Let's break down the problem shown in the image: ### **Problem Statement** - A wooden cube of length **0.5 m** and mass **10 kg** is attached to a slender rod of mass **3 kg** (initially at rest). - A **bullet** of mass **10 grams** (**0.01 kg**) is fired with a velocity of **850 m/s** towards the cube and gets embedded into the center of the cube. - The goal is to **calculate the maximum inclination** (angle, θ) that the slender rod will swing due to the impact. ### **Solution Approach** #### **Step 1: Conservation of Angular Momentum** The bullet embeds into the cube, so use conservation of angular momentum about the pivot point: \[ \text{Initial Angular Momentum} = \text{Final Angular Momentum} \] - Initial angular momentum (just as the bullet hits): \[ L_{\text{initial}} = m_{\text{bullet}} \cdot v_{\text{bullet}} \cdot d \] where \( d = 0.5\, \text{m} \) (center of cube to pivot). \[ L_{\text{initial}} = 0.01 \times 850 \times 0.5 = 4.25\, \text{kg}\cdot\text{m}^2/\text{s} \] - Final angular momentum: \[ L_{\text{final}} = I \cdot \omega \] where \( I \) is the total moment of inertia about the pivot. #### **Step 2: Total Moment of Inertia (I)** - Cube (about edge): \( I_{\text{cube}} = \frac{1}{3} M_{\text{cube}} l^2 = \frac{1}{3} \times 10 \times (0.5)^2 = \frac{1}{3} \times 10 \times 0.25 = 0.833\, \text{kg}\cdot\text{m}^2 \) - Rod (about end): \( I_{\text{rod}} = \frac{1}{3} M_{\text{rod}} l^2 = \frac{1}{3} \times 3 \times (1)^2 = 1\, \text{kg}\cdot\text{m}^2 \) - Bullet (treated as point mass at 0.5 m): \( I_{\text{bullet}} = m_{\text{bullet}} \cdot d^2 = 0.01 \times (0.5)^2 = 0.0025\, \text{kg}\cdot\text{m}^2 \) \[ I = 0.833 + 1 + 0.0025 = 1.8355\, \text{kg}\cdot\text{m}^2 \] #### **Step 3: Angular Velocity after Collision (\(\omega\))** \[ I \omega = L_{\text{initial}} \] \[ \omega = \frac{L_{\text{initial}}}{I} = \frac{4.25}{1.8355} \approx 2.315\, \text{rad/s} \] #### **Step 4: Conservation of Energy (Convert Kinetic to Potential at Max Height)** At max inclination, all rotational kinetic energy is converted into potential energy: \[ \frac{1}{2} I \omega^2 = (M_{\text{cube}} + M_{\text{rod}} + m_{\text{bullet}}) g h \] Let the length of the rod + cube = **1 m** (since rod is 1 m and cube is attached at the end). - **h** is the vertical height gained by the center of mass (CM). Let θ be the angle with the vertical at max height. The vertical rise \( h \) of the CM is: \[ h = L_{\text{CM}} (1 - \cos\theta) \] We need to find \( L_{\text{CM}} \): \[ L_{\text{CM}} = \frac{M_{\text{rod}} \cdot \frac{1}{2}l_{\text{rod}} + (M_{\text{cube}} + m_{\text{bullet}}) \cdot l_{\text{rod}}}{M_{\text{rod}} + M_{\text{cube}} + m_{\text{bullet}}} \] \[ = \frac{3 \times 0.5 + 10.01 \times 1}{3 + 10 + 0.01} = \frac{1.5 + 10.01}{13.01} = \frac{11.51}{13.01} \approx 0.885 \text{ m} \] Now substitute: \[ \frac{1}{2} I \omega^2 = (M_{\text{cube}} + M_{\text{rod}} + m_{\text{bullet}}) g L_{CM} (1 - \cos\theta) \] Values: - \( I = 1.8355 \) - \( \omega = 2.315 \) - \( M_{\text{cube}} + M_{\text{rod}} + m_{\text{bullet}} = 13.01 \) - \( g = 9.8 \) - \( L_{CM} = 0.885 \) \[ \frac{1}{2} \times 1.8355 \times (2.315)^2 = 13.01 \times 9.8 \times 0.885 \times (1 - \cos\theta) \] \[ \frac{1}{2} \times 1.8355 \times 5.364 = 13.01 \times 9.8 \times 0.885 \times (1 - \cos\theta) \] \[ 4.921 \approx 113.01 \times (1 - \cos\theta) \] \[ 1 - \cos\theta = \frac{4.921}{113.01} \approx 0.0435 \] \[ \cos\theta = 1 - 0.0435 = 0.9565 \] \[ \theta = \cos^{-1}(0.9565) \approx 16.5^\circ \] ### **Final Answer** The closest value to this in the options is **13.67°**. So the correct answer is: \[ \boxed{13.67^\circ} \]