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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 and then stick in the collision btw)

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Don’t you need the distance between the two carts to even first calculate the Ug? I would say that Conservation of energy ( Ki+Ui=Kf+Ur) does not apply here because there is an unbalanced external force acting on the system of only the two carts; that force is the force of gravity due to the earth, which is very significant. But how would you know/prove that it increased or decreased? Because technically, as the carts get closer, their U between each other decreases and their K increases, which makes it very fuzzy and intuitively ambiguous.

****A lot of people misunderstand and say U=mgh. I know for sure this is NOT what this question asks though, because there is another of the exact same question with the two cart - AND earth system, where that U=mgh is true. But here it is regular Ug=Gm1m2/r equation I think.

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  • $\begingroup$ Related question by same OP: physics.stackexchange.com/questions/542479/… $\endgroup$ – G. Smith Apr 8 at 21:14
  • $\begingroup$ @G.Smith the thread didn’t have answer and stopped commenting. But if you set the zero point at infinity, is the gravitational PE still zero at a distance infinity plus/minus r? $\endgroup$ – Alexander Ye Apr 8 at 21:15
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    $\begingroup$ Typically in problems like these we don't consider the gravitational interaction between the two carts, as it will be negligible. Why are you wanting to do this? Did your instructor tell you to do this? $\endgroup$ – BioPhysicist Apr 8 at 22:07
  • $\begingroup$ @AaronStevens yep! He’s very direct though and I can’t imagine that this doesn’t have a clear answer $\endgroup$ – Alexander Ye Apr 8 at 22:32
  • $\begingroup$ Just to be super clear, your instructor specifically told you to consider the gravitational interaction between the two carts? $\endgroup$ – BioPhysicist Apr 8 at 22:35
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The issues you are considering (distance between the carts and their possible gravitational interaction) are irrelevant. Internal conservative forces cannot change the total mechanical energy of the system, they can only cause conversions between kinetic and potential energy. The gravitational interaction between the carts is probably negligible, but that doesn't necessarily matter here.

Since the system is just the two carts, the gravitational force of the Earth on the system is an external force. i.e. we do not consider it in our potential energy. This means that we have an external force doing work on the system. I will leave it to you to conclude what this means for energy conservation of the system.

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  • $\begingroup$ But does it Increase or Decrease the energy of the system? So far I have: $\endgroup$ – Alexander Ye Apr 8 at 22:42
  • $\begingroup$ Because for all we know the mass of the carts could be really really huge, meaning there is significant gravitational PE when they are apart and pulled closer by earth’s gravity... $\endgroup$ – Alexander Ye Apr 8 at 22:44
  • $\begingroup$ @AlexanderYe I cannot tell you the answer. The attraction between the carts is irrelevant. Think about what happens to the energy of a system when an external force does work on it. $\endgroup$ – BioPhysicist Apr 8 at 22:45
  • $\begingroup$ Oh, it increases. Though the direction that the carts are pushed also increases PE, so the total ME does not necessarily change (so why wouldn’t the attraction be important)? $\endgroup$ – Alexander Ye Apr 8 at 22:53
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    $\begingroup$ @AlexanderYe Internal conservative forces cannot change the total mechanical energy. It can only convert potential energy to kinetic energy or vice versa $\endgroup$ – BioPhysicist Apr 8 at 22:57
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The distinction your text makes, between a system consisting of only the two carts versus a system consisting of the two carts plus Earth's gravitational field, is not about imagining the two carts colliding due to their mutual gravitational attraction. (Not least because that interaction is tiny; if they started a meter apart, total mass of a kilogram, it would take more than a day for them to touch.)

The distinction is an accounting difference: whether the work done by the Earth on the carts is an internal process, to be accounted as a change in potential energy, or an external process to be accounted as work done from outside.

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  • $\begingroup$ Okay, the system GAINS energy from gravity outside the system from Earth; but doesn’t the system LOSE kinetic energy after the collision too? Rigorously doing this gets no where without the exact angle because it uses trigonometric square values ... $\endgroup$ – Alexander Ye Apr 9 at 1:40
  • $\begingroup$ In energy conservation problems, angles are often red herrings. The current version of the question (v2) does not suggest whether to model the collision as perfectly inelastic (stick together), perfectly elastic (no mechanical energy change), or a more realistic approximation that is somewhere between. $\endgroup$ – rob Apr 9 at 2:02
  • $\begingroup$ sorry, left out the part where it says that they stick together (they have different masses, so it would be an inelastic) $\endgroup$ – Alexander Ye Apr 9 at 2:06
  • $\begingroup$ I think I figured a proof out, but it seems too messy numerically so I’m guessing there should be a cleaner conceptual one. The trigonometric functions are on the denominator so the contribute to the increase in ME. $\endgroup$ – Alexander Ye Apr 9 at 2:07

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