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

Let mass Amoves 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 momentum of the system
(before collision) = $mv\hat{i}$ -------(i)
Final momentum of thesystem
(after collision) = $m\frac{v}{\sqrt{3}}\hat{j}+\overrightarrow{p}$-------- (ii)
From the conservation  linear momentum
$$\begin{gathered} mv\hat{i}=m\frac{v}{\sqrt{3}}\hat{j}+\overrightarrow{p} \\ \overrightarrow{p}=mv\hat{i}-m\frac{v}{\sqrt{3}}\hat{j} \\ \left|\overrightarrow{p}\right|=\sqrt{m\left(v^2+\frac{v^2}{3}\right)}=m\frac{2v}{\sqrt{3}} ;velocity =\frac{2v}{\sqrt{3}} \end{gathered}$$