The explosion is an internal process, so the total momentum of the system is conserved. The energy released is the change in the total kinetic energy of the system. Let $M=6.0 \text{ kg}$ be the initial mass and $\vec{v}_i = 50\hat{i} \text{ m/s}$ be the initial velocity. The pieces are $m_1=2.0 \text{ kg}$, $m_2=1.0 \text{ kg}$. The third piece has mass $m_3 = M - m_1 - m_2 = 3.0 \text{ kg}$. The velocities are $\vec{v}_1 = 500\hat{j} \text{ m/s}$ and $\vec{v}_2 = -400\hat{i} \text{ m/s}$.

[Q1] Conservation of momentum: $M\vec{v}_i = m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3$. We solve for $\vec{v}_3$:

$$m_3\vec{v}_3 = M\vec{v}_i - m_1\vec{v}_1 - m_2\vec{v}_2$$ $$(3.0)\vec{v}_3 = (6.0)(50\hat{i}) - (2.0)(500\hat{j}) - (1.0)(-400\hat{i})$$ $$(3.0)\vec{v}_3 = (300\hat{i}) - (1000\hat{j}) + (400\hat{i}) = 700\hat{i} - 1000\hat{j}$$ $$\vec{v}_3 = (\frac{700}{3}\hat{i} - \frac{1000}{3}\hat{j}) \text{ m/s}$$

[Q2] The energy released is $\Delta E = K_{final} - K_{initial}$.

$$K_{initial} = \frac{1}{2}Mv_i^2 = \frac{1}{2}(6.0)(50)^2 = 7500 \text{ J}$$ $$K_{final} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$$ $$K_{final} = \frac{1}{2}(2.0)(500)^2 + \frac{1}{2}(1.0)(400)^2 + \frac{1}{2}(3.0)\left[\left(\frac{700}{3}\right)^2 + \left(-\frac{1000}{3}\right)^2\right]$$ $$K_{final} = 250000 + 80000 + \frac{3}{2}\frac{1490000}{9} = 330000 + \frac{745000}{3} = \frac{1735000}{3} \text{ J}$$ $$\Delta E = \frac{1735000}{3} - 7500 = \frac{1712500}{3} \text{ J} \approx 5.71 \times 10^5 \text{ J}$$