A ball is dropped from a height `h` onto a floor and rebounds to a height `h//6`. The coefficient of restitution between the ball and the floor is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(2)/(3)`
D. `(1)/sqrt(6)`

Explanation

To find the coefficient of restitution (E) between the ball and the floor, we can follow these steps:

### Step 1: Understand the Problem
The ball is dropped from a height \( h \) and rebounds to a height \( \frac{h}{6} \). We need to find the coefficient of restitution, which is defined as the ratio of the relative speed after the collision to the relative speed before the collision.

### Step 2: Identify Initial and Final Velocities
- When the ball is dropped from height \( h \), its initial velocity \( U_1 \) just before hitting the ground can be calculated using the equation of motion:
\[ V_1^2 = U_1^2 + 2g h \]
Since the initial velocity \( U_1 = 0 \):
\[ V_1^2 = 2gh \quad \Rightarrow \quad V_1 = \sqrt{2gh} \]

- After rebounding to a height \( \frac{h}{6} \), the final velocity \( V_2 \) just after the ball leaves the ground can be calculated similarly:
\[ V_2^2 = U_2^2 - 2g \left(\frac{h}{6}\right) \]
Here, the final velocity \( U_2 = 0 \) at the peak height:
\[ 0 = U_2^2 - 2g \left(\frac{h}{6}\right) \quad \Rightarrow \quad U_2^2 = \frac{2gh}{6} \quad \Rightarrow \quad U_2 = \sqrt{\frac{2gh}{6}} = \frac{1}{\sqrt{3}} \sqrt{2gh} \]

### Step 3: Calculate the Coefficient of Restitution
The coefficient of restitution \( E \) is given by:
\[ E = \frac{V_2}{V_1} \]
Substituting the values we found:
\[ E = \frac{\frac{1}{\sqrt{3}} \sqrt{2gh}}{\sqrt{2gh}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the coefficient of restitution \( E \) between the ball and the floor is:
\[ E = \frac{1}{\sqrt{3}} \]