A) \[v\sqrt{\frac{m}{2k}}\]
B) \[m\sqrt{\frac{v}{2k}}\]
C) \[\sqrt{\frac{mv}{k}}\]
D) \[\frac{mv}{2k}\]
Correct Answer: A
Solution :
Initial momentum of the system (block C) = mv 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 maximum. By the law of conservation of linear momentum \[mv=(m+m)V\Rightarrow V=\frac{v}{2}\] By the law of conservation of energy KE of block C = KE of system + PE of system \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m){{V}^{2}}+\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m){{\left( \frac{v}{2} \right)}^{2}}+\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \] \[k{{x}^{2}}=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow x=v\sqrt{\frac{m}{2k}}.\]You need to login to perform this action.
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