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 \[{{x}_{0}}\].                    [JEE ONLINE 12-05-2012]

The spring constant k will be

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 :

[d]

Initial momentum of the system block \[=m{{v}_{0}}.\]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

\[\frac{1}{2}mv_{0}^{2}=\frac{1}{2}(3m){{\left( \frac{{{v}_{0}}}{3} \right)}^{2}}+\frac{1}{2}kx_{0}^{2}\]

\[\Rightarrow \]\[\frac{1}{2}mv_{0}^{2}=\frac{1}{2}mv_{0}^{2}+\frac{1}{2}kx_{0}^{2}\]

\[\Rightarrow \]\[\frac{1}{2}kx_{0}^{2}=\frac{1}{2}mv_{0}^{2}-\frac{1}{6}mv_{0}^{2}=\frac{mv_{0}^{2}}{3}\]

\[\therefore \]\[k=\frac{2}{3}m{{\left( \frac{{{v}_{0}}}{{{x}_{0}}} \right)}^{2}}\]