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# Potential Energy. watch

1. I think C should be the answer, because when half of the water will go into tank Y, then both tanks will be at equal height. As potential energy = mgh, as height is decreased to half, so potential energy should also decrease to half the original value. But the answer in mark scheme is B. Can anyone explain that where am I doing mistake?
2. Won't it regain some when it rises to the halfway point in Y though, making the answer mgh/4 ?
3. (Original post by The Randy Panda)
Won't it regain some when it rises to the halfway point in Y though, making the answer mgh/4 ?
Nope. How can it regain? Experience tells us that heights of water in both tanks will be equal when water in X will decrease to half the original height and water in Y will increase half the original height. As the mass of total water will remain constant, and height decreased by half, so previous PE = mgh and new PE = mg
4. The value of "h" that you need to use is the height of the centre of gravity of the water, not the highest point of the water. Does the answer now follow from that?
5. (Original post by Pangol)
The value of &quot;h&quot; that you need to use is the height of the centre of gravity of the water, not the highest point of the water. Does the answer now follow from that?
How can that be? The question clearly indicates that height "h" is from base of the tank to the surface of water...
6. (Original post by Zishi)
How can that be? The question clearly indicates that height "h" is from base of the tank to the surface of water...
OK, what is your GPE when you are standing straight up? Your mass x g x your total height, or your mass x g x the distance from the ground to your centre of gravity?
7. It's mgh/4. In the initial tube, energy is calculated by mgh, but since the mass is spread out over that height (and not just at the top) you need to multiply by 0.5, which has the same effect as finding the average GPE of the body of water (effectively from the centre of mass).

The water level will then halve again, so divide mgh/2 by 2 and you get mgh/4. Simples.
8. the height and mass have decreased by half.
9. (Original post by Pangol)
OK, what is your GPE when you are standing straight up? Your mass x g x your total height, or your mass x g x the distance from the ground to your centre of gravity?

(Original post by JayTeeKay)
It's mgh/4. In the initial tube, energy is calculated by mgh, but since the mass is spread out over that height (and not just at the top) you need to multiply by 0.5, which has the same effect as finding the average GPE of the body of water (effectively from the centre of mass).

The water level will then halve again, so divide mgh/2 by 2 and you get mgh/4. Simples.
Oh, now It seems that I get it. I think I never noticed that the objects were small particles, etc while calculating their gravitational potential energy. So for GPE, do we always take the height from the centre of gravity of that particular body?
10. (Original post by Zishi)
Oh, now I think I get it. I think I never noticed that the objects were small particles, etc while calculating their gravitational potential energy. So for GPE, do we always take the height from the centre of gravity of that particular body?
Yes, always.

(Note, however, that the centre of gravity is only 'in the middle' for a uniform body. This applies here, since one particle of water is just like another. You probably don't need to worry about this though. They'd have to be horrible to set you any question involving something non-uniform.)
11. (Original post by Zishi)
Oh, now It seems that I get it. I think I never noticed that the objects were small particles, etc while calculating their gravitational potential energy. So for GPE, do we always take the height from the centre of gravity of that particular body?
Yes. Because we often model bodies as point particles, this is not always obvious. If you model yourself as a particle resting on the floor, you have zero GPE as you are a height of zero above the ground. But in actuality, your centre of gravity is somewhere in your midrift, so it would be more accurate to model yourself as a particle acting from this point.

In the case of the water in your example, you need to take this into account when considering the "before" and "after" GPE of the water. Although I thought that the explaination in terms of the height and the mass both being halved was very elegant.

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