1. Use conservation of momentum

Since the bomb was initially at rest, its initial momentum was zero.

After explosion, total momentum must remain zero: $m_1 v_1 + m_2 v_2 = 0$

Let:

• $m_1 = 4\,\text{kg}$
• $m_2 = 12\,\text{kg}$
• $v_2 = 4\,\text{m s}^{-1}$

So, $4v_1 + 12(4) = 0$ $4v_1 = -48$ $v_1 = -12\,\text{m s}^{-1}$

Thus, the $4\,\text{kg}$ piece has speed $12\,\text{m s}^{-1}$.

2. Find its kinetic energy

Kinetic energy of the $4\,\text{kg}$ piece: $K = \frac{1}{2}mv^2$ $K = \frac{1}{2}(4)(12)^2$ $K = 2 \times 144$ $K = 288\,\text{J}$

3. Match with options

$288\,\text{J}$ corresponds to Option B.

4. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the answer agrees with the stored answer.