Collisions | Mechanics 2

FOUNDATION 01

What Is Momentum?

Momentum is a vector quantity associated with the motion of a particle. It depends on both the particle's mass and its velocity.

p = mv momentum = mass × velocity

Because velocity is a vector, momentum is also a vector. Therefore, the direction of momentum is the same as the direction of the particle's velocity.

Symbol Meaning SI unit p momentum kg m s⁻¹ m mass kg v velocity m s⁻¹

Vector warning

Never treat velocities as positive simply because the particles are moving. You must choose a positive direction and assign negative signs to velocities in the opposite direction.

FOUNDATION 02

Choosing a Positive Direction

Most collision questions become much easier once a positive direction has been chosen.

For example, suppose particle A moves to the right and particle B moves to the left. We may choose right as positive.

RIGHT = + LEFT = −

If A has velocity 8 m/s to the right, its velocity is +8 m/s. If B has velocity 5 m/s to the left, its velocity is −5 m/s.

SIGN-CONVENTION EXAMPLE

A particle of mass 3 kg travels at 6 m/s to the left. Take right as positive.

Since left is negative:

v = −6 m/s

Therefore:

p = mv = 3(−6)

p = −18 kg m s⁻¹

Exam habit

Write your chosen positive direction before starting a collision calculation. This prevents most sign errors.

CORE THEORY 03

Conservation of Momentum

When considering a system of particles, the total momentum remains constant if the resultant external force acting on the system is zero.

Total momentum before = Total momentum after

For two particles with masses m₁ and m₂, initial velocities u₁ and u₂, and final velocities v₁ and v₂:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

This equation is one of the central tools for collision questions. It is a vector equation, so signs matter.

Why is momentum conserved?

Forces during a collision are internal forces within the chosen system. By Newton's third law, the forces the particles exert on each other are equal and opposite, so their impulses cancel when the whole system is considered.

When can we use conservation of momentum?

The system must be clearly defined.
The resultant external impulse on the system must be negligible or zero over the time interval being considered.
Internal forces between the particles do not destroy total momentum conservation.

CORE THEORY 04

Impulse and Change in Momentum

Impulse measures the effect of a force acting over a time interval. It is equal to the change in momentum.

Impulse = Change in momentum

I = mv − mu

If the force is constant:

I = Ft

Therefore:

Ft = mv − mu

Quantity	Meaning	Unit
I	Impulse	N s
F	Force	N
t	Time interval	s
mv − mu	Change in momentum	kg m s⁻¹

Important unit equivalence

1 N s = 1 kg m s⁻¹.

CORE THEORY 05

Direct Impact

A collision is called a direct impact when the particles move along the same straight line before and after the collision.

This is the standard collision model used throughout Mechanics 2.

Direct impact

The motion of both particles can be represented using one straight line, so their velocities can be treated using signed one-dimensional quantities.

A typical problem gives the masses and velocities before impact and then asks for one or both velocities after impact.

Choose a positive direction. Record all velocities using signs.

Apply conservation of momentum.

Use Newton's law of restitution if the coefficient of restitution is given.

Solve the simultaneous equations.

Interpret the signs of the final velocities.

CORE THEORY 06

Newton's Law of Restitution

For two particles undergoing a direct collision, Newton's law of restitution relates the relative speed of separation to the relative speed of approach.

e = (relative speed of separation) ÷ (relative speed of approach)

The coefficient of restitution is normally denoted by e. For the usual collision model:

0 ≤ e ≤ 1

If particle 1 is initially moving faster than particle 2 and they are approaching one another, then:

e = (v₂ − v₁) ÷ (u₁ − u₂)

provided the chosen positive direction is such that particle 1 is moving in the positive direction towards particle 2.

Do not memorise the signs blindly.

The safest method is always to identify the actual relative speed of approach and relative speed of separation from the motion described in the question.

CORE THEORY 07

Coefficient of Restitution: Physical Meaning

e = 1

The relative speed of separation equals the relative speed of approach.

This corresponds to a perfectly elastic collision in the idealised model.

e = 0

The relative speed of separation is zero.

The particles have the same velocity immediately after impact and therefore move together.

v_separation = e v_approach

For 0 < e < 1, some kinetic energy is lost during the collision, although momentum is still conserved for an isolated system.

Critical distinction

Momentum conservation does NOT mean kinetic energy conservation. These are different physical principles.

CORE THEORY 08

Elastic and Inelastic Collisions

The coefficient of restitution gives useful information about the nature of a collision.

