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 1.0 kg moving with a speed of 2.0 m s -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 \ \ \ \ \ -.

Physics · System Of Particles

JEE Advanced 2018 — Paper 1 — Question 10

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 1.0 kg moving with a speed of $2.0 \mathrm{~m} \mathrm{~s}^{-1}$ 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

Numerical answer — enter this value.

Step-by-step solution

For collision :

using com $\rightarrow 1 \times 2=1 \times u+2 \times v$

Using $e \rightarrow 2=-u+v$

$u=\frac{-2}{3} m / s \quad v=\frac{4}{3} m / s$

time taken for the block to came to the unstretched position of spring for the first time after the collision

$=\pi \sqrt{\frac{m}{k}}=\pi \mathrm{sec}$

distance between blocks $=\frac{2 \pi}{3} m=2.09 \mathrm{~m}$ (taking $\pi \approx 3.14$ )

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Physics
Chapter: System Of Particles
Topic: Collisions in One Dimension