A player caught a cricket ball of mass `150gm` moving at a rate of `20m//s` . If the catching process be comleted in `0.1` `s` , then the force of the blow exerted by the ball on the hands of the player is.

Options

A. `0.3N`
B. `30N`
C. `3000N`
D. `3000N`

Explanation

To solve the problem step by step, we will follow the principles of Newton's laws of motion, particularly focusing on the concepts of force, mass, acceleration, and the relationship between them.

### Step 1: Convert the mass of the ball to kilograms
The mass of the ball is given as 150 grams. To use SI units, we need to convert this to kilograms.

\[ \text{Mass} (m) = 150 \, \text{g} = 0.150 \, \text{kg} \]

**Hint**: Remember that 1 kg = 1000 g, so to convert grams to kilograms, divide by 1000.

### Step 2: Identify the initial and final velocities
The initial velocity (u) of the ball is given as 20 m/s (the speed at which it is moving before being caught). The final velocity (v) when the ball is caught is 0 m/s (since it comes to a stop).

\[ u = 20 \, \text{m/s}, \quad v = 0 \, \text{m/s} \]

**Hint**: The final velocity is zero because the ball is caught and comes to a stop.

### Step 3: Calculate the acceleration
We can use the formula for acceleration (a) derived from the equations of motion:

\[ a = \frac{v - u}{t} \]

Substituting the known values:

\[ a = \frac{0 - 20}{0.1} = \frac{-20}{0.1} = -200 \, \text{m/s}^2 \]

**Hint**: Acceleration is negative because the ball is decelerating (slowing down) as it is caught.

### Step 4: Calculate the force exerted by the ball
Using Newton's second law of motion, we can calculate the force (F) exerted by the ball on the player's hands:

\[ F = m \cdot a \]

Substituting the values we have:

\[ F = 0.150 \, \text{kg} \cdot (-200 \, \text{m/s}^2) = -30 \, \text{N} \]

**Hint**: The negative sign indicates that the force exerted by the ball is in the opposite direction of its motion.

### Step 5: Determine the force exerted by the ball on the player
According to Newton's third law of motion, the force exerted by the ball on the player's hands is equal in magnitude but opposite in direction to the force exerted by the player on the ball. Therefore, the force exerted by the ball on the player's hands is:

\[ F_{\text{ball on player}} = 30 \, \text{N} \]

**Hint**: Remember that action and reaction forces are equal in magnitude and opposite in direction.

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

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