question_answer

Two identical blocks A and B, each of mass m resting on smooth floor are connected by a light spring of natural length L and spring constant k with the spring at its natural length. A third identical block C (mass m) moving with a speed v along the line joining A and B collides with A. The maximum compression in the spring is

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 \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m){{V}^{2}}+\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m){{\left( \frac{v}{2} \right)}^{2}}+\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \] \[k{{x}^{2}}=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow x=v\sqrt{\frac{m}{2k}}.\]