A particle of mass m moving at a velocity v strikes with the wall at an angle `alpha` with the wall and leaves ball making same angle with the wall. Choose the correct statemet (s) from the following.

Options

A. The impulsive reaction of wall is 2mV sin `alpha`
B. The impulsive reaction of wall is zero
C. The coefficient of restitution `ne 1`
D. There is no force by particle on the wall

Answer

To solve the problem, we need to analyze the collision of a particle of mass \( m \) with a wall at an angle \( \alpha \). The particle strikes the wall with a velocity \( v \) and leaves at the same angle \( \alpha \). We will evaluate the four statements provided in the question.

### Step-by-Step Solution:

1. **Understanding the Collision**:
- The particle strikes the wall at an angle \( \alpha \) and rebounds off at the same angle.
- The wall is stationary, so its initial velocity is \( 0 \).

2. **Components of Velocity**:
- Before the collision, the velocity of the particle can be resolved into two components:
- **Perpendicular to the wall (y-direction)**: \( v_y = v \cos \alpha \)
- **Parallel to the wall (x-direction)**: \( v_x = v \sin \alpha \)

3. **After the Collision**:
- After the collision, the particle leaves the wall with the same angle \( \alpha \), so the components of the velocity after collision are:
- **Perpendicular to the wall (y-direction)**: \( -v_y' = -v \cos \alpha \) (the direction is reversed)
- **Parallel to the wall (x-direction)**: \( v_x' = v \sin \alpha \) (the direction remains the same)

4. **Change in Momentum**:
- The change in momentum along the x-direction (parallel to the wall):
\[ \Delta p_x = m(v_x' - v_x) = m(v \sin \alpha - v \sin \alpha) = 0 \]
- The change in momentum along the y-direction (perpendicular to the wall):
\[ \Delta p_y = m(-v \cos \alpha - m v \cos \alpha) = -2m v \cos \alpha \]

5. **Impulse on the Wall**:
- The impulse imparted by the wall on the particle is equal to the change in momentum:
\[ J = \Delta p_y = -2m v \cos \alpha \]
- The impulse is directed towards the wall, hence the magnitude is \( 2m v \cos \alpha \).

6. **Coefficient of Restitution**:
- The coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
- Here, the velocity of separation is \( v \sin \alpha \) (after collision) and the velocity of approach is \( v \sin \alpha \) (before collision):
\[ e = \frac{v \sin \alpha}{v \sin \alpha} = 1 \]

7. **Conclusion**:
- Now, let's evaluate the statements:
1. **The impulsive reaction of the wall is \( 2mv \sin \alpha \)**: This is incorrect; it should be \( 2mv \cos \alpha \).
2. **The impulsive reaction of the wall is zero**: This is incorrect; there is an impulse.
3. **The coefficient of restitution not equals to one**: This is incorrect; it equals one.
4. **There is no force by the particle on the wall**: This is incorrect; there is a force due to the impulse.

### Final Answer:
None of the statements are correct.