Explanation

(c) Let $m_{1},m_{2}$be the masses of first and second fragments respectively and $v_{1},v_{2}$be their velocities after explosion.
From conservation of momentum
$Mv = m_{1},m_{2} + m_{2}v_{2}$
Where, M is mass of bomb before explosion and vits velocity.
Since, bomb is stationary, hence v=0
Given ,$m_{1} = 1g = 1 \times 10^{- 3}kg = 0.001kg$
$m_{2} = 3g = 3 \times 10^{- 3}kg = 0.003$kg and $E_{k} = 6.4 \times 10^{4}J$
$\therefore 0 = m_{1}v_{1} + m_{2}v_{2}$
$or\ \ \ \ \ \ \ \ \ \ \ \ 0 = 0.001v_{1} + 0.003v_{2}$
Or $\ \ \ \ \ v_{2} = - \frac{v_{1}}{3}$ .............(i)
Total kinetic energy is
${\ \ \ \ \ \ \ \ \ \ \ E}_{K} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}$
$E_{k} = \frac{1}{2} \times (0.001)v_{1}^{2} + \frac{1}{2} \times (0.003)v_{2}^{2}$ …(ii)
$\therefore E_{k} = \frac{1}{2} \times {(0.001)v}_{1}^{2} + \frac{1}{2}(0.003) \times \left( - \frac{v_{1}^{2}}{3} \right)^{2}$
$E_{k} = \frac{1}{2} \times (0.001)\left( v_{1}^{2} + 3 \times \frac{v_{1}^{2}}{9} \right)$
$E_{k} = \frac{1}{2} \times (0.001) \times \frac{4v_{1}^{2}}{3} = \frac{(0.002)v_{1}^{2}}{3}$ …(iii)
$\therefore\ \ \ \ \ \ \ \ 6.4 \times 10^{4} = \frac{(0.002)v_{1}^{2}}{3}$
Or $v_{1}^{2} = \frac{3 \times 6.4 \times 10^{4}}{0.002}$
Or $v_{1}^{2} = \frac{3 \times 6.4 \times 10^{4}}{0.002}$
$or\ \ \ \ \ \ \ v_{1}^{2} = 96 \times 10^{6} = 9.6 \times 10^{7}ms^{- 1}$
$\ \ $Hence, kinetic energy of smaller fragment is
$E_{K}^{'} = \frac{1}{2}m_{1}v_{1}^{2}$
$E_{k}^{'} = \frac{1}{2} \times (0.001) \times 9.6 \times 10^{7}$
$E_{k}^{'}\ = 4.8 \times 10^{4}J.$