Elastic Potential Energy and mass

[Bottle]

Can someone explain to me why is mass not important factor in this question?

Box A is twice as massive as Box B. They are set on the same spring, and we allow the spring to compress and release the two boxes into the air. What is the height difference between the two boxes?

The Review book I'm using says they will reach the same height because mass is not important.

I don't understand this answer because Elastic potential Energy is U = 1/2 m DeltaX^2. Mass is clearly in the equation so why do we ignore it?

Thanks 🙂

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[theonlytycrane]

Does it say A is twice the mass of B? Or that they are the same mass, but A is twice as large?

 

[salemstein]

Elastic potential energy is 1/2 kx^2 cuz its on a spring. No mass in the equation.

 

[j-thoven]

In this situation I would use conservation of energy, in which there are three parts. 1/2mv^2 + mgdeltaH + 1/2kx^2 = 1/2mv^2 + mgdeltaH + 1/2kx^2. In both cases the initial and final velocity is zero, so the equation becomes
mgdeltaH + 1/2kx^2 = mgdeltaH + 1/2kx^2. In this equation the mass cancels out, and the final equation contains no mass. Hope this helps!

 

[BerkReviewTeach]

In this situation I would use conservation of energy, in which there are three parts. 1/2mv^2 + mgdeltaH + 1/2kx^2 = 1/2mv^2 + mgdeltaH + 1/2kx^2. In both cases the initial and final velocity is zero, so the equation becomes
mgdeltaH + 1/2kx^2 = mgdeltaH + 1/2kx^2. In this equation the mass cancels out, and the final equation contains no mass. Hope this helps!

Actually, mass does not cancel out as you have it written. For it to cancel out, it would need to be in all four terms. If this doesn't make sense, then ask yourself if k cancels out and if g cancels out. You should conclude that none of them cancel out.

Conceptually speaking, if you compress a spring to a fixed distance, you give the system a set amount of potential energy. When the spring fires the object into the air, assuming no loss of energy to resistance or friction, at its max height, it will have the same amount of potential energy (mgh) as it had at the start (1/2kxexp2). If m is doubled, the height has to be half. But that is not what this question is asking.

I think the trick in this question comes down to wording. As I'm interpreting it, both boxes are together on the SAME spring and are fired up into the air together. So the system has a mass that is the sum of both boxes. You might consider that both boxes are feeling the same net force of kx-mg at their lowest point, so they both will feel the same acceleration from rest and thus both achieve the same velocity. When they leave the spring, they should have the same velocity, so if you treat this as a kinematics question, you can say that vexp2 = 2gh for both boxes (mass has dropped out).

Hopefully that approach makes sense if the energetics are not clear.