Coefficient of Restitution e

For a HEAD-ON collision (where the colliding bodies move strictly along one common straight line, both before and after impact), the COEFFICIENT OF RESTITUTION e is defined as the negative of the ratio of the relative velocity of separation (after collision) to the relative velocity of approach (before collision).

For two bodies of masses $m_1$ and $m_2$ with initial velocities $u_1$, $u_2$ and final velocities $v_1$, $v_2$ (all along the same line, with proper algebraic signs), the relative velocity of approach is $u_a=u_2-u_1$ and the relative velocity of separation is $v_s=v_2-v_1$, so $e=-\frac{v_s}{u_a}=-\frac{v_2-v_1}{u_2-u_1}$.

For a PERFECTLY INELASTIC collision, the bodies move jointly after impact, $v_1=v_2$, so $v_s=0$ and hence $e=0$. Conversely, $e=0$ for a head-on collision guarantees it is perfectly inelastic.

For a perfectly ELASTIC head-on collision, combining conservation of momentum ($m_1u_1+m_2u_2=m_1v_1+m_2v_2$) with conservation of kinetic energy leads (after algebraic simplification, dividing the KE-conservation equation by the momentum-conservation equation) to $u_1+v_1=u_2+v_2$, i.e. $u_2-u_1=v_1-v_2$ -- meaning the relative velocity of approach exactly equals the relative velocity of separation in magnitude, giving $e=1$ for a perfectly elastic collision. …