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.

Explanation

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

### Step 1: Understand the Energy Conversion
The potential energy stored in the stretched rubber cord is converted into the kinetic energy of the stone when it is released.

### Step 2: Write the Energy Equations
The potential energy (PE) stored in the rubber cord when it is stretched is given by:
\[ PE = \frac{1}{2} F \Delta L \]
where \( F \) is the force exerted by the rubber cord, and \( \Delta L \) is the change in length (stretching).

The kinetic energy (KE) of the stone when it is launched is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the stone and \( v \) is its velocity.

### Step 3: Set the Potential Energy Equal to Kinetic Energy
Since the potential energy is converted into kinetic energy, we can set these two equations equal to each other:
\[ \frac{1}{2} F \Delta L = \frac{1}{2} mv^2 \]
We can cancel the \( \frac{1}{2} \) from both sides:
\[ F \Delta L = mv^2 \]

### Step 4: Solve for Force
Rearranging the equation gives us:
\[ F = \frac{mv^2}{\Delta L} \]

### Step 5: Calculate the Strain
Strain is defined as the change in length divided by the original length:
\[ \text{Strain} = \frac{\Delta L}{L} \]
where \( L \) is the original length of the rubber cord.

### Step 6: Calculate the Stress
Stress is defined as the force per unit area:
\[ \text{Stress} = \frac{F}{A} \]
where \( A \) is the cross-sectional area of the rubber cord. The area can be calculated using:
\[ A = \pi r^2 \]
where \( r \) is the radius of the cord. Given the diameter is \( 6.0 \, \text{mm} \), the radius \( r \) is:
\[ r = \frac{6.0 \, \text{mm}}{2} = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m} \]

### Step 7: Calculate Young's Modulus
Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Substituting the expressions for stress and strain gives:
\[ Y = \frac{F/A}{\Delta L/L} \]
This can be rearranged to:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]

### Step 8: Substitute Values
Now we can substitute the known values:
- Mass \( m = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} \)
- Velocity \( v = 20 \, \text{m/s} \)
- Change in length \( \Delta L = 20 \, \text{cm} = 0.20 \, \text{m} \)
- Original length \( L = 42 \, \text{cm} = 0.42 \, \text{m} \)
- Cross-sectional area \( A = \pi (3.0 \times 10^{-3})^2 \)

### Step 9: Calculate Each Component
1. Calculate \( F \):
\[ F = \frac{(20 \times 10^{-3}) \cdot (20)^2}{0.20} = \frac{(20 \times 10^{-3}) \cdot 400}{0.20} = \frac{8000 \times 10^{-3}}{0.20} = 40 \, \text{N} \]

2. Calculate \( A \):
\[ A = \pi (3.0 \times 10^{-3})^2 = \pi \cdot 9.0 \times 10^{-6} \approx 2.83 \times 10^{-5} \, \text{m}^2 \]

3. Substitute into Young's modulus formula:
\[ Y = \frac{40 \cdot 0.42}{(2.83 \times 10^{-5}) \cdot 0.20} \]
\[ Y = \frac{16.8}{5.66 \times 10^{-6}} \approx 2.97 \times 10^6 \, \text{N/m}^2 \]

### Final Answer
The Young's modulus for rubber is approximately:
\[ Y \approx 2.97 \times 10^6 \, \text{N/m}^2 \]