Answer :
Let's analyze the problem step-by-step to determine the value of the force exerted on the baseball during the collision.
1. Understand the problem:
- The mass of the baseball, [tex]\( m \)[/tex], is 0.15 kilograms.
- The initial velocity, [tex]\( v_i \)[/tex], is 40 meters per second (m/s).
- The final velocity, [tex]\( v_f \)[/tex], after the collision is 50 meters per second (m/s) in the opposite direction.
- The duration of the collision, [tex]\( \Delta t \)[/tex], is [tex]\( 8.0 \times 10^{-3} \)[/tex] seconds.
2. Calculate the change in velocity ([tex]\( \Delta v \)[/tex]):
Since the final velocity is in the opposite direction, we need to consider it as a negative value relative to the initial velocity. Hence:
[tex]\[ \Delta v = v_f - v_i = -50 \, \text{m/s} - 40 \, \text{m/s} = -90 \, \text{m/s} \][/tex]
3. Calculate the impulse ([tex]\( J \)[/tex]):
Impulse is the change in momentum, which is mathematically given by:
[tex]\[ J = m \cdot \Delta v \][/tex]
Substituting the values:
[tex]\[ J = 0.15 \, \text{kg} \times -90 \, \text{m/s} = -13.5 \, \text{kg} \cdot \text{m/s} \][/tex]
4. Calculate the average force ([tex]\( F \)[/tex]):
The impulse experienced by the ball is also equal to the average force applied multiplied by the duration of the collision:
[tex]\[ J = F \cdot \Delta t \implies F = \frac{J}{\Delta t} \][/tex]
Substituting the values:
[tex]\[ F = \frac{-13.5 \, \text{kg} \cdot \text{m/s}}{8.0 \times 10^{-3} \, \text{s}} = -1687.5 \, \text{N} \][/tex]
Here, the result is a negative value, indicating the direction of the force. Since the problem does not require the direction, we take the absolute value for the magnitude of the force:
[tex]\[ F = 1687.5 \, \text{N} \][/tex]
5. Select the correct answer:
Comparing the calculated force with the given options:
[tex]\[ \begin{align*} A. & \ 1.8 \times 10^2 \ \text{N} \\ B. & \ 1.2 \times 10^2 \ \text{N} \\ C. & \ 1.3 \times 10^3 \ \text{N} \\ D. & \ 1.4 \times 10^3 \ \text{N} \end{align*} \][/tex]
None of these exactly matches 1687.5, so there seems to be an error as the closest (if rounding or approximation were allowed) closest value is much less.
Had the correct steps been followed, it indicates:
None exactly match `-1687.5` so likely typo or approximations:
Closest - `expected correct would be None`.
1. Understand the problem:
- The mass of the baseball, [tex]\( m \)[/tex], is 0.15 kilograms.
- The initial velocity, [tex]\( v_i \)[/tex], is 40 meters per second (m/s).
- The final velocity, [tex]\( v_f \)[/tex], after the collision is 50 meters per second (m/s) in the opposite direction.
- The duration of the collision, [tex]\( \Delta t \)[/tex], is [tex]\( 8.0 \times 10^{-3} \)[/tex] seconds.
2. Calculate the change in velocity ([tex]\( \Delta v \)[/tex]):
Since the final velocity is in the opposite direction, we need to consider it as a negative value relative to the initial velocity. Hence:
[tex]\[ \Delta v = v_f - v_i = -50 \, \text{m/s} - 40 \, \text{m/s} = -90 \, \text{m/s} \][/tex]
3. Calculate the impulse ([tex]\( J \)[/tex]):
Impulse is the change in momentum, which is mathematically given by:
[tex]\[ J = m \cdot \Delta v \][/tex]
Substituting the values:
[tex]\[ J = 0.15 \, \text{kg} \times -90 \, \text{m/s} = -13.5 \, \text{kg} \cdot \text{m/s} \][/tex]
4. Calculate the average force ([tex]\( F \)[/tex]):
The impulse experienced by the ball is also equal to the average force applied multiplied by the duration of the collision:
[tex]\[ J = F \cdot \Delta t \implies F = \frac{J}{\Delta t} \][/tex]
Substituting the values:
[tex]\[ F = \frac{-13.5 \, \text{kg} \cdot \text{m/s}}{8.0 \times 10^{-3} \, \text{s}} = -1687.5 \, \text{N} \][/tex]
Here, the result is a negative value, indicating the direction of the force. Since the problem does not require the direction, we take the absolute value for the magnitude of the force:
[tex]\[ F = 1687.5 \, \text{N} \][/tex]
5. Select the correct answer:
Comparing the calculated force with the given options:
[tex]\[ \begin{align*} A. & \ 1.8 \times 10^2 \ \text{N} \\ B. & \ 1.2 \times 10^2 \ \text{N} \\ C. & \ 1.3 \times 10^3 \ \text{N} \\ D. & \ 1.4 \times 10^3 \ \text{N} \end{align*} \][/tex]
None of these exactly matches 1687.5, so there seems to be an error as the closest (if rounding or approximation were allowed) closest value is much less.
Had the correct steps been followed, it indicates:
None exactly match `-1687.5` so likely typo or approximations:
Closest - `expected correct would be None`.