Momentum

Conservation laws, impulse, collisions and explosions

SPEC 3.4.1.6 • MECHANICS AND MATERIALS

What is Momentum?

Definition

Momentum is the product of an object's mass and velocity. It is a measure of how difficult it is to stop a moving object.

p = m v

Key Properties

Vector quantity - has magnitude and direction

Direction same as velocity

Can be positive or negative (depends on direction chosen)

Units

kg m s⁻¹ (kilogram metres per second)

Or equivalently: N s (newton seconds)

Exam Tip

A stationary object has zero momentum (v = 0), regardless of its mass. A moving object always has momentum.

Calculating Momentum

Objects with the same momentum can have very different masses and velocities:

ObjectMass (kg)Velocity (m s⁻¹)Momentum (kg m s⁻¹)

Bullet0.015005

Tennis ball0.057704

Person walking701.5105

Car12003036,000

Important

Since momentum is a vector, you must define a positive direction. Objects moving in opposite directions have opposite signs for their momentum.

Conservation of Momentum

Principle

The total momentum of a closed system remains constant, provided no external forces act on it.

\sum p_{before} = \sum p_{after}

Conditions for Conservation

Closed system - no objects enter or leave

No external forces - only internal forces between objects

Works for collisions, explosions, and separations

COLLISION IN 1D

m₁

u₁ →

m₂

← u₂

m₁

v₁ →

m₂

v₂ →

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

Applying Conservation of Momentum

Method

Define positive direction (usually direction of initial motion)

Calculate total momentum before the interaction

Apply conservation: total momentum after = total momentum before

Solve for unknown velocity or mass

Interpret sign of answer (direction of motion)

Sign Convention

Objects moving in the positive direction have positive momentum. Objects moving in the negative direction have negative momentum. A negative answer means the object moves opposite to your defined positive direction.

Common Scenarios

Objects stick together:

m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v

Explosion from rest:

0 = m_{1} v_{1} + m_{2} v_{2}

Recoil: Gun and bullet move in opposite directions

Worked Example01

A 1200 kg car travelling at 15 m s⁻¹ collides with a stationary 800 kg car. The two cars stick together after the collision. Calculate their common velocity after the collision.

1

Define positive direction: direction of moving car (→)

2

Momentum before:

p_{before} = m_{1} u_{1} + m_{2} u_{2} = 1200 \times 15 + 800 \times 0 = 18000 kg m s^{- 1}

3

After collision (stuck together):

p_{after} = (m_{1} + m_{2}) v = (1200 + 800) v = 2000 v

4

Apply conservation:

18000 = 2000 v

5

Solve:

v = \frac{18000}{2000} = 9 m s^{- 1}

Answer: 9 m s⁻¹ in the original direction of travel

Worked Example02

A 2.0 kg trolley moving at 3.0 m s⁻¹ to the right collides with a 1.5 kg trolley moving at 4.0 m s⁻¹ to the left. After the collision, the 2.0 kg trolley moves at 0.50 m s⁻¹ to the left. Find the velocity of the 1.5 kg trolley after the collision.

1

Define positive direction: right (→)

2

Before:

p_{before} = 2.0 \times 3.0 + 1.5 \times (- 4.0) = 6.0 - 6.0 = 0 kg m s^{- 1}

3

After:

p_{after} = 2.0 \times (- 0.50) + 1.5 \times v_{2} = - 1.0 + 1.5 v_{2}

4

Apply conservation:

0 = - 1.0 + 1.5 v_{2}

5

Solve:

v_{2} = \frac{1.0}{1.5} = 0.67 m s^{- 1}

Answer: 0.67 m s⁻¹ to the right (positive value = right direction)

Worked Example03

A 5.0 kg gun fires a 0.020 kg bullet at 400 m s⁻¹. Calculate the recoil velocity of the gun.

