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MathidiaMathematics Academy
Further mechanics
Lesson 57 of 75
Easy practice

Collisions

Objective. Use conservation and restitution in one-dimensional impacts.

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  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

Collisions between objects are analysed using conservation of momentum, m₁u₁+m₂u₂ = m₁v₁+m₂v₂, combined with Newton's law of restitution, e = (v₂−v₁)/(u₁−u₂), where e is the coefficient of restitution between 0 (perfectly inelastic) and 1 (perfectly elastic). These two equations together allow the final velocities v₁ and v₂ to be found when the initial velocities and masses are known. For example, two equal masses colliding with e=1 (elastic) exchange velocities exactly, while e=0 means both move off together at the same final velocity.

Worked example

A ball of mass 2kg moving at 6 m/s collides directly with a stationary ball of mass 3kg. Given the coefficient of restitution is 0.5, find the velocities of both balls after the collision.

2 … 6
Illustrated card showing 2 … 6
  1. 1Conservation of momentum: 2(6) + 3(0) = 2v₁ + 3v₂, so 12 = 2v₁ + 3v₂
  2. 2Restitution: 0.5 = (v₂ − v₁)/(6 − 0), so v₂ − v₁ = 3
  3. 3From the second equation, v₂ = v₁ + 3; substitute into the first: 12 = 2v₁ + 3(v₁+3) = 5v₁ + 9
  4. 4Solve: 5v₁ = 3, v₁ = 0.6 m/s, and v₂ = 0.6+3 = 3.6 m/s

Answer: v₁ = 0.6 m/s, v₂ = 3.6 m/s

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Common mistake. Students write the restitution equation with the velocities in the wrong order or sign, forgetting it measures separation speed over approach speed, not the reverse. Always define a consistent positive direction and carefully order (v₂−v₁) over (u₁−u₂).

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