Question

The rubber cord of a catapult has cross-section area $2m{m}^2$ and a total unstretched length $15cm$. It is stretched to $18cm$ and then released to project a particle of mass $3g$. Calculate the velocity of projection if $Y$ for rubber is $8\times {10}^{8}N/m^2$.

• A. $56.5{ms}^{-1}$
• B. $10{ms}^{-1}$
• C. $20{ms}^{-1}$
• D. $40{ms}^{-1}$

Answer 1

The correct option is A $56.5{ms}^{-1}$

Given, $Y=8\times {10}^{8}N/m^2$

$A=2{mm}^2=\frac{2}{{\left(1000\right)}^2}{m}^2$

$L=15cm=\frac{15}{100}m$

$l=(18-15)=3cm=\frac{3}{100}m$

Let, velocity will be V m/s.

$\therefore$ Stored potential energy = Kinetic energy

or, $\frac{YA{l}^2}{2L}=\frac{1}{2}m{v}^2$

or, $v^2=\frac{YA{l}^2}{mL}$

or, $v^2=\frac{(8\times {10}^8)\times 2\times 3^2\times 100\times 1000}{{\left(1000\right)}^2\times {\left(100\right)}^2\times (3\times 10)\times 15}$

$\therefore v=56.5m/s$