The rubber cord of a catapult has cross-section area 1 `mm^(2)` and a total unstretched length 10 cm. It is stretched to 12 cm and then released to project a partical of mass 5 g. Calculate the velocity of projection (Y for rubber is `5xx10^(8)N//m^(2)`)

Options

A. 5 `ms^(-1)`
B. 10 `ms^(-1)`
C. 20 `ms^(-1)`
D. 40 `ms^(-1)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a rubber cord of a catapult that is stretched from its original length of 10 cm to 12 cm. We need to calculate the velocity of a particle of mass 5 g that is projected when the cord is released.

### Step 2: Convert Units
- Cross-sectional area (A) = 1 mm² = \(1 \times 10^{-6} \, m^2\)
- Unstretched length (L₀) = 10 cm = \(0.1 \, m\)
- Stretched length (L) = 12 cm = \(0.12 \, m\)
- Change in length (ΔL) = L - L₀ = \(0.12 \, m - 0.1 \, m = 0.02 \, m\)
- Mass (m) = 5 g = \(0.005 \, kg\)

### Step 3: Calculate the Force (F)
Using the formula for force in terms of Young's modulus (Y):
\[ F = Y \cdot A \cdot \frac{\Delta L}{L_0} \]
Where:
- Y = \(5 \times 10^8 \, N/m^2\)
- A = \(1 \times 10^{-6} \, m^2\)
- ΔL = \(0.02 \, m\)
- L₀ = \(0.1 \, m\)

Substituting the values:
\[ F = (5 \times 10^8) \cdot (1 \times 10^{-6}) \cdot \frac{0.02}{0.1} \]
\[ F = (5 \times 10^8) \cdot (1 \times 10^{-6}) \cdot 0.2 \]
\[ F = (5 \times 0.2) \cdot 10^2 = 1 \times 10^2 \, N = 100 \, N \]

### Step 4: Calculate the Elastic Potential Energy (EPE)
The elastic potential energy stored in the stretched cord is given by:
\[ EPE = \frac{1}{2} F \Delta L \]
Substituting the values:
\[ EPE = \frac{1}{2} \cdot 100 \cdot 0.02 \]
\[ EPE = 1 \, J \]

### Step 5: Set EPE equal to Kinetic Energy (KE)
When the cord is released, all the elastic potential energy converts into kinetic energy of the particle:
\[ EPE = KE \]
\[ 1 = \frac{1}{2} m v^2 \]
Where \(v\) is the velocity we need to find. Rearranging the equation to solve for \(v\):
\[ v^2 = \frac{2 \cdot EPE}{m} \]
Substituting the values:
\[ v^2 = \frac{2 \cdot 1}{0.005} \]
\[ v^2 = \frac{2}{0.005} = 400 \]
\[ v = \sqrt{400} = 20 \, m/s \]

### Step 6: Final Answer
The velocity of projection of the particle is \(20 \, m/s\).

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