IIT JEE , AIEEE Forum - Physics , Mechanics

rooney

http://www.goiit.com/posts/list/30/mechanics-challenge-3-32895.htm

Check the question. and here is a great soln given by freneid's prof.

Let us consider that F2>F1.now let m1 be displaced to the ryt by x1 and m2 to the ryt by x2.therefore net displacement of m1 wrt m2=x1-x2.this is also the elongation of the spring.so we get the concept that the relative disp. of the 2 blocks is the elongation of the spring.let us denote it by Srel.then dSrel/dt=Vrel of the two blocks.in position of max.elongation the Vrel=0.now consider the fbds of the 2 blocks.
T-F1=m1a1.
F2-T=m2a2.
solving a1 and a2 we get a1=T-F1/m1
a2=F2-T/m2
therefore rel accn.=a2-a1.
now write above accn.=VdVrel/dSrel and put T=kSrel....integrate with limtis Srel=0, Vrel=0 and Srel=x,Vrel=0 where x is max elongation.

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