Collisions

CIE A-Level Physics· Unit 3: Dynamics, Topic 4: Collisions· 20 min read

1. Classification of Collisions★★☆☆☆⏱ 10 min

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📘 Definition

Elastic collision

A collision where both total linear momentum and total kinetic energy are conserved across the collision. No kinetic energy is converted to other forms (heat, sound, deformation).

Example:

Collisions between ideal gas molecules

📘 Definition

Inelastic collision

A collision where total momentum is conserved, but total kinetic energy is not. Some kinetic energy is converted to other energy forms during the collision.

Example:

A car crash

📘 Definition

Perfectly inelastic collision

A special case of inelastic collision where the two colliding objects stick together after impact and move with a single common final velocity.

Example:

A bullet embedding in a wooden block

📐 Worked Example

A 2 kg mass moving at 3 m/s collides with a stationary 1 kg mass. After collision, the 2 kg mass moves at 1 m/s in the same direction, and the 1 kg mass moves at 4 m/s. Classify the collision.

1
Calculate total initial kinetic energy:
2
$KE_{initial} = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}(2)(3)^2 + 0 = 9 \text{ J}$
3
Calculate total final kinetic energy:
4
$KE_{final} = \frac{1}{2}(2)(1)^2 + \frac{1}{2}(1)(4)^2 = 1 + 8 = 9 \text{ J}$
5
Check momentum is conserved: Initial momentum = $2 \times 3 = 6$ kg m/s, final momentum = $(2 \times 1) + (1 \times 4) = 6$ kg m/s. Both quantities are conserved.
6
Conclusion: the collision is elastic.

2. Solving One-Dimensional Elastic Collisions★★★☆☆⏱ 15 min

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For all elastic collisions, we have two conservation laws to work with: conservation of momentum and conservation of kinetic energy. This gives us two equations to solve for two unknown final velocities:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \quad (\text{conservation of momentum})

\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \quad (\text{conservation of KE})

Rearranging these equations gives a very useful shortcut for exams: the speed of approach equals the speed of separation:

u_1 - u_2 = v_2 - v_1

📐 Worked Example

A 0.5 kg ball moving at 6 m/s right collides elastically with a 0.25 kg ball moving at 2 m/s right. Find the final velocities of both balls.

1
Define variables: $u_{1} = 6$ m/s, $u_{2} = 2$ m/s, $m_{1} = 0.5$ kg, $m_{2} = 0.25$ kg. We need $v_{1}$ and $v_{2}$ .
2
Apply conservation of momentum:
3
$(0.5)(6) + (0.25)(2) = 0.5v_1 + 0.25v_2 \implies 14 = 2v_1 + v_2$
4
Apply the speed of approach/separation relation:
5
$u_1 - u_2 = v_2 - v_1 \implies 6 - 2 = v_2 - v_1 \implies 4 = v_2 - v_1$
6
Solve the system: substitute $v_{2} = v_{1} + 4$ into the first equation: $14 = 2 v_{1} + v_{1} + 4 ⟹ 3 v_{1} = 10 ⟹ v_{1} = 3.33$ m/s. Then $v_{2} = 3.33 + 4 = 7.33$ m/s.

Exam tip:

Use the relative speed relation instead of rearranging kinetic energy from scratch to save 2+ minutes in exams.

3. Perfectly Inelastic Collisions★★☆☆☆⏱ 12 min

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In perfectly inelastic collisions, the two objects stick together after collision, so they share the same final velocity. This simplifies calculations, as we only have one unknown velocity, which we can find from conservation of momentum alone.

📘 Definition

Common final velocity

The single velocity of the combined mass of two objects that stick together after a perfectly inelastic collision.

📐 Worked Example

A 10 g bullet moving at 400 m/s is fired into a 1 kg stationary wooden block. Calculate the common velocity after collision, and the kinetic energy lost.

