A bullet of mass 120 grams is fired into a target of mass 10 kg with velocity \( 400 \mathrm{~m} / \mathrm{s} \). If the target is free to move, the loss of kinetic energy would be approximately \( \begin{array}{ll}\text { (1) } 9.5 \mathrm{kNm} & \text { (2) } 4.75 \mathrm{kNm} \\ \text { (3) } 19 \mathrm{kNm} & \text { (4) } 0\end{array} \)

To find the loss of kinetic energy when the bullet strikes and embeds itself into the target, we can start by calculating the initial kinetic energy of the bullet and the combined system after the collision. 1. Initial kinetic energy (KE_initial) of the bullet is given by the formula: \[ KE = \frac{1}{2} mv^2 \] For the bullet: \[ KE_{\text{bullet}} = \frac{1}{2} \times 0.12 \, \text{kg} \times (400 \, \text{m/s})^2 = 9600 \, \text{J} = 9.6 \, \text{kJ} \] Since the target is initially at rest, its kinetic energy is zero. 2. After the collision, the bullet and the target move together (perfectly inelastic collision). To find the final velocity \( V_f \), we use conservation of momentum: \[ m_{\text{bullet}} \cdot v_{\text{bullet}} + m_{\text{target}} \cdot v_{\text{target}} = (m_{\text{bullet}} + m_{\text{target}}) \cdot V_f \] Substituting in the values: \[ 0.12 \, \text{kg} \cdot 400 \, \text{m/s} + 10 \, \text{kg} \cdot 0 = (0.12 + 10) \cdot V_f \] \[ 48 = 10.12 \cdot V_f \implies V_f \approx 4.75 \, \text{m/s} \] 3. Now, we can calculate the combined kinetic energy (KE_final) after the collision: \[ KE_{\text{final}} = \frac{1}{2} (m_{\text{bullet}} + m_{\text{target}}) V_f^2 \] \[ KE_{\text{final}} = \frac{1}{2} \cdot 10.12 \cdot (4.75)^2 \approx 114.07 \, \text{J} \approx 0.114 \, \text{kJ} \] 4. Finally, the loss in kinetic energy is: \[ \text{Loss} = KE_{\text{initial}} - KE_{\text{final}} = 9.6 \, \text{kJ} - 0.114 \, \text{kJ} \approx 9.48 \, \text{kJ} \] Thus, the loss of kinetic energy would be approximately **9.5 kJ** (option 1). So, the answer is: (1) 9.5 kNm