Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

Two blocks A and B of mass m and 2m are connected by a massless spring of force constant k. They are placed on smooth horizontal plane . Spring is streched by an amount x and then released . the relative velocity of the blocks when the spring come its natural length is


$ a)\bigg(\sqrt {\large\frac{3k}{2m}}\bigg)\\ b)\bigg(\sqrt {\large\frac{2k}{3m}}\bigg)\\ c)\sqrt {\large\frac{2km}{m}}\\ d)\sqrt {\large\frac{3km}{2x}} $
Can you answer this question?

1 Answer

0 votes
Let $v_r$ be the relative velocity pf the two.
Let the reduced mass of the 2 block system be $\mu$
From conservation of mechnical energy we have
$\large\frac{1}{2}$$kx^2=\large\frac{1}{2}$$\mu v_r^2$
$kx^2=\mu v_r^2$
$\qquad= \large\frac{2}{3} $$mv_r^2$
$v_r= \bigg(\sqrt {\large\frac{3k}{2m}}\bigg)$$x$
Hence a is the correct answer.


answered Aug 6, 2013 by meena.p
edited Feb 10, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App