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                [EAMCET 2003]

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 Þ \[V=\frac{v}{2}\]                 By the law of conservation of energy                 K.E. of block C = K.E. of system + P.E. of system                 \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m)\,{{V}^{2}}+\frac{1}{2}k{{x}^{2}}\]                 Þ \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}(2m)\ {{\left( \frac{v}{2} \right)}^{2}}+\frac{1}{2}k{{x}^{2}}\]                 Þ \[k{{x}^{2}}=\frac{1}{2}m{{v}^{2}}\]                 Þ \[x=v\sqrt{\frac{m}{2k}}\]