A trolley of mass 2.0\ kg moving at 3.0\ m s^-1 collides and sticks to a stationary trolley of mass 1.0\ kg. Find their common velocity afterwards. [3 marks]

A car accelerates uniformly from rest to 24\ m s^-1 in 8.0\ s. Find the distance travelled. [2 marks]

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How do Newton's laws and momentum describe and predict motion?

The equations of motion for uniform acceleration, Newton's three laws, linear momentum and impulse, and the conservation of momentum in collisions.

A CCEA A-Level Physics answer on the equations of motion for uniform acceleration, Newton's three laws, linear momentum and impulse, and the conservation of momentum in elastic and inelastic collisions.

Generated by Claude Opus 4.811 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

CCEA wants you to use the equations of motion for uniform acceleration, state and apply Newton's three laws, define linear momentum and impulse, and apply the conservation of momentum to collisions and explosions. Collision calculations and impulse problems appear in nearly every paper.

The answer

The equations of motion

For projectiles, treat the horizontal and vertical motions separately: the horizontal velocity is constant (no horizontal force) and the vertical motion has acceleration $g$ downward.

Newton's three laws

Momentum, impulse and conservation

In an elastic collision kinetic energy is also conserved; in an inelastic collision momentum is conserved but some kinetic energy is transferred to other forms (heat, sound, deformation). An explosion is the reverse of a perfectly inelastic collision: total momentum stays zero if it started at rest.

Worked example: an explosion (recoil)

Recoil of a trolley spring system

A $0.80\ \text{kg}$ trolley and a $0.20\ \text{kg}$ trolley are held together against a compressed spring, at rest. When released they fly apart. The lighter trolley moves off at $4.0\ \text{m s}^{-1}$ . Find the speed of the heavier trolley.

step Apply conservation of momentum

The total momentum before release is zero, so it must be zero afterwards:

0 = m_1 v_1 + m_2 v_2.

step Substitute the values

Taking the light trolley's direction as positive:

0 = (0.20)(4.0) + (0.80)v_2 \Rightarrow 0 = 0.80 + 0.80\, v_2.

step Solve

v_2 = -\frac{0.80}{0.80} = -1.0\ \text{m s}^{-1}.

The heavier trolley recoils at

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

in the opposite direction, four times slower than the light one because it is four times heavier.

Examples in context

Example 1. A car crumple zone. Crumple zones and airbags increase the time over which a passenger's momentum is brought to zero. Because $F = \Delta p / \Delta t$ , a longer contact time means a smaller average force for the same change in momentum, reducing injury, an everyday application of the impulse relation.

Example 2. A rocket in space. A rocket ejects exhaust gas backwards at high speed; conservation of momentum means the rocket gains forward momentum equal and opposite to the gas. With no external force in space, this is the only way to accelerate, and it is the same physics as the recoiling trolley scaled up.

Try this

Q1. A trolley of mass $2.0\ \text{kg}$ moving at $3.0\ \text{m s}^{-1}$ collides and sticks to a stationary trolley of mass $1.0\ \text{kg}$ . Find their common velocity afterwards. [3 marks]

Cue. $2.0 \times 3.0 = (2.0 + 1.0)v$ , so $v = 2.0\ \text{m s}^{-1}$ .

Q2. State Newton's second law in terms of momentum. [1 mark]

Cue. The resultant force equals the rate of change of momentum.

Q3. A car accelerates uniformly from rest to $24\ \text{m s}^{-1}$ in $8.0\ \text{s}$ . Find the distance travelled. [2 marks]

Cue. $s = \tfrac{1}{2}(u + v)t = \tfrac{1}{2}(0 + 24)(8.0) = 96\ \text{m}$ .

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 20206 marksA ball of mass 0.15 kg moving at 8.0 m per second strikes a wall at right angles and rebounds at 6.0 m per second in the opposite direction. The contact lasts 0.025 s. Calculate the change in momentum of the ball and the average force exerted by the wall on the ball, and state the force exerted by the ball on the wall.

Show worked answer →

Take the initial direction as positive. The change in momentum is

$\Delta p = m v - m u = 0.15 \times (-6.0) - 0.15 \times (8.0) = -0.90 - 1.20 = -2.10$ kg m per second.

The magnitude of the change is 2.10 kg m per second (directed away from the wall).

The average force from impulse equals change in momentum:

$F = \frac{\Delta p}{\Delta t} = \frac{2.10}{0.025} = 84$ N.

By Newton's third law the ball exerts an equal and opposite force of 84 N on the wall.

Markers reward correct signs in the momentum change (the rebound reverses direction), the impulse equation, and the third-law statement.

CCEA 20185 marksA 1500 kg car travelling at 20 m per second collides head-on with a stationary 1000 kg car. The two lock together. Calculate their common velocity immediately after the collision, and determine the kinetic energy lost. State whether the collision is elastic or inelastic.

Show worked answer →

Conserving momentum:

$m_1 u_1 = (m_1 + m_2)v$ , so $1500 \times 20 = (1500 + 1000)v$ , giving $v = \frac{30000}{2500} = 12$ m per second.

Kinetic energy before: $\tfrac{1}{2} \times 1500 \times 20^2 = 3.0 \times 10^{5}$ J.

Kinetic energy after: $\tfrac{1}{2} \times 2500 \times 12^2 = 1.8 \times 10^{5}$ J.

Kinetic energy lost: $3.0 \times 10^{5} - 1.8 \times 10^{5} = 1.2 \times 10^{5}$ J.

The collision is inelastic because kinetic energy is not conserved (momentum still is).

Markers reward conservation of momentum, both kinetic energies, the energy lost, and the inelastic classification.

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Sources & how we know this

CCEA GCE Physics specification — CCEA (2016)