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

detailed solution

Correct option is A

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 θ with the horizontal with velocity v. Initial horizontal momentum of system(before collision) = mv                         ....(i)Final horizontal momentum of system (after collision) = mV cosθ                  ....(ii)From the conservation of horizontal linear momentum          mv = mV cosθ ⇒ v = V cosθ           ...(iii)Initial vertical momentum of system (before collision) is zero.Final vertical momentum of system mv3−mVsinθFrom the conservation of vertical linear momentum  mv3−mVsinθ=0⇒v3=Vsinθ              ...(iv)By solving (iii) and (iv) v2+v23=V2(sin2θ+cos2θ)⇒4v23=V2  ⇒V=23v