The bodies A and B of masses m and 2m respectively are put on a smooth floor. They are connected by a spring . A third body C of mass m moves with a velocity `v_0` along the line joining A and B and collides elastically with A as shown in fig1.18.At a certain instant of time `t_0` after the collision, it is found that the instantaneous velocities of A and B are the same. Further,at this instant the compression of the spring is found to be `x_0`. FInd out the force constant of the spring.

Explanation

According to the principle of conservation of energy,
`1/2mc_(0^2)=1/2mv^2+1/2times2mv^2+1/2kx_(0^2)`
or,`mv_(0^2)=3mv^2+kx_(0^2)`
[k=force constant of the spring]
or, `mv_(0^2)=3m(v_0/3)^2+kx_(0^2)or,k=2/3(mv_(0^2))/x_(0^2)`