Elastic collision

Momentum is conserved.

Kinetic energy is also conserved.

For the ideal one-dimensional case: e = 1.

Inelastic collision

Momentum is conserved for an isolated system.

Kinetic energy is not conserved.

For 0 ≤ e < 1, relative separation is reduced.

KE = ½mv²

The change in kinetic energy can be found by comparing the total kinetic energy before and after the collision.

Exam warning

Do not automatically write a kinetic-energy equation in every collision question. Momentum is the conservation equation. Kinetic energy is only conserved when the collision is elastic or when the question explicitly gives enough information to use it.

CORE THEORY 09

The Standard Collision System

Suppose two particles have masses m₁ and m₂. Before impact their velocities are u₁ and u₂, and after impact their velocities are v₁ and v₂.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

If the particles approach each other and then separate:

e(via relative separation) = relative separation ÷ relative approach

The two equations usually provide the two unknown final velocities.

Typical exam structure

Write the conservation of momentum equation.

Write Newton's law of restitution.

Substitute known values.

Solve the simultaneous equations.

Check whether the resulting directions make physical sense.

WORKED EXAMPLE 01

Finding Both Velocities After Impact

FULL EXAM-STYLE SOLUTION

A particle A of mass 2 kg moves with velocity 6 m/s towards particle B, which has mass 3 kg and is initially at rest. The coefficient of restitution between the particles is 0.5. Find the velocities of A and B after the collision.

Take the direction of A's initial motion as positive.

Therefore:

u_A = 6, u_B = 0

Step 1 — Conservation of momentum

2(6) + 3(0) = 2v_A + 3v_B

Therefore:

12 = 2v_A + 3v_B

Step 2 — Newton's law of restitution

Relative speed of approach:

6 − 0 = 6

Relative speed of separation:

v_B − v_A

Therefore:

0.5 = (v_B − v_A) / 6

So:

v_B − v_A = 3

Step 3 — Solve simultaneously

v_B = v_A + 3

Substitute into momentum:

12 = 2v_A + 3(v_A + 3)

12 = 5v_A + 9

3 = 5v_A

v_A = 0.6

Therefore:

v_B = 3.6

A moves at 0.6 m/s and B moves at 3.6 m/s in the positive direction.

WORKED EXAMPLE 02

A Collision With Opposite Initial Directions

SIGN CONVENTION MATTERS

Particle A of mass 4 kg moves at 5 m/s to the right. Particle B of mass 2 kg moves at 3 m/s to the left. The coefficient of restitution is 0.5. Find their velocities after impact.

Take right as positive.

u_A = 5

u_B = −3

Momentum

4(5) + 2(−3) = 4v_A + 2v_B

20 − 6 = 4v_A + 2v_B

14 = 4v_A + 2v_B

Restitution

Before impact, the particles approach each other at:

5 − (−3) = 8 m/s

So the separation speed is:

0.5 × 8 = 4 m/s

After impact A is expected to be moving to the left relative to B, so:

v_B − v_A = 4

Therefore:

v_B = v_A + 4

Substitute:

14 = 4v_A + 2(v_A + 4)

14 = 6v_A + 8

6v_A = 6

v_A = 1

Therefore:

v_B = 5

A moves at 1 m/s to the right and B moves at 5 m/s to the right.

WORKED EXAMPLE 03

Finding the Coefficient of Restitution

FIND e

Two particles approach each other with speeds 7 m/s and 2 m/s. After collision they separate with speeds 1 m/s and 3 m/s. Find the coefficient of restitution.

The relative speed of approach is:

7 + 2 = 9 m/s

The relative speed of separation is:

3 + 1 = 4 m/s

Therefore:

e = 4 / 9

e = 4/9

Key idea

When two particles move towards each other, their relative speed of approach is usually the sum of their speeds if those speeds are stated as positive magnitudes.

WORKED EXAMPLE 04

Impulse During a Collision

MOMENTUM CHANGE

A 3 kg particle moves at 8 m/s to the right before collision and at 2 m/s to the left after collision. Find the impulse acting on the particle.

Take right as positive.

u = 8

v = −2

Impulse = mv − mu

= 3(−2) − 3(8)

= −6 − 24

Impulse = −30 N s

The negative sign means the impulse acts to the left.

Do not use speed in the impulse formula.

Impulse depends on the change in velocity, so direction must be included.

CORE THEORY 14

Collision and Kinetic Energy

Momentum is conserved in an isolated collision, but kinetic energy may decrease.

