PhysicsCore20 min read

Momentum

The quantity that survives every collision, and why it makes cars safer

This topic appears in:

9th Class Physics · ch.31st Year Physics · ch.2O Level Physics (5054) · ch.1A Level Physics (9702) · ch.3A Level Mathematics (9709) · ch.4
01

Momentum: mass in motion

Definition

Momentum — The product of an object's mass and its velocity, p = mv. A vector, measured in kg m s⁻¹.

Momentum measures how hard something is to stop. It depends on both how heavy the object is and how fast it is going, which is why a slow-moving lorry and a fast-moving motorcycle can be equally difficult to bring to rest.

It is a vector, so direction is part of the answer. In one-dimensional problems that means choosing a positive direction at the start and giving anything travelling the other way a negative value. Most collision questions that go wrong go wrong here rather than in the arithmetic.

The unit has no special name: it is simply the kilogram metre per second, kg m s⁻¹.

p = m va vector — take one direction as positive and keep to it
p
momentumkg m s⁻¹
m
masskg
v
velocitym s⁻¹
02

Force as the rate of change of momentum

Newton did not originally write his second law as F = ma. He wrote it in terms of momentum: the resultant force equals the rate of change of momentum. That form is more general and more useful, because it still works when the mass is changing — for a rocket burning fuel, for instance.

Rearranged, it gives impulse: force multiplied by the time it acts equals the change in momentum produced. This is the single most practically useful equation in the topic, because it explains every piece of safety engineering you will be asked about.

The change in momentum in a crash is fixed — the car is going to stop, whatever happens. What can be changed is the time over which it happens. Spread the same change in momentum over a longer time and the force must be smaller.

F = Δp / Δtimpulse = F Δt = Δp = m(v − u)same change in momentum over a longer time means a smaller force
F
resultant forceN
Δp
change in momentumkg m s⁻¹
Δt
time takens
Worked example 15 marks

A 60 kg passenger in a car travelling at 20 m s⁻¹ is brought to rest in a crash. Find the force on them if they stop in 0.10 s with a seat belt, and in 0.010 s without one.

  1. Change in momentum = m(v − u) = 60 × (0 − 20) = −1200 kg m s⁻¹.The same for both cases — the passenger stops either way.
  2. With belt: F = Δp/Δt = 1200 / 0.10.Taking the magnitude for the size of the force.
  3. F = 12 000 N.
  4. Without belt: F = 1200 / 0.010.Hitting the windscreen stops you ten times faster.
  5. F = 120 000 N, ten times greater.Ten times the force, for the same change in momentum.

12 kN with the belt, 120 kN without

Every safety feature is this one equation

Crumple zones, airbags, seat belts, cycle helmets, crash mats, bending your knees when you land, moving your hands back to catch a ball — all of them increase the time over which the momentum change happens, so the force is smaller. Name the equation in your answer and the marks follow.

03

Conservation of momentum

Definition

Principle of conservation of momentum — In the absence of a resultant external force, the total momentum of a system before an interaction equals the total momentum after it.

When two objects interact, they push on each other with equal and opposite forces for exactly the same length of time — Newton's third law. Equal and opposite forces acting for equal times produce equal and opposite impulses, so whatever momentum one object gains, the other loses. The total is unchanged.

The condition matters and is worth a mark on its own: there must be no resultant external force. Friction from the ground or air resistance would remove momentum from the system, so questions specify smooth surfaces or short interaction times to make the principle apply.

In practice this turns every collision problem into a single equation: total momentum before equals total momentum after. Write both sides carefully, keeping the signs right, and solve.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂u for before, v for after; velocities in the negative direction take a minus sign
m
masskg
u
velocity beforem s⁻¹
v
velocity afterm s⁻¹
Worked example 25 marks

A 2.0 kg trolley moving at 3.0 m s⁻¹ collides head-on with a 1.0 kg trolley moving at 2.0 m s⁻¹ towards it. They stick together. Find their common velocity afterwards.

  1. Take motion of the 2.0 kg trolley as positive.Choose a positive direction first and state it — this is the mark most often lost.
  2. Momentum before = (2.0 × 3.0) + (1.0 × −2.0).The second trolley moves the other way, so its velocity is negative.
  3. = 6.0 − 2.0 = 4.0 kg m s⁻¹.
  4. They stick, so the combined mass is 3.0 kg: 4.0 = 3.0 × v.
  5. v = 1.3 m s⁻¹, in the original direction of the 2.0 kg trolley.Positive, so it keeps that direction — state it.

1.3 m s⁻¹ in the direction of the heavier trolley

Set any masses and any speeds you like. The two bars — total momentum before and after — stay identical every time. Switch between bouncing apart and sticking together and the individual velocities change completely, while the total does not move.

04

Explosions and recoil

An explosion is a collision run backwards. Two objects start at rest, so the total momentum before is zero, and it must still be zero afterwards. The only way for that to happen is for the two to move in opposite directions with equal and opposite momenta.

This is why a rifle recoils when fired, and why a rocket works. The bullet carries momentum forward, so the rifle must carry the same amount backward. The rifle is far heavier, so its velocity is correspondingly smaller — which is fortunate for the shoulder behind it.

It is also why a rocket can accelerate in the vacuum of space with nothing to push against. It does not need anything to push against: it throws exhaust gas backwards, and gains forward momentum equal and opposite to the momentum of the gas.

