Solution:

Initial momentum is zero. Two pieces of mass \(m\) move with velocity \(v\) perpendicular to each other. Their momenta are \(m\vec{v}_1 = mv\hat{i}\, m\vec{v}_2 = mv\hat{j}\). The third piece has mass \(m_3 = 4m - m - m = 2m\). By momentum conservation, \(m_3\vec{v}_3 = -(mv\hat{i} + mv\hat{j})\), so \(|\vec{v}_3| = \frac{\sqrt{(mv)^2 + (mv)^2}}{2m} = \frac{\sqrt{2}mv}{2m} = \frac{v}{\sqrt{2}}\). Total KE = \(\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(2m)(\frac{v}{\sqrt{2}})^2 = mv^2 + \frac{1}{2}mv^2 = \frac{3}{2}mv^2\).