Let's break down the problem: **Given:** - Wooden cube: Length = 0.5 m, Mass = 10 kg - Slender rod: Mass = 3 kg, initially at rest as shown in the figure - Bullet: Mass = 10 grams = 0.01 kg, Velocity = 850 m/s, fired towards the center of the cube and gets embedded **Asked:** Calculate the **maximum inclination** (angle) that the slender rod will swing due to this impact. --- ### Step 1: **Conservation of Angular Momentum** Since the bullet gets embedded, it's a perfectly inelastic collision. **Total Angular Momentum before impact:** The system is initially at rest, except the bullet. The bullet hits the center of the cube, which is at one end of the rod, a distance of \( L = 0.5 \) m from the axis of rotation. Angular momentum of the bullet about the axis: \[ L_{\text{bullet}} = m_{\text{bullet}} \cdot v_{\text{bullet}} \cdot r \] Where: - \( m_{\text{bullet}} = 0.01 \) kg - \( v_{\text{bullet}} = 850 \) m/s - \( r = 0.5 \) m \[ L_{\text{bullet}} = 0.01 \times 850 \times 0.5 = 4.25 \, \text{kg} \cdot \text{m}^2/\text{s} \] --- ### Step 2: **Moment of Inertia** After the collision, the bullet is embedded in the cube. The moment of inertia about the axis for the whole system (rod + cube + bullet): - **Rod (pivoted at one end, length \( L \)):** \[ I_{\text{rod}} = \frac{1}{3} M_{\text{rod}} L^2 = \frac{1}{3} \times 3 \times (0.5)^2 = \frac{1}{3} \times 3 \times 0.25 = 0.25 \, \text{kg} \cdot \text{m}^2 \] - **Cube (as a point mass at end):** \[ I_{\text{cube}} = M_{\text{cube}} L^2 = 10 \times (0.5)^2 = 2.5 \, \text{kg} \cdot \text{m}^2 \] - **Bullet (as a point mass at end):** \[ I_{\text{bullet}} = 0.01 \times (0.5)^2 = 0.0025 \, \text{kg} \cdot \text{m}^2 \] Total moment of inertia: \[ I_{\text{total}} = 0.25 + 2.5 + 0.0025 = 2.7525 \, \text{kg} \cdot \text{m}^2 \] --- ### Step 3: **Angular Velocity after Collision** By conservation of angular momentum: \[ L_{\text{initial}} = I_{\text{total}} \omega \] \[ 4.25 = 2.7525 \omega \] \[ \omega = \frac{4.25}{2.7525} \approx 1.544 \, \text{rad/s} \] --- ### Step 4: **Maximum Inclination (Energy Conservation)** After the collision, the system swings up. At its maximum inclination, all kinetic energy is converted to potential energy. - **Initial rotational kinetic energy:** \[ KE = \frac{1}{2} I_{\text{total}} \omega^2 = \frac{1}{2} \times 2.7525 \times (1.544)^2 \approx 3.284 \, \text{J} \] - **Potential energy at maximum height:** (let θ be the maximum angle with the vertical) The center of mass rises. For a rod of length \( L \) and mass \( M \) pivoted at one end, the change in height for a swing to angle θ is \( h = \text{distance from pivot to COM} \times (1 - \cos\theta) \). For rod: COM at \( L/2 \) from pivot For cube and bullet: at \( L \) from pivot Total change in potential energy: \[ \Delta PE = M_{\text{rod}} g \frac{L}{2} (1 - \cos\theta) + (M_{\text{cube}} + M_{\text{bullet}}) g L (1 - \cos\theta) \] Substitute values: - \( M_{\text{rod}} = 3 \) kg - \( M_{\text{cube}} = 10 \) kg - \( M_{\text{bullet}} = 0.01 \) kg - \( g = 9.8 \) m/s² - \( L = 0.5 \) m \[ \Delta PE = 3 \times 9.8 \times 0.25 (1 - \cos\theta) + 10.01 \times 9.8 \times 0.5 (1 - \cos\theta) \] \[ = 7.35 (1 - \cos\theta) + 49.549 (1 - \cos\theta) \] \[ = 56.899 (1 - \cos\theta) \] Set \( KE = \Delta PE \): \[ 3.284 = 56.899 (1 - \cos\theta) \] \[ 1 - \cos\theta = \frac{3.284}{56.899} \approx 0.0577 \] \[ \cos\theta = 1 - 0.0577 = 0.9423 \] \[ \theta = \cos^{-1}(0.9423) \approx 20^\circ \] --- ### Step 5: **Match with Options** The closest option to \( 20^\circ \) is **20.16°**. --- ## **Final Answer:** **20.16°** is the correct answer. Let me know if you need the solution in a more detailed or step-by-step format!