Key points

  1. Momentum is mv and is a vector — choose a positive direction and keep to it.
  2. Force is the rate of change of momentum, F = Δp/Δt.
  3. Increasing the collision time reduces the force — the basis of every safety feature.
  4. Total momentum is conserved when no resultant external force acts.
  5. In an explosion the total momentum stays zero, so the pieces fly apart with equal and opposite momenta.

Practice questions

6 questions · 28 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

2 · 4 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State the principle of conservation of momentum, including the condition under which it applies.
Model answer

The total momentum of a system before an interaction equals the total momentum after it, provided no resultant external force acts on the system.

Examiner tip. The condition is worth a mark on its own. Never state the principle without it.

SQ2[2 marks]
Explain why a cricketer moves their hands backwards while catching a fast ball.
Model answer

It increases the time over which the ball is brought to rest. Since F = Δp/Δt and the change in momentum is fixed, a longer time means a smaller force on the hands.

Examiner tip. Name the equation. "It hurts less" without the physics scores nothing.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A 0.15 kg ball hits a wall at 12 m s⁻¹ and rebounds at 8.0 m s⁻¹. The contact lasts 0.050 s. Calculate the average force on the ball.

Given. m = 0.15 kg, u = 12 m s⁻¹, v = −8.0 m s⁻¹, Δt = 0.050 s

Full working
  1. Takes the initial direction as positive, so the rebound velocity is −8.0 m s⁻¹the sign is the whole question[1]
  2. Δp = m(v − u) = 0.15 × (−8.0 − 12) = −3.0 kg m s⁻¹a change of 20 m s⁻¹, not 4[1]
  3. Uses F = Δp/Δt[1]
  4. F = −3.0 / 0.050 = −60 N, i.e. 60 N away from the wall[1]

60 N, directed away from the wall

Examiner tip. A rebound reverses the velocity, so the change is u + v in size, not u − v. Treating 12 and 8 as a change of 4 is the single commonest error on this topic.

Long questions

1 · 9 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[9 marks]
A 0.045 kg golf ball is struck by a club. The ball leaves the tee at 60 m s⁻¹ and the contact lasts 0.50 ms.
  1. Calculate the change in momentum of the ball. [2]
  2. Calculate the average force exerted by the club. [3]
  3. The club has a mass of 0.30 kg and was moving at 70 m s⁻¹ before impact. Calculate its speed immediately afterwards. [4]
Mark scheme
  1. Uses Δp = m(v − u) with u = 0the ball starts at rest on the tee[1]
  2. Δp = 0.045 × 60 = 2.7 kg m s⁻¹[1]
  3. Converts the time: 0.50 ms = 5.0 × 10⁻⁴ s[1]
  4. Uses F = Δp/Δt[1]
  5. F = 2.7 / 5.0 × 10⁻⁴ = 5400 N[1]
  6. Applies conservation of momentum to club and ball together[1]
  7. Before: 0.30 × 70 = 21 kg m s⁻¹the ball contributes nothing[1]
  8. After: 21 = 0.30 × v + 2.7[1]
  9. v = 18.3 / 0.30 = 61 m s⁻¹[1]

(a) 2.7 kg m s⁻¹ (b) 5400 N (c) 61 m s⁻¹

Examiner tip. The millisecond conversion in (b) is worth a mark by itself, and getting it wrong makes the force a thousand times too small — which looks almost plausible, so it often survives a check.

Exam questions

2 · 11 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
A 1200 kg car travelling at 15 m s⁻¹ collides with a stationary 800 kg car. The two lock together.
  1. Calculate the total momentum before the collision.
  2. Calculate their common velocity immediately after.
  3. State whether kinetic energy is conserved, and name the type of collision.
Mark scheme
  1. Uses p = mv[1]
  2. p = 1200 × 15 = 18 000 kg m s⁻¹the stationary car contributes nothing[1]
  3. Applies conservation: total after = 18 000 kg m s⁻¹[1]
  4. Combined mass 2000 kg, so v = 18 000 / 2000 = 9.0 m s⁻¹[1]
  5. Kinetic energy is not conserved[1]
  6. Inelastic collisionobjects sticking together is always inelastic[1]

(a) 1.8 × 10⁴ kg m s⁻¹ (b) 9.0 m s⁻¹ (c) not conserved; inelastic

Examiner tip. Whenever two objects stick together, the collision is inelastic and kinetic energy has been lost to heat, sound and deformation. Momentum is still conserved — the two facts are independent.

Q2[5 marks]
A rifle of mass 4.0 kg fires a bullet of mass 0.020 kg at 400 m s⁻¹.
  1. Calculate the recoil velocity of the rifle.
  2. Explain, using momentum, why a heavier rifle recoils more slowly.
Mark scheme
  1. Total momentum before is zeronothing is moving[1]
  2. 0 = 0.020 × 400 + 4.0 × v[1]
  3. v = −8.0 / 4.0 = −2.0 m s⁻¹, i.e. 2.0 m s⁻¹ backwardsdirection required[1]
  4. The rifle's momentum must equal the bullet's in size and be opposite in direction[1]
  5. Since p = mv is fixed, a larger m gives a smaller v[1]

2.0 m s⁻¹ backwards

Examiner tip. Explosion questions always start from "total momentum before = 0". Write that line first and the algebra follows.