A ball falls vertically onto a floor with momentum `p` and then bounces repeatedly. If coefficient of restitution is `e`, then the total momentum imparted by the ball to the floor is

Options

A. (a) `p(1+e)`
B. (b) `(p)/(1-e)`
C. (c) `p((1-e)/(1+e))`
D. (d) `p((1+e)/(1-e))`

Explanation

To solve the problem of calculating the total momentum imparted by a ball to the floor after it bounces repeatedly, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Initial Momentum**:
The ball falls with an initial momentum \( p \). This momentum is given by the product of mass \( m \) and velocity \( v \). Therefore, we can express the initial velocity as:
\[ v = \frac{p}{m} \]

2. **Momentum After First Bounce**:
Upon hitting the floor, the ball bounces back with a momentum reduced by the coefficient of restitution \( e \). Thus, the momentum after the first bounce is:
\[ p_1 = e \cdot p \]

3. **Momentum After Subsequent Bounces**:
Each subsequent bounce will also follow the same pattern. The momentum after the second bounce will be:
\[ p_2 = e \cdot p_1 = e^2 \cdot p \]
The momentum after the third bounce will be:
\[ p_3 = e \cdot p_2 = e^3 \cdot p \]
This pattern continues indefinitely.

4. **Total Momentum Imparted to the Floor**:
The total momentum imparted to the floor can be calculated by summing the momentum changes for each bounce. The change in momentum for each bounce can be expressed as:
- For the first impact:
\[ \Delta p_1 = p - p_1 = p - e \cdot p = p(1 - e) \]
- For the second impact:
\[ \Delta p_2 = p_1 - p_2 = e \cdot p - e^2 \cdot p = e \cdot p(1 - e) \]
- For the third impact:
\[ \Delta p_3 = p_2 - p_3 = e^2 \cdot p - e^3 \cdot p = e^2 \cdot p(1 - e) \]
- And so on...

5. **Summing the Infinite Series**:
The total momentum imparted can be expressed as an infinite geometric series:
\[ \Delta P = \Delta p_1 + \Delta p_2 + \Delta p_3 + \ldots = p(1 - e) + e \cdot p(1 - e) + e^2 \cdot p(1 - e) + \ldots \]
Factoring out \( p(1 - e) \):
\[ \Delta P = p(1 - e) \left(1 + e + e^2 + \ldots \right) \]

6. **Using the Formula for Infinite Geometric Series**:
The sum of the infinite series \( 1 + e + e^2 + \ldots \) can be calculated using the formula for the sum of an infinite geometric series:
\[ S = \frac{1}{1 - e} \quad \text{(for } |e| < 1\text{)} \]
Therefore, substituting back, we get:
\[ \Delta P = p(1 - e) \cdot \frac{1}{1 - e} = p(1 + e) \]

7. **Final Result**:
Thus, the total momentum imparted by the ball to the floor is:
\[ \Delta P = \frac{p(1 + e)}{1 - e} \]

### Final Answer:
The total momentum imparted by the ball to the floor is \( \frac{p(1 + e)}{1 - e} \).