In the figure shown, mass `2m` connected with a spring of force constant `k` is at rest and in equilibrium. A partical of mass `m` is released from height `4.5 mg//k` from `2m` . The partical 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.

Answer

`(2 pi)/(3) sqrt((3m)/(K)) + 3 sqrt((m)/(K))`

Explanation

Velocity of the particle just before collision
`u=sqrt(2gxx(4.5mg)/(K))`
`=3gsqrt((m)/(K))`
time taken `(1)/(2)"gt"_(1)^(2)=4.5(mg)/(K)`
`t_1=3sqrt((m)/(K))`
Now it collides with the plate..
Now just after collision velocity of system of plate and particle.
mu`=3mv`
`impliesV=(u)/(3)=gsqrt((m)/(K))`
Now system performs SHM with time perio `T=2pi`
`sqrt((3m)/(K))` and mean position as `(mg)/(K)` distance below the point of collision.
Let the equation of motion be
`y=Asin(omega+phi)` for `t=0` `y=(mg)/(K)`
`(mg)/(K)=Asinphi`
Now for amplitude
`V=omegasqrt(A^2-x^2)impliesgsqrt((m)/(K))=sqrt((K)/(3m))sqrt(A^2-(m^2g^2)/(K^2))`
`(sqrt3(mg)/(K))^2=A^2-(m^2g^2)/(K^2)`
`A=(2mg)/(k)` .(ii)
`x=(A)/(2)` to `x=0impliest=(T)/(12)`
`x=0` to `x=Aimpliest=(T)/(4)`
Total time `=(T)/(12)+(T)/(4)+3sqrt((m)/(K))=(2pi)/(3)=sqrt((3m)/(K))+3sqrt((m)/(K))`