Two equal discs initially at rest are in contact on a smooth horizontal table. A third disc of same mass but of double radius strikes them symmetrically and comes to rest after impact. The coefficient of restitution is 

Let initial speed of big disc is u.
 $\cos \theta =\frac{2\sqrt{2}}{3R}=\frac{2\sqrt{2}}{3}$
$2J\cos \theta =mu\Rightarrow J=\frac{mu}{2\cos \theta }=\frac{mu3}{2\left(2\sqrt{2}\right)}=\frac{3mu}{4\sqrt{2}}$       
Also, J=mv
$\therefore mv=\frac{3mu}{4\sqrt{2}}\Rightarrow v=\frac{3u}{4\sqrt{2}}$  
Along common normal direction, speed of approach $=u\cos \theta$  and speed of separation =v
$\therefore \theta =\frac{V}{u\cos \theta }=\frac{3u/4\sqrt{2}}{u(2\sqrt{2}/3)}=\frac{9}{16}$