A ball of mass $m$ moving with a speed $2v_0$ collides head-on with an identical ball at rest. If $e$ is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision? 

According to the law of conservation of linear momentum, we get 

$m\left(2v_0\right)+m\times 0=m{v}_1+m{v}_2$

where $v_1$ and $v_2$ are the velocities of the balls after collision.

$v_1+v_2=2v_0 ...\left(i\right)$

By definition of coefficient of restitution

$$\begin{gathered} e=\frac{v_2-v_1}{u_1-u_2}=\frac{v_2-v_1}{2v_0} \left(\because u_1=2v_0,u_2=0\right) \\ {v}_2-v_1=2e{v}_0 ...\left(ii\right) \end{gathered}$$

Adding $\left(i\right)$ and $\left(ii\right)$ ,we get $2v_2=2v_0+2e{v}_0;v_2=\left(1+e\right)v_0$

Subtract $\left(ii\right)$ from $\left(i\right)$, we get 

$2v_1=2v_0-2e{v}_0;v_1=v_0(1-e)$

Their corresponding ratio is $\frac{v_1}{v_2}=\frac{1-e}{1+e}$