A rubber cord has a cross -sectional area `1 mm^(2)`and total unstretched length`10.0cm`. It is streched to `12.0 cm` and then released to project a missile of mass 5.0 g.Taking young's modulus Y for rubber as `5.0xx10^(8) N//m^(2)`.Calculate the velocity of projection .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We need to calculate the velocity of a missile projected by a rubber cord after it is stretched and then released. The given parameters include the cross-sectional area, unstretched length, stretched length, mass of the missile, and Young's modulus for rubber.

### Step 2: Identify the Given Data
- Cross-sectional area (A) = 1 mm² = \(1 \times 10^{-6} \, m^2\)
- Unstretched length (L) = 10 cm = 0.1 m
- Stretched length = 12 cm = 0.12 m
- Mass of the missile (m) = 5 g = \(5 \times 10^{-3} \, kg\)
- Young's modulus (Y) = \(5.0 \times 10^{8} \, N/m^2\)

### Step 3: Calculate the Extension (Δx)
The extension (Δx) of the rubber cord is the difference between the stretched length and the unstretched length:
\[ \Delta x = \text{Stretched length} - \text{Unstretched length} = 0.12 \, m - 0.1 \, m = 0.02 \, m \]

### Step 4: Calculate the Spring Constant (k)
The spring constant (k) can be calculated using the formula:
\[ k = \frac{Y \cdot A}{L} \]
Substituting the values:
\[ k = \frac{(5.0 \times 10^{8} \, N/m^2) \cdot (1 \times 10^{-6} \, m^2)}{0.1 \, m} = \frac{5.0 \times 10^{2} \, N}{0.1} = 5000 \, N/m \]

### Step 5: Calculate the Elastic Potential Energy (EPE)
The elastic potential energy stored in the rubber cord when it is stretched is given by:
\[ EPE = \frac{1}{2} k (\Delta x)^2 \]
Substituting the values:
\[ EPE = \frac{1}{2} \cdot 5000 \, N/m \cdot (0.02 \, m)^2 = \frac{1}{2} \cdot 5000 \cdot 0.0004 = 1 \, J \]

### Step 6: Relate Elastic Potential Energy to Kinetic Energy
When the missile is projected, the elastic potential energy is converted into kinetic energy (KE):
\[ EPE = KE = \frac{1}{2} m v^2 \]
Setting the equations equal gives:
\[ 1 \, J = \frac{1}{2} \cdot (5 \times 10^{-3} \, kg) \cdot v^2 \]

### Step 7: Solve for Velocity (v)
Rearranging the equation to solve for v:
\[ 1 = \frac{1}{2} \cdot (5 \times 10^{-3}) \cdot v^2 \]
\[ 2 = (5 \times 10^{-3}) \cdot v^2 \]
\[ v^2 = \frac{2}{5 \times 10^{-3}} = 400 \]
\[ v = \sqrt{400} = 20 \, m/s \]

### Final Answer
The velocity of projection of the missile is \(20 \, m/s\).

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