1

Define positive direction: direction bullet travels (→)

2

Before firing (at rest):

p_{before} = 0

3

After firing:

p_{after} = m_{gun} v_{gun} + m_{bullet} v_{bullet}

4

Apply conservation:

0 = 5.0 \times v_{gun} + 0.020 \times 400

5

Solve:

0 = 5.0 v_{gun} + 8.0

v_{gun} = \frac{- 8.0}{5.0} = - 1.6 m s^{- 1}

Answer: 1.6 m s⁻¹ backwards (negative sign indicates opposite direction to bullet)

Impulse and Force

Newton's Second Law (Full Form)

Force equals the rate of change of momentum:

F = \frac{Δ ( m v )}{Δ t}

Rearranging gives the impulse equation:

Impulse

F Δ t = Δ (m v)

Impulse = Change in momentum

Impulse Properties

Vector quantity

Units: N s or kg m s⁻¹

Equal to area under F-t graph

Key Relationships

F Δ t = m v - m u

F Δ t = m (v - u)

F Δ t = m Δ v

Force-Time Graphs

The area under a force-time graph equals the impulse (change in momentum).

Constant Force

Impulse = F × Δt (rectangle area)

Variable Force

Impulse = area under curve (estimate or integrate)

Worked Example04

A 0.45 kg football is kicked from rest. The foot is in contact with the ball for 0.12 s and the ball leaves at 20 m s⁻¹. Calculate: (a) the impulse on the ball, (b) the average force on the ball.

(a)

Impulse = change in momentum =

m (v - u)

Impulse =

0.45 \times (20 - 0) = 9.0 N s

(b)

F Δ t = Δ (m v)

F = \frac{Δ ( m v )}{Δ t} = \frac{9.0}{0.12} = 75 N

Answers: (a) 9.0 N s, (b) 75 N

Worked Example05

A force acts on a 2.0 kg object initially at rest. The force-time graph is triangular, rising from 0 to 60 N over 0.10 s, then falling back to 0 N over the next 0.10 s. Calculate the final velocity of the object.

1

The graph forms a triangle with base 0.20 s and height 60 N

2

Area under graph = Impulse = ½ × base × height

Impulse = ½ × 0.20 × 60 = 6.0 N s

3

Impulse = change in momentum =

m (v - u)

6.0 = 2.0 × (v - 0)

4

v = \frac{6.0}{2.0} = 3.0 m s^{- 1}

Answer: 3.0 m s⁻¹

Impact Forces and Contact Times

Key Relationship

From

F Δ t = Δ (m v)

, for a fixed change in momentum:

F = \frac{Δ ( m v )}{Δ t}

Longer contact time → Smaller force

This principle is used in many safety applications:

Car Safety

Crumple zones: Extend collision time, reducing force on passengers

Airbags: Increase stopping time for head/body

Seatbelts: Allow controlled deceleration

Other Applications

Packaging: Foam/bubble wrap extends impact time

Sports: Following through when catching extends contact time

Helmets: Crush zones extend impact time

Worked Example06

A 70 kg person in a car travelling at 15 m s⁻¹ comes to rest in a collision. Calculate the average force on the person if they are stopped in: (a) 0.10 s (no seatbelt), (b) 0.80 s (with seatbelt and crumple zone).

1

Change in momentum:

Δ (m v) = 70 \times (0 - 15) = - 1050 kg m s^{- 1}

(a)

F = \frac{Δ ( m v )}{Δ t} = \frac{- 1050}{0.10} = - 10500 N

Force magnitude = 10,500 N (about 15× body weight!)

(b)

F = \frac{- 1050}{0.80} = - 1313 N

Force magnitude = 1,310 N (about 1.9× body weight)

The crumple zone reduces the force by a factor of 8!

Elastic and Inelastic Collisions

Momentum is ALWAYS conserved in collisions (if no external forces). But what about kinetic energy?

Elastic Collision

Total KE is conserved

Objects bounce apart

Rare in real life (approx. in atomic collisions)

KE_{before} = KE_{after}

Inelastic Collision

Total KE is NOT conserved

Some KE converted to heat, sound, deformation

Most real collisions are inelastic

KE_{before} > KE_{after}

Perfectly Inelastic

When objects stick together after collision, this is called perfectly inelastic. Maximum KE is lost (but momentum still conserved).

Special Case: Equal Masses

Equal Mass Elastic Collision

When two objects of equal mass collide elastically, their velocities are exchanged.

BEFORE COLLISION

u →

← v

AFTER COLLISION

← v

u →

The velocities swap over!

Classic Example

Newton's cradle: When one ball swings in and hits stationary balls of equal mass, only the end ball swings out with the same speed. The velocities are transferred through the line of balls.

