Question

A particle is fired with velocity $u$ at an angle $\theta$ with horizontal. At the high point of its trajectory, it splits up into three segments of masses $m,m$ and $2m$. First part falls vertically downward with zero initial velocity and second part return via the same path to the point of projection. The velocity of the third part of mass $2m$ just after explosion will be

• A. $u\cos \theta$
• B. $\frac{3}{2}u\cos \theta$
• C. $2u\cos \theta$
• D. $\frac{5}{2}u\cos \theta$

Answer 1

The correct option is D $\frac{5}{2}u\cos \theta$

upon conserving momentum in the horizontal direction,

we have, $P_i=P_f$

$4mu\cos \theta =-mu\cos \theta +2mv$

$5mu\cos \theta =2mv$

$\therefore v=\frac{5}{2}u\cos \theta$ in the right direction.