A 0.20\,kg ball moving at 5.0\,m s^-1 is brought to rest in 0.10\,s. Find the average force on it. [3 marks]

WJEC WJEC 2020-style practice: A railway truck of mass 2000\,kg moving at 3.0\,m s^-1 collides with and couples to a stationary truck of mass 3000\,kg. Calculate the common velocity after the collision and determine whether the collision is elastic.

Apply conservation of momentum, then compare the kinetic energy before and after. Momentum before: p = m_1 u_1 = 2000 × 3.0 = 6000\,kg m s^-1 (the second truck is stationary). After coupling, the combined mass is 5000\,kg moving at v: 6000 = 5000 v, so v = 1.2\,m s^-1. Kinetic energy before: 12(2000)(3.0)^2 = 9000\,J. Kinetic energy after: 12(5000)(1.2)^2 = 3600\,J. Kinetic energy has fallen from 9000\,J to 3600\,J, so 5400\,J has transferred to heat and sound. Kinetic energy is not conserved, so the collision is inelastic. Markers reward the momentum calculation, the comparison of kinetic energies, and the conclusion that it is inelastic.

WalesPhysicsSyllabus dot point

How do forces change motion, and why is momentum conserved in a collision?

Newton's three laws of motion, force as the rate of change of momentum, conservation of momentum, and elastic and inelastic collisions.

A focused answer to WJEC A-Level Physics Unit 1 dynamics, covering Newton's three laws of motion, force as the rate of change of momentum, the principle of conservation of momentum, and the difference between 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

WJEC wants you to state and apply Newton's three laws, express force as the rate of change of momentum, use the principle of conservation of momentum, and distinguish elastic from inelastic collisions. Collision and explosion problems are a reliable source of marks, and the momentum form of the second law underpins the safety-engineering questions the examiners enjoy setting.

The answer

Newton's three laws

For constant mass the second law reduces to $F = ma$ , because $\dfrac{\Delta p}{\Delta t} = \dfrac{\Delta(mv)}{\Delta t} = m\dfrac{\Delta v}{\Delta t} = ma$ . The momentum form is the more general statement and is the one WJEC expects when the mass is changing, as in a rocket.

Momentum and impulse

Momentum is $p = mv$ , a vector measured in $\text{kg m s}^{-1}$ . The impulse of a force is $F\Delta t = \Delta p$ , the change in momentum, which equals the area under a force-time graph. Impulse explains why follow-through in sport matters: keeping the force on the ball for longer increases its change of momentum.

Conservation of momentum

It follows directly from Newton's third law: the equal and opposite forces during a collision act for the same time, so they produce equal and opposite impulses, and the momentum gained by one body equals the momentum lost by the other.

Elastic and inelastic collisions

In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved but kinetic energy is not, as some transfers to other forms such as heat, sound or deformation. A collision in which the bodies stick together is perfectly inelastic, which loses the most kinetic energy consistent with conserving momentum.

An explosion releasing two fragments

A stationary firework shell of mass $1.2\,\text{kg}$ explodes into two pieces. A $0.40\,\text{kg}$ fragment flies off at $30\,\text{m s}^{-1}$ . Find the velocity of the other fragment.

step Apply conservation of momentum

Total momentum before the explosion is zero (the shell is at rest), so the total after must also be zero.

step Set up the equation

The second fragment has mass $1.2 - 0.40 = 0.80\,\text{kg}$ and unknown velocity $v$ .

$0 = (0.40)(30) + (0.80)v$ , so $0.80 v = -12$ .

step Solve

$v = -15\,\text{m s}^{-1}$ . The minus sign means the larger fragment moves at $15\,\text{m s}^{-1}$ in the opposite direction, conserving the zero total momentum.

Examples in context

Example 1. Recoil of a rifle. A $4.0\,\text{kg}$ rifle fires a $0.020\,\text{kg}$ bullet at $400\,\text{m s}^{-1}$ . Momentum is conserved from rest, so $0 = (0.020)(400) + (4.0)v$ , giving a recoil velocity of $v = -2.0\,\text{m s}^{-1}$ . The rifle moves back slowly because its mass is much larger, which is why a shoulder can absorb the recoil.

Example 2. Pile driver. A $500\,\text{kg}$ hammer hits a pile at $4.0\,\text{m s}^{-1}$ and they move together afterward. With a $1500\,\text{kg}$ pile, momentum gives $(500)(4.0) = (2000)v$ , so $v = 1.0\,\text{m s}^{-1}$ . Kinetic energy drops from $4000\,\text{J}$ to $1000\,\text{J}$ , and the missing $3000\,\text{J}$ does the work of driving the pile into the ground, which is exactly the point of an inelastic impact here.

Try this

Q1. A $0.20\,\text{kg}$ ball moving at $5.0\,\text{m s}^{-1}$ is brought to rest in $0.10\,\text{s}$ . Find the average force on it. [3 marks]

Cue. $\Delta p = 0.20\times5.0 = 1.0\,\text{kg m s}^{-1}$ ; $F = \frac{\Delta p}{\Delta t} = \frac{1.0}{0.10} = 10\,\text{N}$ .

Q2. State what is conserved in an inelastic collision and what is not. [2 marks]

Cue. Momentum is conserved; kinetic energy is not.

Exam-style practice questions

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

WJEC 20205 marksA railway truck of mass

2000\,\text{kg}

moving at

3.0\,\text{m s}^{-1}

collides with and couples to a stationary truck of mass

3000\,\text{kg}

. Calculate the common velocity after the collision and determine whether the collision is elastic.

Show worked answer →

Apply conservation of momentum, then compare the kinetic energy before and after.

Momentum before: $p = m_1 u_1 = 2000 \times 3.0 = 6000\,\text{kg m s}^{-1}$ (the second truck is stationary).

After coupling, the combined mass is $5000\,\text{kg}$ moving at $v$ :

$6000 = 5000 v$ , so $v = 1.2\,\text{m s}^{-1}$ .

Kinetic energy before: $\tfrac{1}{2}(2000)(3.0)^2 = 9000\,\text{J}$ .

Kinetic energy after: $\tfrac{1}{2}(5000)(1.2)^2 = 3600\,\text{J}$ .

Kinetic energy has fallen from $9000\,\text{J}$ to $3600\,\text{J}$ , so $5400\,\text{J}$ has transferred to heat and sound. Kinetic energy is not conserved, so the collision is inelastic. Markers reward the momentum calculation, the comparison of kinetic energies, and the conclusion that it is inelastic.

WJEC 20173 marksState Newton's second law of motion in terms of momentum, and use it to explain why a longer collision time reduces the force on a car occupant.

Show worked answer →

Newton's second law states that the resultant force on a body equals the rate of change of its momentum, $F = \dfrac{\Delta p}{\Delta t}$ .

In a crash the occupant's change of momentum $\Delta p$ is fixed by their mass and the speed change. Because $F = \Delta p / \Delta t$ , increasing the time $\Delta t$ over which the momentum changes reduces the force $F$ .

Crumple zones, airbags and seat belts all increase the collision time, so the same change in momentum is spread over a longer time and the peak force on the occupant is lower. Markers reward the momentum form of the law and the inverse link between $\Delta t$ and $F$ .

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)