A body of mass 3kg moving with speed `5 ms^(-1)` , hits a wall at an angle of `60^(@)` and return at the same angle. The impact time was 0.2 sec. calculate the force exerted on the wall `:`

Options

A. `150 sqrt( 3) N`
B. `50 sqrt( 3) N`
C. `100N`
D. `75 sqrt( 3) N`

Answer

To solve the problem, we need to calculate the force exerted on the wall by a body of mass 3 kg moving at a speed of 5 m/s that hits the wall at an angle of 60 degrees and returns at the same angle. The impact time is given as 0.2 seconds.

### Step-by-Step Solution:

1. **Identify the Components of Velocity:**
- The body hits the wall at an angle of 60 degrees. We need to resolve the velocity into horizontal and vertical components.
- The horizontal component \( v_x \) is given by:
\[ v_x = v \cdot \cos(60^\circ) = 5 \cdot \cos(60^\circ) = 5 \cdot \frac{1}{2} = 2.5 \, \text{m/s} \]
- The vertical component \( v_y \) is given by:
\[ v_y = v \cdot \sin(60^\circ) = 5 \cdot \sin(60^\circ) = 5 \cdot \frac{\sqrt{3}}{2} = 2.5\sqrt{3} \, \text{m/s} \]

2. **Calculate Initial and Final Momentum:**
- The initial momentum \( \vec{p_i} \) when the body hits the wall:
\[ \vec{p_i} = m \cdot \vec{v} = 3 \, \text{kg} \cdot (2.5 \hat{i} + 2.5\sqrt{3} \hat{j}) \, \text{m/s} = (7.5 \hat{i} + 7.5\sqrt{3} \hat{j}) \, \text{kg m/s} \]
- After hitting the wall, the horizontal component of the velocity reverses, while the vertical component remains the same. Thus, the final momentum \( \vec{p_f} \) is:
\[ \vec{p_f} = m \cdot (-\vec{v}) = 3 \, \text{kg} \cdot (-2.5 \hat{i} + 2.5\sqrt{3} \hat{j}) \, \text{m/s} = (-7.5 \hat{i} + 7.5\sqrt{3} \hat{j}) \, \text{kg m/s} \]

3. **Calculate Change in Momentum:**
- The change in momentum \( \Delta \vec{p} \) is given by:
\[ \Delta \vec{p} = \vec{p_f} - \vec{p_i} = (-7.5 \hat{i} + 7.5\sqrt{3} \hat{j}) - (7.5 \hat{i} + 7.5\sqrt{3} \hat{j}) \]
- Simplifying this gives:
\[ \Delta \vec{p} = (-7.5 - 7.5) \hat{i} + (7.5\sqrt{3} - 7.5\sqrt{3}) \hat{j} = -15 \hat{i} \, \text{kg m/s} \]

4. **Calculate the Force Exerted on the Wall:**
- The force \( \vec{F} \) exerted on the wall can be calculated using the formula:
\[ \vec{F} = \frac{\Delta \vec{p}}{\Delta t} \]
- Substituting the change in momentum and the impact time:
\[ \vec{F} = \frac{-15 \hat{i}}{0.2} = -75 \hat{i} \, \text{N} \]
- The magnitude of the force is:
\[ F = 75 \, \text{N} \]

5. **Final Result:**
- The force exerted on the wall is:
\[ F = 75 \, \text{N} \]