A rubber cord catapult has cross-sectional area \(25 \mathrm{~mm}^{2}\) and initial length of cord is \(10 \mathrm{~cm} .\) It is stretched to $5 \mathrm{~cm}\( and then released to project a missile of mass \)5 \mathrm{gm}$. Taking $\mathrm{Y}_{\text {rubber }}=5 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\( velocity of projected missile is \)\ldots \ldots .$ (A) \(20(\mathrm{~m} / \mathrm{s})\) (B) \(100(\mathrm{~m} / \mathrm{s})\) (C) \(250(\mathrm{~m} / \mathrm{s})\) (D) \(200(\mathrm{~m} / \mathrm{s})\)

Short Answer

Expert verified
The velocity of the projected missile is \(20 \mathrm{~m} / \mathrm{s}\) (A).

Step by step solution

01

Calculating the stress in the rubber cord

The stress in the rubber cord can be calculated using Hooke's law, which states that the stress is directly proportional to the strain: Stress \(= Y \times \) Strain where Y is Young's modulus. First, let's find the strain in the rubber cord when it is stretched by 5 cm. Strain \(= \frac{\Delta L}{L}\), where \(\Delta L\) is the change in length, and L is the initial length. Strain = \(\frac{5 \mathrm{~cm}}{10 \mathrm{~cm}} = \frac{1}{2}\) Now, we can find the stress in the rubber cord: Stress \(= Y \times\) Strain = \(5 \times 10^8 \times \frac{1}{2}\) N/m².
02

Finding the Force applied by the rubber cord

We can find the force applied by the rubber cord using the formula: Force = Stress × Cross-sectional Area Cross-sectional Area = \(25 \mathrm{~mm}^{2}\) = \(25 \times 10^{-6} \mathrm{m}^{2}\) Now, let's find the force: Force = \(5 \times 10^8 \times \frac{1}{2} \times 25 \times 10^{-6}\) N.
03

Calculating the work done by the rubber cord

To calculate the work done by the rubber cord, we need to find the potential energy stored in it when it is stretched. The formula for the potential energy stored in a stretched linear elastic material is: Potential Energy \(= \frac{1}{2} \times\) Force × Change in Length Substituting the values we derived earlier, we can find the potential energy: Potential Energy \(= \frac{1}{2} \times (5 \times 10^8 \times \frac{1}{2} \times 25 \times 10^{-6}) \times 5 \times 10^{-2}\) J.
04

Calculating the velocity of the projected missile

According to the conservation of energy principle, the potential energy stored in the stretched rubber cord is completely transferred to the kinetic energy of the missile. So, the kinetic energy of the missile can be calculated as: Kinetic Energy \(= \frac{1}{2} \times\) Mass × Velocity². We know the mass of the missile is 5 gm = \(5 \times 10^{-3}\) kg. Now, equating the potential energy and kinetic energy: \(\frac{1}{2} \times (5 \times 10^8 \times \frac{1}{2} \times 25 \times 10^{-6}) \times 5 \times 10^{-2} = \frac{1}{2} \times 5 \times 10^{-3} \times\) Velocity² Solving for the velocity: Velocity = 20 m/s So, the velocity of the projected missile is 20 m/s, which corresponds to option (A).

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