Question

A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section $10^{-6}$ m$^2$ stretched by an amount 0.04 m. The velocity of the projected stone is _________ m/s. (Young's modulus of rubber = $0.5 \times 10^9$ N/m$^2$)

Hint:

The key to this problem is the principle of conservation of energy. The potential energy stored in the elastic material is fully converted to the kinetic energy of the projectile. Ensure all units are in the SI system before calculation.

Answer 1

Correct Answer: 20

The potential energy stored in the stretched rubber catapult is converted into the kinetic energy of the stone.
The formula for the elastic potential energy (U) stored in a stretched material is:
$U = \frac{1}{2} \times \frac{YA}{L} (\Delta L)^2$
where Y is Young's modulus, A is the cross-sectional area, L is the original length, and $\Delta L$ is the extension.
Given values:
$Y = 0.5 \times 10^9$ N/m$^2$
$A = 10^{-6}$ m$^2$
$L = 0.1$ m
$\Delta L = 0.04$ m
Substitute these values to find the stored energy U:
$U = \frac{1}{2} \times \frac{(0.5 \times 10^9) \times (10^{-6})}{0.1} \times (0.04)^2$
$U = \frac{1}{2} \times (5 \times 10^3) \times (1.6 \times 10^{-3})$
$U = \frac{1}{2} \times 5 \times 1.6 = 4$ J.
This stored energy is converted to the kinetic energy (KE) of the stone.
$KE = \frac{1}{2}mv^2$
Given mass of the stone, $m = 20$ g = $0.02$ kg.
$4 = \frac{1}{2} \times (0.02) \times v^2$
$4 = 0.01 \times v^2$
$v^2 = \frac{4}{0.01} = 400$
$v = \sqrt{400} = 20$ m/s.