Momentum

CIE A-Level Mathematics· 15 min read

1. Momentum and Impulse★★☆☆☆⏱ 5 min

📘 Definition

Linear Momentum

$p = m v$

A vector quantity with the same direction as the velocity of the object

Example:

A 2 kg mass moving at $3 m s^{- 1}$ has momentum $6 kg m s^{- 1}$

Impulse is defined as the change in momentum of an object, caused by a net force acting over a period of time. For a force-time graph, impulse equals the area under the graph.

I = \Delta p = F_{\text{avg}} \Delta t = m(v - u)

📐 Worked Example

A 5 kg ball moving at $8 m s^{- 1}$ is hit by a bat, reversing its direction to move at $12 m s^{- 1}$ . Calculate the impulse exerted on the ball.

1
1. Choose positive direction to match the final direction of the ball. Initial velocity is negative because it points opposite.
2
1. Calculate initial and final momentum:
3
$p_{\text{initial}} = mu = 5(-8) = -40 \text{ kg m s}^{-1}, \quad p_{\text{final}} = mv = 5(12) = 60 \text{ kg m s}^{-1}$
4
1. Calculate impulse as change in momentum:
5
$I = \Delta p = 60 - (-40) = 100 \text{ N s}$
6
The impulse exerted on the ball is 100 N s in the direction of the final velocity.

Exam tip:

Always define your positive direction first to avoid sign errors in momentum and impulse calculations.

2. Principle of Conservation of Momentum★★☆☆☆⏱ 6 min

📘 Definition

Conservation of Linear Momentum

For a closed system with no external resultant force, total momentum before a collision equals total momentum after a collision.

This principle applies to all collisions and explosions in one or two dimensions. For two colliding objects, the mathematical form is:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

📐 Worked Example

A 2 kg mass moving at $5 m s^{- 1}$ collides with a stationary 3 kg mass. After collision, the 2 kg mass moves at $2 m s^{- 1}$ in the same direction. Find the velocity of the 3 kg mass after collision.

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1. Define positive direction as the original direction of the 2 kg mass. Initial velocity of the 3 kg mass is 0.
2
1. Substitute values into the conservation of momentum formula:
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$(2)(5) + (3)(0) = (2)(2) + (3)v_2$
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1. Simplify and solve for $v_{2}$ :
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$10 = 4 + 3v_2 \implies 3v_2 = 6 \implies v_2 = 2 \text{ m s}^{-1}$
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The positive sign confirms the velocity is in the original direction, so the final velocity is $2 m s^{- 1}$ .

3. Problems involving Explosions★★★☆☆⏱ 4 min

Explosions are a common application of conservation of momentum. Before the explosion, the object is stationary so total initial momentum is zero. By conservation of momentum, total final momentum must also equal zero, so:

m_1v_1 + m_2v_2 = 0 \implies m_1v_1 = -m_2v_2

This means fragments move in opposite directions with equal magnitude momentum.

📐 Worked Example

A stationary shell of mass 4 kg explodes into two fragments, one of mass 1 kg and one of mass 3 kg. The 1 kg fragment moves away at $12 m s^{- 1}$ . Find the velocity of the 3 kg fragment.

1
1. Total initial momentum is zero, so total final momentum is zero. Let positive direction be the direction of the 1 kg fragment.
2
1. Substitute into the momentum equation:
3
$(1)(12) + (3)v = 0$
4
1. Solve for v:
5
$3v = -12 \implies v = -4 \text{ m s}^{-1}$
6
The negative sign means the 3 kg fragment moves in the opposite direction to the 1 kg fragment at $4 m s^{- 1}$ .

4. Elastic and Inelastic Collisions★★★☆☆⏱ 5 min

📘 Definition

Elastic vs Inelastic Collisions

Elastic: both momentum and kinetic energy are conserved. Inelastic: momentum is conserved, kinetic energy is not. Objects that stick together after collision are perfectly inelastic.

CIE exams frequently ask to determine if a collision is elastic by comparing total kinetic energy before and after the collision.

📐 Worked Example

For the collision in Section 2 (2 kg at 5 m/s hits stationary 3 kg, 2 kg moves at 2 m/s after, 3 kg moves at 2 m/s after), determine if the collision is elastic.

1
1. Calculate total kinetic energy before collision:
2
$KE_{\text{initial}} = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}(2)(5)^2 + 0 = 25 \text{ J}$
3
1. Calculate total kinetic energy after collision:
4
$KE_{\text{final}} = \frac{1}{2}(2)(2)^2 + \frac{1}{2}(3)(2)^2 = 4 + 6 = 10 \text{ J}$
5
1. Compare the two values: $K E_{final} < K E_{initial}$ , so kinetic energy is not conserved.
6
Conclusion: The collision is inelastic.

Exam tip:

Never assume a collision is elastic unless the question explicitly states it, or asks you to check.

5. Common Pitfalls

Wrong move:

Forgetting momentum is a vector, not using negative signs for opposite directions.

Why:

This leads to incorrect impulse and final velocity calculations, costing significant marks.

Correct move:

Always explicitly define a positive direction at the start of any momentum problem, then assign signs to velocities.

Wrong move:

Using conservation of kinetic energy for inelastic collisions.

Why:

Kinetic energy is only conserved in elastic collisions, so results will be wrong.

Correct move:

Only use conservation of kinetic energy if the collision is stated as elastic, or you are asked to check if it is elastic.

Wrong move:

Calculating impulse as final momentum instead of change in momentum.

Why:

This causes large errors when velocity reverses direction, as in the bat and ball example.

Correct move:

Always use $I = m v - m u$ , including the correct sign for initial velocity $u$ .

Wrong move:

Setting initial momentum to zero for collisions where objects are moving before impact.

Why:

Only stationary exploding objects have zero total initial momentum. This does not apply to general collisions.

Correct move:

Sum the momentum of all objects before collision with correct signs, do not assume it is zero.

Wrong move:

Forgetting to add masses when objects stick together after a perfectly inelastic collision.

Why:

After sticking, the combined mass is the sum of the two individual masses, not just one.

Correct move:

Write final momentum as $(m_{1} + m_{2}) v$ when two objects stick together after collision.

6. Quick Reference Cheatsheet

Concept	Formula	Key Note
Linear Momentum	$p = m v$	Vector: sign = direction
Impulse	$I = Δ p = F_{avg} Δ t$	Equals area under F-t graph
Conservation of Momentum	$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$	Requires no external force
Elastic Collision	p and KE both conserved	Check by comparing KE before/after
Stationary Explosion	$m_{1} v_{1} + m_{2} v_{2} = 0$	Fragments move opposite directions

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.

2024 · 3
Impulse and collision calculation
2023 · 3
Explosion momentum problem
2022 · 3
Check collision elasticity

What's Next

Momentum is a core foundation for further topics in advanced CIE A-level mechanics, including two-dimensional collisions, connected particles, and energy analysis of interacting systems. Mastering sign conventions for momentum and impulse will help you avoid common errors in multi-topic questions that frequently appear in Paper 3. Most longer exam problem questions combine momentum concepts with energy considerations, so building a strong understanding of this sub-topic is critical for achieving high marks on your exam.