The coefficient of restitution for a body is `e=(1)/(3)`. At what angle the body must be incident on a perfectly hard plane so that the angle between the direction before and after the impact be at right angles:

Options

A. `37^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Explanation

To solve the problem, we need to find the angle of incidence (θ₁) such that the angle between the direction of the body before and after the impact is 90 degrees, given that the coefficient of restitution (e) is 1/3.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a body colliding with a perfectly hard plane.
- The coefficient of restitution (e) is given as \( e = \frac{1}{3} \).
- We need to find the angle of incidence (θ₁) such that the angle between the direction before and after the impact is 90 degrees.

2. **Setting Up the Angles**:
- Let θ₁ be the angle of incidence (the angle the incoming path makes with the normal).
- Let θ₂ be the angle of reflection (the angle the outgoing path makes with the normal).
- According to the problem, \( θ₁ + θ₂ = 90^\circ \).

3. **Using the Coefficient of Restitution**:
- The coefficient of restitution is defined as:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
- For our case, the velocities along the line of impact are:
- Velocity of approach before the collision: \( u \cos θ₁ \)
- Velocity of separation after the collision: \( v \cos θ₂ \)
- Therefore, we can write:
\[ v \cos θ₂ = e \cdot u \cos θ₁ \]
- Substituting \( e = \frac{1}{3} \):
\[ v \cos θ₂ = \frac{1}{3} u \cos θ₁ \]

4. **Using the Relationship Between Angles**:
- Since \( θ₂ = 90^\circ - θ₁ \), we can express \( \cos θ₂ \) as:
\[ \cos θ₂ = \sin θ₁ \]
- Substituting this into the equation gives:
\[ v \sin θ₁ = \frac{1}{3} u \cos θ₁ \]

5. **Conservation of Momentum in the x-direction**:
- The momentum in the x-direction before and after the collision can be expressed as:
\[ u \sin θ₁ = v \sin θ₂ \]
- Substituting \( θ₂ = 90^\circ - θ₁ \) gives:
\[ u \sin θ₁ = v \cos θ₁ \]

6. **Dividing the Two Equations**:
- We now have two equations:
1. \( v \sin θ₁ = \frac{1}{3} u \cos θ₁ \)
2. \( u \sin θ₁ = v \cos θ₁ \)
- Dividing these two equations:
\[ \frac{v \sin θ₁}{u \sin θ₁} = \frac{\frac{1}{3} u \cos θ₁}{v \cos θ₁} \]
- This simplifies to:
\[ \frac{v}{u} = \frac{1}{3} \cdot \frac{u \cos θ₁}{v \cos θ₁} \]
- Rearranging gives:
\[ \tan θ₁ = e \tan θ₂ \]

7. **Substituting for θ₂**:
- Since \( θ₂ = 90^\circ - θ₁ \):
\[ \tan θ₂ = \cot θ₁ \]
- Therefore, we can write:
\[ \tan θ₁ = e \cdot \cot θ₁ \]
- This leads to:
\[ \tan^2 θ₁ = e \]

8. **Finding θ₁**:
- Substituting \( e = \frac{1}{3} \):
\[ \tan^2 θ₁ = \frac{1}{3} \]
- Thus:
\[ \tan θ₁ = \frac{1}{\sqrt{3}} \]
- This implies:
\[ θ₁ = 30^\circ \]

### Final Answer:
The angle of incidence (θ₁) is \( 30^\circ \).