A ball of mass 50 gm is dropped from a height h=10m. It rebounds losing 75 percent of its kinetic energy. If it remains in contanct with the ground for `Deltat=0.01` sec. the impulse of the impact force is (take `g=10 m//s^(2)`)

Options

A. 1.3 N-s
B. 1.06 N-s
C. 1300 N-s
D. 1.05

Explanation

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

### Step 1: Calculate the velocity just before impact (V1)
When the ball is dropped from a height \( h = 10 \, m \), we can use the equation of motion to find the velocity just before it hits the ground. The formula is:

\[ V_1 = \sqrt{2gh} \]

Substituting the values \( g = 10 \, m/s^2 \) and \( h = 10 \, m \):

\[ V_1 = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2} \, m/s \]

### Step 2: Calculate the kinetic energy before impact (K1)
The kinetic energy (K1) just before impact can be calculated using the formula:

\[ K_1 = \frac{1}{2} m V_1^2 \]

First, convert the mass from grams to kilograms:

\[ m = 50 \, gm = 0.05 \, kg \]

Now, substituting the values:

\[ K_1 = \frac{1}{2} \times 0.05 \times (10\sqrt{2})^2 = \frac{1}{2} \times 0.05 \times 200 = 5 \, J \]

### Step 3: Calculate the kinetic energy after rebound (K2)
The problem states that the ball loses 75% of its kinetic energy upon rebounding. Therefore, the remaining kinetic energy (K2) is:

\[ K_2 = K_1 - 0.75 K_1 = 0.25 K_1 \]

Substituting \( K_1 = 5 \, J \):

\[ K_2 = 0.25 \times 5 = 1.25 \, J \]

### Step 4: Calculate the velocity after rebound (V2)
Using the kinetic energy after rebound to find the velocity (V2):

\[ K_2 = \frac{1}{2} m V_2^2 \]

Rearranging gives:

\[ V_2 = \sqrt{\frac{2K_2}{m}} \]

Substituting the values:

\[ V_2 = \sqrt{\frac{2 \times 1.25}{0.05}} = \sqrt{50} = 5\sqrt{2} \, m/s \]

### Step 5: Calculate the change in momentum (ΔP)
The change in momentum (ΔP) can be calculated as:

\[ \Delta P = P_{\text{final}} - P_{\text{initial}} = mV_2 - m(-V_1) \]

Substituting the values:

\[ \Delta P = 0.05 \times (5\sqrt{2}) - 0.05 \times (10\sqrt{2}) = 0.05(5\sqrt{2} + 10\sqrt{2}) = 0.05 \times 15\sqrt{2} \]

Calculating this gives:

\[ \Delta P = 0.75\sqrt{2} \, kg \cdot m/s \]

### Step 6: Calculate the impulse of the impact force
The impulse (I) is equal to the change in momentum and can be calculated as:

\[ I = \Delta P \]

Substituting the value:

\[ I = 0.75\sqrt{2} \, kg \cdot m/s \]

Numerically calculating \( \sqrt{2} \approx 1.414 \):

\[ I \approx 0.75 \times 1.414 \approx 1.06 \, N \cdot s \]

### Final Answer
The impulse of the impact force is approximately \( 1.06 \, N \cdot s \).

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