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A spring - block system is resting on ...

A spring - block system is resting on a frictionless floor as shown in the figure . The spring constant is `2.0 Nm^(-1)` and the mass of the block is 2.0 kg . Ignore the mass of the spring . Initially the spring is in normal length . Another block of mass 1.0 kg moving with a speed of `2.0 " m "s^(1-)` collides elastically with the first block . The collision elastically with thef irst block . The collision is such that the 2.0 kg block does not hot the wall . the disatnce , in metres between the two blocks when the spring retrns to its normal state for the first time after the collision is _______ .

Text Solution

Verified by Experts

The correct Answer is:
B
Just before Collision

Conservation of linear momentum
Just After Collision

`2V_(1) -V_(2) = 2 " " ….(ii)`
On adding the above two equations
`3V_(1)=4 rArr V_(1) = 4/3 `m/s
`V_(2) = 2 - 4/3 = 2/3 ` m/s
Seperation between the bodies will be equal to the distance travelled by 1 kg block in half oscillation of spring block . Time taken by 2 kg block to complete half oscillation to come back to initial position is (T is time period of spring )
`Deltat=T/2 = (2pi)/2 sqrt(m/K) = (2pi)/2 sqrt(2/2) = pi `sec .
Distance ` = piV_(2) = (22/7) (2/3) m = 44/21 m = 2.09 m `
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