## Step 1: Conservation of Momentum During Collision When the bullet lodges in the block, the system (bullet + block) moves together. This is a perfectly inelastic collision, so we use conservation of linear momentum: Let: - \( m_b \) = mass of bullet = .20 lb = \( \frac{.20}{32.2} \) slugs ≈ .0062 slugs - \( m_w \) = mass of wood = 30 lb = \( \frac{30}{32.2} \) slugs ≈ .932 slugs - \( v_b \) = speed of bullet before impact (unknown) - \( v' \) = common speed of block+bullet immediately after collision (unknown) **Equation:** \[ m_b v_b = (m_b + m_w) v' \] **Explanation:** Linear momentum is conserved because the external forces (gravity and tension) act perpendicular to the direction of the bullet's motion at the instant of collision. --- ## Step 2: Energy Conservation for Swing After the collision, the block and bullet swing upward to the dashed position, reaching a maximum angle of 64.8°. At the highest point, all kinetic energy is converted to gravitational potential energy. **Let \( L \) = 10 ft (length of string):** - Height gained (\( h \)) by the center of mass is: \[ h = L (1 - \cos \theta) = 10 (1 - \cos 64.8^\circ) \approx 10(1 - .425) \approx 5.75 \text{ ft} \] **Potential energy gained:** \[ PE = (m_b + m_w) g h \] where \( g = 32.2 \) ft/s² **Set \( KE = PE \):** \[ \frac{1}{2} (m_b + m_w) v'^2 = (m_b + m_w) g h \Rightarrow v' = \sqrt{2gh} \] \[ v' = \sqrt{2 \cdot 32.2 \cdot 5.75} \approx \sqrt{370.3} \approx 19.24 \text{ ft/s} \] Now, plug \( v' \) into the first equation to solve for \( v_b \): \[ .0062\,v_b = (.0062 + .932) \cdot 19.24 \] \[ .0062\,v_b = .9382 \cdot 19.24 \approx 18.06 \] \[ v_b = \frac{18.06}{.0062} \approx 2,913 \text{ ft/s} \] **Explanation:** We used conservation of energy to relate the after-collision speed to the swing height, and then conservation of momentum to find the bullet's original speed. --- ## Final Answers **1. Speed of the bullet before hitting the block:** \[ \boxed{2,913 \text{ ft/s}} \] **2. How much of the bullet's energy was absorbed by the block?** Initial kinetic energy of the bullet: \[ KE_{bullet,\,initial} = \frac{1}{2} m_b v_b^2 = \frac{1}{2} \cdot .0062 \cdot (2913)^2 \approx 26,281 \text{ ft-lb} \] Kinetic energy right after collision (all goes to PE at max height): \[ KE_{after} = \frac{1}{2} (.9382) \cdot (19.24)^2 \approx 173.68 \text{ ft-lb} \] **Energy absorbed by the block:** \[ \Delta KE = KE_{bullet,\,initial} - KE_{after} \approx 26,281 - 173.68 \approx 26,107 \text{ ft-lb} \] --- ### **Summary Table** | Quantity | Value | |--------------------------------------------------|--------------------| | Speed of bullet before hitting block | 2,913 ft/s | | Energy absorbed by the block | 26,107 ft-lb |