Physics Elasticity Question 159
Question: A rubber cord catapult has cross-sectional area $ 25mm^{2} $ and initial length of rubber cord is $ 10cm. $ It is stretched to $ 5,cm. $ and then released to project a missile of mass $ 5gm. $ Taking $ Y _{rubber}=5\times 10^{8}N/m^{2} $ velocity of projected missile is
[CPMT 2002]
Options:
A) $ 20,m{{s}^{-1}} $
B)$ 100,m{{s}^{-1}} $
C) $ 250,m{{s}^{-1}} $
D)$ 200,m{{s}^{-1}} $
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Answer:
Correct Answer: C
Solution:
Potential energy stored in the rubber cord catapult will be converted into kinetic energy of mass. $ \frac{1}{2}mv^{2}=\frac{1}{2}\frac{YAl^{2}}{L} $
Therefore $ v=\sqrt{\frac{YAl^{2}}{mL}} $
$ =\sqrt{\frac{5\times 10^{8}\times 25\times {{10}^{-6}}\times {{(5\times {{10}^{-2}})}^{2}}}{5\times {{10}^{-3}}\times 10\times {{10}^{-2}}}}=250\ m/s $
BETA
