Collisions (Mechanics 2)

Q: What value of e do I use for perfectly elastic collisions?

Use for perfectly elastic collisions, where no kinetic energy is lost. For perfectly inelastic collisions where particles stick together after impact, use .

Edexcel International A-Level Mathematics· 2018 Issue 3, M2 §4.1-4.3· 25 min read

1. Vector Momentum and Impulse-Momentum Principle★★☆☆☆⏱ 5 min

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📘 Definition

Linear Momentum

\vec{p} = m\vec{v}

Vector quantity equal to the product of a particle's mass and velocity, measured in kg m s⁻¹.

Example:

A 2 kg particle moving at 3 m s⁻¹ right has momentum $6 i$ kg m s⁻¹.

For any collision, the impulse exerted on a particle equals its change in momentum (the impulse-momentum principle). For a closed system of colliding particles with no external forces acting, total linear momentum is conserved: total momentum before impact equals total momentum after impact.

📐 Worked Example

A 2 kg particle moving at $3 i$ m s⁻¹ collides with a 3 kg particle moving at $- 1 i$ m s⁻¹. After collision, the 2 kg particle moves at $- 1.5 i$ m s⁻¹. Find the velocity of the 3 kg particle after collision, and the impulse exerted on the 2 kg particle.

1
Apply conservation of linear momentum: total momentum before = total momentum after
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
2
Substitute known values into the equation
$2(3\vec{i}) + 3(-1\vec{i}) = 2(-1.5\vec{i}) + 3v_2$
3
$6\vec{i} - 3\vec{i} = -3\vec{i} + 3v_2 \implies 3\vec{i} + 3\vec{i} = 3v_2 \implies v_2 = 2\vec{i} \text{ m s}^{-1}$
4
Calculate impulse on the 2 kg particle as its change in momentum
$\vec{I} = m_1(v_1 - u_1) = 2(-1.5\vec{i} - 3\vec{i}) = -9\vec{i} \text{ N s}$

Exam tip:

Always assign a clear positive direction at the start of every problem to avoid sign errors in momentum calculations.

2. Newton's Law of Restitution and Energy Loss★★★☆☆⏱ 7 min

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📘 Definition

Coefficient of Restitution

e

Dimensionless ratio of separation speed to approach speed for two colliding objects, with values $0 \leq e \leq 1$ .

Example:

$e = 1$ = perfectly elastic (no KE loss), $e = 0$ = perfectly inelastic (particles stick together).

For direct impacts, Newton's Law of Restitution states: $Speed of separation = e \times Speed of approach$ . The loss of mechanical energy during impact equals the difference between total kinetic energy before and after the collision.

📐 Worked Example

Particles A (1 kg) and B (2 kg) move directly towards each other at 4 m s⁻¹ and 1 m s⁻¹ respectively, with $e = 0.6$ . Find their speeds after collision and the total kinetic energy lost.

1
Set positive direction as A's initial motion. Approach speed = $4 - (- 1) = 5$ m s⁻¹
$v_B - v_A = e \times \text{approach speed} = 0.6 \times 5 = 3$
2
Write conservation of momentum equation
$1(4) + 2(-1) = v_A + 2v_B \implies v_A + 2v_B = 2$
3
Solve simultaneously: substitute $v_{A} = v_{B} - 3$ into momentum equation
$(v_B - 3) + 2v_B = 2 \implies 3v_B = 5 \implies v_B = \frac{5}{3} \approx 1.67 \text{ m s}^{-1}$
4
$v_A = \frac{5}{3} - 3 = -\frac{4}{3} \approx -1.33 \text{ m s}^{-1} (A moves backwards at 1.33 m s⁻¹)$
5
Calculate total KE before collision
$KE_{before} = 0.5(1)(4)^2 + 0.5(2)(1)^2 = 8 + 1 = 9 \text{ J}$
6
Calculate total KE after collision
$KE_{after} = 0.5(1)(\frac{16}{9}) + 0.5(2)(\frac{25}{9}) = \frac{11}{3} \approx 3.67 \text{ J}$
7
Calculate energy loss
$KE_{loss} = 9 - \frac{11}{3} = \frac{16}{3} \approx 5.33 \text{ J}$

Exam tip:

If you calculate a negative energy loss, you have mixed up separation and approach speeds in your restitution equation.

3. Collisions with a Smooth Fixed Plane★★★☆☆⏱ 6 min

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When a particle collides directly with a smooth fixed plane, only the velocity component perpendicular to the plane changes (the parallel component remains constant as no friction acts). The restitution equation simplifies to $v_{r e b o u n d} = e \times v_{a pp r o a c h}$ , where $v_{a pp r o a c h}$ is the speed of the particle hitting the plane.

📐 Worked Example

A 0.5 kg ball is dropped from rest 2 m above a smooth horizontal floor, with $e = 0.8$ . Find the maximum height it reaches after the first bounce, and the impulse exerted on the ball by the floor.

