A 2.0\ kg trolley at 3.0\ m s^-1 collides with and sticks to a stationary 1.0\ kg trolley. Find the common velocity. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A 0.045\ kg golf ball is struck and leaves the club at 48\ m s^-1. The club is in contact with the ball for 0.50\ ms. Calculate the impulse on the ball and the average force exerted by the club.

Impulse equals the change in momentum: p = m v = 0.045 × 48 = 2.16\ N s (the ball starts from rest). Average force from impulse and contact time: F = p t = 2.160.50 × 10^-3 = 4\,320\ N, about 4.3\ kN. Markers reward the impulse 2.16\ N s, using F = ( p)/( t) with the time in seconds, and the force about 4.3\ kN.

WJEC Eduqas Eduqas 2021-style practice: A 1200\ kg car moving at 15\ m s^-1 collides head-on with an 1800\ kg van moving at 10\ m s^-1 in the opposite direction. They lock together after impact. Calculate their common velocity and determine whether the collision is elastic.

Take the car's direction as positive. Conservation of momentum: 1200(15) + 1800(-10) = (1200 + 1800)v. 18\,000 - 18\,000 = 3000v, so v = 0\ m s^-1: the wreckage is momentarily stationary. Kinetic energy before: 12(1200)(15)^2 + 12(1800)(10)^2 = 135\,000 + 90\,000 = 225\,000\ J. Kinetic energy after: 0\ J. Since kinetic energy is not conserved, the collision is inelastic. Markers reward correct signs, the common velocity 0\ m s^-1, comparing the kinetic energy before and after, and concluding the collision is inelastic.

EnglandPhysicsSyllabus dot point

Why is momentum conserved in collisions and explosions, and how does impulse explain the forces involved?

Momentum: linear momentum and its conservation, Newton's second law as rate of change of momentum, impulse and the force-time graph, and elastic versus inelastic collisions.

A focused answer to the Eduqas A-Level Physics Component 1 momentum content, covering linear momentum and its conservation, Newton's second law as the rate of change of momentum, impulse and the area under a force-time graph, and the distinction between elastic and inelastic collisions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Eduqas wants you to define linear momentum, apply the principle of conservation of momentum to collisions and explosions, express Newton's second law as the rate of change of momentum, define impulse and find it from a force-time graph, and distinguish elastic from inelastic collisions by testing whether kinetic energy is conserved.

The answer

Linear momentum

Conservation of momentum

Newton's second law as rate of change of momentum

Impulse and the force-time graph

Elastic and inelastic collisions

Recoil of a cannon

A $1500\ \text{kg}$ cannon fires a $12\ \text{kg}$ shell horizontally at $250\ \text{m s}^{-1}$ . Find the recoil speed of the cannon and the impulse on the shell.

step 1: Apply conservation of momentum

The system is at rest before firing, so total momentum is zero. Taking the shell's direction as positive: $0 = (12)(250) + (1500)(-v)$ .

step 2: Solve for the recoil speed

$1500v = 12 \times 250 = 3000$ , so $v = \dfrac{3000}{1500} = 2.0\ \text{m s}^{-1}$ backwards.

step 3: Impulse on the shell

Impulse equals the shell's change in momentum: $\Delta p = 12 \times 250 = 3000\ \text{N s}$ in the firing direction.

Examples in context

Conservation of momentum explains rocket and jet propulsion, the recoil of firearms, and the analysis of subatomic collisions in particle accelerators. Impulse reasoning drives vehicle safety design: crumple zones, airbags and seatbelts all extend the collision time to cut the peak force on occupants. In sport, a long follow-through in golf, tennis or cricket maximises the impulse and so the speed imparted to the ball.

Try this

Q1. Define linear momentum and state its unit. [2 marks]

Cue. Momentum is mass times velocity, $p = mv$ , measured in $\text{kg m s}^{-1}$ (equivalently $\text{N s}$ ).

Q2. A $2.0\ \text{kg}$ trolley at $3.0\ \text{m s}^{-1}$ collides with and sticks to a stationary $1.0\ \text{kg}$ trolley. Find the common velocity. [2 marks]

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

Q3. State what the area under a force-time graph represents. [1 mark]

Cue. The impulse, equal to the change in momentum.

Exam-style practice questions

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

Eduqas 20194 marksA

0.045\ \text{kg}

golf ball is struck and leaves the club at

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

. The club is in contact with the ball for

0.50\ \text{ms}

. Calculate the impulse on the ball and the average force exerted by the club.

Show worked answer →

Impulse equals the change in momentum: $\Delta p = m\Delta v = 0.045 \times 48 = 2.16\ \text{N s}$ (the ball starts from rest).

Average force from impulse and contact time: $F = \dfrac{\Delta p}{\Delta t} = \dfrac{2.16}{0.50 \times 10^{-3}} = 4\,320\ \text{N}$ , about $4.3\ \text{kN}$ .

Markers reward the impulse $2.16\ \text{N s}$ , using $F = \frac{\Delta p}{\Delta t}$ with the time in seconds, and the force about $4.3\ \text{kN}$ .

Eduqas 20215 marksA

1200\ \text{kg}

car moving at

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

collides head-on with an

1800\ \text{kg}

van moving at

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

in the opposite direction. They lock together after impact. Calculate their common velocity and determine whether the collision is elastic.

Show worked answer →

Take the car's direction as positive. Conservation of momentum: $1200(15) + 1800(-10) = (1200 + 1800)v$ .

$18\,000 - 18\,000 = 3000v$ , so $v = 0\ \text{m s}^{-1}$ : the wreckage is momentarily stationary.

Kinetic energy before: $\tfrac{1}{2}(1200)(15)^2 + \tfrac{1}{2}(1800)(10)^2 = 135\,000 + 90\,000 = 225\,000\ \text{J}$ . Kinetic energy after: $0\ \text{J}$ . Since kinetic energy is not conserved, the collision is inelastic.

Markers reward correct signs, the common velocity $0\ \text{m s}^{-1}$ , comparing the kinetic energy before and after, and concluding the collision is inelastic.

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)