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

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. It covers section 3.2.6 (Momentum), which is shared International AS and A-level content, so everything here is relevant to both qualifications. Read it with the course hub and tick off each point on the printable checklist.

Companion pages: the revision notes and the practice questions. Newton’s laws are covered in motion and Newton’s laws. To find your weak spots first, try the free diagnostics.

What 3.2.6 asks you to do

Specification statement	What you must be able to do
Momentum, p = mv	Calculate momentum and treat it as a vector
Conservation of linear momentum	Apply it quantitatively to problems in one dimension
Elastic and inelastic collisions; explosions	Classify an interaction using kinetic energy and analyse explosions
F = Δ(mv)/Δt, force as the rate of change of momentum	Use it, including when mass changes
Impulse = change in momentum; FΔt = Δ(mv) for constant F	Calculate impulse and the resulting change in velocity
Area under a force-time graph	Explain and use it, including for forces that vary with time
Impact forces and contact times	Explain football kicks, crumple zones and packaging

Momentum

Momentum is the product of mass and velocity:

p = mv

The unit is kg m s⁻¹. Momentum points the way the body moves, so it is a vector. For motion along a line, fix a positive direction first and attach a sign to every velocity. A change in direction is a change in momentum even if the speed stays the same.

Worked example 1. A 0.85 kg trolley moves towards a spring buffer at 1.4 m s⁻¹ and rebounds at 0.60 m s⁻¹. Find its change in momentum.

Let the direction of approach be positive. Initial momentum = 0.85 × 1.4 = +1.19 kg m s⁻¹ Final momentum = 0.85 × (−0.60) = −0.51 kg m s⁻¹ Change = −0.51 − 1.19 = −1.7 kg m s⁻¹

The size is 1.7 kg m s⁻¹, directed away from the buffer. Subtracting the speeds (1.4 − 0.60) is the classic wrong answer.

Conservation of linear momentum

Principle. The total momentum of a closed system stays constant, provided no resultant external force acts on it. Momentum is conserved in elastic collisions, inelastic collisions and explosions.

Why it holds: when two bodies interact, Newton’s third law gives equal and opposite forces acting for the same time. Each body receives equal and opposite impulses, so the changes in momentum cancel and the total is unchanged.

Method for any one-dimensional problem

1. Sketch the situation before and after, and mark which way is positive.
2. Add the momenta of all bodies before, using signs.
3. Add the momenta afterwards, with the unknown included.
4. Set the totals equal, solve, and give the direction in words.

Collisions where the bodies join

Worked example 2. A railway truck of mass 14 000 kg moves at 2.2 m s⁻¹ and couples to a stationary truck of mass 9 000 kg. Find the common speed and the kinetic energy lost.

Momentum before = 14 000 × 2.2 = 30 800 kg m s⁻¹ After: (14 000 + 9 000)v = 30 800 v = 30 800 / 23 000 = 1.3 m s⁻¹ (1.34 before rounding)

KE before = ½ × 14 000 × 2.2² = 33 900 J KE after = ½ × 23 000 × 1.339² = 20 600 J Lost = 13 300 J (3 s.f.), about 39% of the original.

Worked example 3. A 1.5 kg trolley moves right at 4.0 m s⁻¹ and meets a 2.5 kg trolley moving left at 1.0 m s⁻¹. They stick together. Find their velocity.

Right is positive. Before: 1.5 × 4.0 + 2.5 × (−1.0) = 6.0 − 2.5 = 3.5 kg m s⁻¹ After: 4.0v = 3.5 v = +0.88 m s⁻¹, to the right.

Kinetic energy falls from 13.25 J to 1.53 J. The opposite momenta partly cancel in the sum, but the total of 3.5 kg m s⁻¹ is unchanged by the collision.

Elastic and inelastic collisions

• Elastic: no kinetic energy is converted to other forms, so the total is unchanged. Collisions between gas molecules are a good model of this.
• Inelastic: some kinetic energy becomes other forms (thermal energy, sound, deformation). Momentum is still conserved.
• Perfectly inelastic: the bodies move together after the collision, which is the greatest kinetic energy loss possible for the given momentum.

Test: calculate total KE before and after. If they match, the collision is elastic. For an elastic collision in one dimension, the bodies separate at the same relative speed at which they approached. That gives a quick second equation alongside momentum.

Worked example 4. A 0.30 kg cart moving at 2.0 m s⁻¹ hits a stationary 0.10 kg cart. The first cart continues at 1.0 m s⁻¹. Find the speed of the second cart and decide whether the collision is elastic.

Momentum: 0.30 × 2.0 = 0.30 × 1.0 + 0.10 × v 0.60 = 0.30 + 0.10v, so v = 3.0 m s⁻¹

KE before = ½ × 0.30 × 2.0² = 0.60 J KE after = ½ × 0.30 × 1.0² + ½ × 0.10 × 3.0² = 0.15 + 0.45 = 0.60 J

KE is unchanged, so the collision is elastic. Check: they approached at 2.0 m s⁻¹ and separate at 3.0 − 1.0 = 2.0 m s⁻¹.

Explosions

An explosion is a collision run backwards. Bodies start together (often at rest), so total momentum is zero, and they fly apart with momenta that add to zero. Kinetic energy increases, supplied by stored chemical or elastic energy, so explosions are not elastic in the KE sense.

