What is the difference between elastic and inelastic collisions?

What is the difference between elastic and inelastic collisions?

In every collision the total momentum is conserved. What differs is kinetic energy: in an elastic collision it is conserved too, while in an inelastic one some of it is turned into heat, sound or deformation. A perfectly inelastic collision is the extreme case where the bodies stick together and move with a common velocity.

What is always conserved?

Two bodies pushing on each other exert equal and opposite forces for the same time, so the momentum one gains the other loses. With initial velocities $u$ and final velocities $v$:

This holds for a car crash, two billiard balls or a hammer on a nail — as long as no outside force acts during the collision. It is one equation with two unknowns ($v_1$, $v_2$), so you need one more piece of information: how elastic the collision is.

What happens in an elastic collision?

Requiring kinetic energy to be conserved as well gives, for a head-on collision:

Two special cases are worth remembering: equal masses swap velocities, and a light body hitting a very heavy one bounces back with almost its own speed while the heavy body barely moves.

What happens in an inelastic collision?

In a perfectly inelastic collision the bodies stick together, so there is one final velocity $v = (m_1u_1 + m_2u_2)/(m_1 + m_2)$. In between lies the general case, described by the coefficient of restitution:

Combining it with momentum conservation gives $v_1 = \dfrac{m_1u_1 + m_2u_2 - m_2e\,(u_1 - u_2)}{m_1 + m_2}$ and $v_2 = \dfrac{m_1u_1 + m_2u_2 + m_1e\,(u_1 - u_2)}{m_1 + m_2}$. Setting $e = 1$ recovers the elastic formulas; $e = 0$ makes both velocities equal.

A worked example

A $2$ kg body moving at $3$ m/s hits a $1$ kg body at rest. Total momentum is $6\ \text{kg·m/s}$ and the initial kinetic energy is $9$ J.

Check the momentum in each row: $2\times1 + 1\times4 = 6$, $2\times1.5 + 1\times3 = 6$, $3\times2 = 6$. It never changes — only the energy does.

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