Initial mass of particle $=4 M$
Initial velocity of particle $=0$.
After explosion into three pieces of masses $M, M$ and $2 M$ particle of equal masses move along $X$ and $Y$ axes with velocities $4 \mathrm{~m} / \mathrm{s}$ and $6 \mathrm{~m} / \mathrm{s}$, respectively. So, according to law of conservation of momentum mass $2 M$ particle will move in the direction shown in the figure.


Law of conservation of momentum, Momentum before collision $=$ Momentum after collision.
$$
\begin{array}{rlrl}
& & 4 M \times 0 & =2 M \mathbf{v}+M \mathbf{v}_x+M \mathbf{v}_y \\
\Rightarrow & & -2 M \mathbf{v} & =M\left(\mathbf{v}_x+\mathbf{v}_y\right) \\
\Rightarrow & & -2 \mathbf{v} & =4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}} \Rightarrow \mathbf{v}=-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \\
\text { So, }|\mathbf{v}|=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13} \mathrm{~ms}^{-1}
\end{array}
$$