Two blocks `A` and `B` of masses in and `2m` respectively placed on a smooth floor are connected by a spring. A third body `C` of mass `m` moves with velocity `v_(0)` along the line joining `A` and `B` and collides elastically with `A`. At a certain instant of time after collision it is found that the instantaneous velocities of `A` and `B` are same then:

Answer

`((v_0)/(3) , (2 mv_0^2)/(3x_0^2))`

Explanation

As the collision between C and A is elastic and their masses are equal and A was initially at rest, therefore the result of collision will be that C will come to rest and A will initially start moving with a velocity `v_0`. But since A is connected to B with a spring, the spring will get compressed.
At `t = t_0` the velocities of A and B become the same. Applying conservation of mechanical energy.
`1/2 mv_0^2 = 1/2 m v^('2) + 1/2 2 mv^('2) + 1/2 kx_0^2`
Where `x_0` is the compression in the string at `t = t_0`
`:. v_0^2 = 3v^('2) + k/m x_0^2 " ".....(i)`
Applying conservation of linear momentum, we get `m v_0 = m v' + 2 mv'`
`:. v' = (v_0)/3 " ".....(ii)`
From (i) and (ii) `v_0^2 = 3 xx (v_0^2)/(9) + (k)/(m) x_0^2 implies k = (2 mv_0^2)/(3 x_0^2)`