Momentum & Impulse — Science Sage

Momentum

Momentum (p) is the product of an object's mass and its velocity. It is a vector — its direction is the same as the direction of velocity.

p = mv

Symbol	Meaning	Unit
p	Momentum (vector)	kg·m·s⁻¹
m	Mass	kg
v	Velocity (vector)	m·s⁻¹
Δp	Change in momentum	kg·m·s⁻¹

Change in momentum: Δp = mvf − mvi = m(vf − vi). Choose a positive direction and stick to it throughout a problem.

Always define a positive direction first. Velocities opposite to it are negative.

Impulse and the Impulse-Momentum Theorem

Impulse (J) is the product of the net force and the time over which it acts. Impulse equals the change in momentum.

J = F_net Δt = Δp = m(v_f − v_i)

Unit of impulse: N·s = kg·m·s⁻¹ (same as momentum — they are numerically and dimensionally identical).

Practical applications:
• Airbags and crumple zones increase the collision time Δt. Since J = FΔt = Δp (fixed), a larger Δt means a smaller F — reducing injury.
• Catching a ball: pulling your hands back as you catch increases Δt, reducing the impact force on your hands.

Conservation of Momentum

In a closed (isolated) system (no net external forces), the total momentum is constant (conserved).

Σp_before = Σp_after
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

Conservation of momentum is valid for ALL collision types — elastic, inelastic, and perfectly inelastic.

Types of Collisions

Elastic collision: both momentum AND kinetic energy are conserved. KE before = KE after. (Ideal; billiard balls approximate this.)
Inelastic collision: momentum is conserved; kinetic energy is NOT conserved (some converted to heat/sound/deformation).
Perfectly inelastic collision: objects stick together after collision (v₁f = v₂f = vf); maximum kinetic energy is lost.

Perfectly inelastic: (m₁ + m₂)v_f = m₁v₁ᵢ + m₂v₂ᵢ

Explosions

In an explosion, the system starts at rest (total momentum = 0). After the explosion, fragments move apart with equal and opposite momenta so the total remains zero.

0 = m₁v₁f + m₂v₂f ⟹ m₁v₁f = −m₂v₂f

Worked Example — Car Collision

Problem: Car A (1200 kg) moves at 20 m·s⁻¹ East. Car B (900 kg) is stationary. They collide and stick together. Find vf.

Positive direction: East
Σpᵢ = 1200×20 + 900×0 = 24 000 kg·m·s⁻¹
Σpf = (1200+900)×vf
24 000 = 2100×vf
vf = 24 000/2100 = 11.4 m·s⁻¹ East

IEB Extension — 2D Collisions

In a glancing (2D) collision, momentum is conserved independently in the x and y directions:

Σpₓ before = Σpₓ after AND Σpᵧ before = Σpᵧ after

Resolve all momenta into components before and after the collision, then apply conservation in each direction separately.

IEB Extension — Coefficient of Restitution

The coefficient of restitution (e) measures the "bounciness" of a collision:

e = (v₂f − v₁f) / (v₁ᵢ − v₂ᵢ)

e = 1: perfectly elastic. e = 0: perfectly inelastic (objects stick). 0 < e < 1: partially inelastic. e is always between 0 and 1 for real collisions.

Collision Simulator

Collision Type

Block 1 ● (left)

Mass m₁ (kg)4 kg

Velocity v₁ᵢ+5 m·s⁻¹

Block 2 ● (right)

Mass m₂ (kg)2 kg

Velocity v₂ᵢ−2 m·s⁻¹

p total (before)

—kg·m·s⁻¹

p total (after)

—kg·m·s⁻¹

KE before

—J

KE after

—J

KE lost

—J

Status

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Question 1 of 8

A 0.5 kg ball moves at 10 m·s⁻¹ East. Its momentum is:

Question 2 of 8

A net force of 40 N acts on a 2 kg object for 0.5 s. The change in momentum is:

Question 3 of 8

A 1500 kg car (30 m·s⁻¹ East) collides with a 1000 kg stationary car and they stick together. Their combined velocity after is:

Question 4 of 8

Which statement is true about a perfectly inelastic collision?

