A boy has a catapult made of a rubber cord of length `42 cm` and diameter `6.0 mm`. The boy stretches the cord by `20 cm` to catapult a stone of mass `20 g`. The stone flies off with a speed of `20 ms^(-1)`. Find Young's modulus for rubber. Ignore the change in the cross section of the cord in stretching.

Options

A. `10^6m^(-2)`
B. `10^3Nm^(-2)`
C. `10^8Nm^(-2)`
D. `10^4Nm^(-2)`

Explanation

To find Young's modulus for the rubber cord used in the catapult, we will follow these steps:

### Step 1: Gather the given data
- Length of the rubber cord, \( L = 42 \, \text{cm} = 0.42 \, \text{m} \)
- Diameter of the rubber cord, \( d = 6.0 \, \text{mm} = 6.0 \times 10^{-3} \, \text{m} \)
- Radius of the rubber cord, \( r = \frac{d}{2} = 3.0 \times 10^{-3} \, \text{m} \)
- Change in length (stretch), \( \Delta L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Mass of the stone, \( m = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} \)
- Velocity of the stone, \( v = 20 \, \text{m/s} \)

### Step 2: Calculate the kinetic energy of the stone
The kinetic energy (KE) of the stone can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times (20 \times 10^{-3}) \times (20)^2 = \frac{1}{2} \times 0.02 \times 400 = 4 \, \text{J} \]

### Step 3: Calculate the volume of the rubber cord
The volume \( V \) of the rubber cord can be calculated using the formula for the volume of a cylinder:
\[ V = \pi r^2 L \]
Substituting the values:
\[ V = \pi \times (3.0 \times 10^{-3})^2 \times 0.42 \]
Calculating \( r^2 \):
\[ r^2 = (3.0 \times 10^{-3})^2 = 9.0 \times 10^{-6} \, \text{m}^2 \]
Now substituting back:
\[ V = \pi \times 9.0 \times 10^{-6} \times 0.42 \approx 1.19 \times 10^{-5} \, \text{m}^3 \]

### Step 4: Calculate the strain
Strain is defined as:
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{0.2}{0.42} \approx 0.4762 \]

### Step 5: Relate the energy stored in the rubber to the kinetic energy of the stone
The energy stored in the rubber can be expressed as:
\[ \text{Energy stored} = \frac{1}{2} \times \text{Young's Modulus} \times \text{Strain}^2 \times V \]
Let \( Y \) be Young's modulus. Setting the energy stored equal to the kinetic energy:
\[ \frac{1}{2} Y \left(\frac{\Delta L}{L}\right)^2 V = KE \]
Substituting the known values:
\[ \frac{1}{2} Y \left(0.4762\right)^2 (1.19 \times 10^{-5}) = 4 \]

### Step 6: Solve for Young's modulus \( Y \)
First, calculate \( \left(0.4762\right)^2 \):
\[ (0.4762)^2 \approx 0.2275 \]
Now substituting this back into the equation:
\[ \frac{1}{2} Y \times 0.2275 \times 1.19 \times 10^{-5} = 4 \]
This simplifies to:
\[ Y \times 0.2275 \times 1.19 \times 10^{-5} = 8 \]
Now solving for \( Y \):
\[ Y = \frac{8}{0.2275 \times 1.19 \times 10^{-5}} \approx 3.0 \times 10^6 \, \text{N/m}^2 \]

### Conclusion
Thus, the Young's modulus for rubber is approximately \( 3.0 \times 10^6 \, \text{N/m}^2 \).