Q. A bomb of mass $3.0\ kg$ explodes in air into two pieces of masses $2.0\ kg$ and $1.0\ kg$. The smaller mass goes at a speed of $80\ m/s$. The total energy imparted to the two fragments is
A) $1.07\ kJ$
B) $2.14\ kJ$
C) $2.4\ kJ$
D) $4.8\ kJ$
Explanation
(d) Both fragments will possess the equal linear momentum
$m_{1}v_{1} = m_{2}v_{2} \Rightarrow 1 \times 80 = 2 \times v_{2} \Rightarrow v_{2} = 40\ m/s$
$\therefore$ Total energy of system $= \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}$
$= \frac{1}{2} \times 1 \times (80)^{2} + \frac{1}{2} \times 2 \times (40)^{2} = 4800\ J = 4.8\ kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.