Elastic potential energy and equilibrium position when mass changes in vertical mass-spring system

Question

I'm a bit stuck with this situation: Suppose that an object $A$ with mass M=8 kg at 10 mt above the floor falls from rest tied to an 2 mt elastic rope, which does not exert any force since $A$ falls 2 (and I will model it as a spring with elastic constant $k$). I know it will bounce at a minimum height - doesn't matter where it is, but let it be $y_1$-, and at that moment the object decrease its mass in 3 kg, so that new mass is m=5 kg.
I want to know the maximum height it will reach after bouncing, so I set up conservation of energy (assuming there is no friction due to the air). Now, which is the elongation of the spring? I mean, we measure it respect to the equilibrium position. Initially, but it changes when mass changes (it is a a higher position when $A$ weighs less). Although, if elongation is higher after losing mass, I think it implies system has more potential than before but no force has done work to change it. It seems like system has gained energy, which looks absurd.

So, what elongation should I use when I set up the eq: $$Mgy_1 + \frac{1}{2}k(y_\rm{eq}-y_1)^2 = mgy_2 + \frac{1}{2}k(\hat{y_\rm{eq}}-y_2) ^2 $$

Here some images to illustrate this situation.

Where $y_2$ will be the maximum height after bouncing.
Remember $y_\rm{eq}$ is the equilibrium position when $A$ has mass M, and $\hat{y_\rm{eq}}$ is the same when $A$ has mass m.

I hope you could help me to see it more clear. Please, correct me if I'm wrong.

Answer 1

You cannot equate the energy before the mass decrease to the energy after the mass decrease: When you decrease the mass, you take away potential energy, thus the total mechanical energy of the system changes.

What you need to do is to find the height at which $M$ stops (by using the conservation of energy) and then use that height to find where the mass $m$ stops (again, by using the conservation of energy). Note that these two energies, as I have said in the beginning, are different.

Lastly, be careful about the energy expression you have written for the elastic rope. You should use the difference between the height of the object and the height where the spring force is zero, not the height where the total force on the object is zero.

Answer 2

Pretty much, you just can't set it up in the way that you have done, as mass changes, E=mc^2 is lost, I will leave it at that. Best way to approach is to break it into pieces, let K of spring be an unknown variable, in terms of K find max elongation in terms of K, now, just replace the masses and put x = calculated value and find max height by applying energy conservation again. Again, importantly, take reference points as simple as possible like, for spring put origin at the point where its unstrtched and do the same for mass.