A plastic ball falls from a height of 4.9 metre and rebounds several times from the floor. What is the coefficient of restitution during the impact with the floor if 1.3 seconds pass from the first impact to the second one ?

Answer

To solve the problem of finding the coefficient of restitution (e) for a plastic ball that falls from a height of 4.9 meters and rebounds several times, we can follow these steps:

### Step 1: Determine the initial velocity (v0) just before the first impact.
Using the formula for free fall, we can find the velocity just before the impact:
\[ v_0 = \sqrt{2gh_0} \]
where:
- \( h_0 = 4.9 \, \text{m} \)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating \( v_0 \):
\[ v_0 = \sqrt{2 \times 9.8 \times 4.9} = \sqrt{96.04} \approx 9.8 \, \text{m/s} \]

### Step 2: Understand the time intervals.
The total time from the first impact to the second impact is given as 1.3 seconds. This time can be divided into three parts:
- \( t_1 \): time taken to rise to the maximum height after the first impact.
- \( t_2 \): time taken to fall back down to the floor after reaching the maximum height.
- \( t_3 \): time taken to rise to the maximum height after the second impact.

### Step 3: Express the time intervals in terms of the coefficient of restitution (e).
After the first impact, the velocity with which the ball rebounds is:
\[ v_1 = e \cdot v_0 \]

The maximum height after the first rebound (h1) can be expressed as:
\[ h_1 = \frac{(e \cdot v_0)^2}{2g} \]

The time taken to rise to this height (t1) is:
\[ t_1 = \frac{v_1}{g} = \frac{e \cdot v_0}{g} \]

The time taken to fall back down to the floor (t2) is the same as t1:
\[ t_2 = \sqrt{\frac{2h_1}{g}} = \sqrt{\frac{(e \cdot v_0)^2}{g^2}} = \frac{e \cdot v_0}{g} \]

Thus, the total time from the first impact to the second impact is:
\[ t_1 + t_2 + t_3 = 1.3 \]

### Step 4: Substitute the expressions for time into the total time equation.
Since \( t_3 \) is the same as \( t_1 \) (the ball takes the same time to rise after the second impact):
\[ t_1 + t_2 + t_3 = t_1 + t_1 + t_1 = 3t_1 = 1.3 \]

Thus:
\[ 3 \cdot \frac{e \cdot v_0}{g} = 1.3 \]

### Step 5: Solve for the coefficient of restitution (e).
Substituting \( v_0 \) into the equation:
\[ 3 \cdot \frac{e \cdot 9.8}{9.8} = 1.3 \]
This simplifies to:
\[ 3e = 1.3 \]
\[ e = \frac{1.3}{3} \approx 0.4333 \]

### Final Answer:
The coefficient of restitution during the impact with the floor is approximately \( e \approx 0.433 \).

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