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What is momentum, and why is it conserved in collisions?

Momentum: the momentum equation, conservation of momentum in collisions and explosions, and the link between force and rate of change of momentum (higher and separate).

A focused answer to AQA GCSE Physics 4.5.7, covering the momentum equation, the conservation of momentum in collisions and explosions, and how force relates to the rate of change of momentum and to road-safety features.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Momentum
  3. Conservation of momentum
  4. Force and rate of change of momentum
  5. Try this

What this dot point is asking

AQA wants you to use the momentum equation, apply the conservation of momentum to collisions and explosions, and (higher and separate physics) link force to the rate of change of momentum and to safety features.

Momentum

Momentum being a vector is central to every calculation. When two objects move in opposite directions you must give their velocities opposite signs (for example, take rightwards as positive and leftwards as negative) before adding their momenta. A common slip is to add the sizes and ignore the directions, which gives the wrong total. The unit, kilograms metres per second, comes straight from multiplying mass (kg) by velocity (m/s).

Conservation of momentum

Conservation of momentum is a direct consequence of Newton's third law: in a collision the two objects exert equal and opposite forces on each other for the same time, so they receive equal and opposite changes of momentum, leaving the total unchanged. The explosion case is worth practising: a stationary firework or a recoiling rifle starts with zero total momentum, so after the event the forward and backward momenta must still add to zero, which is why a rifle recoils backwards as the bullet goes forwards. Note that while momentum is always conserved in a closed system, kinetic energy usually is not, because some is transferred to other stores such as thermal and sound during the collision.

Force and rate of change of momentum

This idea explains a wide range of everyday safety designs, and AQA often asks you to apply it to a new example. A gymnast lands on a thick crash mat, which squashes and lengthens the time taken to stop, so the force on the gymnast is smaller. A cricketer moves their hands back while catching a fast ball, lengthening the time to stop it and reducing the force on their hands. In every case the change in momentum is the same (the object is brought to rest), so the only way to reduce the force is to spread that momentum change over a longer time. The same equation also explains why being hit by a fast, heavy object hurts more: a larger momentum change in the same short time means a larger force.

Try this

Q1. State the equation for momentum and its unit. [2 marks]

  • Cue. p=mvp = mv, measured in kgm/skg\,m/s.

Q2. A 1500kg1500\,kg car moves at 20m/s20\,m/s. Calculate its momentum. [2 marks]

  • Cue. p=mv=1500×20=30000kgm/sp = mv = 1500 \times 20 = 30000\,kg\,m/s.

Exam-style practice questions

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

AQA 20205 marksA 1.0kg1.0\,\text{kg} trolley moving at 4.0m/s4.0\,\text{m/s} collides with and sticks to a stationary 3.0kg3.0\,\text{kg} trolley. Calculate the velocity of the combined trolleys immediately after the collision, and state the principle you have used.
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Use conservation of momentum: in a closed system the total momentum before equals the total momentum after (1 mark for stating the principle). Momentum before =(1.0×4.0)+(3.0×0)=4.0kg m/s= (1.0 \times 4.0) + (3.0 \times 0) = 4.0\,\text{kg m/s} (1 mark). After the collision the combined mass is 1.0+3.0=4.0kg1.0 + 3.0 = 4.0\,\text{kg} moving at velocity vv, so total momentum after =4.0v= 4.0v (1 mark). Setting before equal to after: 4.0v=4.04.0v = 4.0, so v=1.0m/sv = 1.0\,\text{m/s} (2 marks). Markers reward stating conservation of momentum, finding the momentum before, and solving for the common velocity. A frequent error is to forget the stationary trolley contributes zero momentum before.

AQA 20214 marksExplain, in terms of force and the rate of change of momentum, how a car's crumple zone reduces the force on the passengers during a collision.
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Force is equal to the rate of change of momentum, F=mΔvtF = \frac{m\Delta v}{t} (1 mark). In a collision the passengers' momentum changes by the same amount whether or not there is a crumple zone, because they are brought to rest (1 mark). A crumple zone crushes on impact, which increases the time over which the momentum change happens (1 mark). Because the same change of momentum now happens over a longer time, the rate of change of momentum is smaller, so the force on the passengers is smaller, reducing injury (1 mark). Markers reward citing F=mΔv/tF = m\Delta v / t, recognising the momentum change is fixed, and the link between longer time and smaller force. A common error is to say the crumple zone reduces the momentum change rather than extending the time.

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