Two identical balls are dropped from the same height onto a hard surface, the second ball being released exactly when the first ball bollides with the surface. If the first ball has made two more collisions by the time the second one collides. Then the coefficient of restitution between the ball and the surface satisfies :

Options

A. `e gt 0.5`
B. `e = 0.5`
C. `e = (sqrt(3) - 1)/(2)`
D. `e lt (sqrt(3) - 1)/(2)`

Explanation

To solve the problem, we need to analyze the motion of the two balls and the collisions they undergo. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the time taken by the first ball to hit the ground
When the first ball is dropped from a height \( h \), we can use the second equation of motion to find the time \( t_0 \) it takes to hit the ground. The equation is:

\[ h = \frac{1}{2} g t_0^2 \]

Rearranging gives:

\[ t_0 = \sqrt{\frac{2h}{g}} \]

### Step 2: Calculate the velocity of the first ball just before impact
Using the third equation of motion, we can find the velocity \( v_0 \) of the first ball just before it strikes the ground:

\[ v_0 = \sqrt{2gh} \]

### Step 3: Analyze the first collision
Upon colliding with the ground, the first ball will rebound with a velocity reduced by the coefficient of restitution \( e \):

\[ v_1 = e v_0 = e \sqrt{2gh} \]

### Step 4: Determine the time taken for the first ball to collide again
The time taken for the first ball to rise back to the maximum height and fall again can be calculated. The time to rise to the maximum height is given by:

\[ t_{up} = \frac{v_1}{g} = \frac{e \sqrt{2gh}}{g} \]

The time to fall back down from that height is the same, so the total time for the first collision and return is:

\[ t_1 = 2 \cdot t_{up} = 2 \cdot \frac{e \sqrt{2h}}{g} \]

### Step 5: Analyze the second collision
After the first collision, the ball will again collide with the ground. The velocity just before the second collision will be:

\[ v_2 = e v_1 = e^2 v_0 = e^2 \sqrt{2gh} \]

The time taken for the ball to rise and fall again is:

\[ t_{down} = 2 \cdot \frac{v_2}{g} = 2 \cdot \frac{e^2 \sqrt{2h}}{g} \]

### Step 6: Total time for two collisions
The total time taken for the first ball to collide twice is:

\[ t_{total} = t_1 + t_2 = 2 \cdot \frac{e \sqrt{2h}}{g} + 2 \cdot \frac{e^2 \sqrt{2h}}{g} \]

Factoring out the common terms gives:

\[ t_{total} = \frac{2\sqrt{2h}}{g} (e + e^2) \]

### Step 7: Set up the inequality
According to the problem, the time taken for the second ball to hit the ground \( t_0 \) must be less than the total time for the first ball to collide twice:

\[ t_0 < t_{total} \]

Substituting the expressions we derived:

\[ \sqrt{\frac{2h}{g}} < \frac{2\sqrt{2h}}{g} (e + e^2) \]

### Step 8: Simplifying the inequality
Dividing both sides by \( \sqrt{\frac{2h}{g}} \):

\[ 1 < 2(e + e^2) \]

Rearranging gives:

\[ 2e^2 + 2e - 1 > 0 \]

### Step 9: Solving the quadratic inequality
To find the roots of the quadratic equation \( 2e^2 + 2e - 1 = 0 \):

Using the quadratic formula:

\[ e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]

Calculating gives:

\[ e = \frac{-2 \pm \sqrt{4 + 8}}{4} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2} \]

### Step 10: Determine the valid range for \( e \)
Since \( e \) must be positive, we take the positive root:

\[ e < \frac{-1 + \sqrt{3}}{2} \]

### Conclusion
The coefficient of restitution \( e \) must satisfy:

\[ e < \frac{-1 + \sqrt{3}}{2} \]