p = mv · Σmv constant

Momentum calculator

Momentum and kinetic energy for one object, or the velocities after a head-on collision with momentum conserved and the energy loss reported.

One moving object

Head-on collision

Short answer: 10 kg at 5 m/s has momentum 50 kg·m/s and 125 J of kinetic energy. A 2 kg mass at 3 m/s hitting a stationary 1 kg mass perfectly elastically leaves them at 1 m/s and 4 m/s — momentum 6 kg·m/s before and after, with no energy lost.

The two conservation statements

Momentum is conserved in a collision with no outside force, which is what makes the problem solvable. Energy is not conserved unless the collision is perfectly elastic, and the page shows the loss so the difference between the two statements is visible in the numbers rather than only in prose.

Worked example

A 1,200 kg car at 15 m/s hits a stationary 1,500 kg car and they lock together: v = (1200 × 15) ÷ (1200 + 1500) = 6.67 m/s, and 135 kJ becomes 60 kJ — over half the energy goes into crumpling metal. That is why crumple zones exist and why the restitution 0 case is worth calculating.

Momentum questions

What is momentum?

p = mv — mass times velocity, in kg·m/s. It is a vector, so a ball moving left has negative momentum, and that sign is what makes the collision arithmetic come out right.

Why is kinetic energy separate?

Both come from the same two numbers, but they scale differently: doubling the velocity doubles the momentum and quadruples the energy. That is why a crash at twice the speed does four times the damage, and why both are reported.

What does the restitution setting mean?

It is how bouncy the collision is, from 0 (the two masses stick together) to 1 (perfectly elastic, no energy lost). Real collisions sit between: a car crash is near 0, a steel ball bearing on steel is close to 0.9.

Is momentum really always conserved?

In a collision with no external force, yes — that is why it solves the problem. Energy is not: some always becomes heat, sound and deformation, and the page prints the loss so that a perfectly elastic assumption is never smuggled in.

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