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Two blocks of masses $m_1=2 kg$ and $ m_2=4kg$ are moving in the same direction with speeds $V_1=6 m/s$ and $V_2=3 m/s$, respectively on a frictional less surface. An ideal spring with spring constant $k= 30000\;n/m$ is attached to the bakside of $m_2.$ Then the maxiimum compression of the spring after collision will be

$\begin{array}{1 1} (a)\;0.06\;m & \quad (b)\;0.04\;m \\ (c)\;0.02\;m & \quad  (d)\;none\;of\;these \end{array}$

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At maximum compression velocity of both bodies are same and equal to velocity of centre of mass
By momentum conservation
$m_1V_1+m_2V_2=(m_1+m_2)V$
$V= \large\frac{12+12}{6}$$=4\;m/s$
By energy conservation:
$\large\frac{1}{2}$$m_1v_{1}^2+\large\frac{1}{2}$$m_2v_{2}^2=\large\frac{1}{2} $$(m_1+m_2)V^2+\large\frac{1}{2} $$k x^2$
$\frac{1}{2}kx^2=\large\frac{1}{2}. $$2 \times 36 +\large\frac{1}{2} $$. 4 \times 9-\large\frac{1}{2} $$(6) 16$
$\qquad=54 - 48$
$x^2=\large\frac{12}{k}$
$x=\sqrt {\large\frac{12}{30000}}$
$\qquad= \sqrt {\large\frac{1}{2500}}$
$\qquad=\large\frac{1}{50}$
$\qquad = 0.02\;m$
answered Nov 22, 2013 by meena.p
edited Jun 17, 2014 by lmohan717
 

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