To determine the average force acting on the ball during the collision, we first need to calculate the change in momentum of the ball.
Step 1: Calculate initial and final momentum
Mass of the ball (m):
\( m = 607 , \text{g} = 0.607 , \text{kg} \)
Initial velocity (u):
The ball strikes the wall (let's take this as negative)
\( u = -13.3 , \text{m/s} \)
Final velocity (v):
The ball rebounds (let's take this as positive)
\( v = 10.4 , \text{m/s} \)
Step 2: Calculate change in momentum
The momentum before the collision:
\( p_{\text{initial}} = m \cdot u = 0.607 , \text{kg} \cdot (-13.3 , \text{m/s}) = -8.0651 , \text{kg m/s} \)
The momentum after the collision:
\( p_{\text{final}} = m \cdot v = 0.607 , \text{kg} \cdot 10.4 , \text{m/s} = 6.3138 , \text{kg m/s} \)
Step 3: Calculate change in momentum
Change in momentum (\( \Delta p \)):
\[
\Delta p = p_{\text{final}} - p_{\text{initial}} = 6.3138 , \text{kg m/s} - (-8.0651 , \text{kg m/s})
\]
\[
\Delta p = 6.3138 + 8.0651 = 14.3789 , \text{kg m/s}
\]
Step 4: Calculate average force
The average force (\( F_{\text{avg}} \)) acting on the ball can be calculated using the impulse-momentum theorem, which states that: \[ F_{\text{avg}} = \frac{\Delta p}{\Delta t} \] where \( \Delta t \) is the time of contact.
Given \( \Delta t = 0.046 , \text{s} \), we can find \( F_{\text{avg}} \):
\[ F_{\text{avg}} = \frac{14.3789 , \text{kg m/s}}{0.046 , \text{s}} \approx 312.08 , \text{N} \]
Final Answer:
The magnitude of the average force acting on the ball during the collision is approximately 312.1 N.