A block of mass `m` is dropped onto a spring of constant `k` from a height `h`. The second end of the spring is attached to a second block of mass M as shown in figure. Find the minimum value of h so that the block M bounces off the ground. If the block of mass m sticks to the spring immediately after it comes into contact with it.

Answer

`((M^2+2mM)g)/(2km)`

Explanation

The block `m` drops a height h from A compresses the spring and then goes up to `Bx` above the initial level of spring. Applying work energy theorem between A and B
`W=DeltaPE_s+DeltaPE_g+DeltaKE+w_(NC)`
`0=-mg(h-x)+1/2kx^2+0+0`
`mg(h-x)=1/2kx^2`
The minimum value of x required so that the block bounces is when the N will become zero.
`N+kx=Mg`, Here, `N=0`

`impliesx=(Mg)/(k)`
`mg(h-(Mg)/(k))=1/2k*(M^2g^2)/(k^2)`
`impliesmgh=(M^2g^2)/(2k)+(mMg^2)/(k)`
`h=((M^2+2mM)g)/(2km)`
This is the required minimum value of h.