In the figure shown a massless spring of stiffness `k` and natural length `l_(0)` is rigidly attached to a block of mass `m` and is in vertical position. A wooden ball of mass `m` is released from rest to fall under gravity. Having fallen a height `h` the ball strikes the spring and gets stuck up in the spring at the top. What should be the minimum value of `h` so that the lower block will just lose contact with the ground later on ? Assume that `l_(0) gt gt (4mg)/(k)`. Neglect any loss of energy. (Given `k=mg//2`)

Explanation

The minimum force needed to lift the lower block is equal to its weight. During upward motion the spring will get elongated. If elongation in the spring for just lifting the block is `x_(0)` then
`kx_(0)=mg`
`rArrx_(0)=(mg)/(k)`……`(i)`
From `COE`
`mg(l_(0)+h)=mg(l_(0)+x_(0))+(1)/(2)kx_(0)^(2)`
`rArr mgh=mgx_(0)+(1)/(2)kx_(0)^(2)rArr mgh=((mg)^(2))/(k)+(1)/(2)(m^(2)g^(2))/(k)rArrh=(3mg)/(2k)`
During down ward motion, suppose maximum compression in the spring is `x`
From `COE`
`mg(l_(0)+h)=mg(l_(0)-x)+(1)/(2)kx^(2)`
`rArrmgh= -mgx+(1)/(2)kx^(2)rArrmg"(3mg)/(2k)= -mgx+(1)/(2)kx^(2)`
`rArr 3(mg)^(2)= -2mgkx+k^(2)x^(2)`
`rArr k^(2)x^(2)-2mgkx-3(mg)^(2)=0`
`rArr x=(2mgk+-sqrt(4(mgk)^(2)+12k^(2)(mg)^(2)))/(2k^(2))=(2mgk+-4mgk)/(2k^(2))rArrx=(3mg)/(k)`