A ball reaches a racket at `60m//s` along +X direction and leaves the racket in the opposite direction 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. 300 N along + X direction
B. 300 N along -X direction
C. 300 kN along +X direction
D. 300 kN along - X direction

Answer

To solve the problem, we need to calculate the force exerted by the racket on the ball using the concepts of momentum and Newton's second law of motion. Here’s a step-by-step solution:

### Step 1: Identify the given values
- 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 -X direction)
- Mass of the ball, \( m = 50 \, \text{g} = 0.05 \, \text{kg} \) (convert grams to kilograms)
- Time of contact, \( t = 0.02 \, \text{s} \)

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

### Step 3: Calculate the acceleration
Acceleration (\( a \)) 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 net force using Newton's second law
According to Newton's second law, the net force (\( F \)) is given by:
\[ F = m \cdot a \]
Substituting the values:
\[ F = 0.05 \, \text{kg} \cdot (-6000 \, \text{m/s}^2) = -300 \, \text{N} \]

### Step 5: Interpret the result
The negative sign indicates that the force is in the opposite direction to the positive X direction, which is consistent with the ball leaving the racket in the opposite direction.

### Final Answer
The force exerted by the racket on the ball is \( 300 \, \text{N} \) in the negative X direction.

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