A) \[m\frac{v_{0}^{2}}{x_{0}^{2}}\]
B) \[m\frac{{{v}_{0}}}{2{{x}_{0}}}\]
C) \[2m\frac{{{v}_{0}}}{{{x}_{0}}}\]
D) \[\frac{2}{3}m{{\left( \frac{{{v}_{0}}}{{{x}_{0}}} \right)}^{2}}\]
Correct Answer: D
Solution :
Initial momentum of the system block \[(C)=m{{v}_{0}}.\]After striking with A, the block C comes to rest and now both block A and B moves with velocity v when compression in spring is \[{{x}_{0}}.\] By the law of conservation of linear momentum\[m{{v}_{0}}=(m+2m)v\Rightarrow v=\frac{{{v}_{0}}}{3}\] By the law of conservation of energy K.E. of block C= K.E. of system + P.E. of system \[\frac{1}{2}mv_{0}^{2}=\frac{1}{2}(3m){{\left( \frac{{{v}_{0}}}{3} \right)}^{2}}+\frac{1}{2}kx_{0}^{2}\] \[\Rightarrow \]\[\frac{1}{2}mv_{0}^{2}=\frac{1}{2}mv_{0}^{2}+\frac{1}{2}kx_{0}^{2}\] \[\Rightarrow \]\[\frac{1}{2}kx_{0}^{2}=\frac{1}{2}mv_{0}^{2}-\frac{1}{6}mv_{0}^{2}=\frac{mv_{0}^{2}}{3}\] \[\therefore \]\[k=\frac{2}{3}m{{\left( \frac{{{v}_{0}}}{{{x}_{0}}} \right)}^{2}}\]You need to login to perform this action.
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