A bomb of 12 kg divides in two parts whose ratio of masses is 1 : 3. If kinetic energy of smaller part is 216 J, then momentum of bigger part in kg-m/sec will be

The bomb of mass 12kg divides into two masses 
m₁ and m₂ then ${\mathrm{m}}_1+{\mathrm{m}}_2=12$      …(i)
and  $\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{1}{3}$                              …(ii)
by solving we get ${\mathrm{m}}_1=3\mathrm{kg}$ and  ${\mathrm{m}}_2=9\mathrm{kg}$
Kinetic energy of smaller part =  $\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2=216\mathrm{J}$
$\therefore {\mathrm{v}}_1^2=\frac{216\times 2}{3}\Rightarrow {\mathrm{v}}_1=12\mathrm{m}\text{}/\text{}\mathrm{s}$
So its momentum =  ${\mathrm{m}}_1{\mathrm{v}}_1=3\times 12=36\mathrm{kg}\text{-m}/\text{}\mathrm{s}$
As both parts possess same momentum therefore momentum of each part is  $36\mathrm{kg}\text{-m}/\text{}\mathrm{s}$