A ball of mass 10 gm dropped from a height of 5m hits the floor and rebounds to a height of 1.25m. If the ball is in contact with the ground for 0.1s, the force exerted by the ground on the ball is `(g=10 m//s^(2))`

Options

A. 0.5 N
B. 1.5N
C. 0.15 N
D. 2.5 N

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Convert mass from grams to kilograms
The mass of the ball is given as 10 grams. To convert this to kilograms, we use the conversion factor \(1 \text{ kg} = 1000 \text{ g}\).

\[ m = \frac{10 \text{ g}}{1000} = 0.01 \text{ kg} \]

**Hint:** Remember to convert grams to kilograms when using SI units.

### Step 2: Calculate the velocity just before hitting the ground
We can use the formula for the velocity of a freely falling object:

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

Where:
- \(g = 10 \text{ m/s}^2\) (acceleration due to gravity)
- \(h = 5 \text{ m}\) (height from which the ball is dropped)

Substituting the values:

\[ v_1 = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \text{ m/s} \]

**Hint:** Use the formula for free fall to find the velocity just before impact.

### Step 3: Calculate the velocity just after rebounding
Using the same formula for the rebound height of 1.25 m, we can find the velocity just after the ball rebounds:

\[ v_2 = \sqrt{2gh} \]

Where:
- \(h = 1.25 \text{ m}\)

Substituting the values:

\[ v_2 = \sqrt{2 \times 10 \times 1.25} = \sqrt{25} = 5 \text{ m/s} \]

**Hint:** The rebound height will give you the velocity just after the ball hits the ground.

### Step 4: Calculate the change in momentum
The change in momentum (\(\Delta p\)) can be calculated as:

\[ \Delta p = m(v_2 - (-v_1)) = m(v_2 + v_1) \]

Substituting the values:

\[ \Delta p = 0.01 \text{ kg} \times (5 + 10) = 0.01 \times 15 = 0.15 \text{ kg m/s} \]

**Hint:** Remember that the initial downward velocity is negative, hence we add it to the rebound velocity.

### Step 5: Calculate the force exerted by the ground
The force can be calculated using the formula:

\[ F = \frac{\Delta p}{\Delta t} \]

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

Substituting the values:

\[ F = \frac{0.15 \text{ kg m/s}}{0.1 \text{ s}} = 1.5 \text{ N} \]

**Hint:** Use the formula for force as the rate of change of momentum over time.

### Final Answer
The force exerted by the ground on the ball is **1.5 N**.

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