What is conserved, and what is not

In every collision the total momentum is conserved: the forces the two bodies exchange are internal to the system and, by the third law, equal and opposite, so their sum changes nothing. This holds whether the impact is elastic, inelastic or anything between, which is why it is always the starting point: it is the one equation that is guaranteed.

Kinetic energy, by contrast, is conserved only in a perfectly elastic collision. In the others some of it goes into permanent deformation, heat and sound, and that share is often the answer the exercise wants. In the perfectly inelastic case the bodies stay together and the loss is greatest: they travel on at the centre-of-mass velocity, the only velocity compatible with the momentum they started with.

The coefficient of restitution unifies the two cases textbooks present separately. It is the ratio of the speed at which the bodies separate to the speed at which they approached: 1 for an elastic impact, 0 when they stick, and something between for a tennis ball or two cars. One formula, with the two familiar cases at its ends.

Common mistakes

  • Forgetting the signs: two bodies coming towards each other have velocities of opposite sign, and adding them as positive numbers gets the total momentum wrong.
  • Applying conservation of kinetic energy to an inelastic collision: it is not conserved there, and the energy lost is precisely the number being asked for.
  • Confusing approach speed with the sum of the speeds: it is the difference, because what matters is the relative motion.
  • Believing momentum is lost in an inelastic collision: energy is lost, momentum never.

Frequently asked questions

Why is momentum always conserved and energy not?

Because momentum changes only under external forces, and the ones the two bodies exchange are internal. Kinetic energy, on the other hand, can turn into other forms — deformation, heat, sound — with no external force involved at all.

What is the coefficient of restitution?

The ratio of the separation speed after the impact to the approach speed before it. It is 1 for a perfectly elastic collision, 0 when the bodies stick together, and somewhere between in reality: about 0.75 for a basketball on a wooden floor, far less for two car panels.

What happens if the two bodies have equal masses?

In an elastic collision they simply swap velocities. It is the case of billiard balls and of Newton's cradle, and it falls out of the general formula by setting m₁ = m₂.

Why does the centre-of-mass velocity not change?

Because it is the total momentum divided by the total mass, and neither changes during the impact. Seen from the centre of mass the collision is always symmetric, which is often the fastest way to solve it.

Does this work for oblique collisions?

No, this page handles head-on collisions along a line. In two dimensions momentum is conserved component by component and two equations are needed instead of one.

How this calculation works

From conservation of momentum m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂' and the definition of the coefficient of restitution e = (v₂' − v₁')/(v₁ − v₂) comes a system of two equations in the two final velocities, whose solution is v₁' = ((m₁ − e·m₂)v₁ + (1 + e)m₂v₂)/(m₁ + m₂) and v₂' = ((m₂ − e·m₁)v₂ + (1 + e)m₁v₁)/(m₁ + m₂). At e = 1 the elastic formulas reappear; at e = 0 both velocities collapse onto the centre-of-mass velocity. Kinetic energy is ½mv² before and after, and the difference is what the collision dissipated. The impulse shown is m₁(v₁' − v₁); the second body receives exactly its opposite.