Question

A catapult consists of two parallel rubber strings, each of length $10\text{cm}$ and cross-sectional area $10{\text{mm}}^2$ . When stretched by $5\text{cm}$, it can throw a stone of mass $100\text{gm}$ to a vertical height of $25\text{m}$. Determine the Young's modulus of elasticity of rubber.

• A. ${10}^7{\text{N/m}}^2$
• B. ${10}^8{\text{N/m}}^2$
• C. ${10}^9{\text{N/m}}^2$
• D. ${10}^{10}{\text{N/m}}^2$

Answer 1

The correct option is B ${10}^8{\text{N/m}}^2$

Given, for the catapult
$L=10\text{cm},\Delta L=5\text{cm},A=10{\text{mm}}^2$
For the stone, $m=100\text{gm},h=25\text{m}$

A stretched catapult has elastic potential energy stored in it (Strain energy stored in both the rubber strings)
$U=2\times (\frac{1}{2}\times \text{Stress}\times \text{Strain}\times \text{Volume})$
$=2\times (\frac{1}{2}\times (\text{Strain}\times Y)\times \text{Strain}\times \text{Volume})$
$\Rightarrow U=2\times (\frac{1}{2}\times Y\times {\left(\frac{\Delta L}{L}\right)}^2\times (A\times L\left)\right)$
$\Rightarrow U=2\times \left(\frac{1}{2}\frac{YA\Delta L^2}{L}\right)$

This energy, when imparted to the stone, it flies off to a height $25\text{m}$. Energy possessed by the stone $=mgh$.
Now,
$U=mgh$
$\Rightarrow \frac{YA\Delta L^2}{L}=mgh$
$\Rightarrow Y=\frac{mghL}{A\Delta L^2}=\frac{\left({10}^{-1}\text{kg}\right)\times 10\left({\text{m/s}}^2\right)\times \left(25\text{m}\right)\times (10\times {10}^{-2}\text{m})}{(10\times {10}^{-6}{\text{m}}^2)\times (5\times {10}^{-2}\text{m}{)}^2}$
$\therefore Y={10}^8{\text{N/m}}^2$