A cricket ball of mass `150g` moving with a speed of `126km//h` hits at the middle of the bat, held firmly at its position by the batman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for `0.001s`, the force that the batsman had to apply to hold the bat firmly at its place would be

Options

A. 10.5 N
B. 21 N
C. `1.05 xx 10^4` N
D. `2.1 xx 10^4` N

Explanation

To solve the problem step by step, we will follow the physics principles of momentum, impulse, and force.

### Step-by-Step Solution:

1. **Convert the mass of the cricket ball from grams to kilograms:**
\[ \text{Mass} (m) = 150 \text{ g} = \frac{150}{1000} \text{ kg} = 0.15 \text{ kg} \]

**Hint:** Remember that to convert grams to kilograms, divide by 1000.

2. **Convert the speed of the cricket ball from km/h to m/s:**
\[ \text{Speed} (v) = 126 \text{ km/h} = 126 \times \frac{5}{18} \text{ m/s} = 35 \text{ m/s} \]

**Hint:** Use the conversion factor \( \frac{5}{18} \) to convert km/h to m/s.

3. **Determine the initial and final velocities:**
- Initial velocity (u) = 35 m/s (towards the bat)
- Final velocity (v) = -35 m/s (after bouncing back, in the opposite direction)

**Hint:** In elastic collisions, the speed remains the same, but the direction changes.

4. **Calculate the change in momentum:**
\[ \Delta p = m(v - u) = 0.15 \text{ kg} \times (-35 \text{ m/s} - 35 \text{ m/s}) = 0.15 \text{ kg} \times (-70 \text{ m/s}) = -10.5 \text{ kg m/s} \]

**Hint:** The change in momentum is calculated by finding the difference between final and initial velocities.

5. **Use the impulse-momentum theorem to find the force:**
\[ \text{Impulse} = F \times t = \Delta p \]
Rearranging gives:
\[ F = \frac{\Delta p}{t} \]
Given that the time of contact (t) = 0.001 s:
\[ F = \frac{-10.5 \text{ kg m/s}}{0.001 \text{ s}} = -10500 \text{ N} \]

**Hint:** Impulse is equal to the change in momentum, and force can be found by dividing the change in momentum by the time of contact.

6. **Interpret the result:**
The negative sign indicates that the force exerted by the batsman is in the opposite direction to the motion of the ball.

**Final Answer:** The force that the batsman had to apply to hold the bat firmly at its place is \( 10500 \text{ N} \) in the direction opposite to that of the ball.

### Summary of Steps:
1. Convert mass from grams to kilograms.
2. Convert speed from km/h to m/s.
3. Identify initial and final velocities.
4. Calculate change in momentum.
5. Use impulse-momentum theorem to find force.
6. Interpret the direction of the force.