Browse Questions

# 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

$a)\bigg(\sqrt {\large\frac{3k}{2m}}\bigg)\\ b)\bigg(\sqrt {\large\frac{2k}{3m}}\bigg)\\ c)\sqrt {\large\frac{2km}{m}}\\ d)\sqrt {\large\frac{3km}{2x}}$

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.

edited Feb 10, 2014 by meena.p

+1 vote