Two balls of same mass are dropped from the same height onto the floor. The first ball bounces upwards from the floor elastically. The second ball stricks to the floor. The first applies an impulse to the floor of `I_(1)` and the second applies an imupulse `I_(2)`. The impulses obey :-

Options

A. `I_(2) = 2I_(1)`
B. `I_(2) = (I_(1))/(2)`
C. `I_(2) = 4I_(1)`
D. `I_(2) = (I_(1))/(4)`

Explanation

To solve the problem, we will analyze the situation of two balls dropped from the same height and their respective impulses on the floor after they collide.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Both balls have the same mass \( m \).
- Both are dropped from the same height \( H \).
- The initial velocity of both balls before dropping is \( 0 \).

2. **Calculate the Velocity Just Before Impact**:
- Using the equation of motion under gravity, the velocity \( v \) just before hitting the ground can be calculated as:
\[ v = \sqrt{2gh} \]
- Here, \( g \) is the acceleration due to gravity.

3. **Analyze the First Ball (Elastic Collision)**:
- The first ball bounces back elastically. This means it rebounds with the same speed but in the opposite direction.
- Initial momentum before impact: \( p_{initial1} = -mv \) (downward direction considered negative).
- Final momentum after impact: \( p_{final1} = mv \) (upward direction considered positive).
- The change in momentum (impulse \( I_1 \)) is given by:
\[ I_1 = p_{final1} - p_{initial1} = mv - (-mv) = mv + mv = 2mv \]

4. **Analyze the Second Ball (Inelastic Collision)**:
- The second ball sticks to the floor, meaning it does not bounce back.
- Initial momentum before impact: \( p_{initial2} = -mv \).
- Final momentum after impact: \( p_{final2} = 0 \) (since it sticks to the floor).
- The change in momentum (impulse \( I_2 \)) is given by:
\[ I_2 = p_{final2} - p_{initial2} = 0 - (-mv) = mv \]

5. **Establish the Relationship Between Impulses**:
- Now we have:
\[ I_1 = 2mv \]
\[ I_2 = mv \]
- To find the ratio of the impulses:
\[ \frac{I_1}{I_2} = \frac{2mv}{mv} = 2 \]
- Thus, we can conclude:
\[ I_1 = 2I_2 \]

### Conclusion:
The relationship between the impulses applied by the two balls on the floor is:
\[ I_1 = 2I_2 \]