A 140g ball, in horizontal flight with a speed `v_(1)` of 39.0 m/s. is sturck by a bat. After leaving the bat, the ball travel in the opposite direction with speed `v_(2)=39.0 m//s`. If the impact time `Deltat` for the ball-bat collision is 1.20ms, what average net force acts on the ball

Options

A. 1308N
B. 1090N
C. 9100N
D. 980N

Explanation

To solve the problem, we need to find the average net force acting on the ball during the collision with the bat. We will use the concept of impulse and momentum.

### Step-by-Step Solution:

1. **Convert Mass to Kilograms:**
The mass of the ball is given as 140 grams. We need to convert this to kilograms since the standard unit of mass in physics is kilograms.
\[ m = 140 \, \text{g} = 0.140 \, \text{kg} \]

**Hint:** Remember that 1 kg = 1000 g.

2. **Identify Initial and Final Velocities:**
The initial velocity \( v_1 \) of the ball is 39.0 m/s (in the positive direction), and after being struck by the bat, the ball travels in the opposite direction with a final velocity \( v_2 \) of -39.0 m/s.
\[ v_1 = 39.0 \, \text{m/s}, \quad v_2 = -39.0 \, \text{m/s} \]

**Hint:** Note that the direction of velocity changes after the collision, which is why \( v_2 \) is negative.

3. **Calculate Change in Momentum:**
The change in momentum (\( \Delta p \)) can be calculated using the formula:
\[ \Delta p = m(v_2 - v_1) \]
Substituting the values:
\[ \Delta p = 0.140 \, \text{kg} \times (-39.0 \, \text{m/s} - 39.0 \, \text{m/s}) = 0.140 \, \text{kg} \times (-78.0 \, \text{m/s}) = -10.92 \, \text{kg m/s} \]

**Hint:** The change in momentum is negative because the ball's direction has reversed.

4. **Convert Impact Time to Seconds:**
The impact time \( \Delta t \) is given as 1.20 ms. We need to convert this to seconds:
\[ \Delta t = 1.20 \, \text{ms} = 1.20 \times 10^{-3} \, \text{s} \]

**Hint:** Remember that 1 ms = \( 10^{-3} \) s.

5. **Calculate Average Net Force:**
The average net force (\( F \)) can be calculated using the formula:
\[ F = \frac{\Delta p}{\Delta t} \]
Substituting the values:
\[ F = \frac{-10.92 \, \text{kg m/s}}{1.20 \times 10^{-3} \, \text{s}} \approx -9100 \, \text{N} \]
Since we are interested in the magnitude of the force, we can take the absolute value:
\[ |F| \approx 9100 \, \text{N} \]

**Hint:** The negative sign indicates the direction of the force, which is opposite to the initial direction of the ball.

### Final Answer:
The average net force acting on the ball is approximately **9100 N**.