Conservation of momentum

CIE A-Level Physics· Unit 3: Dynamics· 25 min read

1. The Principle of Conservation of Momentum★★☆☆☆⏱ 8 min

📘 Definition

Principle of Conservation of Linear Momentum

For a closed system of interacting objects, the total linear momentum in any given direction remains constant, provided no net external force acts on the system.

This principle is derived directly from Newton's Third Law. When two objects interact, the force each exerts on the other is equal and opposite, so the total change in momentum of the system is zero.

\sum p_{initial} = \sum p_{final} \quad \text{where} \quad p = mv

📐 Worked Example

A 2 kg mass moving at 3 m s⁻¹ to the right collides with a stationary 1 kg mass. What is the total momentum after the collision?

1
Assign right as the positive direction
2
Calculate total initial momentum:
3
$p_{initial} = m_1 v_1 + m_2 v_2 = (2 \times 3) + (1 \times 0) = 6 \text{ kg m s}^{-1}$
4
By conservation of momentum, total final momentum equals total initial momentum
5
Final total momentum = 6 kg m s⁻¹

Exam tip:

Always state the full principle when answering structured CIE questions to gain full marks.

2. Elastic and Inelastic Collisions★★★☆☆⏱ 9 min

All collisions obey conservation of momentum (for closed systems), but they are classified based on whether kinetic energy is conserved.

📘 Definition

Elastic Collision

A collision where both total momentum and total kinetic energy are conserved

Example:

Ideal billiard ball collisions, collisions between gas molecules

📘 Definition

Inelastic Collision

A collision where total momentum is conserved, but kinetic energy is not. Kinetic energy is converted to heat, sound, or deformation energy

Example:

A car crash, a ball sticking to a surface after impact

📐 Worked Example

A 0.5 kg ball moving at 4 m s⁻¹ collides head-on with a stationary 0.5 kg ball. After collision, the first ball stops, and the second moves at 4 m s⁻¹. Show the collision is elastic.

1
Check momentum is conserved:
2
$p_{initial} = (0.5 \times 4) + (0.5 \times 0) = 2 \text{ kg m s}^{-1} \\ p_{final} = (0.5 \times 0) + (0.5 \times 4) = 2 \text{ kg m s}^{-1}$
3
Calculate initial and final kinetic energy:
4
$KE_{initial} = \frac{1}{2}(0.5)(4)^2 + 0 = 4 \text{ J} \\ KE_{final} = 0 + \frac{1}{2}(0.5)(4)^2 = 4 \text{ J}$
5
Kinetic energy is unchanged, so the collision is elastic

Exam tip:

CIE almost always asks to distinguish between elastic and inelastic collisions, so memorize both definitions.

3. Explosions and Recoil★★★☆☆⏱ 8 min

Explosions are the reverse of collisions: a single stationary object splits into two or more parts. Total initial momentum is zero, so by conservation of momentum, total final momentum must also be zero. The parts move in opposite directions with equal magnitude momentum, which causes recoil of the larger object.

📐 Worked Example

A stationary cannon of mass 1000 kg fires a 10 kg cannonball at 150 m s⁻¹. Calculate the recoil speed of the cannon.

1
Assign the cannonball's direction as positive. Total initial momentum = 0, since the cannon is stationary.
2
Apply conservation of momentum:
3
$0 = m_{ball} v_{ball} + m_{cannon} v_{cannon}$
4
Rearrange to solve for recoil velocity:
5
$v_{cannon} = - \frac{m_{ball} v_{ball}}{m_{cannon}} = - \frac{10 \times 150}{1000} = -1.5 \text{ m s}^{-1}$
6
The negative sign indicates the cannon moves opposite to the cannonball. Recoil speed is 1.5 m s⁻¹

4. Common Pitfalls

Wrong move:

Ignoring that momentum is a vector, so signs for opposite directions are omitted

Why:

This leads to incorrect total momentum calculations for objects moving in opposite directions

Correct move:

Assign a positive direction at the start of the problem, and use negative signs for velocity in the opposite direction

Wrong move:

Claiming kinetic energy is conserved in all collisions

Why:

Only elastic collisions conserve kinetic energy; nearly all real collisions are inelastic

Correct move:

Only state kinetic energy is conserved if the question explicitly identifies the collision as elastic

Wrong move:

Applying conservation of momentum to systems with external net forces

Why:

The principle only holds for closed systems with no net external force

Correct move:

Check for unbalanced external forces (e.g. friction) before applying the principle

Wrong move:

Confusing mass and velocity when calculating recoil speed after an explosion

Why:

Recoil speed is inversely proportional to mass for a fixed momentum change

Correct move:

Set total initial momentum to zero, then solve for the unknown speed using equal and opposite momentum

5. Quick Reference Cheatsheet

Concept	Key Rule	Formula
Conservation of momentum	Closed system, no external force	$\sum p_{ini t ia l} = \sum p_{f ina l}$
Elastic collision	Momentum and KE conserved	$K E_{ini t ia l} = K E_{f ina l}$
Inelastic collision	Only momentum conserved	$K E_{ini t ia l} > K E_{f ina l}$
Stationary explosion	Total initial $p = 0$	$m_{1} v_{1} = - m_{2} v_{2}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 11
Conservation of momentum collision
2023 · 22
Explosion recoil calculation
2024 · 13
Elastic vs inelastic collision

Going deeper

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What's Next

Conservation of momentum is a fundamental conservation law that extends to two-dimensional problems and is used across many areas of physics, including nuclear reactions, where it helps calculate the momentum of emitted particles such as alpha particles. It also connects back to Newton's laws of motion and underpins concepts like impulse and force interactions. Mastery of one-dimensional momentum problems is essential for all dynamics questions in the CIE A-Level exam, and prepares you for more complex problems in further topics.