1
Convert to SI units: $m_{1} = 10$ g = $0.01$ kg, $m_{2} = 1$ kg, $u_{1} = 400$ m/s, $u_{2} = 0$ .
2
Apply conservation of momentum, where final mass is $m_{1} + m_{2}$ :
3
$m_1u_1 + m_2u_2 = (m_1 + m_2)v$
4
$(0.01)(400) + 0 = (1.01)v \implies v = \frac{4}{1.01} \approx 3.96 \text{ m/s}$
5
Calculate initial and final kinetic energy:
6
$KE_{initial} = \frac{1}{2}(0.01)(400)^2 = 800 \text{ J}$
7
$KE_{final} = \frac{1}{2}(1.01)(3.96)^2 \approx 7.92 \text{ J}$
8
Kinetic energy lost = $800 - 7.92 \approx 792$ J, converted to heat and deformation.

4. Two-Dimensional Collisions★★★★☆⏱ 15 min

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For oblique collisions in two dimensions, conservation of momentum applies separately to the perpendicular x and y components of momentum. If the collision is elastic, kinetic energy is also conserved.

📐 Worked Example

A stationary identical ball is struck by an identical ball moving at 5 m/s. After elastic collision, the stationary ball moves at 3 m/s at 90° to the original direction of the incident ball. Find the final speed of the incident ball.

1
Set original direction of the incident ball as the x-axis. Both masses are equal ( $m$ ).
2
Total y-momentum before collision is 0, so after collision:
3
$0 = m v_{1y} + m (3) \implies v_{1y} = -3 \text{ m/s}$
4
Total x-momentum before collision is $5 m$ , the stationary ball has 0 x-velocity after collision, so:
5
$5m = m v_{1x} + 0 \implies v_{1x} = 5 \text{ m/s}$
6
Use conservation of kinetic energy for elastic collision:
7
$\frac{1}{2}m (5)^2 = \frac{1}{2}m v_1^2 + \frac{1}{2}m (3)^2 \implies 25 = v_1^2 + 9 \implies v_1 = 4 \text{ m/s}$

Exam tip:

Always resolve momentum into perpendicular x and y components for 2D collision problems.

5. Common Pitfalls

Wrong move:

Assuming all collisions conserve kinetic energy

Why:

Only elastic collisions conserve kinetic energy; all inelastic collisions have kinetic energy loss

Correct move:

Only assume kinetic energy is conserved if the collision is explicitly stated as elastic, or you are asked to prove it is elastic

Wrong move:

Forgetting to convert masses from grams to kilograms for calculations

Why:

CIE commonly uses grams for bullet/block problems, leading to answers that are off by orders of magnitude

Correct move:

Always convert all quantities to SI units (kg for mass, m/s for velocity) before starting calculations

Wrong move:

Messing up the sign in the relative speed formula $u_{2} - u_{1} = v_{1} - v_{2}$

Why:

Direction sign errors lead to incorrect final velocities

Correct move:

Remember: speed of approach (initial relative speed) equals speed of separation (final relative speed), both are positive values

Wrong move:

Adding momentum magnitudes directly for 2D collisions

Why:

Momentum is a vector, so magnitudes do not add algebraically when directions differ

Correct move:

Always split momentum into perpendicular x and y components, apply conservation to each component separately

Wrong move:

Treating final velocities as separate unknowns for perfectly inelastic collisions

Why:

Wastes time solving unnecessary simultaneous equations

Correct move:

Use a single combined mass and one common final velocity, calculated from conservation of momentum alone

6. Quick Reference Cheatsheet

Collision Type	Momentum Conserved	Kinetic Energy Conserved	Key Rule
Elastic	Yes	Yes	Speed of approach = Speed of separation
Inelastic	Yes	No	Use momentum conservation only
Perfectly Inelastic	Yes	No	Objects stick, $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v$
2D Collision	Yes (per component)	Only if elastic	Resolve into x/y components first

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 · 12
Classify collision type MCQ
2023 · 22
1D elastic collision calculation
2021 · 11
2D collision energy check

Going deeper

syllabusOfficial CIE 9702 2025-2027 SyllabusSee page 28 for learning outcomes

What's Next

Collisions are a core application of momentum conservation, the foundational principle for many advanced topics in CIE A-Level Physics, including explosions, rocket propulsion, and nuclear particle interactions. Mastering collision problem solving builds the vector analysis and conservation law skills you need for all mechanics and modern physics topics. After completing this sub-topic, you can move on to related topics to extend your knowledge and prepare for exam questions.