A ball of mass 1 kg is dropped from height 9.8 m, strikes with ground and rebounds at height 4.9 m, if the time of contact between ball and ground is 0.1 s then find impulse and average force acting on ball.

Options

A. 23.52 N-s,235.2 N
B. 235.2 N-s,23.53 N
C. 42.5 N-s,525 N
D. 52.5 N-s,525N

Explanation

To solve the problem, we need to calculate the impulse and the average force acting on the ball when it strikes the ground and rebounds. Here’s the step-by-step solution:

### Step 1: Calculate the velocity just before impact
When the ball is dropped from a height \( h = 9.8 \, \text{m} \), we can use the equation of motion to find the velocity just before it strikes the ground. The equation is given by:

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

Where:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 9.8 \, \text{m} \)

Substituting the values:

\[ v = \sqrt{2 \times 9.8 \times 9.8} = \sqrt{192.08} \approx 13.86 \, \text{m/s} \]

### Step 2: Calculate the velocity just after rebound
The ball rebounds to a height of \( 4.9 \, \text{m} \). We can again use the equation of motion to find the velocity just after it rebounds:

\[ v' = \sqrt{2gh'} \]

Where:
- \( h' = 4.9 \, \text{m} \)

Substituting the values:

\[ v' = \sqrt{2 \times 9.8 \times 4.9} = \sqrt{96.04} \approx 9.80 \, \text{m/s} \]

### Step 3: Calculate the change in momentum
The momentum before impact (downward) and after impact (upward) can be calculated as follows:

- Momentum before impact:
\[ p_{before} = m \cdot v = 1 \cdot (-13.86) = -13.86 \, \text{kg m/s} \]
(Note: The negative sign indicates the downward direction.)

- Momentum after impact:
\[ p_{after} = m \cdot v' = 1 \cdot 9.80 = 9.80 \, \text{kg m/s} \]

Now, the change in momentum (\( \Delta p \)) is given by:

\[ \Delta p = p_{after} - p_{before} = 9.80 - (-13.86) = 9.80 + 13.86 = 23.66 \, \text{kg m/s} \]

### Step 4: Calculate the impulse
Impulse is equal to the change in momentum:

\[ \text{Impulse} = \Delta p = 23.66 \, \text{kg m/s} \]

### Step 5: Calculate the average force
The average force (\( F_{avg} \)) can be calculated using the formula:

\[ F_{avg} = \frac{\Delta p}{\Delta t} \]

Where \( \Delta t = 0.1 \, \text{s} \):

\[ F_{avg} = \frac{23.66}{0.1} = 236.6 \, \text{N} \]

### Final Results
- Impulse = \( 23.66 \, \text{Ns} \)
- Average Force = \( 236.6 \, \text{N} \)