Animated Solution for Physics - Properties of Solids and Liquids: A stone of mass $20\text{ g}$ is projected from a rubber catapult of length $0.1\text{ m}$ and area of cross-section $10^{-6}{\text{ m}}^2$ stretched by an amount $0.04\text{ m}$. The velocity of the projected stone is ............ $\text{m/s}$. (Take, Young's modulus of rubber $=0.5\times 10^9{\text{ N/m}}^2$)

Imagine you are holding a catapult. You place a small stone in the pouch, pull the rubber band back, and hold it there, feeling the tension. What you are actually doing is doing mechanical work against the restoring force of the rubber band. According to the law of conservation of energy, this work doesn't just disappear; it gets stored within the molecular structure of the rubber as Elastic Potential Energy.

When you let go, the rubber band snaps back to its original shape, and all that stored potential energy is instantaneously transferred to the stone, manifesting as Kinetic Energy. This beautiful interplay of energies is the core principle behind this problem.

To find out exactly how fast the stone will fly, we first need to calculate how much energy is stored in the stretched rubber band. The formula for the elastic potential energy $U$ stored in a stretched material is given by:

We know from Hooke's Law that $\text{Stress}=Y\times \text{Strain}$, where $Y$ is the Young's modulus of the material. Substituting this into our energy equation, we get a much more useful form:

Let's break down the components based on the given data: Young's Modulus ($Y$): $0.5\times 10^9{\text{ N/m}}^2$ Strain: This is the ratio of change in length to the original length, $\frac{\Delta l}{l}=\frac{0.04}{0.1}=0.4$ Volume:* The volume of the rubber band is its cross-sectional area multiplied by its original length, $A\times l=10^{-6}\times 0.1=10^{-7}{\text{ m}}^3$

Now, we apply the conservation of energy. The elastic potential energy $U$ becomes the kinetic energy $K$ of the stone. The formula for kinetic energy is $K=\frac{1}{2}m{v}^2$. Equating the two, we get our master equation:

Notice how the $\frac{1}{2}$ on both sides elegantly cancels out. Before we plug in the numbers, there is a classic trap we must avoid: units. The mass of the stone is given as $20\text{ g}$. In physics, we must always work in standard SI units (kilograms, meters, seconds) to ensure our final answer is correct. Therefore, we must convert the mass: $m=20\text{ g}=0.02\text{ kg}$.

Let's substitute all our carefully prepared values into the master equation:

Now, we execute the arithmetic step-by-step. First, square the strain:

Next, group the powers of 10 together ($10^9\times 10^{-7}=10^2=100$) and multiply the decimals ($0.5\times 0.16=0.08$):

Finally, isolate $v^2$ and solve for the velocity:

The stone leaves the catapult at a blazing speed of $20\text{ m/s}$. This problem perfectly demonstrates how macroscopic kinetic phenomena are directly governed by the microscopic elastic properties of materials!