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A catapult uses a rubber cord 42 cm long and 6 mm in diameter (negligible mass). A 0.02 kg stone is placed on it and the cord is stretched by 20 cm with a constant applied force. On release the stone leaves with a speed of 20 m/s. Ignoring any change in cross-sectional area on stretching, the Young's modulus of the rubber is closest to:
- 10⁴ N m⁻²
- 10⁸ N m⁻²
- 10⁶ N m⁻²
- 10³ N m⁻²
Correct answer: 10⁶ N m⁻²
Solution
Equating elastic PE to KE gives Y = m v² L /(A x²); plugging in numbers gives about 3*10⁶, closest to 10⁶ N/m².
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