a
(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 = \(\frac{1}{2}{m_1}\) \(v_1^2 = 216J\)

\(v_1^2 = \frac{{216 \times 2}}{3}\)

==> \({v_1} = 12m/s\)

So its momentum = \({m_1}{v_1} = 3 \times 12 = 36\;kg{\rm{ - }}{\rm{m}}/s\)

As both parts possess same momentum therefore momentum of each part is \(36\;kg{\rm{ - }}m/s\)