12 kg નો સ્થિર બોમ્બ ફૂટતાં 1:3 દળના 2 ટુક

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 $36\;kg{\rm{ - }}m/s$