Two balls of same mass are dropped from the same height h, on to the floor. The first ball bounces to a height `h//4`, after the collision & the second ball to a height `h//16`. The impulse applied by the first & second ball on the floor are `I_(1)` and `I_(2)` respectively. Then :-

Options

A. `5I_(1) = 6I_(2)`
B. `6I_(1) = 5I_(1)`
C. `I_(1) = 2I_(2)`
D. `2I_(1) = I_(2)`

Explanation

To solve the problem, we need to calculate the impulse applied by each ball on the floor and then find the ratio of these impulses.

### Step-by-Step Solution:

1. **Determine the initial velocity before impact for both balls:**
The initial velocity \( V_i \) just before hitting the ground can be calculated using the formula:
\[ V_i = \sqrt{2gh} \]

2. **Calculate the final velocity after bouncing for the first ball:**
The first ball bounces to a height of \( \frac{h}{4} \). The final velocity \( V_{f1} \) after bouncing can be calculated using:
\[ V_{f1} = \sqrt{2g \left(\frac{h}{4}\right)} = \sqrt{\frac{2gh}{4}} = \frac{1}{2} \sqrt{2gh} = \frac{1}{2} V_i \]

3. **Calculate the impulse \( I_1 \) for the first ball:**
Impulse is defined as the change in momentum. The initial momentum \( P_i \) and final momentum \( P_f \) for the first ball are:
\[ P_i = M V_i \quad \text{(downward)} \]
\[ P_f = -M V_{f1} \quad \text{(upward)} \]
Therefore, the impulse \( I_1 \) is:
\[ I_1 = P_f - P_i = -M V_{f1} - M V_i = -M \left(\frac{1}{2} V_i\right) - M V_i = -\frac{3}{2} M V_i \]

4. **Calculate the final velocity after bouncing for the second ball:**
The second ball bounces to a height of \( \frac{h}{16} \). The final velocity \( V_{f2} \) after bouncing can be calculated using:
\[ V_{f2} = \sqrt{2g \left(\frac{h}{16}\right)} = \sqrt{\frac{2gh}{16}} = \frac{1}{4} \sqrt{2gh} = \frac{1}{4} V_i \]

5. **Calculate the impulse \( I_2 \) for the second ball:**
Similarly, the initial and final momentum for the second ball are:
\[ P_i = M V_i \quad \text{(downward)} \]
\[ P_f = -M V_{f2} \quad \text{(upward)} \]
Therefore, the impulse \( I_2 \) is:
\[ I_2 = P_f - P_i = -M V_{f2} - M V_i = -M \left(\frac{1}{4} V_i\right) - M V_i = -\frac{5}{4} M V_i \]

6. **Find the ratio of impulses \( \frac{I_1}{I_2} \):**
Now we can find the ratio of the impulses:
\[ \frac{I_1}{I_2} = \frac{-\frac{3}{2} M V_i}{-\frac{5}{4} M V_i} = \frac{\frac{3}{2}}{\frac{5}{4}} = \frac{3}{2} \cdot \frac{4}{5} = \frac{12}{10} = \frac{6}{5} \]

### Final Result:
The ratio of the impulses is:
\[ \frac{I_1}{I_2} = \frac{6}{5} \]