How this calculator works
If external impulse during a collision is negligible, the combined momentum of the two bodies is conserved. Momentum alone leaves one free final velocity, so a collision mode adds a restitution condition. A value of one preserves relative approach speed; zero makes the bodies share a velocity. In unknown-quantity modes, supplied final velocities may represent an energy-releasing event, so kinetic energy loss may be negative.
Formula
m1 u1 + m2 u2 = m1 v1 + m2 v2; v2-v1 = e(u1-u2); K = (m1 v1^2 + m2 v2^2)/2.
Worked example
Two equal 2 kg bodies exchange velocities in an elastic collision: 3 and 0 m/s become 0 and 3 m/s. If a 2 kg body at 3 m/s sticks to a stationary 1 kg body, both leave at 2 m/s and lose 3 J of kinetic energy.
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Frequently asked questions
Is kinetic energy always conserved?
Only the perfectly elastic mode conserves total kinetic energy. Momentum conservation does not require that.
How do I describe a head-on collision?
Choose one positive axis, then enter the velocity of the body travelling the other way as negative.
Can this solve a two-dimensional collision?
No. The tool uses a single line of motion and no rotational degrees of freedom.