Momentum

Further Maths · Further Mechanics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

In a collision between two spheres, total momentum is conserved, but one equation cannot give two unknown velocities. The second equation is Newton's experimental law: the speed of separation is a fixed fraction e of the speed of approach. The number e is the coefficient of restitution and lies between 0 and 1.

If e = 1 the collision is perfectly elastic and no kinetic energy is lost. If e = 0 it is inelastic and the spheres move on together.

For smooth spheres the impulse acts along the line of centres only. In an oblique impact, split each velocity into a part along this line and a part perpendicular to it. The perpendicular parts do not change. The same idea works for a smooth fixed wall.

Rules to remember

Momentum = mass × velocity; total momentum before = total momentum after (m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂), with one direction taken as positive
Newton's experimental law: speed of separation = e × speed of approach (v₂ − v₁ = e(u₁ − u₂)), with 0 ≤ e ≤ 1
e = 1: perfectly elastic, kinetic energy conserved. e = 0: inelastic, the spheres have the same velocity after impact
Direct impact with a fixed surface: rebound speed = e × impact speed; a ball dropped from height h rebounds to height e²h
Oblique impact with a smooth fixed wall: the component parallel to the wall is unchanged; the component perpendicular to the wall is reversed and multiplied by e
If the path makes angle α with the wall before impact and β after, tan β = e tan α
Oblique impact of two smooth spheres: components perpendicular to the line of centres are unchanged; use momentum and Newton's law along the line of centres

Common mistake

Sign errors in Newton's experimental law are the usual cause of lost marks. Draw a before and after diagram, mark every velocity in the same positive direction, and write v₂ − v₁ = e(u₁ − u₂). A negative answer then simply means the sphere moves the other way.

Worked example

A smooth sphere A of mass 4 kg moving at 5 m s⁻¹ collides directly with a stationary smooth sphere B of mass 1 kg. The coefficient of restitution is 0.6. Find the speeds of A and B after the collision.

Let the velocities after impact be a and b in the original direction of A. Momentum: 4 × 5 + 0 = 4a + b, so 4a + b = 20.
Newton's experimental law: b − a = 0.6 × (5 − 0) = 3, so b = a + 3.
Substitute: 4a + a + 3 = 20, so a = 3.4 and b = 6.4.
Check momentum: 4 × 3.4 + 6.4 = 13.6 + 6.4 = 20.

Answer: A: 3.4 m s⁻¹, B: 6.4 m s⁻¹, both in the original direction of A

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