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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9231Paper 3 · Further MechanicsMomentumMomentum and its conservation

Momentum · Subtopic 1 of 2

Momentum and its Conservation

Momentum is mass times velocity, and in any collision the total stays the same. One equation, "momentum before $=$ momentum after", solves the whole topic.

Cambridge 9709 Paper 4Cambridge 9231 Paper 3Edexcel YMA01 WME02Edexcel 9FM0 OptionAQA 7367 Paper 3≈ 30 minRevision 5 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Calculate momentum $mv$ , with its sign showing direction.
Find a change in momentum, including when a particle rebounds.
Use conservation of momentum to find an unknown speed after a collision.
Apply conservation to explosions and recoil, where the total starts at zero.

Pause and think

A lorry and a bicycle are both moving at $5\ \text{m s}^{-1}$ . Which is harder to stop, and why does "speed" not capture the difference?

Section 2 of 14

Core idea

Start herePool balls

A cue ball slams into a stationary ball and stops dead; the other rolls off at the cue ball's speed. Something was handed over completely: momentum. In any collision the total momentum stays the same.

The big idea

Momentum $=mv$ (kg m/s, a vector). With no external force, total momentum before = total after: $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ .

1. Why momentum is conserved

From Newton's laws

$F_{A\text{ on }B}=-F_{B\text{ on }A}$
Third law.
$m_B\Delta v_B=F\,\Delta t,\quad m_A\Delta v_A=-F\,\Delta t$
Second law: $F=m\frac{\Delta v}{\Delta t}$ , same contact time.
$m_A\Delta v_A+m_B\Delta v_B=0$
Add: total change zero.

Before and after a collision

Before:

3(6)+2(1)=20

. After:

3(2.5)+2v=20

, so

v=6.25

m/s.

Worked example3 kg at 6 m/s hits 2 kg at 1 m/s

After, the 3 kg moves at 2.5 m/s. Find the 2 kg's velocity.

Goal: $v$ .
Strategy: Conservation.

$3(6)+2(1)=20$
Before.
$3(2.5)+2v=20\Rightarrow v=6.25$
After.

Answer

6.25

m/s.

✓ CheckThe front ball now moves faster than the back one ✓ (they separate).

Worked exampleBullet into a block

A 0.05 kg bullet at 400 m/s embeds in a 2 kg block at rest on smooth ice.

Goal: Common speed.
Strategy: Combined mass after.

$0.05\times400=20$
Before.
$2.05v=20\Rightarrow v=9.76$
After.

Answer

9.76

m/s.

✓ CheckMuch slower than the bullet, much faster than rest ✓.

Pause and think

A 60 kg skater pushes a 40 kg skater from rest; the 40 kg moves at 3 m/s. The 60 kg?

Summary

$p=mv$ .
Total conserved (Newton's third law).

Momentum: Mass × velocity.

HistoryDescartes nearly had it. In 1644 René Descartes proposed that the total "quantity of motion" in the universe never changes, measuring it as mass times speed. That ignored direction, and his collision rules were wrong. In 1668 John Wallis, Christopher Wren and Christiaan Huygens independently sent the Royal Society the correct rules, with direction taken into account.

Fun factNewton's cradle. The desk toy with a row of hanging steel balls (named "Newton's cradle" by its British designer in 1967) shows conservation beautifully: pull back one ball and exactly one flies out the other end. Two out would also conserve momentum, but would break conservation of energy, so it never happens.

Section 3 of 14

Key formulae

Momentum

p=mv

Change in momentum

\Delta p=mv-mu

Conservation of momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

Explosion or recoil from rest

0=m_1v_1+m_2v_2

TrickBefore-and-after sketch. Draw two rows: "before" and "after". Put each particle as a circle with its mass inside and an arrow with its velocity above. Unknown velocities point in the positive direction; if the answer comes out negative, the particle is moving the other way.

Section 4 of 14

Worked examples

Example 1 · Rebound0 / 2 steps

A ball of mass

0.5

kg hits the floor at

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

and rebounds at

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

. Find the magnitude of its change in momentum.

Example 2 · Collision0 / 2 steps

Particle

A

(

0.4

kg) moving at

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

catches up with particle

B

(

0.6

kg) moving at

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

in the same direction. After the collision

A

moves at

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

. Find

B

's speed.

Example 3 · Recoil0 / 2 steps

A gun of mass

4

kg fires a bullet of mass

0.02

kg at

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

. Find the recoil speed of the gun.

Section 5 of 14

Explore the graph

Every outcome that conserves momentum

Loading the interactive graph…

Try this

Particle $A$ ( $2$ kg) at $u\ \text{m s}^{-1}$ hits $B$ ( $3$ kg) at rest. The horizontal axis is $A$ 's velocity after, the vertical axis $B$ 's. Every point on the line conserves momentum.
The point on the axis is "no collision": $A$ keeps going and $B$ stays put. The other point is where they stick together, both moving at $\tfrac{2u}{5}$ . Check it lies on the line.
Points with $x>y$ would mean $A$ passes through $B$ . Which part of the line is physically possible?

Conservation of momentum gives one equation, so on its own it allows a whole line of outcomes. The question always supplies one more fact.

