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Gravitational U = -Gm1m2/r, thus the distance (r) the two carts are apart affects how much U you end up with, no? Thus it also affects whether or no ME is ultimately conserved, because you can just set the distance so that it is. Can you even freely set the distance/ where h=0 is, even though it is not an earth-object system? So because of such variation, I feel like distance does matter in these cases, and you need to be given the distance between the objects to answer the question.

Consider the time interval from when the two carts are released until just after they collide. For the system consisting of only the two carts, indicate whether the total mechanical energy increases, decreases, or remains the same. (The carts start out at rest btw)

Cart A has mass M and is released from rest at a height 2H on a ramp making an angle 2θ with the horizontal, as shown. Cart B has mass 2M and is released from rest at a height H on a ramp making an angle θ with the horizontal. The carts roll toward each other, have a head-on collision on the horizontal portion of the ramp, and stick together. The masses of the carts’ wheels are negligible, as are any frictional or drag forces.

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Mechanical energy is the sum of potential energy and kinetic energy.

Potential energy is the gravitational potential each cart has due to being raised distances $2H$ and $H$ from the surface.

This is not referring to the very very small amount of potential energy they would have if considered to be mini planets at a distance $r$ apart, as your equation suggests. This "celestial-body potential energy" is negligible here as they have such small masses so we do not take into account when working with mechanical energy.

So, yes, you are right that for a true proof of conservation of all types of energy you would need to be told the distance apart that they started. This would allow you to calculate the loss in cart-to-cart gravitational potential energy lost due to work done by the very small gravitational force of attraction between the carts. However, all questions of this sort are presented in an abstract framework where things are considered to be "ideal" and negligible effects, like this tiny potential energy, are ignored.

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  • $\begingroup$ Note though that it says the system of ONLY the two carts. There is another part to this question that includes the system of the two carts AND the earth specifically (for which I chose to include the minute U between the carts)... I am pretty sure it did canceled out though. So basically this question is ONLY the two objects’ potential energy, even though Earth’s gravity is involved. $\endgroup$
    – Alexander Ye
    Commented Apr 8, 2020 at 20:32
  • $\begingroup$ If there was no Earth in this setup (which raises questions like what is the ramp sitting on!), then what incentives would there by for the carts to even roll down the slopes at all? $\endgroup$
    – Joe Iddon
    Commented Apr 8, 2020 at 20:36
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    $\begingroup$ True. In this question though, it is clear that they only want you to consider the potential energy an object gains from being lifted away from the surface of the earth: $U = mgh$. This "approximate" potential energy formula assumes that the only objects in the universe is the object, of mass $m$, and the Earth (or any planet), which has a gravitational field strength, $g$ at its surface. It breaks down when the height lifted, $h$, is very large since the gravitational potential field can no longer be approximated as linear. $\endgroup$
    – Joe Iddon
    Commented Apr 8, 2020 at 20:40
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    $\begingroup$ Hmm, I'm not sure that makes sense... I'm struggling to understand your justification for why ME is not conserved: "the external force of the earth’s gravitational force acting on the objects". Consider a simple example: you drop a ball from a height of $1m$. Is ME conserved? Yes, it is under the Earth's gravitational force and it is conserved as the potential is converted to kinetic. $\endgroup$
    – Joe Iddon
    Commented Apr 8, 2020 at 20:48
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    $\begingroup$ @AlexanderYe It's convenient when calculating stuff on Earth to set the zero point for gravitational PE at $h=0$. But in orbital mechanics the usual convention is to set the gravitational PE zero point at infinite distance from the centre of the main body, or from the barycentre of a system of bodies. $\endgroup$
    – PM 2Ring
    Commented Apr 8, 2020 at 20:49

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