At time t=0 string A is cut, then what will be the maximum compression in the spring if spring constant is K.(Assume surfaces are smooth and initially spring was at its natural length)

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4 Answers
2gk
 Akash Anand Wrong answer I guessUpvote·0· Reply ·20140321 23:08:25
At the mean position, Kx=2g; x=2gK
2gx=12Kx^{2} +K.E (Using energy conservation)
K.E is the kinetic energy of the two block system ,individually they each have kinetic energy=K.E2
Just after that the string would become loose,So the upper block would continue to move with that velocity,
So at max. compression(h) the kinetic energy of system=K.E2
2gh=12Kh^{2} +K.E2
Solving we get,h=(2+√2)gK
IS IT CORRECT?
 Himanshu Giria At mean position x wu 0 i guess d b = to
 Himanshu Giria at mean position i guess x wud b =0...
 Niraj kumar Jha After cutting A ,"mean position" means when the net force on lower block is zero.
 Akash Anand After the mean position, by COE loss in P.E + Loss in K.E would be equal to the P.E stored in the spring na??
 Niraj kumar Jha Sir, we can also do that and the obtained ans. should be addded to 2g/K to get max. compression.but I took initial point as point of reference and as initially there was no K.E Loss in P.E=K.E at max. comp.+P.E stored in spring.the result here directly gives max. comp.
Niraj..I am not able to find the fallacy, but the given answer is far different from this. Given answer is 0.45m .
 Niraj kumar Jha Sir, answer isn't in even in terms of "K"?
 Akash Anand Actually K is given 100N/m.