A particle of mass m collides with another stationary particle of mass M. If the particle m stops just after the collision, the coefficient of restitution of collision is equal to:

The net horizontal force acting on the system (M+m) is zero. Therefore the momentum of the system just before and after the collision remains constant. Let M move with the velocity v'.
$$\begin{gathered} \Rightarrow \mathrm{m}\overrightarrow{\mathrm{v}}+0=\mathrm{m}\left(0\right)+\mathrm{M}\overrightarrow{{\mathrm{v}}^{,}} \\ \Rightarrow \frac{{\mathrm{v}}^{,}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}} ...\left(1\right) \end{gathered}$$
Newton’s experimental formula $\mathrm{e}=-\frac{\left({\mathrm{v}}^{,}-0\right)}{0-\mathrm{v}}$
$\Rightarrow \mathrm{e}=\frac{{\mathrm{v}}^{,}}{\mathrm{v}} ...\left(2\right)$
Equating (1) and (2),

e = (m/M)