LearnYMA01WME01 · Mechanics 1Dynamics of a particle in a straight lineMomentum and impulse

Dynamics of a particle in a straight line · Subtopic 4 of 4

Momentum and Impulse

Momentum is mass × velocity, so it has a direction. An impulse changes momentum: $I=mv-mu=Ft$ . When two particles collide, the total momentum before equals the total after, as long as no external impulse acts.

Edexcel YMA01 WME01Edexcel 9FM0 OptionAQA 7367 Paper 3≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Calculate momentum, with direction.
Use impulse $=$ change in momentum $=Ft$ .
Use conservation of momentum in collisions.
Handle rebounds, coalescence and particles moving towards each other.

Pause and think

A ball hits a wall at $15$ m s⁻¹ and bounces back at $10$ m s⁻¹. Is the change in velocity $5$ m s⁻¹?

Section 2 of 14

Core idea

Start hereCatching an egg

Catch an egg by stopping your hand dead and it breaks; let your hand move back with it and it survives. Either way the egg loses the same momentum. What changes is how long that takes, and so the size of the force.

The big idea

Momentum $mv$ . Impulse $I=Ft=mv-mu$ . In collisions total momentum is conserved; impulses on the two bodies are equal and opposite.

1. Impulse

Impulse from F = ma

$F=ma=m\frac{v-u}t$
Constant acceleration.
$Ft=mv-mu$
Multiply by $t$ .

Impulse (N s) = change in momentum (kg m s⁻¹).

Worked exampleBall hits a wall

0.5 kg at 8 m s $^{-1}$ rebounds at 6 m s $^{-1}$ . Impulse; average force if contact lasts 0.05 s.

Goal: $I$ , $F$ .
Strategy: Signs: towards the wall positive.

$u=8,\ v=-6$
Rebound is negative.
$I=0.5(-6-8)=-7\text{ N s}$
Magnitude $7$ N s, away from the wall.
$F=7/0.05=140\text{ N}$
$I=Ft$ .

Answer

7

N s;

140

✓ CheckIgnoring the sign would give

1

N s: far too small ✓.

2. Conservation

Before and after: momentum is conserved

Right positive. Before:

3(4)+2(-2)=8

. After:

3v_1+2(3)=8

, so

v_1=\tfrac23

m/s. It is positive, so the 3 kg ball keeps moving to the right, more slowly.

Why momentum is conserved

$I_{A}=-I_{B}$
Newton's third law: equal and opposite forces for the same time.
$\Delta(m_Av_A)+\Delta(m_Bv_B)=0$
Changes cancel.

m_Au_A+m_Bu_B=m_Av_A+m_Bv_B

Worked exampleHead-on collision

3 kg at 4 m s $^{-1}$ meets 2 kg at 2 m s $^{-1}$ coming the other way; the 2 kg rebounds at 3 m s $^{-1}$ . Find the 3 kg's velocity and the impulse.

Goal: $v_1$ , $I$ .
Strategy: Conservation with signs.

$3(4)+2(-2)=8$
Before.
$3v_1+2(3)=8\Rightarrow v_1=\tfrac23$
After.
$I=2(3-(-2))=10\text{ N s}$
On the 2 kg.

Answer

\frac23

m s

^{-1}

(same direction);

10

N s.

✓ CheckOn the 3 kg:

3(\frac23-4)=-10

✓ equal and opposite.

Common misconceptionForgetting direction Velocities in opposite directions need opposite signs.

Pause and think

Two particles stick together: 2 kg at 6 and 4 kg at rest. Common speed?

Summary

$I=Ft=mv-mu$ .
Total momentum conserved.

Impulse: Change in momentum.
Momentum: $mv$ .

HistoryDescartes' "quantity of motion". René Descartes proposed in 1644 that the universe's total "quantity of motion" is conserved, but he used speed rather than velocity, so his version was wrong. Christiaan Huygens and later Newton fixed it by giving motion a direction.

Fun factNewton's cradle. Lift one ball of a Newton's cradle and let go: one ball flies out the far side. Two in, two out. Only that outcome conserves both momentum and kinetic energy for equal steel balls.

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