Total KE before = ½m₁u₁² + ½m₂u₂²

Total KE after = ½m₁v₁² + ½m₂v₂²

For an elastic collision:

½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²

For an inelastic collision, the final kinetic energy is smaller than the initial kinetic energy.

Where does the lost kinetic energy go?

It may be transformed into internal energy, deformation, sound, heat or other forms of energy. Total energy is still conserved; it is kinetic energy that is not necessarily conserved.

CORE THEORY 15

Perfectly Inelastic Collision

A perfectly inelastic collision is the special case where the particles have zero relative speed of separation.

e = 0

Therefore:

v₁ = v₂

The particles move together immediately after impact.

Conservation of momentum can then be used to find their common velocity.

EXAMPLE — PARTICLES STICK TOGETHER

A 2 kg particle moving at 6 m/s collides with a stationary 4 kg particle. They move together after collision. Find their common velocity.

Momentum before:

2(6) + 4(0) = 12

Momentum after:

(2 + 4)v = 6v

Therefore:

12 = 6v

v = 2 m/s

CORE THEORY 16

When Will Two Particles Actually Collide?

A collision can only occur if the particles are moving towards each other along the same line.

For example, if particle A is behind particle B and both move in the same direction, A will catch B only if A has the greater speed.

Catch-up condition: v_A > v_B

If A is initially a distance d behind B and their velocities remain constant, the time to collision is found using relative speed:

time = distance ÷ relative speed

t = d / (v_A − v_B)

Collision condition

Do not automatically assume that two particles mentioned in the same question will collide. Check their relative positions and velocities.

CORE THEORY 17

Repeated Collisions

Some Mechanics 2 problems involve a particle colliding with another particle, after which the particles move away and may later collide with other objects.

The safest strategy is to treat each collision as a separate event.

Determine the velocities immediately before the collision.

Apply momentum conservation to that collision.

Apply Newton's law of restitution.

Find the new velocities.

Use the new velocities to determine what happens next.

Think in events

Collision 1 → new velocities → motion → Collision 2 → new velocities. Never mix the equations from different collisions.

EXAM METHOD 18

The Universal Collision Method

For most two-particle direct-impact questions, this procedure will keep your working organised.

Draw or visualise the motion. Identify which particle is approaching from which direction.

Choose positive. Write the sign of every velocity.

Write momentum conservation. Keep all velocities signed.

Write restitution. Use relative separation divided by relative approach.

Solve simultaneously. Usually there are two equations in two unknown velocities.

Interpret the answer. A negative velocity means the particle moves opposite to your chosen positive direction.

Check the result. Ask whether the particles are separating after impact and whether the answer is physically reasonable.

COMMON ERRORS 19

Mistakes That Cost Marks

1. Ignoring direction

Momentum is a vector. Writing every velocity as positive can destroy the momentum equation.

2. Using speed instead of velocity

Restitution uses relative speeds, but momentum uses signed velocities.

3. Reversing approach and separation

Always identify what happens before and after impact.

4. Assuming kinetic energy is conserved

Momentum conservation is the standard principle. Kinetic energy requires an elastic collision.

5. Forgetting the mass

p = mv

A heavier particle can have greater momentum even when moving more slowly.

6. Misinterpreting a negative answer

A negative velocity is not automatically a wrong answer. It indicates motion opposite to the chosen positive direction.

7. Using e incorrectly

The coefficient of restitution is a ratio of relative speeds, not simply the ratio of the two individual final speeds.

DEEPER CONNECTION 20

Momentum, Impulse and Force

Momentum and impulse are closely connected to Newton's laws.

I = Δp

If a force acts for a short time during a collision, the impulse is the area under a force-time graph.

Impulse = area under an F–t graph

For a constant force:

I = Ft

For a variable force, the area must be determined using the appropriate geometry or integration where applicable.

Physical interpretation

A large force acting for a very short time can produce the same impulse as a smaller force acting for a longer time.

DEEPER CONNECTION 21

Why Newton's Restitution Law Works

Newton's law of restitution does not say that the velocities after collision are simply a fixed percentage of the velocities before collision.

Instead, it compares the motion of the two particles relative to one another.

e = relative separation speed / relative approach speed

This is important because a collision concerns the relative motion between the particles rather than the motion of either particle by itself.

Best conceptual interpretation

Momentum controls the overall motion of the system, while restitution controls how much relative motion is recovered after impact.

