A ball falls vertically on an inclined plane of inclination `alpha` with speed `v_0` and makes a perfectly elastic collision. What is angle of velocity vector with horizontal after collision.

Explanation

To solve the problem of finding the angle of the velocity vector with the horizontal after a perfectly elastic collision of a ball falling vertically onto an inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The ball falls vertically with an initial speed \( v_0 \).
- The inclined plane makes an angle \( \alpha \) with the horizontal.

2. **Resolve the Initial Velocity**:
- The initial velocity \( v_0 \) can be resolved into two components relative to the inclined plane:
- Perpendicular to the inclined plane: \( v_{0 \perp} = v_0 \cos(\alpha) \)
- Parallel to the inclined plane: \( v_{0 \parallel} = v_0 \sin(\alpha) \)

3. **Understand the Collision**:
- Since the collision is perfectly elastic, the component of velocity perpendicular to the inclined plane will change direction but maintain its magnitude.
- The parallel component of velocity will remain unchanged.

4. **Determine the Components After Collision**:
- After the collision, the new velocity components will be:
- Perpendicular to the inclined plane: \( v_{1 \perp} = -v_0 \cos(\alpha) \) (changes direction)
- Parallel to the inclined plane: \( v_{1 \parallel} = v_0 \sin(\alpha) \) (remains the same)

5. **Combine the Velocity Components**:
- The final velocity vector \( \vec{V_f} \) can be expressed as:
\[ \vec{V_f} = v_{1 \parallel} \hat{p} + v_{1 \perp} \hat{n} \]
- Substituting the values:
\[ \vec{V_f} = v_0 \sin(\alpha) \hat{p} - v_0 \cos(\alpha) \hat{n} \]

6. **Convert to Cartesian Coordinates**:
- The components can be expressed in terms of the horizontal (x) and vertical (y) directions:
- Horizontal component: \( V_{fx} = v_0 \sin(\alpha) \)
- Vertical component: \( V_{fy} = -v_0 \cos(\alpha) \)

7. **Calculate the Angle with the Horizontal**:
- The angle \( \theta \) with the horizontal can be found using the tangent function:
\[ \tan(\theta) = \frac{V_{fy}}{V_{fx}} = \frac{-v_0 \cos(\alpha)}{v_0 \sin(\alpha)} = -\cot(\alpha) \]
- Therefore, the angle \( \theta \) can be expressed as:
\[ \theta = \tan^{-1}(-\cot(\alpha)) \]

8. **Final Result**:
- The angle of the velocity vector with the horizontal after the collision is:
\[ \theta = \tan^{-1}(-\cot(\alpha)) \]