Q. A bullet of mass $m$ hits a target of mass $M$ hanging by a string and gets embedded in it. If the block rises to a height $h$ as a result of this collision, the velocity of the bullet before collision is
A) $v = \sqrt{2gh}$
B) $v = \sqrt{2gh}\left( 1 + \frac{m}{M} \right)$
C) $v = \left( 1 + \frac{M}{m} \right)\sqrt{2gh}$
D) $v = \sqrt{2gh}\left( 1 - \frac{m}{M} \right)$
Explanation
(c) If initial velocity of bullet be $v$ then after collision combined velocity of bullet and target is
$v^{'} = \frac{mv}{(M + v)}$ and $h = \frac{v^{'2}}{2g}$ or $v^{'} = \sqrt{2gh}$
$\therefore\ \ \frac{mv}{(M + m)} = \sqrt{2gh}$
$\Rightarrow \ v = \left( \frac{M + m}{m} \right) \bullet \sqrt{2gh} = \left( 1 + \frac{M}{m} \right)\sqrt{2gh}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.