According to the law of conservation of linear momentum, since the body explodes due to internal forces, the total momentum of the system remains conserved . Initially, the body is at rest, so the initial momentum is zero.

Let the two pieces of mass $M/4$ be moving along the x and y axes. Their momenta are: $\vec{p}_1 = \frac{M}{4} \times 3 \hat{i} = \frac{3M}{4} \hat{i}$ $\vec{p}_2 = \frac{M}{4} \times 4 \hat{j} = M \hat{j}$

The mass of the third piece is $m_3 = M - \frac{M}{4} - \frac{M}{4} = \frac{M}{2}$. Let the velocity of the third piece be $\vec{v}_3$. Its momentum is $\vec{p}_3 = \frac{M}{2} \vec{v}_3$.

By conservation of momentum: $\vec{p}_{\text{initial}} = \vec{p}_{\text{final}}$ $0 = \vec{p}_1 + \vec{p}_2 + \vec{p}_3$ $0 = \frac{3M}{4}\hat{i} + M\hat{j} + \frac{M}{2}\vec{v}_3$ $\frac{M}{2}\vec{v}_3 = -\left(\frac{3M}{4}\hat{i} + M\hat{j}\right)$ $\vec{v}_3 = -1.5\hat{i} - 2\hat{j}$

The magnitude of the velocity is: $|\vec{v}_3| = \sqrt{(-1.5)^2 + (-2)^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5 \text{ m/s}$.