Question 5 of 8

A 5 kg gun fires a 0.01 kg bullet at 400 m·s⁻¹. The recoil speed of the gun is:

Question 6 of 8

Airbags reduce injury during a collision primarily by:

Question 7 of 8

A 60 kg crash-test dummy decelerates from 20 m·s⁻¹ to rest. Without a crumple zone, the collision lasts 0.08 s. With a crumple zone fitted, the same crash instead lasts 0.4 s. By what factor is the average force on the dummy reduced due to the crumple zone?

Question 8 of 8

A 0.02 kg bullet is fired horizontally into a stationary 2 kg wooden block hanging from a string (a ballistic pendulum). The bullet embeds in the block, and the block+bullet swing up to a height of 0.2 m before momentarily stopping. Using energy conservation for the swing (to find the speed just after impact) and momentum conservation for the collision (to find the bullet's speed), calculate the bullet's initial speed. (Use g = 9.8 m·s⁻².)

IEB Extended Questions

IEB Question 1

Ball A (2 kg, 4 m·s⁻¹ East) hits Ball B (2 kg, at rest). After the collision, Ball A moves at 1 m·s⁻¹ East and Ball B at 3 m·s⁻¹ East. The coefficient of restitution is:

IEB Question 2

In a 2D collision, a 3 kg puck moving at 6 m·s⁻¹ North strikes a stationary 3 kg puck. After the collision, the first puck moves at 3 m·s⁻¹ West. Using conservation of momentum, the second puck's northward velocity is:

Define a positive direction at the start of every problem. Momentum is a vector — include direction in all answers. Show formulae, substitutions and units.

Question 1 — Impulse Calculation (5 marks)

A 0.15 kg cricket ball bowled at 30 m·s⁻¹ is hit back by a bat at 40 m·s⁻¹ in the opposite direction. The bat is in contact with the ball for 0.002 s. (a) Define a positive direction. (b) Calculate the change in momentum of the ball. (c) Calculate the impulse on the ball. (d) Calculate the average force exerted by the bat on the ball.

Question 2 — Conservation in a Collision (6 marks)

Trolley A (3 kg) moves at 5 m·s⁻¹ East. Trolley B (2 kg) moves at 3 m·s⁻¹ West. They collide and move separately afterwards: A at 1 m·s⁻¹ East. (a) State the law of conservation of momentum. (b) Find the velocity (magnitude and direction) of B after the collision. (c) Is this an elastic or inelastic collision? Show working.

Question 3 — Explosion (5 marks)

A stationary 80 kg astronaut in space pushes off a 200 kg space module. The astronaut moves at 3 m·s⁻¹ to the right. (a) State why total momentum is conserved here. (b) Calculate the velocity of the space module after the push. (c) Calculate the kinetic energy of each object and comment on where this energy came from.

Question 4 — Kinetic Energy Change in Collision (5 marks)

A 6 kg block moving at 8 m·s⁻¹ collides with a stationary 2 kg block. They stick together after the collision. (a) Find the velocity after the collision. (b) Calculate the kinetic energy before and after the collision. (c) Calculate the KE lost. (d) Where does the lost kinetic energy go?

Question 5 — Airbag Design Reasoning (4 marks)

A car passenger (70 kg) decelerates from 25 m·s⁻¹ to rest. (a) Calculate the change in momentum of the passenger. (b) Without an airbag, the collision takes 0.05 s. Calculate the average force on the passenger. (c) With an airbag, the collision takes 0.4 s. Calculate the new average force. (d) Explain using physics principles why airbags save lives.

Question 6 — Verifying Momentum Conservation from Ticker-Tape Data (6 marks)

In a trolley collision experiment, a ticker-timer records the position of Trolley A (mass 0.5 kg) at equal time intervals as it moves at constant velocity BEFORE colliding with stationary Trolley B (mass 0.5 kg):

Time (s)	Position of A (m)
0.0	0.000
0.1	0.050
0.2	0.100
0.3	0.150

Trolley A then collides with Trolley B and they stick together. The ticker-timer continues recording the position of the combined trolleys AFTER the collision:

Time (s)	Position (m)
0.3	0.150
0.4	0.175
0.5	0.200

(a) Using the gradient of the first table, calculate Trolley A's velocity before the collision. (2 marks)
(b) Using the gradient of the second table, calculate the combined trolleys' velocity after the collision. (2 marks)
(c) Calculate the total momentum before and after the collision, and state whether these results confirm conservation of momentum. (2 marks)