A ball reaches a racket at `60m//s` along +X dirction and leaves the racket in the opposite direaction with the same speed. Assuming that the mass of the ball as `50gm` and the contact time is `0.02` second the force exerted by the racket on the ball is .

Options

A. `300N` along `+X` direction
B. `300N` along `-X` direction
C. `3,00,000 N` along `+X` direction
D. `3,00,000 N` along `-X` direction

Explanation

To solve the problem step by step, we will use the concepts of momentum, force, and acceleration.

### Step 1: Identify the given data
- Initial velocity of the ball, \( u = 60 \, \text{m/s} \) (in the +X direction)
- Final velocity of the ball, \( v = -60 \, \text{m/s} \) (in the opposite direction)
- Mass of the ball, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Contact time, \( t = 0.02 \, \text{s} \)

### Step 2: Calculate the change in velocity (\( \Delta v \))
The change in velocity is given by:
\[ \Delta v = v - u = -60 \, \text{m/s} - 60 \, \text{m/s} = -120 \, \text{m/s} \]

### Step 3: Calculate the acceleration (\( a \))
Acceleration is defined as the change in velocity over time:
\[ a = \frac{\Delta v}{t} = \frac{-120 \, \text{m/s}}{0.02 \, \text{s}} = -6000 \, \text{m/s}^2 \]

### Step 4: Calculate the force (\( F \))
Using Newton's second law, the force exerted by the racket on the ball can be calculated using:
\[ F = m \cdot a \]
Substituting the values we have:
\[ F = 0.05 \, \text{kg} \cdot (-6000 \, \text{m/s}^2) = -300 \, \text{N} \]

### Step 5: Conclusion
The force exerted by the racket on the ball is \( -300 \, \text{N} \). The negative sign indicates that the force is in the opposite direction to the initial motion of the ball.

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