Explanation

(a) The bomb of mass $12kg$ divides into two masses
$m_{1}$ and $m_{2}$ then $m_{1} + m_{2} = 12$ …(i)
And $\frac{m_{1}}{m_{2}} = \frac{1}{3}$ …(ii)
By solving we get $m_{1} = 3kg$ and $m_{2} = 9kg$
Kinetic energy of smaller part $= \frac{1}{2}m_{1}v_{1}^{2} = 216\ J$
$\therefore\ v_{1}^{2} = \frac{216 \times 2}{3} \Rightarrow v_{1} = 12\ m/s$
So its momentum $= m_{1}v_{1} = 3 \times 12 = 36\ kg - m/s$
As both parts possess same momentum therefore momentum of each part is $36\ kg - m/s$