A ball of mass 100 gram falls from 20 m height on floor and rebounds to 5m. Time of contact is 0.02s. Find average force acting on it during impact. (g = 10 m/`s^(2)`)

Options

A. 120 N
B. 100 N
C. 200 N
D. 150 N

Explanation

To find the average force acting on the ball during impact, we can follow these steps:

### Step 1: Convert mass to kilograms
The mass of the ball is given as 100 grams. We need to convert this to kilograms for our calculations.
\[ m = 100 \, \text{grams} = 0.1 \, \text{kg} \]

### Step 2: Calculate the initial velocity just before impact
The ball falls from a height of 20 m. We can use the equation of motion to find the velocity just before impact (using the conservation of energy):
\[ v_i = \sqrt{2gh} \]
where \( g = 10 \, \text{m/s}^2 \) and \( h = 20 \, \text{m} \).
\[ v_i = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \, \text{m/s} \]

### Step 3: Calculate the final velocity just after rebound
The ball rebounds to a height of 5 m. We can again use the conservation of energy to find the velocity just after the rebound:
\[ v_f = \sqrt{2gh} \]
where \( h = 5 \, \text{m} \).
\[ v_f = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \, \text{m/s} \]

### Step 4: Calculate the change in momentum
The change in momentum (\(\Delta P\)) can be calculated as:
\[ \Delta P = mv_f - mv_i \]
Substituting the values:
\[ \Delta P = 0.1 \times 10 - 0.1 \times 20 = 1 - 2 = -1 \, \text{kg m/s} \]

### Step 5: Calculate the average force
The average force (\(F_{\text{avg}}\)) can be calculated using the formula:
\[ F_{\text{avg}} = \frac{\Delta P}{\Delta t} \]
where \(\Delta t = 0.02 \, \text{s}\).
\[ F_{\text{avg}} = \frac{-1}{0.02} = -50 \, \text{N} \]

### Step 6: Interpret the result
The negative sign indicates that the force is acting in the opposite direction of the initial motion (upward). The magnitude of the average force is:
\[ F_{\text{avg}} = 50 \, \text{N} \]

### Final Answer
The average force acting on the ball during impact is **50 N**.

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