According to the law of conservation of linear momentum, the total initial momentum of the system is equal to the total final momentum. Initial momentum, $\vec{P}_i = 0$ (since the particle is at rest) Let the three fragments be $m_1 = m$, $m_2 = m$, and $m_3 = 5m - m - m = 3m$. Given that $m_1$ and $m_2$ move in mutually perpendicular directions with speed $v$. Let $\vec{v}_1 = v\hat{i}$ and $\vec{v}_2 = v\hat{j}$. Let the velocity of the third fragment be $\vec{v}_3$. Final momentum, $\vec{P}_f = m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3$ $\vec{P}_f = m(v\hat{i}) + m(v\hat{j}) + 3m\vec{v}_3$ Since $\vec{P}_i = \vec{P}_f$: $0 = mv\hat{i} + mv\hat{j} + 3m\vec{v}_3$ $3m\vec{v}_3 = -mv(\hat{i} + \hat{j})$ $\vec{v}_3 = -\frac{v}{3}(\hat{i} + \hat{j})$ The magnitude of the velocity of the third fragment is: $|\vec{v}_3| = \sqrt{\left(-\frac{v}{3}\right)^2 + \left(-\frac{v}{3}\right)^2} = \sqrt{\frac{v^2}{9} + \frac{v^2}{9}} = \frac{\sqrt{2}v}{3}$ The energy released during the process is equal to the total final kinetic energy of the fragments (since initial kinetic energy was zero). $K = K_1 + K_2 + K_3$ $K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$ $K = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(3m)\left(\frac{\sqrt{2}v}{3}\right)^2$ $K = mv^2 + \frac{3m}{2} \left(\frac{2v^2}{9}\right)$ $K = mv^2 + \frac{mv^2}{3} = \frac{4}{3}mv^2$