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: Determine the mass of the ball
The mass of the ball is given as 100 grams. To convert this to kilograms:
\[ \text{Mass (m)} = 100 \text{ grams} = 0.1 \text{ kg} \]

### Step 2: Calculate the velocity just before impact
The ball falls from a height of 20 m. We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \) (initial velocity),
- \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( s = -20 \, \text{m} \) (displacement, negative because it falls down).

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 20 \]
\[ v^2 = 400 \implies v = -20 \, \text{m/s} \quad (\text{negative sign indicates downward direction}) \]

### Step 3: Calculate the velocity just after impact
The ball rebounds to a height of 5 m. Using the same kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( u = 0 \) (final velocity after rebound),
- \( a = -10 \, \text{m/s}^2 \) (deceleration due to gravity),
- \( s = 5 \, \text{m} \) (displacement, positive because it goes up).

Substituting the values:
\[ 0 = u^2 + 2 \times (-10) \times 5 \]
\[ 0 = u^2 - 100 \implies u^2 = 100 \implies u = 10 \, \text{m/s} \quad (\text{positive sign indicates upward direction}) \]

### Step 4: Calculate the change in momentum
The change in momentum (\( \Delta p \)) during the impact can be calculated using:
\[ \Delta p = m(v - u) \]
where:
- \( v = 10 \, \text{m/s} \) (upward velocity after rebound),
- \( u = -20 \, \text{m/s} \) (downward velocity just before impact).

Substituting the values:
\[ \Delta p = 0.1 \, \text{kg} \times (10 - (-20)) = 0.1 \, \text{kg} \times (10 + 20) = 0.1 \, \text{kg} \times 30 = 3 \, \text{kg m/s} \]

### Step 5: Calculate the average force
The average force (\( F \)) can be calculated using the formula:
\[ F = \frac{\Delta p}{\Delta t} \]
where \( \Delta t = 0.02 \, \text{s} \).

Substituting the values:
\[ F = \frac{3 \, \text{kg m/s}}{0.02 \, \text{s}} = 150 \, \text{N} \]

### Conclusion
The average force acting on the ball during impact is \( 150 \, \text{N} \).

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