Let's solve the problem step by step: --- ### **Step 1: Find the velocities after impact using the coefficient of restitution** **Given:** - Mass of ball \( A \): \( m_A = 1 \) lb \( = \frac{1}{32.2} \) slugs - Mass of block \( B \): \( m_B = 10 \) lb \( = \frac{10}{32.2} \) slugs - Initial velocity of \( A \): \( v_{A1} = 20 \) ft/s - Initial velocity of \( B \): \( v_{B1} = 0 \) ft/s - Coefficient of restitution: \( e = 0.6 \) #### **Apply conservation of momentum:** \[ m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2} \] \[ \left(\frac{1}{32.2}\right) \times 20 + \left(\frac{10}{32.2}\right) \times 0 = \left(\frac{1}{32.2}\right) v_{A2} + \left(\frac{10}{32.2}\right) v_{B2} \] \[ 20 = v_{A2} + 10v_{B2} \] #### **Apply coefficient of restitution:** \[ e = \frac{v_{B2} - v_{A2}}{v_{A1} - v_{B1}} \] \[ 0.6 = \frac{v_{B2} - v_{A2}}{20} \] \[ v_{B2} - v_{A2} = 12 \] Now, solve these two equations: 1. \( v_{A2} + 10v_{B2} = 20 \) 2. \( v_{B2} - v_{A2} = 12 \) From (2): \( v_{B2} = v_{A2} + 12 \) Substitute into (1): \[ v_{A2} + 10(v_{A2} + 12) = 20 \\ v_{A2} + 10v_{A2} + 120 = 20 \\ 11v_{A2} = 20 - 120 \\ 11v_{A2} = -100 \\ v_{A2} = -9.09 \text{ ft/s} \] \[ v_{B2} = v_{A2} + 12 = -9.09 + 12 = 2.91 \text{ ft/s} \] **Explanation:** We used conservation of momentum and the coefficient of restitution equations to solve for the velocities of both bodies after the collision. --- ### **Step 2: Calculate the average normal force** The average force \( F_{\text{avg}} \) can be found using the impulse-momentum equation for ball \( A \): \[ F_{\text{avg}} \cdot \Delta t = m_A (v_{A2} - v_{A1}) \] \[ F_{\text{avg}} \cdot 0.02 = \frac{1}{32.2} \left(-9.09 - 20\right) \] \[ F_{\text{avg}} \cdot 0.02 = \frac{1}{32.2} \left(-29.09\right) \] \[ F_{\text{avg}} = \frac{-29.09}{32.2 \times 0.02} \] \[ F_{\text{avg}} = \frac{-29.09}{0.644} \] \[ F_{\text{avg}} = -45.18 \text{ lb} \] We take the magnitude for the average normal force: \[ \boxed{F_{\text{avg}} = 45.2 \text{ lb}} \] **Explanation:** We used the impulse-momentum principle for ball \( A \) to find the average force during the impact. The negative sign indicates direction, but the magnitude of the force is \( 45.2 \) lb. --- ## **Final Answer:** \[ \boxed{45.2 \text{ lb}} \] is the average normal force exerted between \( A \) and \( B \) during the impact.