Explanation

(b) Here initial momentum $\overrightarrow{p} = 0$. Since no external force exists, hence momentum must remain conserved $ie$, ${\overrightarrow{p}}_{1} + {\overrightarrow{p}}_{2} + {\overrightarrow{p}}_{3} = 0$
As two fragments of mass mass $m$ each are moving with speed $v$ each at right angles,
$\left| {\overrightarrow{p}}_{1} + {\overrightarrow{p}}_{2} \right| = m\sqrt{v^{2} + v^{2}} = \sqrt{2}mv$
$\therefore\ \left| {\overrightarrow{p}}_{3} \right| = \left| {\overrightarrow{p}}_{1} + {\overrightarrow{p}}_{2} \right| = 2mv$.
The mass of third fragment is 2 m.
$\therefore$ Kinetic energies of three fragments are
$K_{1} = \frac{p_{1}^{2}}{2m} = \frac{1}{2}mv^{2},K_{2} = \frac{p_{2}^{2}}{2m} = \frac{1}{2}mv^{2}$
And $K_{3} = \frac{p_{3}^{2}}{2(2m)} = \frac{1}{2}mv^{2}$
Total kinetic energy released during explosion
$= K_{1} + K_{2} + K_{3} = \frac{3}{2}mv^{2}$