Worked example 5. A rifle of mass 3.8 kg fires a 0.012 kg bullet at 640 m s⁻¹. Find the recoil speed of the rifle (free to move) and compare the kinetic energies.

Bullet direction positive. Total momentum before = 0. 0 = 0.012 × 640 + 3.8 × v v = −7.68 / 3.8 = −2.0 m s⁻¹ (rifle moves backwards at 2.0 m s⁻¹)

KE bullet = ½ × 0.012 × 640² = 2 460 J KE rifle = ½ × 3.8 × 2.02² = 7.8 J

The momenta are equal in size, but the light bullet carries almost all of the kinetic energy because KE = p²/2m.

Worked example 6. A stationary 8.0 kg object splits in two. The 5.0 kg part moves east at 12 m s⁻¹. Find the velocity of the 3.0 kg part and the energy released.

East positive: 0 = 5.0 × 12 + 3.0v, so v = −20 m s⁻¹, i.e. 20 m s⁻¹ west. Energy released = ½ × 5.0 × 12² + ½ × 3.0 × 20² = 360 + 600 = 960 J.

Force as rate of change of momentum

Newton’s second law in its general form is:

F = Δ(mv)/Δt

The resultant force is the rate of change of momentum and acts in the direction of that change. When the mass is constant this reduces to F = mΔv/Δt = ma. The momentum form is the one to use when mass changes, as with a stream of water hitting a surface.

Worked example 7. A hose delivers 1.2 kg of water every second at 14 m s⁻¹ onto a wall. The water stops and runs down the wall. Find the force on the wall.

Momentum lost per second = 1.2 × 14 = 16.8 kg m s⁻¹ per second F = Δ(mv)/Δt = 17 N (2 s.f.)

If the water bounced back it would lose more momentum per second, so the force would be larger.

Impulse

Rearranging gives, for a constant force:

FΔt = Δ(mv)

Impulse is the change in momentum. The unit is N s, equivalent to kg m s⁻¹. Impulse is a vector, so a force that reverses a body’s motion produces an impulse larger than mv.

Worked example 8. A footballer kicks a 0.41 kg ball from rest and it leaves at 22 m s⁻¹. The contact lasts 10 ms. Find the impulse and the average force.

Impulse = Δ(mv) = 0.41 × 22 − 0 = 9.0 N s (9.02 before rounding) F = 9.02 / 0.010 = 9.0 × 10² N

The weight of the ball (0.41 × 9.81 = 4.0 N) is tiny compared with this, so ignoring it is sensible.

Force-time graphs

The area under a force-time graph is the impulse, so it equals the change in momentum. A bigger area means a bigger change in velocity for the same mass.

• Straight-sided graphs: split into triangles, rectangles and trapeziums.
• Curved graphs: count squares, or use the trapezium rule on tabulated data.
• Average force = impulse / contact time.

Worked example 9. The force on a ball rises uniformly from zero to 600 N in 8.0 ms, then falls uniformly to zero at 20 ms. The 0.45 kg ball starts at rest. Find its final speed.

Triangle: area = ½ × base × height = ½ × 0.020 × 600 = 6.0 N s v = impulse / m = 6.0 / 0.45 = 13 m s⁻¹

Worked example 10. A 0.20 kg ball at rest is struck, and the force on it is recorded every 3.0 ms:

t / ms	0	3	6	9	12	15
F / N	0	240	520	380	120	0

Trapezium rule with strip width 0.0030 s: Impulse = 0.0030 × [½(0 + 0) + 240 + 520 + 380 + 120] = 0.0030 × 1260 = 3.8 N s Final speed = 3.78 / 0.20 = 19 m s⁻¹ Average force = 3.78 / 0.015 = 250 N, which is about half the 520 N peak.

Impact forces and contact times

For a fixed change in momentum, F = Δp/Δt shows that a longer contact time means a smaller average force. Crumple zones, seat belts, cushioned packaging, and bending the knees when landing all use this. The momentum change is the same, but it is spread over a longer time, so the force is lower. Kicking a football is the reverse: a hard, short contact gives a large force and a large speed.

Worked example 11. A 2.4 kg instrument falls 0.80 m onto a hard floor and stops in 4.0 ms. Take g = 9.81 m s⁻². Compare this with a foam pad that makes the stop last 40 ms.

Speed on impact: v = √(2gh) = √(2 × 9.81 × 0.80) = 3.96 m s⁻¹ Δp = 2.4 × 3.96 = 9.51 kg m s⁻¹ Hard floor: resultant force = 9.51 / 0.0040 = 2 380 N Foam: resultant force = 9.51 / 0.040 = 238 N

The resultant force is the contact force minus weight. Weight = 2.4 × 9.81 = 23.5 N. So the floor pushes with 2 400 N on the hard floor and 261 N on foam. Ten times the time gives roughly one tenth of the force.

Common errors

• Forgetting signs: velocities in opposite directions must have opposite signs.
• Treating kinetic energy as conserved in every collision (it is not). Only momentum is always conserved when the system is closed.
• Adding speeds rather than subtracting velocities to find a change in momentum on rebound.
• Using time in milliseconds without converting to seconds.
• Saying “force is conserved” or “momentum is used up”.
• Forgetting weight when asked for the contact force during a vertical impact.
• Explaining a crumple zone as “absorbing the force”. Say that it increases the stopping time, which reduces the rate of change of momentum.

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.