
Crumple zone, seat belt, airbag	Longer	Smaller force on occupant
Bending knees on landing	Longer	Smaller force on joints
Foam or air cushion in packaging	Longer	Smaller force on contents
Kicking or hitting a ball	Short, hard contact	Large force, large final speed

Good wording: “The crumple zone increases the time to stop, so the rate of change of momentum, and hence the force, is smaller for the same change in momentum.”

Choosing the right approach

• Speeds before and after a collision, masses given, no time given: conservation of momentum.
• Contact time given and a force wanted: F = Δ(mv)/Δt.
• A force-time graph supplied: find the area first, then divide by the mass for the change in velocity.
• Asked whether energy is conserved: compare kinetic energies before and after (KE is never negative, so signs do not matter).
• A stream of material stopped or deflected: use mass per second multiplied by the velocity change.

Must-know distinctions

• Momentum is a vector and always conserved in a closed system. Kinetic energy is a scalar and only conserved in elastic collisions.
• Impulse is the change in momentum, not momentum itself.
• Peak force and average force are different.
• Inelastic does not mean momentum is lost.
• A collision and an explosion both conserve momentum. Only in an explosion does KE increase.

Quick self-test

1. Find the momentum of a 0.058 kg ball moving at 45 m s⁻¹.
2. Write the unit of impulse in SI base units.
3. A 60 kg skater and a 45 kg skater, both at rest on ice, push apart. The 60 kg skater moves at 1.5 m s⁻¹. How fast does the other skater move?
4. A 0.50 kg trolley moving at 0.80 m s⁻¹ sticks to a 0.30 kg trolley at rest. Find the common speed.
5. A constant 25 N force acts for 0.40 s on a 5.0 kg object at rest. Find the impulse and final speed.
6. A force of 80 N acts for 0.050 s. What is the area under the force-time graph?
7. A 0.45 kg football leaves the foot at 20 m s⁻¹ after 12 ms of contact. Find the average force.
8. A 1.0 kg trolley moving at 4.0 m s⁻¹ hits an identical stationary trolley. Afterwards they move at 1.0 m s⁻¹ and 3.0 m s⁻¹. Is the collision elastic?
9. Explain why an airbag reduces injury.
10. Explain why momentum is conserved when two bodies collide.
11. A hose delivers 0.80 kg of water per second at 6.0 m s⁻¹ onto a wall, and the water stops. What force acts on the wall?

Answers

1. 0.058 × 45 = 2.6 kg m s⁻¹.
2. kg m s⁻¹.
3. 60 × 1.5 = 45 × v, so v = 2.0 m s⁻¹ in the opposite direction.
4. 0.50 × 0.80 = 0.40, then 0.40 / 0.80 = 0.50 m s⁻¹.
5. Impulse = 25 × 0.40 = 10 N s. v = 10 / 5.0 = 2.0 m s⁻¹.
6. 80 × 0.050 = 4.0 N s.
7. F = 0.45 × 20 / 0.012 = 750 N.
8. Momentum: 4.0 before, 1.0 + 3.0 = 4.0 after. KE before 8.0 J, after 0.5 + 4.5 = 5.0 J. KE is lost, so it is inelastic.
9. It increases the time over which the occupant stops. The momentum change is the same, so the average force is smaller.
10. By Newton’s third law the forces on the two bodies are equal and opposite and last equally long, so their impulses are equal and opposite and total momentum is unchanged.
11. F = 0.80 × 6.0 = 4.8 N.

Where marks are usually lost

• Missing or inconsistent signs when bodies move in opposite directions.
• Quoting only a speed where the question wants velocity, with its direction.
• Using time in ms without converting to seconds.
• Assuming a collision is elastic without calculating kinetic energy.
• Writing “force is absorbed” instead of “time of contact is increased, so rate of change of momentum is reduced”.
• Reading the peak value of a force-time graph as the average force.
• Leaving out weight in vertical impacts when the question asks for the contact force.
• Dropping units, or writing N s and kg m s⁻¹ as if they were different quantities.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Section 3.2.6 Momentum.