Q. A bullet of mas $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be
A) $\frac{1}{2}mv^{2} \times \frac{m}{(m + M)}$
B) $\frac{1}{2}mv^{2} \times \frac{M}{(m + M)}$
C) $\frac{1}{2}mv^{2} \times \frac{(M + m)}{(M)}$
D) $\frac{1}{2}Mv^{2} \times \frac{m}{(m + M)}$
Explanation
(a) By conservation of momentum, $mv + M \times 0 = (m + M)V$
Velocity of composite block $V = \left( \frac{m}{m + M} \right)v$
K.E. of composite block $= \frac{1}{2}\ (M + m)V^{2}$
$= \frac{1}{2}\ (M + m)\left( \frac{m}{M + m} \right)^{2}v^{2} = \frac{1}{2}\ mv^{2}\left( \frac{m}{m + M} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.