A `1.5` kg ball drops vertically on a floor hitting with a speed of 25 `ms^(1)`. It rebounds with an initial speed of `15 ms^(-1)`. If the ball was in contact for only `0.03`, the force exerted on the floor by the ball is

Options

A. 2000 N
B. 3000 N
C. 3500 N
D. 4000 N

Answer

To solve the problem of finding the force exerted on the floor by the ball, we can follow these steps:

### Step 1: Identify the given data
- Mass of the ball, \( m = 1.5 \, \text{kg} \)
- Initial velocity before hitting the floor, \( u = -25 \, \text{m/s} \) (negative because it's downward)
- Final velocity after rebounding, \( v = 15 \, \text{m/s} \) (positive because it's upward)
- Time of contact with the floor, \( \Delta t = 0.03 \, \text{s} \)

### Step 2: Calculate the change in momentum
The change in momentum (\( \Delta p \)) can be calculated using the formula:
\[ \Delta p = m(v - u) \]
Substituting the values:
\[ \Delta p = 1.5 \, \text{kg} \times (15 \, \text{m/s} - (-25 \, \text{m/s})) \]
\[ \Delta p = 1.5 \, \text{kg} \times (15 + 25) \, \text{m/s} \]
\[ \Delta p = 1.5 \, \text{kg} \times 40 \, \text{m/s} \]
\[ \Delta p = 60 \, \text{kg m/s} \]

### Step 3: Calculate the average force exerted on the floor
The average force (\( F \)) exerted can be calculated using the impulse-momentum theorem:
\[ F = \frac{\Delta p}{\Delta t} \]
Substituting the values:
\[ F = \frac{60 \, \text{kg m/s}}{0.03 \, \text{s}} \]
\[ F = 2000 \, \text{N} \]

### Step 4: Conclusion
The force exerted on the floor by the ball is \( 2000 \, \text{N} \).

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