Two masses on a single spring how far will each go when released?

[michaelw]

This is a strange question and I am stumped how to solve it
Objects A and B are placed on a spring, object A has twice as much mass as object B. If the spring is depressed and released, how high will object A go compared to object B? (the masses are both on the spring at once, and when released will move upward)

I thought that object A would go 1/2 the distance as object B, but this is incorrect.. (mgh = kx^2/2) because the total elastic potential energy is split among the two masses.. each mass does not receive all the energy or an equal proportion of elastic potential energy

The answer in the book states they will rise to the same height

Anyone know why this is?

 

[pete worthington]

Perhaps if you break up the energy transformation into each step you will see it.

1) the springs potential energy is transformed into the kinetic energy of the two masses (minus the gravitational potential energy gained by the masses due to the springs extension).
2) The kinetic energy of the two masses is transformed into gravitational potential energy of the two masses (excluding any dissipative forces i.e. air resistance).

Set up the second scenario into an equation and notice what happens to the masses.

 

[Doc Al]

The answer in the book states they will rise to the same height

Anyone know why this is?

Hint: When the masses leave the spring, how do their speeds compare?

 

[michaelw]

Is this what you are getting at?
Total kinetic energy of both masses at bottom = Total potential energy of both at top = Spring potential energy of spring
3m*v^2/2 = 3mgh = kx^2/2
The 3m cancel and we get v^2 = gh (mass independent)

 

[pete worthington]

Right ! Perhaps I'm overcomplicating this qualitative problem but just be aware of the fact that gravitational potential energy is increasing as the spring decompresses and the mass accelerates. assume there is three positions (like you have described above): 1) mass at rest on spring compressed, 2 mass at max velocity on spring at equillibrium and 3) mass at rest at highest point in air:

PE (spring)1 = KE (mass)2 + PE (grav)2 = PE (grav)3

Where PE (grav)2 = 3mgx and PE (grav)3 = 3mg(x+h)

 

[jdavel]

This is a strange problem.

As the spring expands, it initially stays in contact with both masses (or so it would seem). So the two masses must have the same acceleration. But during this time the spring exerts the same force on each mass, namely -kx. So F = -kx = ma = 2ma.

Hmmm...

Edit: The above is incorrect. Doc Al's post #8 explains why.

 

[michaelw]

hmm
if both masses were not on the spring at once, how high will each go?

 

[Doc Al]

As the spring expands, it initially stays in contact with both masses (or so it would seem). So the two masses must have the same acceleration. But during this time the spring exerts the same force on each mass, namely -kx. So F = -kx = ma = 2ma.

Consider this: Are the masses side by side, so they are both in contact with the spring? (For example, placed on a plate attached to the spring.) Or are they placed one atop the other?

In any case, the net force exerted by the spring is given by F = kx. This does not mean that the spring exerts the same force on each mass (assuming both masses are in contact with the spring).

 

[ramollari]

I think the problem is far simpler than discussed here. The spring will be in an oscillation, so the speeds of the two bodies will be given by the SHM equation:

$$v = -\omega A sin\phi$$

This indicates that the speeds of the bodies are independent of their masses. Since they leave the spring at the same phase, their speeds will be equal, meaning that they will rise to the same heights.

 

[jdavel]

Consider this: Are the masses side by side, so they are both in contact with the spring? (For example, placed on a plate attached to the spring.) Or are they placed one atop the other?

In any case, the net force exerted by the spring is given by F = kx. This does not mean that the spring exerts the same force on each mass (assuming both masses are in contact with the spring).

Doc Al,

Oops, you're right! Thanks for straightening me out.

jdl

 

[jdavel]

I think the problem is far simpler than discussed here. The spring will be in an oscillation, so the speeds of the two bodies will be given by the SHM equation:

$$v = -\omega A sin\phi$$

This indicates that the speeds of the bodies are independent of their masses. Since they leave the spring at the same phase, their speeds will be equal, meaning that they will rise to the same heights.

ramollari,

I think you're right, the masses leave the spring at the equal speed. But isn't omega in your equation for speed a function of the two masses? And just intuitively, if the masses were huge it seems they might not leave the spring at all, and if they're tiny, they going flying off really fast.

 

[ramollari]

ramollari,

I think you're right, the masses leave the spring at the equal speed. But isn't omega in your equation for speed a function of the two masses? And just intuitively, if the masses were huge it seems they might not leave the spring at all, and if they're tiny, they going flying off really fast.

I meant that omega is the same for both bodies even though one is lighter than the other, because it is defined for the spring. The important thing is that the two bodies will reach equal heights.
On the other hand, as you correctly said, omega depends on the combined mass of the two bodies, but that's not important in the question.