Question

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

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

Hint:

For stretched elastic materials, use potential energy stored as \( \frac{1}{2}YAx^2/L \). Equate to kinetic energy to find launch velocity.

Answer 1

The Correct Option is D

Step 1: Use elastic potential energy: \[ U = \frac{1}{2} Y \cdot A \cdot \frac{x^2}{L} \] Where: - \( Y = 5 \times 10^8\,\text{Nm}^{-2} \) - \( A = 5\,\text{mm}^2 = 5 \times 10^{-6}\,\text{m}^2 \) - \( x = 2\,\text{cm} = 0.02\,\text{m} \) - \( L = 0.02\,\text{m} \) \[ U = \frac{1}{2} \cdot 5 \times 10^8 \cdot 5 \times 10^{-6} \cdot \frac{(0.02)^2}{0.02} = 2.5\,\text{J} \] Step 2: Equating to kinetic energy: \[ \frac{1}{2} mv^2 = U \Rightarrow \frac{1}{2} \cdot 0.02 \cdot v^2 = 2.5 \Rightarrow v^2 = 250 \Rightarrow v = \boxed{50\,\text{ms}^{-1}} \]