According to the law of conservation of linear momentum, the total momentum of an isolated system remains constant. Since the bomb is initially at rest, its initial momentum is zero . Let the mass of the first piece be $m_1 = 8 \text{ kg}$ and its velocity be $\vec{v}_1 = 6 \text{ m/s}$. Let the mass of the second piece be $m_2 = 4 \text{ kg}$ and its velocity be $\vec{v}_2$. $\text{Initial momentum} = \text{Final momentum}$ $0 = m_1\vec{v}_1 + m_2\vec{v}_2$ $0 = (8 \times 6) + (4 \times \vec{v}_2)$ $4\vec{v}_2 = -48$ $\vec{v}_2 = -12 \text{ m/s}$ The negative sign indicates that the second piece moves in the opposite direction . The kinetic energy ($K$) of the second mass ($m_2$) is given by: $K = \frac{1}{2}m_2v_2^2$ $K = \frac{1}{2} \times 4 \times (-12)^2$ $K = 2 \times 144 = 288 \text{ J}$