Two identical spheres B and C, each of mass m, are in contact on a smooth horizontal table. A third sphere A of same size as that of B and C but mass $m/2$  impinges symmetrically on them with a velocity $u$ and is itself brought to rest. The coefficient of restitution between the two spheres $A$ and $B$ ( or between $A$ and C ) is

U= velocity of sphere A before impact. As the spheres are identical, the triangle ABC formed by joining their centres is equilateral. The spheres B and C will move in direction AB and AC after impact making an angle of  ${30}^0$  with the original lines of motion of ball A.
Let v be the speed of the ball B and C after impact
Momentum conservation gives
 $\begin{array}{l}\left(\frac{m}{2}\right)u=mv\cos {30}^0+mv\cos {30}^0\\ u=2\sqrt{3}v\Rightarrow v=\frac{u}{2\sqrt{3}}\left(i\right)\end{array}$
From Newton’s experimental law, for an oblique collision, we have to take components along normal, i.e., along AB for balls A and B
$$\begin{gathered} v_B-v_A=-e(u_B-u_A) \\ v-0=-(0-u\cos {30}^0) \\ v=eu\cos {30}^0 \end{gathered}$$
Combining Eqs. (i) and (ii), we get  $e=1/3$.