A ball of mass `1 kg` drops vertically on to the floor wit a speed of `25 m//s`. It rebounds with an initial velocity of `10 m//s`. What impulse acts on the ball during contact ?

Options

A. `35 kg m//s` downwards
B. `35 kg m//s` upwards
C. `30 kg m//s` downwards
D. `30kg m//s` upwards

Explanation

To solve the problem, we need to calculate the impulse acting on the ball during its contact with the floor. Impulse can be defined as the change in momentum of an object. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the ball, \( m = 1 \, \text{kg} \)
- Initial velocity before the impact (downward), \( u = 25 \, \text{m/s} \)
- Final velocity after the impact (upward), \( v = -10 \, \text{m/s} \) (Note: We take upward direction as negative because the ball is moving in the opposite direction after the rebound)

2. **Calculate the Change in Momentum:**
- The change in momentum (\( \Delta p \)) can be calculated using the formula:
\[ \Delta p = m(v - u) \]
- Substitute the values into the formula:
\[ \Delta p = 1 \, \text{kg} \left(-10 \, \text{m/s} - 25 \, \text{m/s}\right) \]
- Simplifying the equation:
\[ \Delta p = 1 \, \text{kg} \left(-35 \, \text{m/s}\right) = -35 \, \text{kg m/s} \]

3. **Interpret the Result:**
- The negative sign indicates that the impulse acts in the downward direction. The magnitude of the impulse is \( 35 \, \text{kg m/s} \).

4. **Conclusion:**
- Therefore, the impulse acting on the ball during contact with the floor is \( 35 \, \text{kg m/s} \) downward.

### Final Answer:
The impulse acting on the ball during contact is \( 35 \, \text{kg m/s} \) downward.

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