1
Calculate speed of ball just before impact using $v^{2} = u^{2} + 2 a s$ , $g = 9.8$ m s⁻²
$v^2 = 0 + 2(9.8)(2) = 39.2 \implies v = \sqrt{39.2} \approx 6.26 \text{ m s}^{-1} (downwards)$
2
Apply restitution to find rebound speed
$v_{rebound} = 0.8 \times 6.26 \approx 5.01 \text{ m s}^{-1} (upwards)$
3
Calculate maximum height after bounce using $v^{2} = u^{2} + 2 a s$ with final speed = 0
$0 = (5.01)^2 + 2(-9.8)s \implies s = \frac{25.1}{19.6} \approx 1.28 \text{ m}$
4
Calculate impulse (change in momentum, upwards as positive)
$I = 0.5(5.01 - (-6.26)) = 0.5(11.27) \approx 5.64 \text{ N s}$

Exam tip:

Do not use conservation of momentum for particle-plane collisions: the plane has effectively infinite mass so it does not move.

4. Successive Impacts Between Multiple Particles★★★★☆⏱ 7 min

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For successive impacts between up to 3 particles, solve each collision sequentially: use the final velocities from the first collision as initial velocities for the next collision. Apply both conservation of momentum and restitution for each pair of colliding particles.

📐 Worked Example

Three particles A (1 kg), B (1 kg), C (2 kg) are at rest in a straight line. A is projected towards B at 4 m s⁻¹. $e_{A B} = 1$ , $e_{B C} = 0.5$ . Find the speed of C after all collisions are complete.

1
First collision: A and B, perfectly elastic ( $e = 1$ ). Conservation of momentum
$1(4) + 1(0) = v_A + v_B \implies v_A + v_B = 4$
2
Restitution equation for A and B
$v_B - v_A = 1 \times 4 = 4$
3
Solve: add equations to get $2 v_{B} = 8 ⟹ v_{B} = 4$ m s⁻¹, $v_{A} = 0$ m s⁻¹. A stops, B moves towards C at 4 m s⁻¹.
4
Second collision: B and C, $e = 0.5$ . Conservation of momentum
$1(4) + 2(0) = v_B' + 2v_C \implies v_B' + 2v_C = 4$
5
Restitution equation for B and C
$v_C - v_B' = 0.5 \times 4 = 2$
6
Solve: substitute $v_{B}^{'} = v_{C} - 2$ into momentum equation
$(v_C - 2) + 2v_C = 4 \implies 3v_C = 6 \implies v_C = 2 \text{ m s}^{-1}$

Exam tip:

After each collision, check relative velocities: if a particle behind is moving slower than the one in front, no further collision occurs.

5. Common Pitfalls

Wrong move:

Using scalar momentum instead of signed/vector values, ignoring direction of motion

Why:

Momentum is a vector, so direction changes alter the value of momentum used in calculations

Correct move:

Assign a clear positive direction at the start of every problem, and use consistent signs for all velocities

Wrong move:

Swapping separation and approach speeds in the restitution equation

Why:

This leads to invalid negative e values or negative energy loss, which are physically impossible

Correct move:

Write restitution as $v_{2} - v_{1} = e (u_{1} - u_{2})$ with a consistent sign convention for velocities

Wrong move:

Applying conservation of momentum to collisions between a particle and fixed plane

Why:

The plane has effectively infinite mass, so total momentum of the system cannot be calculated with this method

Correct move:

For particle-plane collisions, only use the restitution rule $v_{r e b o u n d} = e u$ and kinematics as needed

Wrong move:

Calculating energy loss for only one particle instead of the whole system

Why:

Energy loss is the total reduction in kinetic energy of all colliding objects

Correct move:

Sum the KE of all particles before collision, sum their KE after collision, subtract the after value from the before value

Wrong move:

Assuming all possible successive collisions occur without checking relative velocities

Why:

If a particle behind is moving slower than the particle in front, they will not collide again

Correct move:

After each collision, check relative speeds to confirm if further collisions are possible before continuing calculations

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Notes
Conservation of Momentum	$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$	Use signed/vector velocities; applies only to closed systems
Newton's Restitution Law	$Separation speed = e \times Approach speed$	$0 \leq e \leq 1$ ; $e = 1$ elastic, $e = 0$ inelastic
Collision Energy Loss	$K E_{l oss} = \sum K E_{b e f or e} - \sum K E_{a f t er}$	Always non-negative; ignore potential energy unless height changes
Particle-Plane Collision	$v_{r e b o u n d} = e \times v_{a pp r o a c h}$	Parallel velocity unchanged for smooth planes; no momentum conservation needed
Successive Impacts	Solve each collision sequentially	Check relative velocities after each step to confirm further collisions

7. Frequently Asked

What value of e do I use for perfectly elastic collisions?

Use $e = 1$ for perfectly elastic collisions, where no kinetic energy is lost. For perfectly inelastic collisions where particles stick together after impact, use $e = 0$ .

Do I need to consider friction for collisions with smooth planes?

No, smooth planes have zero friction, so only the velocity component perpendicular to the plane changes during collision; the parallel velocity component remains constant.

Can energy loss in a collision be negative?

No, energy loss is always non-negative. If you calculate a negative value, you have mixed up separation and approach speeds in your restitution equation.

Going deeper

What's Next

Now that you have mastered collision problems for Edexcel IAL M2, you can move on to more advanced mechanics topics. Collision principles are often combined with work, energy and power concepts in 8-12 mark exam questions, so revising those will help you score full marks on problem-solving tasks. You should also practice past paper collision questions to familiarize yourself with Edexcel's marking scheme requirements. Next, you can learn about centre of mass, another core M2 topic frequently tested alongside collisions in multi-part exam questions.