Q46 / 54 · JEE Advanced 2018 (Paper 1) · Physics

A spring - block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 N m-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 1.0kg moving with a speed of 2.0 m s-1 collides elastically with the first block. The collision is such that the 2.0kg 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 _________. Figure

Correct answer: 2.09

Solution

Given here: m=2.0 kg and k=2.0 N m-1.

Time period of spring is given as T=2πmk=2π sec
Block returns to original position in T2=π sec

The speed of block after collision is v=2×22+1=43 m/s and v'=22+1=23 m/s.
Figure
The distance between the two blocks is d=23π=233.14=2.0933m
d=2.09 m

Hence, the correct answer is 2.09.

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