The rubber cord of catapult has a cross-sectional area 1 ${\mathrm{mm}}^2$ and total unstretched length 10.0 cm. It is stretched to 12.0 cm and then released to project a missile of mass 5.0 g. Taking young's modulus Y for rubber as $5.0\times {10}^8 {\mathrm{Nm}}^2$. Total elastic energy of catapult is converted into kinetic energy of missile without any heat loss. The velocity of projection is

 Total elastic energy of catapult is converted into kinetic energy of missile without any heat loss.

Elastic energy = $\frac{1}{2}\times \mathrm{load}\times \mathrm{extension}$

Here extension,

$∆\mathrm{L} = 12.0 - 10.0 = 2.0 \mathrm{cm} = 2.0 \times {10}^{-2} \mathrm{m}$

From relation $\mathrm{Y} = \frac{\mathrm{FL}}{\mathrm{A}∆\mathrm{L}}$

Load, $\mathrm{F} = \frac{\mathrm{YA}∆\mathrm{L}}{\mathrm{L}}$

              $= \frac{5.0\times {10}^8\times 1\times {10}^{-6}\times 2.0\times {10}^{-2}}{10.0 \times {10}^{-2}} = 100 \mathrm{N}$

If v is velocity of projection, we have

Elastic energy of catapult = kinetic energy of missile

i.e., $\frac{1}{2}\times \mathrm{load}\times \mathrm{extension} = \frac{1}{2}{\mathrm{mv}}^2$

     $\frac{1}{2}\times 100\times 2.0\times {10}^{-2} = \frac{1}{2}\times (5.0\times {10}^{-3}){\mathrm{v}}^2$

This gives ${\mathrm{v}}^2 = \frac{100\times {10}^{-2}\times 2}{5\times {10}^{-3}} = 400$

v = $\sqrt{400} = 20 \mathrm{m}/\mathrm{s}$