Let's break down the problem using the information in the image. ### **Given Data:** - Wooden cube: - Length = 0.5 m - Mass = 10 kg - Slender rod: - Mass = 3 kg - Length = 3 m - Bullet: - Mass = 10 grams = 0.01 kg - Velocity = 850 m/s - The bullet gets embedded into the center of the cube. ### **Required:** Calculate the **maximum inclination** (angle) that the slender rod will swing due to the impact. --- ## **Step 1: Conservation of Angular Momentum (Just after impact)** The bullet is embedded in the cube at the end of the rod, so it's an inelastic collision. Let’s calculate the angular velocity just after the collision. The bullet’s linear momentum is converted into angular momentum about the pivot. - Distance from pivot to impact: \( L = 3 + 0.25 = 3.25 \) m (center of cube is at half its length, so 0.25 m added to the rod's length) - Initial angular momentum (bullet): \( L_{initial} = m_{bullet} \cdot v_{bullet} \cdot L \) --- ### **Step 2: Moment of Inertia of the System** - **Rod (about one end):** \( I_{rod} = \frac{1}{3} m_{rod} L_{rod}^2 = \frac{1}{3} \times 3 \times 3^2 = 9 \) kg·m² - **Cube (about pivot):** \( I_{cube} = m_{cube} \cdot r^2 \) \( r = 3 + 0.25 = 3.25 \) m \( I_{cube} = 10 \times (3.25)^2 = 10 \times 10.5625 = 105.625 \) kg·m² - **Bullet (now part of cube, at same point):** \( I_{bullet} = 0.01 \times (3.25)^2 = 0.01 \times 10.5625 = 0.105625 \) kg·m² - **Total I:** \( I_{total} = 9 + 105.625 + 0.105625 = 114.730625 \) kg·m² --- ### **Step 3: Angular Velocity after Impact (\( \omega \))** By conservation of angular momentum: \[ m_{bullet} \cdot v_{bullet} \cdot L = I_{total} \cdot \omega \] \[ \omega = \frac{m_{bullet} \cdot v_{bullet} \cdot L}{I_{total}} \] \[ \omega = \frac{0.01 \cdot 850 \cdot 3.25}{114.730625} \] \[ \omega = \frac{27.625}{114.730625} = 0.2408\, \text{rad/s} \] --- ### **Step 4: Conservation of Energy (Swing to Maximum Height)** All rotational kinetic energy is converted to gravitational potential energy at the maximum angle (\( \theta \)): \[ \frac{1}{2} I_{total} \omega^2 = (M_{cube} + M_{rod} + M_{bullet}) g h \] But, more accurately, the rise in potential energy of the center of mass is: \[ \Delta PE = (M_{rod} h_{rod} + M_{cube} h_{cube} + M_{bullet} h_{bullet}) g \Delta h \] But since all masses are at the same distance from the pivot, you can sum them as one point mass at distance \( r_{com} \): - \( h = r_{com} (1 - \cos\theta) \) - \( r_{com} = \frac{M_{rod} \cdot 1.5 + (M_{cube} + M_{bullet}) \cdot 3.25}{M_{rod} + M_{cube} + M_{bullet}} \) Calculate \( r_{com} \): \[ M_{total} = 3 + 10 + 0.01 = 13.01 \text{ kg} \] \[ r_{com} = \frac{3 \times 1.5 + 10.01 \times 3.25}{13.01} \] \[ = \frac{4.5 + 32.5325}{13.01} \] \[ = \frac{37.0325}{13.01} = 2.846\, \text{m} \] So, \[ \frac{1}{2} I_{total} \omega^2 = M_{total} g r_{com} (1-\cos\theta) \] Plug in the values: - \( I_{total} = 114.73 \) - \( \omega = 0.2408 \) - \( M_{total} = 13.01 \) - \( r_{com} = 2.846 \) - \( g = 9.8 \) \[ \frac{1}{2} \cdot 114.73 \cdot (0.2408)^2 = 13.01 \cdot 9.8 \cdot 2.846 \cdot (1-\cos\theta) \] \[ \frac{1}{2} \cdot 114.73 \cdot 0.0580 = 362.72 \cdot (1-\cos\theta) \] \[ 3.327 = 362.72 \cdot (1-\cos\theta) \] \[ 1-\cos\theta = \frac{3.327}{362.72} = 0.00918 \] \[ \cos\theta = 1 - 0.00918 = 0.99082 \] \[ \theta = \arccos(0.99082) \approx 7.75^\circ \] --- ## **Final Answer:** **The closest answer is:** \[ \boxed{7.75^\circ} \] --- If you need the detailed step-by-step calculations with any clarification, let me know!