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# 2 masses on a pulley - height of lighter mass is what? Watch

1. 2 masses are hung using a long (but massless?) cord on a frictionless pulley. They are both level, thus are both the same height (h) above the ground. The pulley is attached to the ceiling (not sure of distance from ground to ceiling). Mass A is heavier than mass B.

What maximum height does mass B reach when the masses are allowed to move (ie are not held at the same level at rest)?

I tried to use energy to solve this.

I figured that at the initial stage (stage 1), the masses have 0 kinetic energy, and each have a potential energy of mgh.

So the total energy of the system intitially is:
mAgh + mBgh

When the masses are allowed to move, mass A hits the ground and thus has 0 energy (no potential energy or kinetic energy if it is on the ground). mass B still moves upwards (at a constant velocity I think as opposed to accelerating up???), then it reaches it's max. height, and then it comes down.

When mass B reaches max. height, surely it has no kinetic energy (it is stationary for that tiny period of time that it is at max. height, surely?)but it does have a new potential energy of mgh (where h is greater than in the initial stage).

So to work out mass B's max. height, I thought I do this:

Total initial energy = Total final energy
mAgh + mBgh = mBgh2(where h2 is a new height than h)

So the total energy of the system divided by mass of B, and by g, should give max. height of B, no?

It does not work.

Any useful help would be awesome, thank you!

EXAMPLE 5.95 9P. 171 of Uni Physics 13th Edition Y & F book):
Two objects with masses 5.00kg and 2.00kg hang 0.600m above the floor from the ends of a cord 6.00m long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the 2.00kg object.
2. Use energy considerations to find the speed of the mass that's rising at the point when the other mass hits the ground. (Or the speed of the mass as it hits the ground as this is the same.)
Then the mass that's rising carries on for a small distance. You can work out how much further it travels using a straightforward suvat equation using the velocity you found as u, g, and v=0.
3. (Original post by Stonebridge)
Use energy considerations to find the speed of the mass that's rising at the point when the other mass hits the ground. (Or the speed of the mass as it hits the ground as this is the same.)
Then the mass that's rising carries on for a small distance. You can work out how much further it travels using a straightforward suvat equation using the velocity you found as u, g, and v=0.

That does seem to work.

I just don't understand why you can't just use total energy at the start = total energy at the end to work this out :/ Why would that not work/what have I missed out when using that method?

Thank you
4. (Original post by PhysicsGal)
That does seem to work.

I just don't understand why you can't just use total energy at the start = total energy at the end to work this out :/ Why would that not work/what have I missed out when using that method?

Thank you
Because the kinetic energy of the system is not conserved when the falling mass collides with the ground. That's an inelestic collision.
5. (Original post by Stonebridge)
Because the kinetic energy of the system is not conserved when the falling mass collides with the ground. That's an inelestic collision.
Dude, you're awesome!

That sounds so simple now, no idea why I didn't think of that :/ So some energy would be transferred into sound etc...hmm!
Thanks

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