A mass m moves with a velocity v and collides | vedclass

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Question

(A) Let mass $A$ move with velocity $v$ and collide inelastically with mass $B$,which is initially at rest.According to the problem,mass $A$ moves in

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let mass $A$ move with velocity $v$ and collide inelastically with mass $B$,which is initially at rest.According to the problem,mass $A$ moves in a perpendicular direction after the collision with velocity $\frac{v}{\sqrt{3}}$. Let mass $B$ move at an angle $	heta$ with the horizontal with velocity $V$.Initial horizontal momentum of the system (before collision) $= mv$.Final horizontal momentum of the system (after collision) $= m \left( \frac{v}{\sqrt{3}} ight) \cos(90^{\circ}) + mV \cos 	heta = mV \cos 	heta$.From the conservation of horizontal linear momentum: $mv = mV \cos 	heta \implies v = V \cos 	heta$ $...(i)$.Initial vertical momentum of the system (before collision) $= 0$.Final vertical momentum of the system (after collision) $= m \left( \frac{v}{\sqrt{3}} ight) - mV \sin 	heta$.From the conservation of vertical linear momentum: $m \left( \frac{v}{\sqrt{3}} ight) - mV \sin 	heta = 0 \implies \frac{v}{\sqrt{3}} = V \sin 	heta$ $...(ii)$.Squaring and adding $(i)$ and $(ii)$:$v^2 + \frac{v^2}{3} = V^2 (\cos^2 	heta + \sin^2 	heta)$.$\frac{4v^2}{3} = V^2$.$V = \frac{2}{\sqrt{3}}v$.

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