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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Physics
Chapter
Oscillations

JEE Advanced 2018 · Paper 1 · Physics · Oscillations

A spring - block system is resting on a frictionless floor as shown...

A spring - block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0Nm−12.0N{m}^{- 1} and the mass of the block is 2.0kg2.0kg . 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 2.0ms−12.0m{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 _________. A spring - block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0N m ^ - 1

Correct answer

0

Step-by-step solution

Given here: m=2.0kgm = 2 . 0kg and k=2.0Nm−1k = 2 . 0N{m}^{- 1}.

Time period of spring is given as T=2πmk=2πsecT = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi sec
Block returns to original position in T2=πsec\dfrac{T}{2} = \pi sec

The speed of block after collision is v=2×22+1=43m/sv = \dfrac{2 \times 2}{2 + 1} = \dfrac{4}{3}m / s and v′=22+1=23m/sv ' = \dfrac{2}{2 + 1} = \dfrac{2}{3}m / s.
A spring - block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0N m ^ - 1
The distance between the two blocks is d=23(π)=23(3.14)=2.0933md = \dfrac{2}{3}\left(\pi\right) = \dfrac{2}{3}\left(3.14\right) = 2.0933m
d=2.09md = 2.09m

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