A) \[\frac{m{{\text{v}}_{0}}^{2}}{{{x}_{0}}^{2}}\]
B) \[\frac{m{{\text{v}}_{0}}^{2}}{2{{x}_{0}}^{2}}\]
C) \[\frac{3}{2}\frac{m{{\text{v}}_{0}}^{2}}{{{x}_{0}}^{2}}\]
D) \[\frac{2}{3}\frac{m{{\text{v}}_{0}}^{2}}{{{x}_{0}}^{2}}\]
Correct Answer: D
Solution :
When C strikes A \[\frac{1}{2}m{{v}_{0}}^{2}=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}k{{x}_{0}}^{2}\] (v=? = velocity of A) \[k{{x}_{0}}^{2}=m({{v}_{0}}^{2}-v{{'}^{2}})\] ..?..(i) \[\frac{1}{2}2mv{{'}^{2}}=\frac{1}{2}k{{x}_{0}}^{2}\] (When A and B Block attains K.E.) \[\therefore \,\,\,\frac{1}{2}2mv{{'}^{2}}=\frac{1}{2}k{{x}_{0}}^{2}\] ......... (ii) From (i) and (ii), \[k{{x}_{0}}^{2}=m{{v}_{0}}^{2}-mv{{'}^{2}}=m{{v}_{0}}^{2}-\frac{k}{2}{{x}_{0}}^{2}\] \[\Rightarrow \,\,\,k{{x}_{0}}^{2}+\frac{k}{2}{{x}_{0}}^{2}=m{{v}_{0}}^{2}\] \[\frac{3}{2}k{{x}_{0}}^{2}=m{{v}_{0}}^{2}\] \[\therefore \,\,k=\frac{2}{3}m\frac{{{v}_{0}}^{2}}{{{x}_{0}}^{2}}\]You need to login to perform this action.
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