Momentum, impulse and Newton’s second law

Definitions you must know

📖 Definitions

Momentum: the product of an object’s mass and its velocity. p = mv (kg·m·s⁻¹, a vector)

Newton’s second law in terms of momentum: the net (resultant) force acting on an object is equal to the rate of change of momentum of the object (in the direction of the net force). F_net = Δp ÷ Δt

Impulse: the product of the net force acting on an object and the time the net force acts on the object. Impulse = F_netΔt = Δp

Principle of conservation of linear momentum: the total linear momentum of an isolated system remains constant (is conserved).

Isolated system: a system on which the resultant (net) external force is zero.

Elastic collision: a collision in which both total momentum and total kinetic energy are conserved.

Change in momentum and average force

Momentum is a vector, so choose a positive direction and give every velocity a sign.

✏️ Worked example 1: a ball rebounds from a wall

1A 0,15 kg ball moves left at 10 m·s⁻¹, hits a wall and rebounds right at 6 m·s⁻¹. Take right as positive.

2Δp = m(v_f − v_i) = 0,15(6 − (−10)) = 2,4 kg·m·s⁻¹ to the right.

3The ball is in contact with the wall for 0,02 s: F_net = Δp/Δt = 2,4 ÷ 0,02 = 120 N to the right.

On a force–time graph, the area under the graph equals the impulse. On a momentum–time graph, the gradient equals the net force.

Safety applications: airbags, crumple zones and arrestor beds increase the time of the collision. For the same change in momentum, a larger Δt gives a smaller average force, so injuries are reduced.

Conservation of momentum

In an isolated system, Σp_before = Σp_after.

✏️ Worked example 2: trolleys collide and stick together

1Trolley A (2 kg) moves east at 3 m·s⁻¹ and hits trolley B (1 kg) moving west at 1 m·s⁻¹. They stick together. East positive.

2Σp_before = 2(3) + 1(−1) = 5 kg·m·s⁻¹.

3Σp_after = (2 + 1)v → v = 5 ÷ 3 = 1,67 m·s⁻¹ east.

Elastic or inelastic? Prove it with kinetic energy

✏️ Worked example 3: using the trolleys above

1E_k before = ½(2)(3)² + ½(1)(1)² = 9 + 0,5 = 9,5 J.

2E_k after = ½(3)(1,67)² = 4,17 J.

3Kinetic energy is not conserved, so the collision is inelastic. (Momentum is conserved in both types.)

A ball that rebounds with a smaller speed than it arrived with has lost kinetic energy, so the bounce is inelastic.

Answer in your exercise book, then open each memo to mark your work.

Question 1 · (2 marks)

State Newton’s second law of motion in terms of momentum.

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The net (resultant) force acting on an object is equal to the rate of change of momentum of the object ✓✓ (in the direction of the net force).

Question 2 · (5 marks)

A 0,4 kg ball is kicked from rest and leaves the boot at 25 m·s⁻¹. The boot is in contact with the ball for 0,01 s. (a) Calculate the change in momentum. (b) Calculate the average net force on the ball. (c) Define impulse.

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(a) Δp = m(v_f − v_i) = 0,4(25 − 0) ✓ = 10 kg·m·s⁻¹ in the direction of the kick ✓
(b) F_net = Δp/Δt = 10 ÷ 0,01 ✓ = 1 000 N ✓
(c) The product of the net force acting on an object and the time the net force acts ✓ (accept F_netΔt with explanation).

Question 3 · (6 marks)

A 1 200 kg car moving east at 20 m·s⁻¹ collides with a stationary 800 kg car. They move off together. (a) State the principle of conservation of linear momentum. (b) Calculate their common velocity. (c) Prove, with a calculation, whether the collision is elastic.

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(a) The total linear momentum of an isolated system remains constant ✓
(b) 1 200(20) + 0 = (2 000)v ✓ → v = 12 m·s⁻¹ east ✓
(c) E_k before = ½(1 200)(20)² = 240 000 J ✓; E_k after = ½(2 000)(12)² = 144 000 J ✓; E_k not conserved, so inelastic ✓

Question 4 · (3 marks)

Explain, using physics principles, why a car with crumple zones reduces the force on passengers in a collision.

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The change in momentum of the passenger is the same with or without crumple zones ✓. Crumple zones increase the time taken to stop ✓. Since F_net = Δp/Δt, a longer time gives a smaller average force ✓.