Let's break down and **solve the problem step by step**: ### **Given Data** - Mass of ball \( A \): \( m_A = 1\, \text{lb} \) - Mass of block \( B \): \( m_B = 10\, \text{lb} \) - Initial velocity of \( A \): \( v_{A_i} = 20\, \text{ft/s} \) - Initial velocity of \( B \): \( v_{B_i} = 0\, \text{ft/s} \) - Time of impact: \( \Delta t = 0.02\, \text{s} \) - Coefficient of restitution: \( e = 0.6 \) ### **Step 1: Convert Weights to Mass** Recall that \( 1\, \text{lb} = \frac{1}{32.2}\, \text{slugs} \). - \( m_A = \frac{1}{32.2}\, \text{slugs} \) - \( m_B = \frac{10}{32.2}\, \text{slugs} \) --- ### **Step 2: Conservation of Momentum** Let \( v_{A_f} \) and \( v_{B_f} \) be the final velocities after collision. \[ m_A v_{A_i} + m_B v_{B_i} = m_A v_{A_f} + m_B v_{B_f} \] \[ \frac{1}{32.2} \cdot 20 + \frac{10}{32.2} \cdot 0 = \frac{1}{32.2} v_{A_f} + \frac{10}{32.2} v_{B_f} \] \[ 20 = v_{A_f} + 10 v_{B_f} \] --- ### **Step 3: Coefficient of Restitution** \[ e = \frac{v_{B_f} - v_{A_f}}{v_{A_i} - v_{B_i}} \] \[ 0.6 = \frac{v_{B_f} - v_{A_f}}{20 - 0} \] \[ 0.6 \cdot 20 = v_{B_f} - v_{A_f} \] \[ 12 = v_{B_f} - v_{A_f} \] \[ v_{B_f} = v_{A_f} + 12 \] --- ### **Step 4: Solve the System of Equations** Substitute \( v_{B_f} = v_{A_f} + 12 \) into the momentum equation: \[ 20 = v_{A_f} + 10(v_{A_f} + 12) \] \[ 20 = v_{A_f} + 10v_{A_f} + 120 \] \[ 20 = 11v_{A_f} + 120 \] \[ 11v_{A_f} = 20 - 120 = -100 \] \[ v_{A_f} = -\frac{100}{11} \approx -9.09\, \text{ft/s} \] Now, plug into \( v_{B_f} \): \[ v_{B_f} = v_{A_f} + 12 = -9.09 + 12 = 2.91\, \text{ft/s} \] --- ### **Step 5: Impulse-Momentum to Find Force** For ball \( A \): \[ \text{Impulse} = m_A(v_{A_f} - v_{A_i}) = F_\text{avg} \Delta t \] \[ F_\text{avg} = \frac{m_A(v_{A_f} - v_{A_i})}{\Delta t} \] Plug in the numbers: \[ F_\text{avg} = \frac{\frac{1}{32.2}(-9.09 - 20)}{0.02} \] \[ = \frac{\frac{1}{32.2}(-29.09)}{0.02} \] \[ = \frac{-29.09}{32.2 \times 0.02} \] \[ = \frac{-29.09}{0.644} \] \[ = -45.2\, \text{lb} \] --- ### **Step 6: Direction and Magnitude** The negative sign indicates the direction is opposite to the initial velocity of \( A \)—which makes sense since the force acts to decelerate \( A \). #### **Final Answer:** \[ \boxed{45.2\, \text{lb}} \] is the **average normal force** exerted between \( A \) and \( B \) during the impact.