A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is `2.0Nm^(-1)` and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of `2.0ms^(-1)` collides elastically with the first block. The collision is such that the 2.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is

Answer

2.09

Explanation

`2v_(2)-1v_(1)=1(2)` (Momentum conservaation)
Also `(v_(1)+v_(2))1(2)` (Cofficient of restitution)
`3v_(2)=4 implies v_(2)=(4)/(3)` implies `v_(1)=(2)/(3)`
As 2kg is attached to a spring,
Time taken by 2 kg too return to its initial position is `t=(T)/(2)=pisqrt((m)/(k))=pisqrt((2)/(2))=pi sec`.
Distance move by 1 kg in this time `v_(1)t=(2pi)/(3)`