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. 500N
B. 1000N
C. 2000N
D. 3000N

Answer

To solve the problem, we need to calculate the average force exerted by the wall on the ball when the ball strikes the wall and bounces back. We can follow these steps:

### Step 1: Identify the given values
- 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, \( v_f = -2 \, \text{m/s} \) (away from the wall, hence negative)
- Time of contact with the wall, \( t = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \)

### Step 2: Calculate the change in momentum
The change in momentum (\( \Delta p \)) can be calculated using the formula:
\[ \Delta p = p_f - p_i \]
Where:
- \( p_f = m \cdot v_f \)
- \( p_i = m \cdot v_i \)

Calculating \( p_i \) and \( p_f \):
\[ p_i = m \cdot v_i = 0.5 \, \text{kg} \cdot 2 \, \text{m/s} = 1 \, \text{kg m/s} \]
\[ p_f = m \cdot v_f = 0.5 \, \text{kg} \cdot (-2) \, \text{m/s} = -1 \, \text{kg m/s} \]

Now, substituting these values to find \( \Delta p \):
\[ \Delta p = -1 \, \text{kg m/s} - 1 \, \text{kg m/s} = -2 \, \text{kg m/s} \]

### Step 3: Calculate the average force
The average force (\( F \)) can be calculated using the formula:
\[ F = \frac{\Delta p}{t} \]
Substituting the values we found:
\[ F = \frac{-2 \, \text{kg m/s}}{1 \times 10^{-3} \, \text{s}} = -2000 \, \text{N} \]

The negative sign indicates that the force exerted by the wall is in the opposite direction of the ball's initial motion. 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 N**.

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