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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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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

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Let $v_r$ be the relative velocity pf the two.
Let the reduced mass of the 2 block system be $\mu$
$\mu=\large\frac{(m)(2m)}{m+2m}=\frac{2}{3}$$m$
From conservation of mechnical energy we have
$\large\frac{1}{2}$$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
 

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