Q. A bomb of mass $3m\ kg$ explodes into two pieces of mass $m\ kg$ and $2m\ kg$. If the velocity of $m\ kg$ mass is $16\ m/s$, the total kinetic energy released in the explosion is
A) $192\ mJ$
B) $96\ mJ$
C) $384\ mJ$
D) $768\ mJ$
Explanation
(a) By the conservation of momentum, $m_{A}v_{A} = m_{B}v_{B}$
$\Rightarrow m \times 16 = 2m \times v_{B} \Rightarrow v_{B} = 8\ m/s$
Kinetic energy of system $= \frac{1}{2}m_{A}v_{A}^{2} + \frac{1}{2}m_{B}v_{B}^{2}$
$= \frac{1}{2} \times m \times (16)^{2} + \frac{1}{2} \times (2m) \times 8^{2} = 192mJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.