Explosions

Explosion

An explosion is the opposite of a perfectly inelastic collision. Objects initially together separate and move apart.

Key Properties

Momentum is conserved: If system starts at rest, total momentum = 0 before AND after

Kinetic energy INCREASES: Energy comes from chemical energy (explosives), stored elastic PE (spring), etc.

Objects move in opposite directions to conserve momentum

EXPLOSION FROM REST

m₁ + m₂

At rest (p = 0)

m₁

← v₁

m₂

v₂ →

0 = m_{1} (- v_{1}) + m_{2} v_{2}

therefore

m_{1} v_{1} = m_{2} v_{2}

Exam Tip

In an explosion where two fragments are produced, the heavier fragment will have the smaller velocity. Since

m_{1} v_{1} = m_{2} v_{2}

, if

m_{1} > m_{2}

then

v_{1} < v_{2}

.

Worked Example07

Two trolleys of mass 1.5 kg and 2.5 kg collide and stick together. Before the collision, the 1.5 kg trolley was moving at 4.0 m s⁻¹ and the 2.5 kg trolley was stationary. (a) Calculate their common velocity after collision. (b) Show that the collision is inelastic by calculating KE before and after.

(a)

Conservation:

1.5 \times 4.0 + 2.5 \times 0 = (1.5 + 2.5) \times v

6.0 = 4.0 v

v = 1.5 m s^{- 1}

(b)

KE before = ½ × 1.5 × 4.0² + 0 = 12 J

KE after = ½ × 4.0 × 1.5² = 4.5 J

KE lost = 12 - 4.5 = 7.5 J

Since KE is not conserved (KE decreased), the collision is inelastic.

Practice

Question 1: A 0.50 kg ball moving at 6.0 m s⁻¹ collides with a 0.30 kg ball at rest. After the collision, the 0.50 kg ball moves at 2.0 m s⁻¹ in the same direction. Calculate the velocity of the 0.30 kg ball after the collision.

1

Momentum before:

0.50 \times 6.0 + 0.30 \times 0 = 3.0 kg m s^{- 1}

2

Momentum after:

0.50 \times 2.0 + 0.30 \times v = 1.0 + 0.30 v

3

Apply conservation:

3.0 = 1.0 + 0.30 v

v = \frac{2.0}{0.30} = 6.7 m s^{- 1}

Answer: 6.7 m s⁻¹ (in original direction of motion)

Practice

Question 2: A tennis racket exerts an average force of 400 N on a 0.057 kg tennis ball during a serve. The contact time is 5.0 ms. If the ball was initially at rest, calculate the speed of the ball as it leaves the racket.

1

Convert time: 5.0 ms = 5.0 × 10⁻³ s = 0.0050 s

2

Impulse = FΔt = 400 × 0.0050 = 2.0 N s

3

Impulse = change in momentum = mv - mu

2.0 = 0.057 × v - 0

4

v = \frac{2.0}{0.057} = 35 m s^{- 1}

Answer: 35 m s⁻¹

Practice

Question 3: Two ice skaters, A (mass 60 kg) and B (mass 80 kg), push off from each other while stationary. Skater A moves backwards at 2.0 m s⁻¹. Calculate: (a) the velocity of skater B, (b) the total kinetic energy after the push.

(a)

Total momentum before = 0 (both at rest)

After:

60 \times (- 2.0) + 80 \times v_{B} = 0

- 120 + 80 v_{B} = 0

v_{B} = 1.5 m s^{- 1}

(forwards)

(b)

KE of A = ½ × 60 × 2.0² = 120 J

KE of B = ½ × 80 × 1.5² = 90 J

Total KE = 120 + 90 = 210 J

Answers: (a) 1.5 m s⁻¹ forwards, (b) 210 J

Summary

Key Formulas

ConceptFormulaUnits

Momentum

p = m v

kg m s⁻¹ or N s

Conservation

\sum p_{before} = \sum p_{after}

-

Force (momentum)

F = \frac{Δ ( m v )}{Δ t}

N

Impulse

F Δ t = Δ (m v)

N s

Key Exam Skills

Always define a positive direction for momentum problems

Remember: momentum is ALWAYS conserved (no external forces)

Kinetic energy is only conserved in elastic collisions

Use impulse = area under F-t graph for variable forces

Longer contact time means smaller force (for same impulse)