A bomb of mass 9 kg explodes into 2 pieces of mass 3 kg and 6 kg. The velocity of mass 3 kg is 1.6 m/s. The K.E. of mass 6 kg is

As the bomb initially was at rest therefore
Initial momentum of bomb = 0
Final momentum of system = ${\mathrm{m}}_1{\mathrm{v}}_1 + {\mathrm{m}}_2{\mathrm{v}}_2$
As there is no external force
$\therefore {\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2 = 0 \Rightarrow 3\times 1.6+6\times {\mathrm{v}}_2 = 0$

velocity of 6 kg mass ${\mathrm{v}}_2$ = 0.8 m/s (numerically)
Its kinetic energy = $\frac{1}{2}{\mathrm{m}}_2{{\mathrm{v}}^2}_2 = \frac{1}{2}\times 6\times {(0.8)}^{2 } = 1.92 \mathrm{J}$