Question

A 607 g ball strikes a wall at 13.3 m/s and
rebounds at 10.4 m/s. The ball is in contact
with the wall for 0.046 s.
What is the magnitude of the average force
acting on the ball during the collision?
Answer in units of N.

Answer 1

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**.