Learn9702Paper 1 · Multiple ChoiceDynamicsConservation of momentum and collisions

Dynamics · Subtopic 2 of 2

Conservation of Momentum and Collisions

In any interaction with no external resultant force, total momentum is conserved, in every direction. In an elastic collision kinetic energy is also conserved, and the relative speed of approach equals the relative speed of separation. In an inelastic collision some kinetic energy becomes heat, sound or deformation. In two dimensions, conserve momentum separately in perpendicular directions.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH14AQA 7408 Paper 1≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Apply conservation of momentum to collisions and explosions in one dimension.
Distinguish elastic and inelastic collisions using kinetic energy.
Use relative speeds for perfectly elastic collisions.
Apply conservation of momentum in two dimensions.

Pause and think

Two identical cars, each at $20$ m s⁻¹, collide head-on and stop. Momentum before: $+20m-20m=0$ . After: $0$ . Momentum is conserved, but where did all the kinetic energy go?

Section 2 of 14

Core idea

Start hereNewton's cradle

Release one ball of a Newton's cradle and exactly one ball flies off the other end, at the same speed. Why not two balls at half speed? Momentum would be conserved either way; only one choice also conserves kinetic energy.

The big idea

Momentum always conserved (isolated system). Elastic: KE conserved too, and relative speed of approach = relative speed of separation.

1. Elastic collisions

Elastic collision of equal masses: they swap velocities

Newton's cradle works this way. Momentum

m\times5

and kinetic energy

\tfrac12m\times25

both pass entirely to the second ball.

Relative speeds in an elastic collision

$m_1(u_1-v_1)=m_2(v_2-u_2)$
Momentum, rearranged.
$m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)$
Kinetic energy (×2), rearranged.
$u_1+v_1=v_2+u_2$
Divide the second by the first (difference of squares).
$u_1-u_2=v_2-v_1$
Rearrange: approach = separation.

Worked example2 kg at 6 m s⁻¹ hits 1 kg at rest, elastic

Final velocities.

Goal: $v_1$ , $v_2$ .
Strategy: Momentum + relative speeds (two linear equations).

$12=2v_1+v_2$
Momentum.
$6=v_2-v_1$
Relative speeds.
$6=3v_1\Rightarrow v_1=2,\ v_2=8$
Subtract.

Answer

2

m s

^{-1}

and

8

m s

^{-1}

✓ CheckKE:

36

J before;

4+32=36

J after ✓.

2. Inelastic collisions

Kinetic energy before and after: elastic vs perfectly inelastic

Two 2 kg trolleys, one at 5 m/s. Sticking together:

v=2.5

m/s, KE

=\tfrac12(4)(6.25)=12.5

J. Half the KE is lost.

Worked exampleBallistic pendulum

10 g bullet embeds in a 2.0 kg block; block rises 5.0 cm.

Goal: Bullet speed.
Strategy: Work backwards: energy after, momentum during.

$v=\sqrt{2(9.81)(0.050)}=0.990$
KE → PE after impact.
$0.010u=2.010\times0.990$
Momentum in the collision (energy is not conserved here).
$u=199\text{ m s}^{-1}$
Solve.

Answer

199

m s

^{-1}

✓ CheckKE before

198

J, after

0.99

J: very inelastic ✓.

Common misconceptionEnergy across the impact Don't use energy conservation through a sticking collision.

Edge caseTwo dimensions Conserve momentum separately in

x

and

y

Summary

Momentum always; KE only if elastic.
Elastic: approach = separation.

Elastic: Kinetic energy conserved.

HistoryNewton's cradle. The desk toy with swinging steel balls is named after Newton, though it was popularised in the

1960

s. Lift one ball and release it: one ball swings out the other side. Lift two, and two swing out. Only that outcome conserves both momentum and kinetic energy.

Fun factChadwick finds the neutron. In

1932

, James Chadwick used conservation of momentum and energy in collisions to show that a mysterious radiation knocking protons out of wax must be a neutral particle with about the same mass as a proton: the neutron.

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