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.

enter image description here

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.


2 Answers 2


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.

  • $\begingroup$ Thanks for replying! So, when $M$ decrease to $m$, the system has less energy available to do work, and energy of the system with mass $m$ is conserved? On the other side, why do you say I shouldn't use height where total force is zero? Maybe I get confused with SHM, because if I wanted to get the position of $A$ at time $t$, I'd have to measure it respect to that equilibrium position, or not? $\endgroup$
    – Carlens
    Mar 5, 2016 at 18:24
  • $\begingroup$ Well, since the force $mg$ is constant, you can use the equilibrium height (of course, if the gravity force was not constant, the motion would no longer be simple harmonic). There will be an extra, constant term in the energy which won't change the result. If you do that though, you should not include the potential energy due to gravity - it will be contained within the spring energy term. $\endgroup$
    – erenust
    Mar 5, 2016 at 18:52
  • $\begingroup$ You're right! Within the elastic energy appear the terms $-mgh + \frac{1}{2}k(h-y_1)^2$, where $h$ is the height at which spring is unstretched. How did you realize it? It seems like a very particular coincidence. Do you have any interpretation of those terms (the gravitational at least, which is curiosly negative)? $\endgroup$
    – Carlens
    Mar 6, 2016 at 22:39

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.