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^@`

Answer

To solve the problem step by step, we need to analyze the situation involving the coefficient of restitution and the angles of incidence and reflection.

### Step 1: Understand the Coefficient of Restitution
The coefficient of restitution (e) is defined as the ratio of the velocity of separation to the velocity of approach. Mathematically, it can be expressed as:
\[ e = \frac{V_{\text{separation}}}{V_{\text{approach}}} \]

### Step 2: Define the Angles
Let:
- \( \alpha \) be the angle of incidence (the angle at which the body strikes the surface).
- \( \beta \) be the angle of reflection (the angle at which the body leaves the surface).
According to the problem, we want \( \alpha + \beta = 90^\circ \).

### Step 3: Express the Velocities
The velocity components can be expressed as:
- The approach velocity (before impact) is \( V_1 \cos \alpha \) (normal component) and \( V_1 \sin \alpha \) (parallel component).
- The separation velocity (after impact) is \( V_2 \cos \beta \) (normal component) and \( V_2 \sin \beta \) (parallel component).

### Step 4: Set Up the Equations
From the definition of the coefficient of restitution:
\[ e = \frac{V_2 \cos \beta}{V_1 \cos \alpha} \]
Given that \( e = \frac{1}{3} \), we can write:
\[ \frac{V_2 \cos \beta}{V_1 \cos \alpha} = \frac{1}{3} \]
This implies:
\[ V_2 \cos \beta = \frac{1}{3} V_1 \cos \alpha \]

### Step 5: Use the Angle Relationship
Since \( \beta = 90^\circ - \alpha \), we can substitute \( \beta \) into the equation:
\[ V_2 \cos(90^\circ - \alpha) = V_2 \sin \alpha \]
Thus, the equation becomes:
\[ V_2 \sin \alpha = \frac{1}{3} V_1 \cos \alpha \]

### Step 6: Relate the Velocities
Now we can express the ratio of the velocities:
\[ \frac{V_2}{V_1} = \frac{1}{3} \cdot \frac{\cos \alpha}{\sin \alpha} = \frac{1}{3} \cot \alpha \]
This gives us:
\[ V_2 = \frac{1}{3} V_1 \cot \alpha \]

### Step 7: Substitute into the Coefficient of Restitution Equation
Now we can substitute \( V_2 \) back into the coefficient of restitution equation:
\[ \frac{1}{3} \cot \alpha \cdot \cos \beta = \frac{1}{3} V_1 \cos \alpha \]
Substituting \( \beta = 90^\circ - \alpha \):
\[ \frac{1}{3} \cot \alpha \cdot \sin \alpha = \frac{1}{3} V_1 \cos \alpha \]

### Step 8: Solve for \( \tan \alpha \)
From the above equation, we can simplify:
\[ \cot \alpha \cdot \sin \alpha = \cos \alpha \]
This implies:
\[ \frac{\cos \alpha}{\sin \alpha} \cdot \sin \alpha = \cos \alpha \]
Thus:
\[ \tan^2 \alpha = \frac{1}{3} \]
Taking the square root:
\[ \tan \alpha = \frac{1}{\sqrt{3}} \]

### Step 9: Find the Angle \( \alpha \)
Using the inverse tangent function:
\[ \alpha = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ \]

### Conclusion
The angle at which the body must be incident on the perfectly hard plane so that the angle between the direction before and after the impact is at right angles is:
\[ \alpha = 30^\circ \]