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, we can follow these steps:

### Step 1: Calculate the initial velocity (v0) just before impact
We can use the equation of motion:
\[ v_0^2 = u^2 + 2gh_0 \]
Since the ball is dropped from rest, \( u = 0 \). Therefore:
\[ v_0^2 = 0 + 2gh_0 \]
Where:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h_0 = 4.9 \, \text{m} \)

Substituting the values:
\[ v_0^2 = 2 \times 9.8 \times 4.9 \]
\[ v_0^2 = 96.04 \]
Taking the square root:
\[ v_0 = \sqrt{96.04} \]
\[ v_0 \approx 9.8 \, \text{m/s} \]

### Step 2: Determine the time taken for the first drop (t1)
Using the equation of motion:
\[ h_0 = \frac{1}{2}gt_1^2 \]
Rearranging gives:
\[ t_1 = \sqrt{\frac{2h_0}{g}} \]
Substituting the values:
\[ t_1 = \sqrt{\frac{2 \times 4.9}{9.8}} \]
\[ t_1 = \sqrt{1} = 1 \, \text{s} \]

### Step 3: Calculate the time taken for the ball to rebound to height \( h_1 \) and fall back to the floor (t2 + t3)
The height after the first impact is given by:
\[ h_1 = e \cdot h_0 \]
The time to rise to height \( h_1 \) (t2) and fall back down (t3) can be calculated as:
\[ t_2 = t_3 = \sqrt{\frac{2h_1}{g}} = \sqrt{\frac{2(e \cdot h_0)}{g}} \]
Thus:
\[ t_2 + t_3 = 2\sqrt{\frac{2(e \cdot h_0)}{g}} \]

### Step 4: Total time from the first impact to the second impact
The total time from the first impact to the second impact is:
\[ T = t_1 + t_2 + t_3 \]
Substituting the values:
\[ T = 1 + 2\sqrt{\frac{2(e \cdot h_0)}{g}} \]
Given that \( T = 1.3 \, \text{s} \):
\[ 1 + 2\sqrt{\frac{2(e \cdot 4.9)}{9.8}} = 1.3 \]

### Step 5: Solve for the coefficient of restitution (e)
Rearranging the equation:
\[ 2\sqrt{\frac{2(e \cdot 4.9)}{9.8}} = 0.3 \]
Dividing both sides by 2:
\[ \sqrt{\frac{2(e \cdot 4.9)}{9.8}} = 0.15 \]
Squaring both sides:
\[ \frac{2(e \cdot 4.9)}{9.8} = 0.0225 \]
Multiplying both sides by 9.8:
\[ 2(e \cdot 4.9) = 0.0225 \times 9.8 \]
\[ 2(e \cdot 4.9) = 0.2205 \]
Dividing by 2:
\[ e \cdot 4.9 = 0.11025 \]
Finally, solving for e:
\[ e = \frac{0.11025}{4.9} \]
\[ e \approx 0.0225 \]

### Conclusion
The coefficient of restitution (e) during the impact with the floor is approximately \( 0.0225 \).

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