AQA AQA 2018-style practice: A trolley of mass 0.50 kg moving at 4.0 m s^-1 collides with a stationary trolley of mass 1.5 kg and they move off together. Calculate their common velocity and determine whether the collision is elastic.

Apply conservation of momentum, total before equals total after: p_before = (0.50)(4.0) = 2.0 kg m s^-1. After: (0.50 + 1.5)v = 2.0, so v = 2.02.0 = 1.0 m s^-1. Compare kinetic energy: before = 12(0.50)(4.0)^2 = 4.0 J; after = 12(2.0)(1.0)^2 = 1.0 J. Kinetic energy is not conserved (3.0 J is lost), so the collision is inelastic. Markers reward conservation of momentum for the common velocity and comparing kinetic energy before and after to classify the collision.

AQA AQA 2022-style practice: A ball of mass 0.15 kg travelling at 12 m s^-1 strikes a wall and rebounds straight back at 9.0 m s^-1. The contact time is 0.020 s. Calculate the magnitude of the average force exerted by the wall on the ball.

Take the initial direction as positive. The change in momentum is p = m(v - u) = 0.15(-9.0 - 12) = 0.15 × (-21) = -3.15 kg m s^-1. The average force is the rate of change of momentum: F = p t = -3.150.020 = -158 N. The magnitude of the force is 1.6 × 10^2 N, directed away from the wall (the negative sign shows it opposes the initial motion). Markers reward treating velocity as a vector with the rebound negative, the change in momentum, and dividing by the contact time.

EnglandPhysicsSyllabus dot point

What stays constant when objects collide or explode apart?

Momentum as mass times velocity, the principle of conservation of momentum, force as rate of change of momentum, impulse and the area under a force-time graph, and elastic and inelastic collisions.

A focused answer to AQA A-Level Physics 3.4.1.6, covering momentum as mass times velocity, the principle of conservation of momentum, force as the rate of change of momentum, impulse and the area under a force-time graph, and the difference between elastic and inelastic collisions.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

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What this dot point is asking
Momentum and its conservation
Force as rate of change of momentum
Impulse
Elastic and inelastic collisions
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What this dot point is asking

AQA specification point 3.4.1.6 wants you to define momentum, apply conservation of momentum to collisions and explosions in one dimension, use force as the rate of change of momentum, work with impulse and force-time graphs, and distinguish elastic from inelastic collisions.

Momentum and its conservation

For one-dimensional problems, assign one direction as positive and treat velocities in the opposite direction as negative. In an explosion starting from rest, the total momentum stays zero, so the fragments fly off with equal and opposite momenta; this is the principle behind rocket propulsion and recoil. Conservation of momentum follows directly from Newton's third law: the forces the colliding bodies exert on each other are equal and opposite and act for the same time, so the momentum gained by one equals the momentum lost by the other.

Force as rate of change of momentum

Newton's second law in its full form is $F = \dfrac{\Delta(mv)}{\Delta t}$ , so a force is whatever changes an object's momentum. For constant mass this reduces to $F = ma$ , because $\dfrac{\Delta(mv)}{\Delta t} = m\dfrac{\Delta v}{\Delta t} = ma$ . The full form is needed when mass changes, such as for a rocket burning fuel.

Impulse

This is why crumple zones and airbags increase the contact time $\Delta t$ : for the same change in momentum, a longer time means a smaller average force, reducing injury. A force-time graph that is not rectangular still gives the impulse as the area beneath it.

Elastic and inelastic collisions

In an elastic collision both momentum and kinetic energy are conserved (for example, near-elastic collisions between hard spheres or gas molecules). In an inelastic collision momentum is conserved but kinetic energy is not, with some converted to heat, sound or permanent deformation. When objects stick together and move off as one, the collision is perfectly inelastic and the maximum possible kinetic energy is lost.

Recoil of a rifle (an explosion problem)

A rifle of mass $4.0 \text{ kg}$ fires a bullet of mass $0.020 \text{ kg}$ at $400 \text{ m s}^{-1}$ . Find the recoil velocity of the rifle.

step 1: State conservation of momentum

The system starts at rest, so the total momentum after firing is also zero.

step 2: Write the momentum equation

$m_{\text{bullet}} v_{\text{bullet}} + m_{\text{rifle}} v_{\text{rifle}} = 0$ .

step 3: Substitute the values

$(0.020)(400) + (4.0)v_{\text{rifle}} = 0$ , so $8.0 + 4.0 \, v_{\text{rifle}} = 0$ .

step 4: Solve for the recoil velocity

$v_{\text{rifle}} = \dfrac{-8.0}{4.0} = -2.0 \text{ m s}^{-1}$ , that is $2.0 \text{ m s}^{-1}$ backwards.

Try this

Q1. Define the impulse of a force. [1 mark]

Cue. The product of force and the time for which it acts; it equals the change in momentum.

Q2. Explain how an airbag reduces the force on a passenger in a crash. [2 marks]

Cue. It increases the time over which the momentum changes, so for the same change in momentum the force is smaller.

Q3. State what is conserved in every collision. [1 mark]

Cue. Momentum (total momentum of the system).

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 20185 marksA trolley of mass

0.50 \text{ kg}

moving at

4.0 \text{ m s}^{-1}

collides with a stationary trolley of mass

1.5 \text{ kg}

and they move off together. Calculate their common velocity and determine whether the collision is elastic.

Show worked answer →

Apply conservation of momentum, total before equals total after: $p_{\text{before}} = (0.50)(4.0) = 2.0 \text{ kg m s}^{-1}$ .

After: $(0.50 + 1.5)v = 2.0$ , so $v = \dfrac{2.0}{2.0} = 1.0 \text{ m s}^{-1}$ .

Compare kinetic energy: before $= \tfrac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ J}$ ; after $= \tfrac{1}{2}(2.0)(1.0)^2 = 1.0 \text{ J}$ . Kinetic energy is not conserved ( $3.0 \text{ J}$ is lost), so the collision is inelastic.

Markers reward conservation of momentum for the common velocity and comparing kinetic energy before and after to classify the collision.

AQA 20224 marksA ball of mass

0.15 \text{ kg}

travelling at

12 \text{ m s}^{-1}

strikes a wall and rebounds straight back at

9.0 \text{ m s}^{-1}

. The contact time is

0.020 \text{ s}

. Calculate the magnitude of the average force exerted by the wall on the ball.

Show worked answer →

Take the initial direction as positive. The change in momentum is $\Delta p = m(v - u) = 0.15(-9.0 - 12) = 0.15 \times (-21) = -3.15 \text{ kg m s}^{-1}$ .

The average force is the rate of change of momentum: $F = \dfrac{\Delta p}{\Delta t} = \dfrac{-3.15}{0.020} = -158 \text{ N}$ .

The magnitude of the force is $1.6 \times 10^2 \text{ N}$ , directed away from the wall (the negative sign shows it opposes the initial motion).

Markers reward treating velocity as a vector with the rebound negative, the change in momentum, and dividing by the contact time.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)