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. `2.5 xx 10^(4)`N

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass and speed to standard units
- The mass of the cricket ball is given as `150 g`. We convert this to kilograms:
\[ m = 150 \, \text{g} = 0.150 \, \text{kg} \]
- The speed of the ball is given as `126 km/h`. We convert this to meters per second:
\[ u = 126 \, \text{km/h} = \frac{126 \times 1000}{3600} \, \text{m/s} = 35 \, \text{m/s} \]

### Step 2: Understand the collision
- The ball hits the bat and then moves back towards the bowler. Since the collision is elastic, the speed of the ball after the collision will be equal in magnitude but opposite in direction:
\[ v = -u = -35 \, \text{m/s} \]

### Step 3: Calculate the change in momentum
- The change in momentum (\( \Delta p \)) can be calculated using the formula:
\[ \Delta p = m(v - u) \]
Substituting the values we have:
\[ \Delta p = 0.150 \, \text{kg} \times (-35 \, \text{m/s} - 35 \, \text{m/s}) = 0.150 \, \text{kg} \times (-70 \, \text{m/s}) = -10.5 \, \text{kg m/s} \]

### Step 4: Calculate the average force exerted by the batsman
- The average force (\( F \)) can be calculated using the impulse-momentum theorem, which states that the impulse is equal to the change in momentum:
\[ F \Delta t = \Delta p \]
Rearranging gives:
\[ F = \frac{\Delta p}{\Delta t} \]
Given that the time of contact (\( \Delta t \)) is \( 0.001 \, \text{s} \):
\[ F = \frac{-10.5 \, \text{kg m/s}}{0.001 \, \text{s}} = -10500 \, \text{N} \]

### Step 5: 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, which is expected since he is holding the bat against the ball.

### Final Answer
The force that the batsman had to apply to hold the bat firmly at its place is:
\[ F = 10500 \, \text{N} \]

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