A 150 g tenins ball coming at a speed of 40m/s is hit straight back by a bat to speed of 60m/s. The magnitude of the average force F on the ball, when it is in contact for 5ms is,

Options

A. 2500N
B. 3000N
C. 3500N
D. 4000N

Explanation

To solve the problem, we need to calculate the average force exerted on the tennis ball during the time it is in contact with the bat. We will use the formula for average force, which is given by the change in momentum divided by the time duration of contact.

### Step-by-Step Solution:

1. **Convert Mass to Kilograms**:
The mass of the tennis ball is given as 150 grams. We need to convert this to kilograms since the standard unit of mass in physics is kilograms.
\[ \text{Mass} = 150 \, \text{g} = 0.150 \, \text{kg} \]

**Hint**: Remember to convert grams to kilograms by dividing by 1000.

2. **Identify Initial and Final Velocities**:
The initial velocity \( u \) of the ball is given as 40 m/s (towards the bat), and since it is hit back, we consider this direction as negative. Therefore:
\[ u = -40 \, \text{m/s} \]
The final velocity \( v \) after being hit back is 60 m/s (in the opposite direction):
\[ v = 60 \, \text{m/s} \]

**Hint**: Pay attention to the direction of the velocities; assign negative or positive signs accordingly.

3. **Calculate Change in Momentum**:
The change in momentum \( \Delta p \) is given by:
\[ \Delta p = m(v - u) \]
Substituting the values:
\[ \Delta p = 0.150 \, \text{kg} \times (60 \, \text{m/s} - (-40 \, \text{m/s})) \]
\[ \Delta p = 0.150 \, \text{kg} \times (60 + 40) \, \text{m/s} \]
\[ \Delta p = 0.150 \, \text{kg} \times 100 \, \text{m/s} = 15 \, \text{kg m/s} \]

**Hint**: Remember that the change in momentum considers both the final and initial velocities.

4. **Convert Time to Seconds**:
The time duration \( \Delta t \) during which the ball is in contact with the bat is given as 5 milliseconds. We need to convert this to seconds:
\[ \Delta t = 5 \, \text{ms} = 5 \times 10^{-3} \, \text{s} \]

**Hint**: Milliseconds can be converted to seconds by multiplying by \( 10^{-3} \).

5. **Calculate Average Force**:
The average force \( F \) can now be calculated using the formula:
\[ F = \frac{\Delta p}{\Delta t} \]
Substituting the values we have:
\[ F = \frac{15 \, \text{kg m/s}}{5 \times 10^{-3} \, \text{s}} = 3000 \, \text{N} \]

**Hint**: The average force is directly proportional to the change in momentum and inversely proportional to the time duration.

6. **Final Answer**:
The magnitude of the average force \( F \) on the ball is:
\[ F = 3000 \, \text{N} \]

Therefore, the correct answer is \( 3000 \, \text{N} \).