A ball of mass 0.15kg is dropped from a height 10m strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (`g = 10 m/s^2`) nearly:

Options

A. 2.1 kg m/s
B. 1.4 kg m/s
C. 0 kg m/s
D. 4.2 kg m/s

Answer

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

### Step 1: Determine the velocity just before the ball strikes the ground
When the ball is dropped from a height of 10 m, it converts its potential energy into kinetic energy just before it hits the ground. The potential energy (PE) at the height can be calculated using the formula:

\[ PE = mgh \]

Where:
- \( m = 0.15 \, \text{kg} \) (mass of the ball)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 10 \, \text{m} \) (height)

Substituting the values:

\[ PE = 0.15 \times 10 \times 10 = 15 \, \text{J} \]

At the point just before hitting the ground, all this potential energy will have converted into kinetic energy (KE):

\[ KE = \frac{1}{2} mv^2 \]

Setting \( PE = KE \):

\[ 15 = \frac{1}{2} \times 0.15 \times v^2 \]

### Step 2: Solve for \( v \)
Rearranging the equation to solve for \( v^2 \):

\[ 15 = 0.075 v^2 \]

\[ v^2 = \frac{15}{0.075} = 200 \]

Taking the square root:

\[ v = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

### Step 3: Calculate the impulse
Impulse (J) is defined as the change in momentum. The momentum before and after the collision can be calculated as follows:

- Momentum just before hitting the ground (downward direction):
\[ p_{\text{initial}} = mv = 0.15 \times (-10\sqrt{2}) = -1.5\sqrt{2} \, \text{kg m/s} \]

- Momentum just after rebounding (upward direction):
\[ p_{\text{final}} = mv = 0.15 \times (10\sqrt{2}) = 1.5\sqrt{2} \, \text{kg m/s} \]

### Step 4: Calculate the change in momentum
The change in momentum (which is the impulse) is given by:

\[ \Delta p = p_{\text{final}} - p_{\text{initial}} = 1.5\sqrt{2} - (-1.5\sqrt{2}) = 1.5\sqrt{2} + 1.5\sqrt{2} = 3.0\sqrt{2} \, \text{kg m/s} \]

### Step 5: Calculate the numerical value of impulse
Now, we can calculate the numerical value of the impulse:

\[ \Delta p = 3.0 \times 1.414 \approx 4.242 \, \text{kg m/s} \]

### Final Answer
The magnitude of the impulse imparted to the ball is approximately **4.24 kg m/s**.

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