A ball is dropped on a smooth inclined plane and is observed to move horizontally after the impact. The coefficient of restitution between the plane and ball is e. The inclination of the plane is

Options

A. `45^(@)`
B. `tan^(-1)e`
C. `tan^(-1)sqrte`
D. `tan^(-1)((e)/(2))`

Answer

To solve the problem, we need to analyze the motion of the ball as it collides with the inclined plane and determine the angle of inclination (θ) based on the coefficient of restitution (e).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A ball is dropped onto a smooth inclined plane and bounces off, moving horizontally after the impact.
- The coefficient of restitution (e) is defined as the ratio of the velocity of separation to the velocity of approach.

2. **Identify the Velocities**:
- Let the velocity of the ball just before the impact be \( V \).
- The velocity of the ball can be resolved into two components relative to the inclined plane:
- Along the incline: \( V \cos(90^\circ - \theta) = V \sin(\theta) \)
- Perpendicular to the incline: \( V \sin(90^\circ - \theta) = V \cos(\theta) \)

3. **Velocity After Collision**:
- Let \( U \) be the velocity of the ball after the collision.
- The component of the velocity of separation (after collision) perpendicular to the incline is \( U \cos(\theta) \).

4. **Applying the Coefficient of Restitution**:
- The coefficient of restitution is given by:
\[ e = \frac{\text{Velocity of Separation}}{\text{Velocity of Approach}} = \frac{U \cos(\theta)}{V \sin(\theta)} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ U \cos(\theta) = e V \sin(\theta) \]

6. **Considering the Horizontal Motion**:
- After the impact, the ball moves horizontally, which means the vertical component of the velocity after the impact is zero.
- Thus, the vertical component of the velocity before the impact must equal the vertical component after the impact:
\[ V \cos(\theta) = U \sin(\theta) \]

7. **Setting Up the Equations**:
- From the two equations derived:
1. \( U \cos(\theta) = e V \sin(\theta) \)
2. \( V \cos(\theta) = U \sin(\theta) \)

8. **Substituting for U**:
- From the second equation, we can express \( U \):
\[ U = \frac{V \cos(\theta)}{\sin(\theta)} \]

9. **Substituting into the First Equation**:
- Substitute \( U \) into the first equation:
\[ \frac{V \cos(\theta)}{\sin(\theta)} \cos(\theta) = e V \sin(\theta) \]
- Simplifying gives:
\[ V \frac{\cos^2(\theta)}{\sin(\theta)} = e V \sin(\theta) \]
- Canceling \( V \) (assuming \( V \neq 0 \)):
\[ \frac{\cos^2(\theta)}{\sin(\theta)} = e \sin(\theta) \]

10. **Rearranging**:
- Rearranging gives:
\[ \cos^2(\theta) = e \sin^2(\theta) \]

11. **Using Trigonometric Identity**:
- Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ 1 - e \sin^2(\theta) = e \sin^2(\theta) \]
\[ 1 = (1 + e) \sin^2(\theta) \]
\[ \sin^2(\theta) = \frac{1}{1 + e} \]

12. **Finding θ**:
- Therefore, we can find \( \theta \):
\[ \theta = \sin^{-1}\left(\sqrt{\frac{1}{1 + e}}\right) \]
- Alternatively, we can express this in terms of tangent:
\[ \tan(\theta) = \sqrt{e} \]
\[ \theta = \tan^{-1}(\sqrt{e}) \]

### Final Answer:
The angle of inclination \( \theta \) is given by:
\[ \theta = \tan^{-1}(\sqrt{e}) \]