WORKED EXAMPLE 05

Finding e From a Complete Collision

MULTI-STAGE REASONING

A particle A moves at 10 m/s towards a stationary particle B. After impact A moves at 2 m/s in the same direction and B moves at 6 m/s in the same direction. Find the coefficient of restitution.

Before impact, the relative speed of approach is:

10 − 0 = 10 m/s

After impact, the relative speed of separation is:

6 − 2 = 4 m/s

Therefore:

e = 4/10

e = 0.4

EXAM CHECK 23

How to Check Your Final Answer

Before moving to the next part of the question, perform a quick physical check.

Does the momentum equation balance?
Are the particles moving apart after the collision?
Is the coefficient of restitution between 0 and 1 for the usual collision model?
Have you interpreted negative velocities correctly?
If the collision is elastic, is kinetic energy conserved?
Are your units correct?

Fast sanity check

If your calculated coefficient of restitution is greater than 1, do not immediately accept it. Recheck your relative approach and separation speeds and your sign convention.

PRACTICE 24

Collision Practice Bank

1. A 5 kg particle moves at 4 m/s. Find its momentum. FOUNDATION

Show solution

p = mv

= 5 × 4

p = 20 kg m s⁻¹

2. A 3 kg particle changes velocity from 5 m/s to 9 m/s in the same direction. Find its impulse. FOUNDATION

Show solution

I = m(v − u)

= 3(9 − 5)

I = 12 N s

3. A 2 kg particle moving at 8 m/s collides with a stationary 3 kg particle. If they move together, find their common velocity. CORE

Show solution

2(8) + 3(0) = (2 + 3)v

16 = 5v

v = 3.2 m/s

4. Two particles approach each other at 9 m/s and 4 m/s. After collision they separate at 3 m/s and 2 m/s. Find e. CORE

Show solution

Approach speed = 9 + 4 = 13 m/s

Separation speed = 3 + 2 = 5 m/s

e = 5/13

5. A 4 kg particle moving at 7 m/s collides with a stationary 2 kg particle. The coefficient of restitution is 0.5. Find the velocities after impact. EXAM

Show solution

Momentum:

4(7) = 4v₁ + 2v₂

28 = 4v₁ + 2v₂

Restitution:

v₂ − v₁ = 0.5(7)

v₂ − v₁ = 3.5

v₂ = v₁ + 3.5

28 = 4v₁ + 2(v₁ + 3.5)

28 = 6v₁ + 7

v₁ = 3.5

v₂ = 7

v₁ = 3.5 m/s, v₂ = 7 m/s

6. A particle reverses direction during a collision. Explain why its impulse cannot be found by using only the difference between the two speeds. CONCEPT

Show solution

Impulse depends on the change in momentum, and momentum depends on velocity including direction. Therefore the final velocity must be assigned the correct sign before calculating mv − mu.

7. Explain why momentum can be conserved while kinetic energy decreases during a collision. THEORY

Show solution

Momentum is conserved because the internal forces during the collision provide equal and opposite impulses. Kinetic energy, however, may be transformed into other forms such as heat, sound and deformation.

8. A particle A is 12 m behind particle B. A travels at 8 m/s and B travels at 5 m/s in the same direction. Find the time before A catches B. RELATIVE MOTION

Show solution

Relative speed = 8 − 5 = 3 m/s

t = distance / relative speed

t = 12/3

t = 4 s

MASTER SUMMARY 25

Collision Formula Atlas

MOMENTUM p = mv

CONSERVATION Σp before = Σp after

IMPULSE I = Δp

CONSTANT FORCE I = Ft

RESTITUTION e = separation / approach

RANGE 0 ≤ e ≤ 1

PERFECTLY INELASTIC e = 0

PERFECTLY ELASTIC e = 1

KINETIC ENERGY KE = ½mv²

RELATIVE SPEED distance / time

FINAL EXAM CHECKLIST 26

Before You Submit a Collision Question

Have I chosen a positive direction?
Have I used signed velocities in the momentum equation?
Have I written total momentum before and after impact?
Have I correctly identified the relative speed of approach?
Have I correctly identified the relative speed of separation?
Have I used the coefficient of restitution in the correct direction?
Have I solved the simultaneous equations accurately?
Have I interpreted negative velocities correctly?
If the collision is elastic, have I checked kinetic energy?
Are all final quantities given with appropriate units?
Does the physical situation make sense?

IAL ATLAS principle

Do not memorise collision equations as isolated formulas. Understand the physical story: momentum describes the total motion of the system, while restitution describes how the relative motion changes during impact.