Let's analyze the problem step-by-step: **Given Data:** - Wooden cube: - Side length, \( L = 0.5\, \text{m} \) - Mass, \( m_c = 10\, \text{kg} \) - Rod: - Mass, \( m_r = 3\, \text{kg} \) - Bullet: - Mass, \( m_b = 10\, \text{g} = 0.01\, \text{kg} \) - Velocity, \( v_b = 850\, \text{m/s} \) - The cube is initially at rest, and the bullet strikes the center of the cube and embeds itself. --- ### Step 1: Find the velocity of the combined mass after impact **Conservation of linear momentum:** \[ m_b v_b = (m_c + m_b) v' \] \[ v' = \frac{m_b v_b}{m_c + m_b} \] \[ v' = \frac{0.01 \times 850}{10 + 0.01} \approx \frac{8.5}{10.01} \approx 0.849\, \text{m/s} \] **So, the combined cube + embedded bullet moves with velocity \( v' \approx 0.849\, \text{m/s} \).** --- ### Step 2: Find the initial kinetic energy transferred to the cube + bullet \[ KE_{initial} = \frac{1}{2} (m_c + m_b) v'^2 \] \[ KE_{initial} = \frac{1}{2} \times 10.01 \times (0.849)^2 \approx 5.005 \times 0.721 \approx 3.61\, \text{J} \] --- ### Step 3: Determine the energy transferred to the rod during swinging The question asks for the maximum inclination, which occurs when the kinetic energy is fully converted into potential energy at the highest point of swing. Assuming no energy loss: \[ KE_{initial} = PE_{max} \] The potential energy at maximum inclination angle \( \theta \): \[ PE = (m_c + m_b) g h \] where \( h \) is the vertical rise of the center of mass. --- ### Step 4: Find the vertical rise \( h \) The center of mass of the system (cube + embedded bullet + rod) will rise as the system swings. - The initial velocity imparts rotational motion about the pivot point (assumed at the attachment point of the rod). - The initial kinetic energy translates into rotational potential energy. **However, for simplicity** (assuming the rod swings about one end and the entire system pivots): \[ KE = (m_c + m_b + m_r) g h \] But more precisely, the maximum inclination angle is related to the initial velocity: \[ \frac{1}{2} M v'^2 = M g h \] with \( M = m_c + m_b + m_r \). **Mass total:** \[ M = 10 + 0.01 + 3 = 13.01\, \text{kg} \] **Vertical height \( h \):** \[ h = L_{cm} (1 - \cos \theta) \] But since the entire system is swinging, the maximum height depends on the length of the rod and how the system pivots. --- ### Step 5: Estimating maximum inclination \( \theta \) Given the options, and based on typical pendulum energy calculations: \[ \frac{1}{2} M v'^2 = M g L_{cm} (1 - \cos \theta) \] assuming the center of mass is roughly at the midpoint of the cube + embedded bullet (approximate to 0.25 m from the pivot): \[ h \approx 0.25 (1 - \cos \theta) \] Rearranged: \[ \frac{1}{2} v'^2 = g \times 0.25 (1 - \cos \theta) \] \[ 0.5 \times (0.849)^2 = 9.81 \times 0.25 (1 - \cos \theta) \] \[ 0.5 \times 0.721 = 2.4525 (1 - \cos \theta) \] \[ 0.3605 = 2.4525 (1 - \cos \theta) \] \[ 1 - \cos \theta = \frac{0.3605}{2.4525} \approx 0.147 \] \[ \cos \theta = 1 - 0.147 = 0.853 \] \[ \theta = \cos^{-1}(0.853) \approx 31.8^\circ \] This is higher than the options provided, but considering approximations, the closest options are: - 20.16° - 13.60° - 7.75° - 27.16° Our calculated value (~31.8°) suggests the maximum inclination is close to **27.16°**. --- ### **Final Answer:** **The correct option is:** ### **27.16°** --- **Note:** The approximation is based on simplified assumptions, but it aligns best with option 27.16° given the calculations.