A ball of mass `0.5 kg` moving with a velocity of `2 m//s` strikes a wall normally and bounces back with the same speed . If the time of contact between the ball and the wall is 1 millisecond , the average force exerted by the wall on the ball is

Options

A. 2000N
B. 1000N
C. 5000N
D. 125N

Explanation

To solve the problem, we will calculate the average force exerted by the wall on the ball using the concepts of momentum and Newton's second law.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the ball, \( m = 0.5 \, \text{kg} \)
- Initial velocity of the ball, \( v_i = 2 \, \text{m/s} \) (towards the wall)
- Final velocity of the ball after bouncing back, \( v_f = -2 \, \text{m/s} \) (away from the wall)
- Time of contact, \( \Delta t = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \)

2. **Calculate Initial Momentum:**
\[ p_i = m \cdot v_i = 0.5 \, \text{kg} \cdot 2 \, \text{m/s} = 1 \, \text{kg m/s} \]

3. **Calculate Final Momentum:**
\[ p_f = m \cdot v_f = 0.5 \, \text{kg} \cdot (-2 \, \text{m/s}) = -1 \, \text{kg m/s} \]

4. **Calculate Change in Momentum (\( \Delta p \)):**
\[ \Delta p = p_f - p_i = -1 \, \text{kg m/s} - 1 \, \text{kg m/s} = -2 \, \text{kg m/s} \]

5. **Calculate Average Force (\( F \)):**
According to Newton's second law, the average force is given by:
\[ F = \frac{\Delta p}{\Delta t} \]
Substituting the values:
\[ F = \frac{-2 \, \text{kg m/s}}{1 \times 10^{-3} \, \text{s}} = -2000 \, \text{N} \]

6. **Interpret the Result:**
The negative sign indicates that the force exerted by the wall is in the opposite direction to the initial motion of the ball. Therefore, the magnitude of the average force exerted by the wall on the ball is:
\[ |F| = 2000 \, \text{N} \]

### Final Answer:
The average force exerted by the wall on the ball is \( 2000 \, \text{N} \) in the direction opposite to the ball's initial motion.