Height mass 1 travels after mass 2 is released from spring

[Emilsen]

Hi,
I'm a mechanical engineer from Norway who has an work related query.
Please see the attached picture describing the problem.

I'm trying to figure out how heigh mass 1 travels if mass 2 is quick released when being suspended in a spring. I'm also trying to figure out a the maximum acceleration, but that's 2. priority.

I have tried to work out an answer my self, but I don't think it's correct.
The thing I can't get my head around is how to calculate the height mass 1 travels above it's equilibrium due to the speed and spring tension. If the mass 2 had been release slowly, the answer could easily be found using the energy forumula(E=1/2*k*x^2), as far as I can understand.

Any bright minds who could spare a few minutes explaining me this?

Help is much appreciated!

Have a nice sunday.

 

[tiny-tim]

Hi Emilsen! Welcome to PF!

I think "quick release" simply means that M₂ starts going down, and M₁ starts going up.

So you can use conservation of energy (for M₁ and the spring).

 

[Emilsen]

Hi tiny-tim,
Thanks for your reply.

I have tried the following equation setup:
E1spring=E2spring+E2kinetic
1/2kx₁²=1/2kx²₂+1/2mv²

I used F=kx to find x₁ and x₂ for mass 1+2 and mass 1.

I then tried the following formula to calculate how heigh the mass 1 went above the E2spring equilibrium:
v₂=v₁-gs/v₁=0, to find s.

Is this the right path?

 

[tiny-tim]

Hi Emilsen!

I'm trying to figure out how heigh mass 1 travels if mass 2 is quick released when being suspended in a spring.

I have tried the following equation setup:
E1spring=E2spring+E2kinetic
1/2kx₁²=1/2kx²₂+1/2mv²

nooo …

v is irrelevant, since the system both starts and finishes with v = 0

and what about gravity? ​