In the real worldHow rockets move in empty space. A rocket in space has nothing to push against. It throws exhaust gas backwards at high speed, and conservation of momentum means the rocket gains the same momentum forwards. Recoil from a gun and a rocket engine are the same idea.

Section 6 of 14

Common mistakes

IncorrectChange in momentum on rebound

=m(v-u)

with both positive

Correct

m(v+u)

in size

The velocities have opposite signs.

IncorrectForgetting a particle that is moving before the collision

CorrectInclude every particle's

mu

Momentum before is the total of all of them.

IncorrectSpeeds instead of velocities

CorrectSigned velocities

Momentum is a vector.

IncorrectKinetic energy is conserved too

CorrectUsually not

Only momentum is guaranteed to be conserved.

Expectation

Momentum is conserved, easy

Reality

Forgot one particle was moving left, so its velocity is negative

Section 7 of 14

Exam tip

Exam tipSay "by conservation of momentum". Write the principle you are using before the equation. Then set out "before" on the left and "after" on the right, each term as mass

\times

velocity, with a clear positive direction. That layout earns the method mark even if a sign slips.

Section 8 of 14

Guided practice

SA-9709-P4-MO-G011 markL1 FoundationCalculatorMomentum

Find the momentum of a

0.5

kg ball moving at

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

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P4-MO1-0012 marksL1 FoundationCalculatorMomentum

Find the momentum of a car of mass

1200

kg travelling at

72

km/h.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-MO-0082 marksL2 StandardCalculatorMomentum

A ball of mass

0.2

kg hits a wall at

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

and rebounds along the same line at

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

. Find the magnitude of its change in momentum.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-MO1-0042 marksL2 StandardCalculatorMomentum

A ball of mass

0.5

kg hits the floor at

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

and rebounds vertically at

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

. Find the magnitude of its change in momentum.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-MO-0022 marksL1 FoundationCalculatorConservation

A (

2

kg) moving at

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

hits B (

3

kg) at rest. Afterwards A moves at

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

in the same direction. Find B's speed.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-MO1-0023 marksL1 FoundationCalculatorConservation

Particle

A

of mass

5

kg moving at

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

hits particle

B

of mass

2

kg, which is at rest. After the collision

B

moves at

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

in the direction

A

was moving. Find the speed of

A

after the collision.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-MO1-0033 marksL2 StandardCalculatorConservation

Particle

A

(

0.4

kg) moving at

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

catches up with particle

B

(

0.6

kg) moving at

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

in the same direction. After the collision

A

moves at

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

in its original direction. Find the speed of

B

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-MO1-0053 marksL2 StandardCalculatorConservation

A gun of mass

4

kg, initially at rest, fires a bullet of mass

0.02

kg horizontally at

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

. Find the speed with which the gun recoils.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-MO1-0063 marksL2 StandardCalculatorConservation

Two ice skaters, of mass

60

kg and

40

kg, stand at rest facing each other and push apart. The

40

kg skater moves off at

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

. Find the speed of the

60

kg skater.

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P4-MO-0093 marksL2 StandardCalculatorConservation

A (

4

kg,

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

) hits a particle B of mass

m

kg at rest. Afterwards A moves at

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

and B at

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

, both in A's original direction. Find

m

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Momentum $=mv$ , in N s; it carries a sign for direction.
Change in momentum $=mv-mu$ ; a rebound adds the speeds.
In any collision or explosion, total momentum before $=$ total after.
From rest, the total is zero, so the pieces move apart in opposite directions.

Section 12 of 14

Quick quiz

Q1 · SA-9709-P4-MO-Q011 markL1 FoundationCalculatorMomentum

The units of momentum are

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P4-MO-Q031 markL1 FoundationCalculatorMomentum

Momentum is

Salam Academy Original Practice · Proposed mark scheme

Q3 · SA-9709-P4-MO-Q021 markL1 FoundationCalculatorConservation

In a collision between two particles, which quantity is always conserved?

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 9709 AS & A Level Mathematics
Where: Paper 4, Mechanics · 4.4 Linear momentum
Expected: Use the definition of linear momentum and show understanding of its vector nature (in one dimension only); use conservation of linear momentum to solve problems that may be modelled as the direct impact of two bodies.
Calculator: Permitted. Use g = 10 m s⁻² unless told otherwise.
Typical style: Two particles collide; one velocity after the collision is given and the other must be found.
Open in: 9709 · Paper 4 9231 · Paper 3

Section 14 of 14

Past paper questions

5 questions (33 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

9709/42 · Feb/March 2025Question 57 marks
Collisions on a smooth plane, including particles that coalesce.
Question paper, page 8 Mark scheme
9709/42 · Oct/Nov 2024Question 59 marks
Two particles moving vertically, their speeds before colliding, and a speed at the ground.
Question paper, page 8 Mark scheme
9709/42 · Oct/Nov 2023Question 58 marks
Vertical motion, an impulse in a collision, and the time between two landings.
Question paper, page 6 Mark scheme
9709/42 · May/June 2023Question 24 marks
Collision on a smooth table: unknown speed and kinetic energy lost.
Question paper, page 3 Mark scheme
9709/42 · May/June 2022Question 15 marks
Speeds after a collision and the kinetic energy lost.
Question paper, page 2 Mark scheme

All 9709 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9231, Current syllabus (official specification).