Calculate the momentum of a 1200\,kg car moving at 15\,m/s. [2 marks]

WJEC WJEC 2022-style practice: A 2.0\,kg trolley moving at 3.0\,m/s collides with a stationary 1.0\,kg trolley and they move off together. Calculate their common velocity.

A topic 2.4 Calculate question using conservation of momentum. Total momentum before = (2.0 × 3.0) + (1.0 × 0) = 6.0\,kg m/s (1 mark). After the collision the combined mass is 2.0 + 1.0 = 3.0\,kg (1 mark), and momentum is conserved, so 6.0 = 3.0 × v (1 mark), giving v = 2.0\,m/s (1 mark). Markers reward momentum before, the combined mass and the final velocity. A common error is to forget that the trolleys join together.

WalesPhysicsSyllabus dot point

What is momentum, and how is it conserved in collisions and explosions?

Momentum, the equation momentum equals mass times velocity, and conservation of momentum in collisions and explosions.

A focused answer to WJEC GCSE Physics topic 2.4 on momentum, covering the momentum equation, the conservation of momentum, how momentum applies to collisions and explosions, and the link between force, change in momentum and safety features.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-14

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

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What this topic is asking
Momentum
Conservation of momentum
Force and change in momentum
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What this topic is asking

WJEC wants you to define and calculate momentum and apply the conservation of momentum to collisions and explosions. This is topic 2.4 Further motion concepts in Unit 2 of WJEC GCSE Physics (3420).

Momentum

Because velocity carries a direction, momentum does too. When you add momenta in a collision, take one direction as positive and the opposite as negative. This matters when two objects approach from opposite directions: their momenta partly cancel, so the total momentum can be small even when both objects are moving fast.

Momentum is closely linked to Newton's second law. A resultant force acting for a time changes an object's momentum, and the bigger the force or the longer it acts, the bigger the change in momentum. This is why a small force applied over a long time can produce the same change in motion as a large force applied briefly, an idea that returns in the safety section below.

Conservation of momentum

In an explosion (such as a gun firing or two trolleys pushed apart by a spring), the total momentum before is zero, so the total momentum after must also be zero: the two parts move apart with equal and opposite momenta. The lighter part moves off faster, because for the momenta to be equal in size the smaller mass needs the larger velocity. This explains the recoil (kick back) of a gun: the bullet has a small mass and high velocity, the gun a large mass and small backward velocity, and the two momenta are equal and opposite.

Conservation of momentum is one of the most powerful tools in Unit 2 because it lets you find an unknown velocity without knowing anything about the forces during the collision. Set the total momentum before equal to the total momentum after, substitute the masses and velocities you know, and solve for the one you do not.

Force and change in momentum

Conservation of momentum in a collision

A $4.0\,\text{kg}$ trolley moving at $5.0\,\text{m/s}$ hits a stationary $1.0\,\text{kg}$ trolley and the two stick together. Find their velocity after the collision.

step 1: Find the total momentum before

$p_\text{before} = (4.0 \times 5.0) + (1.0 \times 0) = 20\,\text{kg m/s}$ .

step 2: Write the momentum after

The trolleys join, so the combined mass is $4.0 + 1.0 = 5.0\,\text{kg}$ , moving at velocity $v$ : $p_\text{after} = 5.0 \times v$ .

step 3: Apply conservation of momentum

Momentum is conserved, so $20 = 5.0 \times v$ , giving $v = \dfrac{20}{5.0}$ .

step 4: State the answer

$v = 4.0\,\text{m/s}$ in the original direction. Always set total momentum before equal to total momentum after.

Try this

Q1. Calculate the momentum of a $1200\,\text{kg}$ car moving at $15\,\text{m/s}$ . [2 marks]

Cue. $p = mv = 1200 \times 15 = 18\,000\,\text{kg m/s}$ .

Q2. State the principle of conservation of momentum. [1 mark]

Cue. The total momentum before an event equals the total momentum after, if no external force acts.

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 20203 marksA

0.50\,\text{kg}

ball moves at

4.0\,\text{m/s}

. Calculate its momentum.

Show worked answer →

A topic 2.4 Calculate question. Use $p = mv$ (1 mark). Substitute $m = 0.50\,\text{kg}$ and $v = 4.0\,\text{m/s}$ : $p = 0.50 \times 4.0$ (1 mark) $= 2.0\,\text{kg m/s}$ (1 mark for the answer with units). Markers reward the equation, the substitution and the unit kilogram metres per second. A common error is to omit the unit or use $\text{N}$ .

WJEC 20224 marksA

2.0\,\text{kg}

trolley moving at

3.0\,\text{m/s}

collides with a stationary

1.0\,\text{kg}

trolley and they move off together. Calculate their common velocity.

Show worked answer →

A topic 2.4 Calculate question using conservation of momentum. Total momentum before $= (2.0 \times 3.0) + (1.0 \times 0) = 6.0\,\text{kg m/s}$ (1 mark). After the collision the combined mass is $2.0 + 1.0 = 3.0\,\text{kg}$ (1 mark), and momentum is conserved, so $6.0 = 3.0 \times v$ (1 mark), giving $v = 2.0\,\text{m/s}$ (1 mark). Markers reward momentum before, the combined mass and the final velocity. A common error is to forget that the trolleys join together.

Related dot points

Sources & how we know this

WJEC GCSE Physics specification (3420) from 2016 — WJEC (2016)