$\begin{array}{l}
Let\,\bar v'\,be\,velocity\,of\,third\,piece\,of\,mass\\
2m.\\
Initial\,momentum\,,\,{{\bar p}_i} = 0\left( {As\,the\,body\,is\,at\,rest} \right)\\
Final\,momentum,\,{{\bar P}_i} = mv\hat i + mv\hat j + 2m\bar v'\\
According\,to\,law\,of\,conservation\,of\\
momentum\,\\
\,\,\,\,\,\,\,\,\,\,\,\,{{\bar P}_i} = {{\bar P}_f}\\
0 = mv\hat i + mv\hat j + 2m\bar v'\\
\,\,\,\,\,\,\,\,\,\,\,\,\bar v' =  – \frac{v}{2}\hat i – \frac{v}{2}\,\hat j
\end{array}$

$\begin{array}{l}
The\,magnitude\,of\,v'\,is\\
\,\,\,\,\,\,\,\,\,\,\,v' = \sqrt {{{\left( { – \frac{v}{2}} \right)}^2} + {{\left( { – \frac{v}{2}} \right)}^2}}  = \frac{v}{{\sqrt 2 }}\\
Total\,kinetic\,energy\,generated\,due\,to\\
{\rm{explosion}}\,\\
 = \frac{1}{2}m{v^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {2m} \right)v'2\\
 = \frac{1}{2}m{v^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {2m} \right){\left( { – \frac{v}{{\sqrt 2 }}} \right)^2}\\
 = m{v^2} + \frac{{m{v^2}}}{2} = \frac{3}{2}m{v^2}
\end{array}$