Timeline for Conservation of momentum and conservation of energy
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2018 at 18:47 | answer | added | Maury Markowitz | timeline score: 0 | |
| Feb 4, 2016 at 18:20 | comment | added | Emilio Pisanty | You misunderstand me. Post that question as a coherent, separate post of its own, with its own space for answers. That form is pretty close to what you should probably be asking. | |
| Feb 4, 2016 at 17:46 | comment | added | ergon | The question is simple: you have to same masses one non-deformable and the other plastically deformable. You exert the same force for same time. Will they move the same? Even though one wiill deform first and then move while the other will move immediately? | |
| Feb 4, 2016 at 16:22 | comment | added | Emilio Pisanty | That analysis is too big for a comment, but if you rephrase the question (or post a new one) to make it clear that this is the issue, I'll be happy to post an answer. (I won't shoot a moving target, though: provide a clear question, and don't change it afterwards.) The short of it is that for the front of the car to get crumpled, it needs to be squeezed by the back, i.e. the back provides a forwards force on the front. By Newton's third law, the back exerts exactly that much force on the back. | |
| Feb 4, 2016 at 16:07 | comment | added | ergon | What you are saying is just what the high school books say, without any analysis. You are saying that ALL the force is used to accelerate the masses? Then how plastic deformation takes place? Magically? | |
| Feb 4, 2016 at 0:38 | comment | added | Emilio Pisanty | If you want to be taken seriously, then a slightly more respectful tone would not come amiss, and if you want us to address your (superficial) analysis then you should provide the analysis instead of unproven claims. However much force is used to deform the objects, the net force of one object on the other must still be the same, and equal to the reaction force on the other object. Hence if the masses are different the accelerations will be asymmetric and the joined bodies will not remain at rest. To be clear: there is no part of the forces which does not cause acceleration of the COM. | |
| Feb 3, 2016 at 22:53 | comment | added | ergon | During the collision the acting forces are F (and its equal interaction) to each of the masses, for same time t. So you are saying that this force F cannot cause a plastic deformation X1 to one of the masses and a deformation X2 to the other mass so that all the work done by the forces will be used to cause the plastic deformation and little or no part of the forces will cause acceleration of the masses. You need to go deeper in analysis like I do to understand what I am saying and not take for granted high school ways of analysing problems. | |
| Feb 3, 2016 at 20:58 | comment | added | Emilio Pisanty | If they don't have the same mass, it is impossible for them to come to a complete stop, and the final momentum will be nonzero. | |
| Feb 3, 2016 at 20:35 | comment | added | ergon | They dont have to be the same mass. It's just that their whole kinetic energy was converted into structural deformations. Is this IMPOSSIBLE to happen? | |
| Feb 3, 2016 at 18:05 | comment | added | Emilio Pisanty | So you're thinking of equal-mass cars colliding at equal velocity and opposite directions? That same situation where the total momentum (vector sum of the individual momenta) is zero? How exactly is this a contradiction? | |
| Feb 3, 2016 at 12:42 | history | edited | ergon | CC BY-SA 3.0 |
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| Feb 2, 2016 at 17:28 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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| Feb 2, 2016 at 16:03 | answer | added | Bill N | timeline score: 4 | |
| Feb 2, 2016 at 15:56 | vote | accept | ergon | ||
| Feb 2, 2016 at 15:29 | answer | added | Farcher | timeline score: 2 | |
| Feb 2, 2016 at 15:09 | history | asked | ergon | CC BY-SA 3.0 |