A player caught a criket ball of mass 150 g moving at the rate of `20 ms^(-1)` . If the catching process the completed in 0.1s , the force of the blow exerted by the ball on the hands of the player is

Options

A. (a)0.3 N
B. (b)30 N
C. (c)300 N
D. (d)3000 N

Answer

To solve the problem of finding the force exerted by the cricket ball on the player's hands, we can follow these steps:

### Step 1: Identify the given values
- Mass of the cricket ball, \( m = 150 \, \text{g} = 0.15 \, \text{kg} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \))
- Initial velocity of the ball, \( u = 20 \, \text{m/s} \)
- Final velocity of the ball, \( v = 0 \, \text{m/s} \) (since the ball comes to rest when caught)
- Time taken to catch the ball, \( t = 0.1 \, \text{s} \)

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

### Step 3: Calculate the average force exerted
Using the impulse-momentum theorem, the average force (\( F \)) can be calculated as:
\[ F = \frac{\Delta p}{t} \]
Substituting the values:
\[ F = \frac{-3 \, \text{kg m/s}}{0.1 \, \text{s}} = -30 \, \text{N} \]

### Step 4: Interpret the result
The negative sign indicates that the force exerted by the ball is in the opposite direction of the ball's motion. Therefore, the magnitude of the force exerted by the ball on the player's hands is:
\[ F = 30 \, \text{N} \]

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

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