Question

A rubber cord of cross-sectional area 5 mm²and length 10 cm is stretched to 15 cm and then released to project a stone of mass 5 g. Find the velocity of projection. (Young modulus for rubber = 5 × 10⁸N/m²)

Answer 1

Equate elastic potential energy of the cord with kinetic energy:

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{1}{2}\times stress\times strain\times volume \\ m{v}^2=stress\times strain\times volume \\ m{v}^2={\left(strain\right)}^2\times Y\times volume \\ m{v}^2={\left(\frac{△L}{L}\right)}^2\times Y\times AL \\ v=\sqrt{\frac{{\left(△L\right)}^{2}YA}{Lm}} \end{gathered}$$

Calculation of velocity:

$$\begin{gathered} v=\sqrt{\frac{{\left(0.05\mathrm{m}\right)}^2\left(5\times {10}^8\mathrm{N}/{\mathrm{m}}^2\right)\left(5\times {10}^{-6}{\mathrm{m}}^2\right)}{\left(0.1\mathrm{m}\right)\left(5\times {10}^{-3}\mathrm{kg}\right)}} \\ =111.8\mathrm{m}/\mathrm{s} \end{gathered}$$