Question

A rubber band catapult has initial length \(2 \, \text{cm}\) and cross-sectional area \(5 \, \text{mm}^2\). It is stretched to \(2 \, \text{cm}\) and then released to project a stone of mass \(20 \, \text{g}\). The velocity of projected stone is \((Y=5\times 10^8\,\text{N m}^{-2})\):

• A. \(20\,\text{m s}^{-1}\)
• B. \(50\,\text{m s}^{-1}\)
• C. \(100\,\text{m s}^{-1}\)
• D. \(250\,\text{m s}^{-1}\)

Hint:

For a stretched wire or rubber band, elastic potential energy is given by \[ U=\frac{1}{2}\frac{YA(\Delta L)^2}{L} \] and this energy converts into kinetic energy when released.

Answer 1

The Correct Option is B

Step 1: Calculate elastic potential energy stored in the stretched rubber band.
\( L = 0.02\,\text{m} \), \( \Delta L = 0.02\,\text{m} \), \( A = 5\times10^{-6}\,\text{m}^2 \), \( Y = 5\times10^8\,\text{N m}^{-2} \). \[ U = \frac{YA(\Delta L)^2}{2L} = \frac{5\times10^8\times5\times10^{-6}\times(0.02)^2}{2\times0.02} = 25\,\text{J} \]

Step 2: Equate to kinetic energy of the stone.
\( \frac{1}{2}mv^2 = 25 \Rightarrow v = \sqrt{\frac{50}{0.02}} = 50\,\text{m s}^{-1} \). \[ \boxed{50\,\text{m s}^{-1}} \]