Elastic Collision — Final Velocity of Body 1
Worked example: Equal masses, target at rest → v1 = 0 — press Try an example to run it live, then adjust anything.
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Grade 12Grade 12 Physics
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Elastic Collision — Final Velocity of Body 1 explained
An elastic collision conserves both momentum and kinetic energy, and solving those two equations together gives this closed form for the first body's rebound. Three cases are worth memorising. Equal masses with the target at rest: v₁ = 0 and the bodies swap velocities exactly — the behaviour of a Newton's cradle and of a well-struck cue ball. A light body striking a much heavier one: v₁ ≈ −u₁, a near-perfect bounce back, as when a ball hits a wall. A heavy body striking a light one: v₁ ≈ u₁, it barely notices.
Worked example: a 1 kg ball at 4 m/s hits a stationary 3 kg ball, giving v₁ = ((1 − 3)(4) + 0) ⁄ 4 = −2 m/s — it rebounds at half speed while the heavier ball moves off at 2 m/s. Truly elastic collisions are an idealisation for everyday objects (steel bearings come close, billiard balls reach about 95%), but they are exact for gas molecules and for neutrons scattering in a reactor moderator — which is precisely why moderators use light nuclei like hydrogen or carbon, where each collision strips the most energy.
Elastic Collision — Final Velocity of Body 1 formula
- = Final velocity of body 1 (m/s)
- = Mass 1 (kg)
- = Initial velocity 1 (m/s)
- = Mass 2 (kg)
- = Initial velocity 2 (m/s)
Missing one of these? Work it out first, then come back
- Final velocity of body 1 — Final Velocity (Uniform Acceleration), Velocity-Displacement Relation (v² = v₀² + 2ad)
- Mass 1 — Newton's Second Law, Kinetic Energy
- Initial velocity 1 — Final Velocity (Uniform Acceleration), Displacement (Uniform Acceleration)
- Mass 2 — Newton's Second Law, Kinetic Energy
- Initial velocity 2 — Final Velocity (Uniform Acceleration), Displacement (Uniform Acceleration)