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 a ball dropped on a smooth inclined plane and how it behaves after the impact. The coefficient of restitution (e) plays a crucial role in determining the angle of inclination (θ) of the plane.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the inclined plane make an angle θ with the horizontal.
- The ball is dropped vertically onto the inclined plane and moves horizontally after the impact.

2. **Components of Velocity**:
- When the ball is dropped, it has a vertical velocity component. Let’s denote this initial vertical velocity as \( u \).
- The velocity components just before impact can be resolved into two directions: parallel and perpendicular to the inclined plane.

3. **Velocity Components**:
- The component of the velocity perpendicular to the inclined plane is \( u \cos \theta \).
- The component of the velocity parallel to the inclined plane is \( u \sin \theta \).

4. **Applying the Coefficient of Restitution**:
- According to the definition of the coefficient of restitution, the velocity component after the impact perpendicular to the plane is given by:
\[ V \sin(90 - \theta) = e \cdot (u \cos \theta) \]
- Since \( \sin(90 - \theta) = \cos \theta \), we can rewrite this as:
\[ V \cos \theta = e \cdot (u \cos \theta) \]
- From this, we can simplify to:
\[ V = e \cdot u \]

5. **Parallel Component of Velocity**:
- The velocity component parallel to the inclined plane remains unchanged:
\[ V \sin(90 - \theta) = u \sin \theta \]
- Again, substituting \( \sin(90 - \theta) = \cos \theta \):
\[ V \sin \theta = u \sin \theta \]
- Thus, we have:
\[ V \cos \theta = u \sin \theta \]

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

7. **Dividing the Equations**:
- Dividing the second equation by the first:
\[ \frac{V \cos \theta}{V} = \frac{u \sin \theta}{e \cdot u} \]
- This simplifies to:
\[ \cos \theta = \frac{\sin \theta}{e} \]

8. **Using Trigonometric Identities**:
- Rearranging gives us:
\[ e = \frac{\sin \theta}{\cos \theta} = \tan \theta \]
- Thus, we can express \( e \) in terms of \( \tan \theta \).

9. **Final Expression**:
- Squaring both sides gives:
\[ e = \tan^2 \theta \]
- Therefore, taking the square root:
\[ \tan \theta = \sqrt{e} \]
- Finally, the angle of inclination is:
\[ \theta = \tan^{-1}(\sqrt{e}) \]

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