Q. A bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$, the initial velocity of the block will be
A) $\sqrt{2gh}$
B) $\frac{M + m}{m}\sqrt{2gh}$
C) $\frac{m}{M + m}2gh$
D) $\frac{M + m}{M}\sqrt{2gh}$
Explanation
(a) Initial K.E. of block when bullet strikes to it
$= \frac{1}{2}(m + M)V^{2}$
Due to this K.E. block will rise to a height $h$
Its potential energy $= (m + M)gh$
By the law of conservation of energy
$\frac{1}{2}(m + M)V^{2} = (m + M)gh$ $\therefore V = \sqrt{2gh}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.