Question
A ball of mass 0.05kg strikes a smooth vertical wall normally four times in 2 seconds with a velocity of 10ms^-1. Each time the ball rebounds with the same speed of 10ms^-1. Calculate the average force exerted on the wall
Answers
GPT-4o mini
To calculate the average force exerted on the wall by the ball, we can use the concept of impulse and momentum.
1. **Determine the change in momentum for each collision:**
When the ball strikes the wall, it has a velocity of \(10 \, \text{m/s}\) towards the wall. Upon rebounding, it has a velocity of \(-10 \, \text{m/s}\) (the negative sign indicates the opposite direction).
The change in momentum (\( \Delta p \)) for each collision is calculated as follows:
\[
\Delta p = p_{\text{final}} - p_{\text{initial}}
\]
Where:
- \(p_{\text{initial}} = m \cdot v = 0.05 \, \text{kg} \cdot 10 \, \text{m/s} = 0.5 \, \text{kg m/s}\)
- \(p_{\text{final}} = m \cdot (-v) = 0.05 \, \text{kg} \cdot (-10 \, \text{m/s}) = -0.5 \, \text{kg m/s}\)
So:
\[
\Delta p = -0.5 \, \text{kg m/s} - 0.5 \, \text{kg m/s} = -1.0 \, \text{kg m/s}
\]
This means that the change in momentum for each collision is \(1.0 \, \text{kg m/s}\) (the negative here is just indicating a reversal in direction).
2. **Calculate the total change in momentum over four bounces:**
Since the ball strikes the wall four times:
\[
\Delta p_{\text{total}} = 4 \cdot 1.0 \, \text{kg m/s} = 4.0 \, \text{kg m/s}
\]
3. **Determine the duration over which these changes occur:**
The total time for four collisions is given as 2 seconds.
4. **Calculate the average force:**
The average force \( F \) exerted on the wall can be calculated using the impulse-momentum theorem:
\[
F = \frac{\Delta p_{\text{total}}}{\Delta t}
\]
Where \( \Delta t \) is the time period over which the momentum change occurs.
Substituting the known values:
\[
F = \frac{4.0 \, \text{kg m/s}}{2 \, \text{s}} = 2.0 \, \text{N}
\]
Therefore, the average force exerted on the wall by the ball is **2.0 N**.
1. **Determine the change in momentum for each collision:**
When the ball strikes the wall, it has a velocity of \(10 \, \text{m/s}\) towards the wall. Upon rebounding, it has a velocity of \(-10 \, \text{m/s}\) (the negative sign indicates the opposite direction).
The change in momentum (\( \Delta p \)) for each collision is calculated as follows:
\[
\Delta p = p_{\text{final}} - p_{\text{initial}}
\]
Where:
- \(p_{\text{initial}} = m \cdot v = 0.05 \, \text{kg} \cdot 10 \, \text{m/s} = 0.5 \, \text{kg m/s}\)
- \(p_{\text{final}} = m \cdot (-v) = 0.05 \, \text{kg} \cdot (-10 \, \text{m/s}) = -0.5 \, \text{kg m/s}\)
So:
\[
\Delta p = -0.5 \, \text{kg m/s} - 0.5 \, \text{kg m/s} = -1.0 \, \text{kg m/s}
\]
This means that the change in momentum for each collision is \(1.0 \, \text{kg m/s}\) (the negative here is just indicating a reversal in direction).
2. **Calculate the total change in momentum over four bounces:**
Since the ball strikes the wall four times:
\[
\Delta p_{\text{total}} = 4 \cdot 1.0 \, \text{kg m/s} = 4.0 \, \text{kg m/s}
\]
3. **Determine the duration over which these changes occur:**
The total time for four collisions is given as 2 seconds.
4. **Calculate the average force:**
The average force \( F \) exerted on the wall can be calculated using the impulse-momentum theorem:
\[
F = \frac{\Delta p_{\text{total}}}{\Delta t}
\]
Where \( \Delta t \) is the time period over which the momentum change occurs.
Substituting the known values:
\[
F = \frac{4.0 \, \text{kg m/s}}{2 \, \text{s}} = 2.0 \, \text{N}
\]
Therefore, the average force exerted on the wall by the ball is **2.0 N**.