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Cambridge International AS & A Level Mathematics 9709 · Mechanics

Momentum

Linear momentum as a vector, conservation of momentum in the direct impact of two bodies, bodies that coalesce, and collisions combined with other mechanics.

4.3Paper 4: MechanicsFree lesson

Take $g = 10 {m s}^{- 2}$ wherever the acceleration due to gravity is needed, as in the Paper 4 exam.

What the syllabus says

4.3 (Paper 4)

use the definition of linear momentum and show understanding of its vector nature
use conservation of linear momentum to solve problems that may be modelled as the direct impact of two bodies.

Notes and examples

For motion in one dimension only.
Including direct impact of two bodies where the bodies coalesce on impact.
Knowledge of impulse and the coefficient of restitution is not required.

Quoted from the Cambridge International AS and A Level Mathematics 9709 syllabus (2026-2027 and 2028-2030 editions, which share this content).

Momentum and its vector nature Paper 4

The linear momentum of a particle of mass $m$ moving with velocity $v$ is

$momentum = m v$

Its unit is the newton second, $N s$ (the same as ${kg m s}^{- 1}$ ). Momentum measures how hard it is to stop something: a slow lorry and a fast cricket ball can have the same momentum.

Velocity is a vector, so momentum is a vector, with the same direction as the velocity. In Paper 4 all momentum problems are in one dimension, along a straight line, so the direction is a sign. Choose a positive direction and give every velocity a sign. A ball of mass $0.2 kg$ moving at $15 {m s}^{- 1}$ towards a wall and bouncing back at $10 {m s}^{- 1}$ has momentum $3 N s$ before and $- 2 N s$ after, so its momentum changes by $- 2 - 3 = - 5 N s$ . Ignoring the sign would give a change of $1$ , which is wrong: the wall had to stop the ball and then send it back.

Check yourself Paper 4

Find the momentum of a ball of mass $0.15 kg$ moving at $24 {m s}^{- 1}$ .

Answer it in the portal, where your answer is checked as you type →

Conservation of linear momentum Paper 4

When two bodies collide, and no external force acts along the line of motion, the total momentum before the collision equals the total momentum after it.

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Why it works: during the impact, $A$ pushes $B$ and $B$ pushes $A$ . By Newton's third law these forces are equal and opposite, and they act for exactly the same time. So whatever momentum $A$ gives $B$ , $B$ takes the same amount from $A$ , and the total is unchanged. Gravity and the normal reaction act at right angles to the motion, so on a horizontal surface they do not affect it.

The syllabus calls this the direct impact of two bodies: they move along the same straight line before and after. Your method:

Draw a before-and-after diagram with every mass and velocity, and an arrow for the positive direction.
Write total momentum before $=$ total momentum after, with signs.
Solve, and interpret the sign of any unknown velocity: negative means it moves in the negative direction.

Momentum is conserved in every collision, but kinetic energy is usually not: some is lost as sound, heat and deformation. Questions often ask for the loss of kinetic energy; find the total before and the total after and subtract. Knowledge of impulse and of the coefficient of restitution is not required for this syllabus, so conservation of momentum is the only collision law you need.

Worked example 1 Paper 4

A sphere $A$ of mass $0.5 kg$ moving at $6 {m s}^{- 1}$ collides directly with a sphere $B$ of mass $0.3 kg$ at rest. After the collision $A$ moves at $2 {m s}^{- 1}$ in its original direction. Find the speed of $B$ after the collision.

Take the original direction of $A$ as positive.
Momentum before: $0.5 \times 6 + 0.3 \times 0 = 3 N s$ .
Momentum after: $0.5 \times 2 + 0.3 v = 1 + 0.3 v$ .
Conservation: $1 + 0.3 v = 3$ , so $v = \frac{2}{0.3} = \frac{20}{3} = 6.67 {m s}^{- 1}$ (3 s.f.).
It is reasonable that $B$ , the lighter sphere, moves off faster than $A$ was going.

Answer: $6.67 {m s}^{- 1}$

Directions after impact Paper 4

When the bodies move towards each other, one velocity is negative before the collision. After the collision either body may continue, stop or reverse. Do not guess: call the unknown velocity $v$ in the positive direction and let the algebra decide. A negative answer means the body moves the other way, and the question usually wants the speed and the direction in words.

If a question tells you a body "rebounds" or "reverses its direction", put that sign in yourself.

Worked example 2 Paper 4

Particles $P$ of mass $2 kg$ and $Q$ of mass $3 kg$ move towards each other along a straight line on a smooth horizontal surface, with speeds $4 {m s}^{- 1}$ and $2 {m s}^{- 1}$ . After they collide, $Q$ moves at $1.5 {m s}^{- 1}$ with its direction reversed. Find the speed and direction of $P$ after the collision.

Take the direction of $P$ 's original motion as positive, so $Q$ has velocity $- 2$ before and $+ 1.5$ after.
Momentum before: $2 \times 4 + 3 \times (- 2) = 8 - 6 = 2 N s$ .
Momentum after: $2 v + 3 \times 1.5 = 2 v + 4.5$ .
$2 v + 4.5 = 2$ , so $v = - 1.25$ .
$P$ moves at $1.25 {m s}^{- 1}$ , with its direction reversed. Both particles bounce back.

