Q. A bomb of mass $9\ kg$ explodes into 2 pieces of mass $3kg$ and $6kg$. The velocity of mass $3kg$ is $1.6\ m/s$, the K.E. of mass $6kg$ is
A) $3.84\ J$
B) $9.6\ J$
C) $1.92\ J$
D) $2.92\ J$
Explanation
(c) As the bomb initially was at rest therefore
Initial momentum of bomb = 0
Final momentum of system $= m_{1}v_{1} + m_{2}v_{2}$
As there is no external force
$\therefore m_{1}v_{1} + m_{2}v_{2} = 0 \Rightarrow 3 \times 1.6 + 6 \times v_{2} = 0$
Velocity of $6\ kg$ mass $v_{2} = 0.8\ m/s$ (numerically)
Its kinetic energy $= \frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2} \times 6 \times (0.8)^{2} = 1.92\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.