Find the momentum of a 4\,kg mass moving at 3\,m s^-1. [1 mark]

Two equal masses coalesce; one was at 6\,m s^-1, the other at rest. Find their common speed. [2 marks]

CCEA CCEA A2 2020-style practice: A particle of mass 3\,kg moving at 4\,m s^-1 collides directly with a particle of mass 2\,kg moving in the opposite direction at 1\,m s^-1. They coalesce on impact. Find their common velocity after the collision.

Take the direction of the 3\,kg particle as positive, so the 2\,kg particle has velocity -1\,m s^-1. Conservation of momentum: total momentum before equals total momentum after. Let the common velocity be v: 3(4) + 2(-1) = (3 + 2)v. 12 - 2 = 5v 10 = 5v v = 2\,m s^-1. The combined mass moves at 2\,m s^-1 in the original direction of the 3\,kg particle. Markers reward the sign convention, the conservation-of-momentum equation, and the common velocity of 2\,m s^-1.

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How do momentum and impulse describe collisions, and what does the coefficient of restitution add?

Linear momentum and impulse, conservation of momentum in collisions, Newton's experimental law with the coefficient of restitution, and kinetic energy lost in impacts.

A CCEA A2 Further Maths Mechanics answer on linear momentum and impulse, conservation of momentum in collisions, Newton's experimental law with the coefficient of restitution, and the kinetic energy lost in an impact.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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 linear momentum and impulse, apply the conservation of momentum to direct collisions, use Newton's experimental law with the coefficient of restitution $e$ , and find the kinetic energy lost in an impact.

The answer

Momentum and impulse

Conservation of momentum

Newton's experimental law (restitution)

Kinetic energy lost

Impulse on a struck ball

A ball of mass $0.2\,\text{kg}$ moving at $6\,\text{m s}^{-1}$ is struck and rebounds straight back at $9\,\text{m s}^{-1}$ . Find the magnitude of the impulse on the ball.

step Choose a positive direction

Take the initial direction as positive: $u = 6\,\text{m s}^{-1}$ and $v = -9\,\text{m s}^{-1}$ (rebounds the other way).

step Apply the impulse-momentum principle

J = mv - mu = 0.2(-9) - 0.2(6) = -1.8 - 1.2 = -3.0\,\text{N s}.

step State the magnitude

The impulse has magnitude $3.0\,\text{N s}$ , directed opposite to the original motion. The sign change of velocity is what makes the impulse larger than for a ball that simply stops.

Examples in context

Example 1. Crumple zones in cars. A crumple zone lengthens the time over which a crash brings the car to rest, and since impulse $= Ft =$ change in momentum, a longer time means a smaller force on the occupants. The impulse-momentum principle is a safety-engineering tool.

Example 2. Newton's cradle. The clicking spheres approximate $e = 1$ collisions, so both momentum and kinetic energy are very nearly conserved, which is why one ball out gives one ball out. Real cradles slowly stop because $e$ is just below $1$ and a little energy is lost each impact.

Try this

Q1. Find the momentum of a $4\,\text{kg}$ mass moving at $3\,\text{m s}^{-1}$ . [1 mark]

Cue. $p = mv = 4 \times 3 = 12\,\text{kg m s}^{-1}$ .

Q2. What value of $e$ describes a perfectly elastic collision? [1 mark]

Cue. $e = 1$ .

Q3. Two equal masses coalesce; one was at $6\,\text{m s}^{-1}$ , the other at rest. Find their common speed. [2 marks]

Cue. $m(6) + m(0) = 2mv$ , so $v = 3\,\text{m s}^{-1}$ .

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 A2 20207 marksA particle of mass

3\,\text{kg}

moving at

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

collides directly with a particle of mass

2\,\text{kg}

moving in the opposite direction at

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

. They coalesce on impact. Find their common velocity after the collision.

Show worked answer →

Take the direction of the $3\,\text{kg}$ particle as positive, so the $2\,\text{kg}$ particle has velocity $-1\,\text{m s}^{-1}$ .

Conservation of momentum: total momentum before equals total momentum after. Let the common velocity be $v$ :

$3(4) + 2(-1) = (3 + 2)v.$

$12 - 2 = 5v \Rightarrow 10 = 5v \Rightarrow v = 2\,\text{m s}^{-1}.$

The combined mass moves at $2\,\text{m s}^{-1}$ in the original direction of the $3\,\text{kg}$ particle.

Markers reward the sign convention, the conservation-of-momentum equation, and the common velocity of $2\,\text{m s}^{-1}$ .

CCEA A2 20188 marksA smooth sphere

A

of mass

2\,\text{kg}

moving at

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

collides directly with a stationary smooth sphere

B

of mass

3\,\text{kg}

. The coefficient of restitution is

0.6

. Find the velocities of

A

and

B

after the collision.

Show worked answer →

Let the velocities after be $v_A$ and $v_B$ (positive in the original direction of $A$ ).

Conservation of momentum: $2(5) + 3(0) = 2v_A + 3v_B$ , so $10 = 2v_A + 3v_B$ .

Newton's experimental law: the speed of separation equals $e$ times the speed of approach. Approach speed $= 5 - 0 = 5$ , so separation speed $v_B - v_A = 0.6 \times 5 = 3$ .

From the second equation $v_B = v_A + 3$ . Substitute: $10 = 2v_A + 3(v_A + 3) = 5v_A + 9$ , so $v_A = 0.2\,\text{m s}^{-1}$ and $v_B = 3.2\,\text{m s}^{-1}$ .

Both are positive, so both move in the original direction. Markers reward the momentum equation, the restitution equation, and both final velocities.

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Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)