A player caught a cricket ball of mass 150g moving at a rate of `20m//s`. If the catching process is completed in 0.1s, the force of the blow exerted by the ball on the hand of the player is equal to

Options

A. (a) 150N
B. (b) 3N
C. (c) 30N
D. (d) 300N

Explanation

To solve the problem, we need to calculate the force exerted by the cricket ball on the player's hand when he catches it. We will use the concept of impulse and the formula for force.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the ball, \( m = 150 \, \text{g} = 0.15 \, \text{kg} \) (convert grams to kilograms by dividing by 1000).
- Initial velocity of the ball, \( u = 20 \, \text{m/s} \).
- Final velocity of the ball, \( v = 0 \, \text{m/s} \) (the ball comes to rest when caught).
- Time taken to catch the ball, \( t = 0.1 \, \text{s} \).

2. **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:
\[ \Delta p = 0.15 \, \text{kg} \times (0 - 20 \, \text{m/s}) = 0.15 \, \text{kg} \times (-20 \, \text{m/s}) = -3 \, \text{kg m/s} \]

3. **Calculate the Average Force:**
- The average force (\( F \)) exerted can be calculated using the impulse-momentum theorem, which states:
\[ F = \frac{\Delta p}{t} \]
- Substituting the values:
\[ F = \frac{-3 \, \text{kg m/s}}{0.1 \, \text{s}} = -30 \, \text{N} \]
- The negative sign indicates that the force is in the opposite direction of the ball's initial motion, but we are interested in the magnitude of the force, which is:
\[ F = 30 \, \text{N} \]

4. **Conclusion:**
- The force exerted by the ball on the player's hand is \( 30 \, \text{N} \).

### Final Answer:
The force of the blow exerted by the ball on the hand of the player is **30 N**.