question_answer

Two bodies A and B of mass m and 2m respectively are placed on a smooth floor. They are connected by a spring of negligible mass. A' third body C of mass m is placed on the floor. The body C moves with a velocity \[{{v}_{0}}\] along the line joining A and B and collides elastically with A. At a certain time after the collision it is found that the instantaneous velocities of A and B are same and the compression of the spring is \[{{x}_{0}}\]. The spring constant k will be   JEE Main Online Paper (Held On 12 May 2012)

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 \[\frac{1}{2}mv_{0}^{2}=\frac{1}{2}(3m){{\left( \frac{{{v}_{0}}}{3} \right)}^{2}}+\frac{1}{2}kx_{0}^{2}\] \[\Rightarrow \]\[\frac{1}{2}mv_{0}^{2}=\frac{1}{2}mv_{0}^{2}+\frac{1}{2}kx_{0}^{2}\] \[\Rightarrow \]\[\frac{1}{2}kx_{0}^{2}=\frac{1}{2}mv_{0}^{2}-\frac{1}{6}mv_{0}^{2}=\frac{mv_{0}^{2}}{3}\] \[\therefore \]\[k=\frac{2}{3}m{{\left( \frac{{{v}_{0}}}{{{x}_{0}}} \right)}^{2}}\]