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What is momentum, and how is it conserved in collisions and explosions?

Momentum p = mv, the principle of conservation of momentum, and calculations for one-dimensional collisions and explosions.

A CCEA GCSE Physics answer on momentum p = mv, the principle of conservation of momentum, and how to work out velocities in one-dimensional collisions and explosions.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

CCEA wants you to define momentum, use the equation p = mv, and apply the principle of conservation of momentum to one-dimensional collisions and explosions. These calculations reward careful bookkeeping of momentum before and after an event.

The answer

What momentum is

A heavy, fast object has a large momentum; a stationary object has zero momentum.

Conservation of momentum

Because momentum is a vector, you must give one direction a positive sign and the opposite direction a negative sign before adding.

Collisions and explosions

In a collision, objects may move apart or stick together, but total momentum is conserved. In an explosion the objects start at rest, so the total momentum is zero before and after; the pieces fly apart with equal and opposite momenta.

Momentum is closely linked to force. A resultant force changes an object's momentum, and the bigger the force or the longer it acts, the bigger the change. This is why a small force acting for a long time can produce the same change in momentum as a large force acting briefly, an idea used in vehicle safety: spreading the change in momentum over more time gives a smaller force.

Worked example: a collision with rebound

Examples in context

Example 1. Newton's cradle
When one ball swings in and strikes the row, one ball swings out the far side at the same speed. The incoming momentum is passed through the line, conserving total momentum.
Example 2. A rocket
Hot gas is pushed out backwards with momentum in one direction, so the rocket gains equal and opposite momentum forwards. Total momentum stays zero overall, which is conservation of momentum in action.
Example 3. Catching a cricket ball
A fielder moves their hands back as they catch a fast ball. This lengthens the time over which the ball's momentum falls to zero, so the force on their hands is smaller and the catch hurts less, the same physics that protects passengers in a crash.

Try this

Q1. Calculate the momentum of a 1200 kg1200\ \text{kg} car moving at 15 m/s15\ \text{m/s}. [2 marks]

  • Cue. p=mv=1200×15=18000 kg m/sp = mv = 1200 \times 15 = 18000\ \text{kg m/s}.

Q2. State the principle of conservation of momentum. [1 mark]

  • Cue. In a closed system the total momentum before equals the total momentum after.

Q3. Two 0.50 kg0.50\ \text{kg} trolleys, initially at rest and held together by a compressed spring, are released. One moves left at 2.0 m/s2.0\ \text{m/s}. State the velocity of the other. [2 marks]

  • Cue. Total momentum is zero, so the other moves right at 2.0 m/s2.0\ \text{m/s}.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA style4 marksA 1500 kg car moving at 12 m/s collides with a stationary 1000 kg car and they lock together. Calculate their common velocity immediately after the collision.
Show worked answer →

Total momentum before equals total momentum after.

Before: p=(1500×12)+(1000×0)=18000 kg m/s.p = (1500 \times 12) + (1000 \times 0) = 18000\ \text{kg m/s}.

After, combined mass =2500 kg= 2500\ \text{kg} moving at velocity vv:

2500v=18000⇒v=180002500=7.2 m/s.2500v = 18000 \Rightarrow v = \dfrac{18000}{2500} = 7.2\ \text{m/s}.

Markers reward total momentum before, conservation of momentum, the combined mass, and the value 7.2 m/s.

CCEA style4 marksA 0.060 kg air rifle pellet leaves the barrel at 150 m/s. The rifle has a mass of 3.0 kg. Calculate the recoil velocity of the rifle, explaining your reasoning.
Show worked answer →

Before firing, the total momentum is zero (everything at rest).

By conservation of momentum, the total momentum after firing is also zero, so the pellet's momentum and the rifle's momentum are equal and opposite.

mpvp=mrvr⇒0.060×150=3.0×vr.m_p v_p = m_r v_r \Rightarrow 0.060 \times 150 = 3.0 \times v_r.

vr=9.03.0=3.0 m/sv_r = \dfrac{9.0}{3.0} = 3.0\ \text{m/s} in the opposite direction to the pellet.

Markers reward total momentum zero before and after, equal and opposite momenta, and the recoil value 3.0 m/s.

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