A cricketer catches a ball of mass 150 gm. in 0.1 second moving with speed `20 ms^(–1)`. Then he experiences force of : -

Options

A. 300 N
B. 30 N
C. 3 N
D. 0.3 N

Explanation

To find the force experienced by the cricketer when he catches the ball, we can use the concept of impulse and momentum. The steps to solve this problem are as follows:

### Step 1: Convert the mass of the ball from grams to kilograms.
The mass of the ball is given as 150 grams. To convert grams to kilograms, we divide by 1000.

\[ \text{Mass} (m) = 150 \, \text{g} = \frac{150}{1000} \, \text{kg} = 0.15 \, \text{kg} \]

### Step 2: Calculate the initial momentum of the ball.
Momentum (p) is given by the formula:

\[ p = m \cdot v \]

where \( m \) is the mass and \( v \) is the velocity. The initial velocity of the ball is given as 20 m/s.

\[ p_{\text{initial}} = 0.15 \, \text{kg} \cdot 20 \, \text{m/s} = 3 \, \text{kg m/s} \]

### Step 3: Determine the final momentum of the ball after the catch.
When the cricketer catches the ball, the final velocity of the ball becomes 0 m/s (since it comes to rest). Therefore, the final momentum is:

\[ p_{\text{final}} = 0.15 \, \text{kg} \cdot 0 \, \text{m/s} = 0 \, \text{kg m/s} \]

### Step 4: Calculate the change in momentum.
The change in momentum (Δp) is given by:

\[ \Delta p = p_{\text{final}} - p_{\text{initial}} = 0 - 3 = -3 \, \text{kg m/s} \]

### Step 5: Calculate the average force experienced by the cricketer.
The average force (F) can be calculated using the impulse-momentum theorem, which states that the impulse is equal to the change in momentum. Impulse is also given by the product of force and time (Δt):

\[ F \cdot \Delta t = \Delta p \]

Rearranging this gives:

\[ F = \frac{\Delta p}{\Delta t} \]

Substituting the values we have:

\[ F = \frac{-3 \, \text{kg m/s}}{0.1 \, \text{s}} = -30 \, \text{N} \]

The negative sign indicates that the force is in the opposite direction to the motion of the ball.

### Final Answer:
The cricketer experiences a force of 30 N (in the opposite direction of the ball's motion).

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