Timeline for Problem conserving 4-momentum at CoM frame in an inelastic collision [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2015 at 6:14 | vote | accept | user1581390 | ||
| Aug 13, 2015 at 3:44 | comment | added | Bill N | What is your extra mass $\delta$ doing in your frame 1? You must include it in the analysis of that frame. If you don't, then you are trying to compare two different isolated systems. | |
| Aug 12, 2015 at 9:56 | history | closed |
John Rennie ACuriousMind♦ Kyle Kanos Jim Neuneck |
Needs details or clarity | |
| Aug 12, 2015 at 1:23 | comment | added | dmckee --- ex-moderator kitten | I think that you really need to sit down with an expert and walk them through your thinking (which means rather more than just showing your math), because you seem to have a very strange idea about what is going on in this kind of problem. Again, I don't have a clue what you think you are saying in the second paragraph. It doesn't make any sense, and that may be where your problem lies. | |
| Aug 11, 2015 at 16:02 | answer | added | udrv | timeline score: 1 | |
| Aug 10, 2015 at 14:41 | comment | added | user1581390 | I have added the calculation to show the issue. There is no third object, so I am taking account of an isolated system. This is a mass-energy conversion, and not relativistic. | |
| Aug 10, 2015 at 14:38 | history | edited | user1581390 | CC BY-SA 3.0 |
added 1855 characters in body
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| Aug 9, 2015 at 17:16 | comment | added | Timaeus | Please describe the setup of your problem. Momentum is conserved even if an object throws off some of its particles to have less (rest) mass. And the mass of an isolated system is also conserved when the energy and momentum are conserved (though it isn't the sum of the masses of the parts). | |
| Aug 9, 2015 at 13:27 | comment | added | ragnar | In any frame, the total momentum is conserved, so if you boost to a frame where the initial total momentum is zero, the final total momentum is also zero. You don't give enough details to tell where you're going wrong in your calculation. | |
| Aug 9, 2015 at 9:29 | history | edited | Qmechanic♦ |
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| Aug 9, 2015 at 6:25 | comment | added | user1581390 | Do you have a reference for the frames being the same? I realize CoM here means Center of Momentum. | |
| Aug 9, 2015 at 5:22 | review | Close votes | |||
| Aug 12, 2015 at 9:56 | |||||
| Aug 9, 2015 at 4:37 | comment | added | dmckee --- ex-moderator kitten | "The center of momentum (CoM) frame is not the same before and after the collision." Yes it is; or at least all such frames are at rest with respect to one another. I'm really very uncertain what you are trying to say in the second paragraph or why you think there is a problem. Be sure to notice that---though we us the abbreviation "CoM" for both "center of mass" and "center of momentum" at times---here we mean only center of momentum. | |
| Aug 9, 2015 at 4:18 | history | asked | user1581390 | CC BY-SA 3.0 |