A) 0.01 m
B) 0.02m
C) 0.05m
D) 0.03m
Correct Answer: C
Solution :
As the block\[A\]moves with velocity\[0.15\,\,m{{s}^{-1}}\], it compresses the spring which pushes\[B\]towards right.\[A\]goes on compressing the spring till the velocity acquired by\[B\]becomes equal to the velocity of\[A\],\[ie\],\[0.15\,m{{s}^{-1}}\]. Let the velocity be\[v\]. Now, spring is in a state of maximum compression. Let\[x\]be the maximum compression at this stage. According to the law of conservation of linear momentum, we get \[{{m}_{A}}u=({{m}_{A}}+{{m}_{B}})v\] or \[v=\frac{{{m}_{A}}u}{{{m}_{A}}+{{m}_{B}}}\] \[=\frac{2\times 0.15}{2+3}=0.06\,\,m{{s}^{-1}}\] According to the law of conservation of energy. \[\frac{1}{2}{{m}_{A}}{{u}^{2}}=\frac{1}{2}({{m}_{A}}+{{m}_{B}}){{v}^{2}}=\frac{1}{2}k{{x}^{2}}\] \[\frac{1}{2}{{m}_{A}}{{u}^{2}}-\frac{1}{2}({{m}_{A}}+{{m}_{B}}){{v}^{2}}=\frac{1}{2}k{{x}^{2}}\] \[\frac{1}{2}\times 2\times {{(0.15)}^{2}}-\frac{1}{2}(2+3){{(0.06)}^{2}}=\frac{1}{2}k{{x}^{2}}\] \[0.0225-0.009=\frac{1}{2}k{{x}^{2}}\,\,or\,\,0.0135=\frac{1}{2}k{{x}^{2}}\] or \[x=\sqrt{\frac{0.027}{k}}=\sqrt{\frac{0.027}{10.8}}=0.05m\]You need to login to perform this action.
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