1. Set up momentum conservation

Two identical masses $m$ are involved. Assume the second mass is initially at rest.

• Initial momentum: $\vec p_i = m v\,\hat i$

After collision:

• First mass moves with speed $\dfrac{v}{\sqrt 3}$ perpendicular to the initial direction. Let this be along $\hat j$.
• Let the second mass have velocity components $u_x, u_y$.

So final momentum is $\vec p_f = m\left(\frac{v}{\sqrt 3}\hat j\right) + m(u_x\hat i + u_y\hat j)$

By conservation of momentum, $mv\hat i = m\left(\frac{v}{\sqrt 3}\hat j + u_x\hat i + u_y\hat j\right)$

Cancel $m$: $v\hat i = u_x\hat i + \left(u_y + \frac{v}{\sqrt 3}\right)\hat j$

2. Compare components

Along $x$-direction: $u_x = v$

Along $y$-direction: $u_y + \frac{v}{\sqrt 3} = 0$ $u_y = -\frac{v}{\sqrt 3}$

3. Find speed of the second mass

Its speed is $u = \sqrt{u_x^2 + u_y^2}$ $u = \sqrt{v^2 + \left(\frac{v}{\sqrt 3}\right)^2}$ $u = v\sqrt{1 + \frac{1}{3}}$ $u = v\sqrt{\frac{4}{3}}$ $u = \frac{2v}{\sqrt 3}$

4. Match with the options

$\boxed{\frac{2v}{\sqrt 3}}$

So the correct option is D.