A rubber cord catapult has cross-sectional area `25 mm^(2)` and initial length of rubber cord is `10 cm`. It is stretched to `5 cm`. And then released to protect a missle of mass `5 gm` Taking `Y_("nibber")=5xx10^(7)N//m^(2)` velocity of projected missle is

Options

A. `20 ms^(-1)`
B. `100 ms^(-1)`
C. `250ms^(-1)`
D. `200 ms^(-1)`

Answer

To solve the problem step by step, we will calculate the velocity of the missile projected from the rubber cord catapult using the given data.

### Given Data:
- Cross-sectional area of the rubber cord, \( A = 25 \, \text{mm}^2 = 25 \times 10^{-6} \, \text{m}^2 \)
- Initial length of the rubber cord, \( L = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} \)
- Stretch in the rubber cord, \( \Delta L = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)
- Mass of the missile, \( m = 5 \, \text{g} = 5 \times 10^{-3} \, \text{kg} \)
- Young's modulus of rubber, \( Y = 5 \times 10^{7} \, \text{N/m}^2 \)

### Step 1: Calculate the spring constant \( k \)
The spring constant \( k \) can be calculated using the formula:
\[ k = \frac{Y \cdot A}{L} \]

Substituting the values:
\[ k = \frac{(5 \times 10^{7} \, \text{N/m}^2) \cdot (25 \times 10^{-6} \, \text{m}^2)}{(10 \times 10^{-2} \, \text{m})} \]

Calculating:
\[ k = \frac{(5 \times 10^{7}) \cdot (25 \times 10^{-6})}{(10 \times 10^{-2})} \]
\[ k = \frac{(125 \times 10^{1})}{(10 \times 10^{-2})} = \frac{1250}{0.1} = 12500 \, \text{N/m} \]

### Step 2: Calculate the potential energy stored in the rubber cord
The potential energy \( PE \) stored in the rubber cord when it is stretched is given by:
\[ PE = \frac{1}{2} k (\Delta L)^2 \]

Substituting the values:
\[ PE = \frac{1}{2} (12500) \cdot (5 \times 10^{-2})^2 \]
\[ PE = \frac{1}{2} (12500) \cdot (25 \times 10^{-4}) = \frac{1}{2} (12500 \cdot 0.0025) = \frac{12500 \cdot 2.5}{1000} = 31.25 \, \text{J} \]

### Step 3: Equate potential energy to kinetic energy
The potential energy stored in the rubber cord will be converted into kinetic energy \( KE \) of the missile:
\[ KE = \frac{1}{2} m v^2 \]

Setting \( PE = KE \):
\[ 31.25 = \frac{1}{2} (5 \times 10^{-3}) v^2 \]

### Step 4: Solve for the velocity \( v \)
Rearranging the equation:
\[ v^2 = \frac{31.25 \cdot 2}{5 \times 10^{-3}} \]
\[ v^2 = \frac{62.5}{5 \times 10^{-3}} = \frac{62.5}{0.005} = 12500 \]
\[ v = \sqrt{12500} \approx 111.8 \, \text{m/s} \]

### Final Answer
The velocity of the projected missile is approximately \( v \approx 111.8 \, \text{m/s} \).