Two identical balls in contact on a table are in equilibrium. A third identical ball collides them simultaneously symmetrically and remains at rest after impact. If coefficient of restitution between the balls is  $\frac{2}{n}.$ Find the value of n

Using conservation of momentum in the direction of motion of colliding ball
 
$mu=mv\cos {30}^0+mv\cos {30}^{0}oru=\sqrt{3}v$ 
Coefficient of restitution  $e=-\left(\frac{v_1-v_2}{u_1-u_2}\right)$,
Where velocities of balls are along the line of collision.
$\therefore e=-\frac{(0-v)}{(u\cos {30}^0-0)}=\frac{v}{u\cos {30}^0}=\frac{v}{\sqrt{3}v\times \frac{\sqrt{3}}{2}}=\frac{2}{3}$ .