Explanation

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