Answer: $1.25 {m s}^{- 1}$ , opposite to its original direction

Check yourself Paper 4

A particle $A$ of mass $4 kg$ moving at $5 {m s}^{- 1}$ collides directly with a particle $B$ of mass $1 kg$ moving towards it at $3 {m s}^{- 1}$ . After the collision $B$ moves at $6 {m s}^{- 1}$ in the direction in which $A$ was originally moving. Find the speed of $A$ after the collision.

Answer it in the portal, where your answer is checked as you type →

Particles that coalesce Paper 4

If the two bodies stick together on impact (they coalesce), they move off as one body of mass $m_{1} + m_{2}$ with a common velocity $V$ :

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$

Railway trucks that couple, a bullet that embeds in a block and two lumps of clay are all examples. Coalescing always loses kinetic energy, often a large share of it.

Worked example 3 Paper 4

A trolley of mass $1.2 kg$ moving at $3 {m s}^{- 1}$ runs into a trolley of mass $0.8 kg$ moving at $1 {m s}^{- 1}$ in the same direction, and they couple together. Find their common speed and the loss of kinetic energy.

Momentum before: $1.2 \times 3 + 0.8 \times 1 = 4.4 N s$ .
After: $(1.2 + 0.8) V = 2 V$ , so $V = 2.2 {m s}^{- 1}$ .
Kinetic energy before: $\frac{1}{2} \times 1.2 \times 9 + \frac{1}{2} \times 0.8 \times 1 = 5.4 + 0.4 = 5.8 J$ .
Kinetic energy after: $\frac{1}{2} \times 2 \times {2.2}^{2} = 4.84 J$ .
Loss $= 5.8 - 4.84 = 0.96 J$ .

Answer: $2.2 {m s}^{- 1}$ ; $0.96 J$ lost

Exam-style problems: momentum with other mechanics Paper 4

Paper 4 questions often join a collision to motion before or after it. The collision itself is always handled by conservation of momentum, but the velocities going into it, or the motion coming out of it, use the other topics:

Kinematics: a particle slides to the collision, or slows down afterwards, with constant acceleration.
Newton's laws and friction: after the collision a block moves onto a rough surface and decelerates at $μ g$ .
Energy: the loss of kinetic energy in the collision.

Keep the stages separate: before the impact, the impact, after the impact. Momentum is conserved only during the impact itself, not while friction is acting.

Worked example 4 (exam style) Paper 4

A particle $A$ of mass $0.4 kg$ moves at $5 {m s}^{- 1}$ on a smooth horizontal surface and collides directly with a particle $B$ of mass $0.6 kg$ , which is at rest. After the collision $A$ moves at $0.5 {m s}^{- 1}$ in its original direction.

(a) Find the speed of $B$ after the collision. [2]

(b) Find the loss of kinetic energy in the collision. [2]

(c) $B$ then moves onto a rough horizontal surface, where the coefficient of friction is $0.3$ . Find the distance $B$ travels on the rough surface before it stops. [3]

(a) Before: $0.4 \times 5 = 2 N s$ . After: $0.4 \times 0.5 + 0.6 v = 0.2 + 0.6 v$ . So $0.6 v = 1.8$ and $v = 3 {m s}^{- 1}$ .
(b) Before: $\frac{1}{2} \times 0.4 \times 25 = 5 J$ . After: $\frac{1}{2} \times 0.4 \times 0.25 + \frac{1}{2} \times 0.6 \times 9 = 0.05 + 2.7 = 2.75 J$ . Loss $= 2.25 J$ .
(c) On the rough surface $R = 0.6 \times 10 = 6 N$ and friction $= 0.3 \times 6 = 1.8 N$ , so the deceleration is $\frac{1.8}{0.6} = 3 {m s}^{- 2}$ .
$0 = 3^{2} - 2 \times 3 \times s$ , so $s = 1.5 m$ .

Answer: (a) $3 {m s}^{- 1}$ (b) $2.25 J$ (c) $1.5 m$

Check yourself Paper 4

Two particles, each of mass $2 kg$ , move towards each other with speeds $5 {m s}^{- 1}$ and $3 {m s}^{- 1}$ . They collide and coalesce. Find the speed of the combined particle.

Answer it in the portal, where your answer is checked as you type →

Key points

Momentum $= m v$ , measured in $N s$ ; it is a vector, so in one dimension it carries a sign.
In a direct impact, total momentum before $=$ total momentum after.
Coalescing: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$ .
Let an unknown velocity be positive and interpret a negative answer as a reversal.
Kinetic energy is usually lost in a collision; momentum never is.
Impulse and the coefficient of restitution are not needed for 9709.

Where marks are lost

Giving a body moving the other way a positive velocity.
Reporting a negative velocity as a speed, or not stating the direction when asked.
Using conservation of kinetic energy instead of momentum.
Forgetting to add the masses when the bodies coalesce.
Applying conservation of momentum across a stage where friction acts.
Finding a change in momentum by subtracting speeds rather than signed velocities.

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An original lesson written to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026 to 2027 and 2028 to 2030). The worked examples and questions are our own, not Cambridge questions. Back to every AS & A Level Mathematics 9709 lesson.