Question

The rubber cord of a catapult has a cross-sectional area $1mm$${}^2$ and total unstretched length $10cm$. It is stretched to $12cm$ and then released to project a missile of mass $5gm$. Taking '$Y$' for rubber 5 x 10${}^8$ N/m${}^2$, the velocity of projection is:

• A. $10m/s$
• B. $15m/s$
• C. $20m/s$
• D. $25m/s$

Answer 1

The correct option is C $20m/s$

$F=\frac{YA\Delta l}{l}$

$=\frac{5\times {10}^8\times {10}^{-}6\times 2\times {10}^{-}2}{10\times {10}^{-}2}$

$energy=\frac{1}{2}\times F\times \Delta l=\frac{1}{2}\times \frac{5\times {10}^8\times {10}^{-}6\times 2\times {10}^{-}2}{10\times {10}^{-}2}\times 2\times {10}^{-}2=\frac{1}{2}m{v}^2$

So, $\frac{5\times {10}^8\times {10}^{-}6\times 2\times {10}^{-}2\times 2\times {10}^{-}2}{10\times {10}^{-}2\times 5\times {10}^{-}3}=v^2$

$v=20m/s$