A body of mass 2kg collides with a wall with speed 100 m/s and rebounds with same speed. If the time of contact was 1/50 second, the force exerted on the wall is

Options

A. `10^(4) N`
B. 4N
C. ` 2 xx 10^(4)N`
D. 8N

Explanation

To solve the problem, we need to calculate the force exerted on the wall when a body collides with it and rebounds. We will use the concept of momentum and the formula for force.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the body (m) = 2 kg
- Initial speed (u) = 100 m/s (towards the wall)
- Final speed (v) = -100 m/s (away from the wall, hence negative)
- Time of contact (Δt) = 1/50 seconds

2. **Calculate the initial momentum (p_initial):**
\[ p_{\text{initial}} = m \cdot u = 2 \, \text{kg} \cdot 100 \, \text{m/s} = 200 \, \text{kg m/s} \]

3. **Calculate the final momentum (p_final):**
\[ p_{\text{final}} = m \cdot v = 2 \, \text{kg} \cdot (-100) \, \text{m/s} = -200 \, \text{kg m/s} \]

4. **Calculate the change in momentum (Δp):**
\[ \Delta p = p_{\text{final}} - p_{\text{initial}} = -200 \, \text{kg m/s} - 200 \, \text{kg m/s} = -400 \, \text{kg m/s} \]

5. **Calculate the force (F) using the formula:**
\[ F = \frac{\Delta p}{\Delta t} \]
Substituting the values:
\[ F = \frac{-400 \, \text{kg m/s}}{1/50 \, \text{s}} = -400 \, \text{kg m/s} \cdot 50 = -20000 \, \text{N} \]

6. **Express the force in magnitude:**
The magnitude of the force exerted on the wall is:
\[ |F| = 20000 \, \text{N} = 2 \times 10^4 \, \text{N} \]

### Final Answer:
The force exerted on the wall is \( 2 \times 10^4 \, \text{N} \).

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