Conservation of Momentum Calculator

Linear momentum is always conserved in a collision. There are no exceptions, provided you count everything the collision touches. What differs between collision types is whether kinetic energy is also conserved.

The coefficient of restitution e measures how much relative speed is preserved:

e = (v₂f − v₁f) / (v₁i − v₂i)

e = 1 is elastic: kinetic energy is fully conserved (billiard ball collisions approximate this) e = 0 is perfectly inelastic: the objects stick together after impact (car crashes) 0 < e < 1 is partially inelastic: most real collisions, with energy lost to heat and deformation

Final velocities for a 1D collision:

v₁f = [(m₁ − e·m₂)v₁i + (1 + e)m₂v₂i] / (m₁ + m₂) v₂f = [(1 + e)m₁v₁i + (m₂ − e·m₁)v₂i] / (m₁ + m₂)

The total momentum before and after is the same:

p = m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f

Sign convention: rightward velocities are positive, leftward are negative. A stationary object has velocity 0.

Typical e values: steel on steel ≈ 0.65, rubber ball on concrete ≈ 0.6–0.8, baseball bat ≈ 0.55, car crash ≈ 0.1–0.3.

The special case of e = 1 with equal masses is elegant: the two objects swap velocities. A moving ball hitting an identical stationary ball will stop completely, transferring all its speed. Newton’s cradle demonstrates this.

A worked example

A 2 kg cart moving right at 5 m/s hits a stationary 3 kg cart. Take e = 1, a perfectly elastic bumper.

v₁f = [(2 − 1×3)(5) + (1+1)(3)(0)] / 5 = −5/5 = −1 m/s, so the light cart bounces back at 1 m/s. v₂f = [(1+1)(2)(5) + (3 − 1×2)(0)] / 5 = 20/5 = +4 m/s.

Check the books. Momentum before is 2×5 + 3×0 = 10 kg·m/s; after it is 2×(−1) + 3×4 = 10. Kinetic energy before is ½×2×25 = 25 J; after it is ½×2×1 + ½×3×16 = 25 J. Both conserved, as they must be when e = 1.

Now set e = 0 and rerun it. Both carts leave together at 2 m/s, momentum is still 10, but kinetic energy has dropped from 25 J to 10 J. Those 15 joules went into crumpling, heat and sound. That energy gap is the whole difference between a bounce and a crash, and it is why crumple zones are designed to absorb as much of it as possible.

Where the “no exceptions” claim needs care. Momentum is conserved for the system, so if one object bounces off a wall you have to count the Earth as part of the system. It receives an equal and opposite momentum change; you just never notice, because its mass is 10²⁵ times larger and the resulting velocity is unmeasurable.