Hailstorms are observed to strike the surface of a frozen lake at an angle of `30^(@)` with the vertical and rebound at an angle of `60^(@)` with vertical. Assuming the contact to be smooth, the coefficient of restitution is

Options

A. `e = (1)/(sqrt(3))`
B. `e = (1)/(3)`
C. `e = sqrt(3)`
D. e = 3

Explanation

To find the coefficient of restitution (e) for the hailstones striking and rebounding from the frozen lake, we can follow these steps:

### Step 1: Understanding the Angles
The hailstones strike the surface at an angle of \(30^\circ\) with the vertical and rebound at an angle of \(60^\circ\) with the vertical.

### Step 2: Identify Components of Velocity
Let \(U\) be the initial velocity of the hailstone before impact, and \(V\) be the velocity after the impact. We can break these velocities into components:

- For the initial velocity \(U\):
- Vertical component (perpendicular to the surface): \(U \cos(30^\circ)\)
- Horizontal component (parallel to the surface): \(U \sin(30^\circ)\)

- For the final velocity \(V\):
- Vertical component (perpendicular to the surface): \(V \cos(60^\circ)\)
- Horizontal component (parallel to the surface): \(V \sin(60^\circ)\)

### Step 3: Equating Horizontal Components
Since there is no change in the horizontal component of velocity (the surface is smooth), we can set the horizontal components equal:
\[ U \sin(30^\circ) = V \sin(60^\circ) \]
Substituting the values of sine:
\[ U \cdot \frac{1}{2} = V \cdot \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ U = \sqrt{3} V \]

### Step 4: Applying the Coefficient of Restitution Formula
The coefficient of restitution \(e\) is defined as the ratio of the relative velocity after impact to the relative velocity before impact in the direction perpendicular to the surface:
\[ e = \frac{\text{Relative velocity after impact}}{\text{Relative velocity before impact}} \]
In this case:
- The relative velocity before impact (downward) is \(U \cos(30^\circ)\)
- The relative velocity after impact (upward) is \(V \cos(60^\circ)\)

Substituting the values:
\[ e = \frac{0 - V \cos(60^\circ)}{U \cos(30^\circ)} \]
Since the ground is stationary, the velocity of the ground is 0.

### Step 5: Substitute the Values
Substituting the cosine values:
\[ e = \frac{-V \cdot \frac{1}{2}}{U \cdot \frac{\sqrt{3}}{2}} \]
This simplifies to:
\[ e = \frac{-V/2}{U \cdot \sqrt{3}/2} = \frac{-V}{U \sqrt{3}} \]

### Step 6: Substitute \(U\) from Step 3
From Step 3, we know \(U = \sqrt{3} V\):
\[ e = \frac{-V}{\sqrt{3} V \cdot \sqrt{3}} = \frac{-V}{3V} = -\frac{1}{3} \]
Since we are interested in the magnitude:
\[ e = \frac{1}{3} \]

### Final Answer
The coefficient of restitution is:
\[ \boxed{\frac{1}{3}} \]