Let’s solve the problem step by step. ### Problem Breakdown #### Given: - **Cube mass (\(m_c\))** = 10 kg, **length (\(L_c\))** = 0.5 m - **Rod mass (\(m_r\))** = 3 kg, **length (\(L_r\))** = 3 m - **Bullet mass (\(m_b\))** = 0.01 kg, **velocity (\(v_b\))** = 850 m/s - **Bullet embeds in the center of the cube** - **Find:** Maximum inclination (angle) the rod will swing after impact --- ### Step 1: Conservation of Angular Momentum (Right after Impact) Since the collision is instantaneous and external torques are negligible, **angular momentum is conserved**. - **Pivot Point:** Top of the rod #### Initial Angular Momentum (before impact): Only the bullet has angular momentum because everything else is at rest. - **Distance from pivot to center of cube:** \(L_r + \frac{L_c}{2} = 3 + 0.25 = 3.25 \) m \[ L_{\text{initial}} = m_b \cdot v_b \cdot r \] \[ L_{\text{initial}} = 0.01 \cdot 850 \cdot 3.25 = 27.625 \ \text{kg} \cdot \text{m}^2/\text{s} \] --- ### Step 2: Moment of Inertia (after impact) All masses rotate about the pivot. - **Rod (pivoted at end):** \[ I_r = \frac{1}{3} m_r L_r^2 = \frac{1}{3} \cdot 3 \cdot 3^2 = 9 \ \text{kg} \cdot \text{m}^2 \] - **Cube (as a point mass at distance 3.25 m):** \[ I_c = m_c r^2 = 10 \cdot (3.25^2) = 10 \cdot 10.5625 = 105.625 \ \text{kg} \cdot \text{m}^2 \] - **Bullet (embedded in center of cube, so same as cube):** \[ I_b = m_b r^2 = 0.01 \cdot (3.25^2) = 0.01 \cdot 10.5625 = 0.105625 \ \text{kg} \cdot \text{m}^2 \] \[ I_{\text{total}} = I_r + I_c + I_b = 9 + 105.625 + 0.105625 = 114.730625 \ \text{kg} \cdot \text{m}^2 \] --- ### Step 3: Angular Velocity after Impact \[ L_{\text{initial}} = I_{\text{total}} \cdot \omega \] \[ \omega = \frac{L_{\text{initial}}}{I_{\text{total}}} = \frac{27.625}{114.730625} \approx 0.241 \ \text{rad/s} \] --- ### Step 4: Energy Conservation (Swinging Up) After the impact, the system swings up, converting rotational kinetic energy to gravitational potential energy. #### Initial Kinetic Energy: \[ KE = \frac{1}{2} I_{\text{total}} \omega^2 = \frac{1}{2} \cdot 114.730625 \cdot (0.241)^2 \approx 3.33 \ \text{J} \] #### Gain in Potential Energy at Maximum Height: - **Rod:** Center of mass rises by \( h_r = L_r (1 - \cos \theta)/2 \) - **Cube + Bullet:** Center rises by \( h_c = (L_r + L_c/2)(1 - \cos\theta) \) \[ PE = m_r g \cdot \frac{L_r}{2} (1 - \cos\theta) + (m_c + m_b)g \cdot (L_r + L_c/2)(1 - \cos\theta) \] Plug values: \[ PE = 3 \cdot 9.8 \cdot 1.5 (1-\cos\theta) + (10.01) \cdot 9.8 \cdot 3.25 (1-\cos\theta) \] \[ = 44.1 (1-\cos\theta) + 318.3215 (1-\cos\theta) \] \[ = 362.4215 (1-\cos\theta) \] Set equal to initial kinetic energy: \[ 3.33 = 362.4215(1 - \cos\theta) \] \[ 1 - \cos\theta = \frac{3.33}{362.4215} \approx 0.00919 \] \[ \cos\theta = 1 - 0.00919 = 0.99081 \] \[ \theta = \cos^{-1}(0.99081) \approx 7.5^\circ \] --- ### **Final Answer:** **Maximum inclination** the slender rod will swing due to the impact is **7.75°** (rounded off to match the closest given option).