In the figure shown, mass `2m` connected with a spring of force constant `k` is at rest and in equilibrium. A particle of mass `m` is released from height `4.5 mg//k` from `2m` . The particle stick to the block. Neglecting the duration of collision find time from the release of `m` to the moment when the spring has maximum compression.

Explanation

Velocity of the particle just before collision.
`u=sqrt(2gxx(4.5mg)/(K))impliesu=3gsqrt((m)/(K))`
Now it collides with the plate.
Now just after collision velocity `(V)` of the system of plate `+` particle'
`m u=3mVimpliesV=(u)/(3)=gsqrt((m)/(K))`
Now the system performs SHM with time period
`T=2pisqrt((3m)/(K))omega=sqrt((K)/(3m))` and mean position as `(mg)/(K)`
distance below the point of collision.
Let the equation of motion is
`y=Asin(omegat+phi)` .(i)
`v=(dy)/(dt)=Aomegacos(omegat+phi)` ..(ii)
At `t=0`,`y=(mg)/(K)` and `v=gsqrt((m)/(K))`
From eqs (i) and (ii) `(mg)/(K)=Asinphi` ..(iii)
`A=(2mg)/(K)` ..(iv)
From Eqs. (iii) and (iv), `phi=(5pi)/(K)` `y=(2mg)/(K)sin(sqrt((K)/(3m))t+(5pi)/(6))`
Hence equation of SHM should be
`y=-A=-(2mg)/(K)=(2mg)/(K)sin(sqrt((K)/(2m))t+(5pi)/(6))`
The plate will be at rest again when
`y=-A=-(2mg)/(K)=(2mg)/(K)sin(sqrt((K)/(3m))+(5pi)/(6))`
`impliessin(sqrt((K)/(3m))t+(5pi)/(6))=-1=sin(3pi)/(2)`
`impliessqrt((K)/(3m))t+(5pi)/(6)=(3pi)/(2)implies=(2pi)/(3)sqrt((3m)/(K)`