A partical falls from a height `h` upon a fixed horizontal plane and rebounds. If `e` is the coefficient of restitution, the total distance travelled before rebounding has stopped is

Options

A. `h ((1 + e^(2))/(1 - e^(2)))`
B. `h ((1 - e^(2))/(1 + e^(2)))`
C. `(h)/(2) ((1 - e^(2))/(1 + e^(2)))`
D. `(h)/(2) ((1 + e^(2))/(1 - e^(2)))`

Explanation

To solve the problem of a particle falling from a height \( h \) and rebounding with a coefficient of restitution \( e \), we need to calculate the total distance traveled before the particle stops rebounding. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Motion
When the particle falls from height \( h \), it hits the ground and rebounds to a height \( h_1 \). The coefficient of restitution \( e \) relates the heights before and after the bounce.

### Step 2: Calculate the Height After First Rebound
The height after the first rebound \( h_1 \) can be calculated using the formula:
\[ h_1 = e^2 \cdot h \]
This is derived from the definition of the coefficient of restitution, which states that the ratio of the velocity after the bounce to the velocity before the bounce is \( e \). Since potential energy is proportional to height, the heights are related by the square of the coefficient of restitution.

### Step 3: Calculate the Height After Subsequent Rebounds
After the first rebound, the particle will continue to bounce and reach subsequent heights:
- The height after the second rebound \( h_2 \) is:
\[ h_2 = e^2 \cdot h_1 = e^2 \cdot (e^2 \cdot h) = e^4 \cdot h \]
- The height after the third rebound \( h_3 \) is:
\[ h_3 = e^2 \cdot h_2 = e^2 \cdot (e^4 \cdot h) = e^6 \cdot h \]
- This pattern continues indefinitely.

### Step 4: Total Distance Calculation
The total distance traveled by the particle before it stops rebounding includes:
1. The initial drop from height \( h \).
2. The upward and downward distances for each rebound.

The total distance \( D \) can be expressed as:
\[ D = h + 2(h_1 + h_2 + h_3 + \ldots) \]

### Step 5: Sum the Infinite Series
The series \( h_1 + h_2 + h_3 + \ldots \) can be rewritten as:
\[ h_1 + h_2 + h_3 + \ldots = e^2 \cdot h + e^4 \cdot h + e^6 \cdot h + \ldots \]
This is a geometric series with the first term \( a = e^2 \cdot h \) and common ratio \( r = e^2 \):
\[ \text{Sum} = \frac{a}{1 - r} = \frac{e^2 \cdot h}{1 - e^2} \]

### Step 6: Substitute Back into Total Distance
Now substituting back into the total distance formula:
\[ D = h + 2 \cdot \frac{e^2 \cdot h}{1 - e^2} \]
\[ D = h + \frac{2e^2 \cdot h}{1 - e^2} \]
Factoring out \( h \):
\[ D = h \left(1 + \frac{2e^2}{1 - e^2}\right) \]

### Final Step: Simplify the Expression
Combining the terms gives:
\[ D = h \left(\frac{1 - e^2 + 2e^2}{1 - e^2}\right) = h \left(\frac{1 + e^2}{1 - e^2}\right) \]

### Final Answer
Thus, the total distance traveled before the rebounding stops is:
\[ D = h \left(\frac{1 + e^2}{1 - e^2}\right) \]

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