A) 0.01 m
B) 0.02 m
C) 0.05 m
D) 0.03 m
Correct Answer: C
Solution :
As the block A moves with velocity \[0.15m{{s}^{-1}},\]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, \[0.15m{{s}^{-1}},\] Let this 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\[x=\sqrt{\frac{0.027}{k}}k=\sqrt{\frac{0.027}{10.8}}=0.05m\]You need to login to perform this action.
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