(4)

Let the speed of the third fragment of mass $3m$ be $v\text{'}.$

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From the law of conservation of linear momentum,

$3mv\text{'}=\sqrt{2}mv\Rightarrow v\text{'}=\frac{\sqrt{2}v}{3}$                       ...(i)

$\therefore$   Energy released during the process is,

$\text{K.E.}=2\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}m{v}^2\right)+\frac{1}{2}\left(3m\right)v{\text{'}}^2=m{v}^2+\frac{1}{2}\left(3m\right)\frac{2v^2}{9} \text{(Using eqn. (i))}$

$=m{v}^2+\frac{m{v}^2}{3}=\frac{4}{3}m{v}^2$

(2)

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Let $\overrightarrow{v\text{'}}$ be the velocity of third piece of mass $2m.$

Initial momentum, $\overrightarrow{p_i}=0$  (As the body is at rest)

Final momentum, $\overrightarrow{p_f}=mv\hat{i}+mv\hat{j}+2m\overrightarrow{v\text{'}}$

According to law of conservation of momentum

         $\overrightarrow{p_i}=\overrightarrow{p_f} \text{or,} 0=mv\hat{i}+mv\hat{j}+2m\overrightarrow{v\text{'}}$ $\text{or,} \overrightarrow{v\text{'}}=-\frac{v}{2}\hat{i}-\frac{v}{2}\hat{j}$

The magnitude of $v\text{'}$ is

             $v\text{'}=\sqrt{{(-\frac{{\displaystyle v}}{{\displaystyle 2}})}^2+{(-\frac{v}{2})}^2}=\frac{v}{\sqrt{2}}$

Total kinetic energy generated due to explosion

$=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}\left(2m\right)v{\text{'}}^2$

$=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{1}{2}\left(2m\right){\left(\frac{{\displaystyle v}}{{\displaystyle \sqrt{2}}}\right)}^2$  $=m{v}^2+\frac{m{v}^2}{2}=\frac{3}{2}m{v}^2$