A mass 'm' moves with a velocity 'v' and collides inelastically with another identical mass. After collision the I^st mass moves with velocity $\frac{\mathrm{v}}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. Find the speed of the 2^nd mass after collision

Let mass A moves with velocity v and collides inelastically with mass B, which is at rest. 

According to problem mass A moves in a perpendicular direction and let the mass B moves at angle $\mathrm{\theta }$ with the horizontal with velocity v. 
Initial horizontal momentum of system
(before collision) = mv                         ....(i)
Final horizontal momentum of system 
(after collision) = mV cos$\mathrm{\theta }$                  ....(ii)
From the conservation of horizontal linear momentum          mv = mV cos$\mathrm{\theta }$ $\Rightarrow$ v = V cos$\mathrm{\theta }$           ...(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system $\frac{\mathrm{mv}}{\sqrt{3}}-\mathrm{mVsin\theta }$
From the conservation of vertical linear momentum  $\frac{\mathrm{mv}}{\sqrt{3}}-\mathrm{mVsin\theta }=0\Rightarrow \frac{\mathrm{v}}{\sqrt{3}}=\mathrm{Vsin\theta }$              ...(iv)
By solving (iii) and (iv) 
${\mathrm{v}}^2+\frac{{\mathrm{v}}^2}{3}={\mathrm{V}}^2({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta })$
$\Rightarrow \frac{4{\mathrm{v}}^2}{3}={\mathrm{V}}^2 \Rightarrow \mathrm{V}=\frac{2}{\sqrt{3}}\mathrm{v}$