Packets

This one uses a great technique I've codenamed "packets". (Can you figure it out?)


Two positrons and two protons are kept on four corners of a square of side a = 1 fermi.
The mass of the proton is much larger than the mass of positron. Then determine the kinetic energies of one of the positrons and one of the protons respectively after a very long time.

Note: Its more of a mechanics question than electrostats.


P.S.: Report your answer at the judge in micro-ergs, correct upto 2 decimal places. eg - "0.58,6.50" if ur answer is 0.58 microerg for a positron and 6.50 microerg for a proton.

P.P.S.: You may take K = 9x10^9, e = 1.6x10^(-19) SI Units.

-Shivanker
Click here for the solution.

15 comments:

  1. mai kyu bataunMay 5, 2011 at 11:15 AM

    proton: (ke^2)/2a.root2 + GM^2/2a.root2 + 2ke^2/a(1+M/m) + 2GMm/a(1+M/m) ???
    here M is mass of proton & m dat of positron?
    for positron, just interchange M & m

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  2. mai kyu bataunMay 5, 2011 at 12:22 PM

    shivanker bhaiya answer to batao

    ReplyDelete
  3. poochha kisne hai ??May 5, 2011 at 2:23 PM

    sorry.. i've updated the judge..

    but PLZ, u dont need to tell us, we didnt ask u..

    ReplyDelete
  4. shaanMay 7, 2011 at 1:49 AM

    i can think of only one eqn{conservation of energy}, :(

    wonder what is hidden in 'Packets!!'..

    a hint will be helpful..

    ReplyDelete
  5. AnujMay 7, 2011 at 2:18 AM

    cmon...conservation of momentum must be used

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  6. AnujMay 7, 2011 at 2:19 AM

    and i'd like ppl to use real names/nicknames...

    we've had really awesome ones now::
    "munni badnaam hui..."
    "poocha kisne hai"
    "mai kyun bataun"

    cool.

    ReplyDelete
  7. Sambhav GuptaMay 7, 2011 at 3:32 AM

    anuj bhaiya, conservation of momentum only tells us relation among the two positrons, and among the two protons, which we could tell just by symmetry, it tells nothing about how energy is distributed, try urself :-)

    yes, the question is difficult, and the hint is provided in the line "The mass of the proton is much larger than the mass of positron."

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  8. shaanMay 7, 2011 at 2:24 PM

    yes anuj bhaiya, C O momentum was of no use...

    well due to high mass of P+, do we consider KE of P+ ->0{in comparison to e+)?....(though this seems illogical)

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  9. ShivankerMay 7, 2011 at 2:29 PM

    u'r wrong shaan.. (:

    ReplyDelete
  10. ShivankerMay 7, 2011 at 7:28 PM

    the post has now been updated with the link to the solution..

    ReplyDelete
  11. amritpalMay 9, 2011 at 1:35 PM

    yes soln seems fine but what would be K.E. then?? And question agar kisi aur ka hai to usey credit de dena chahiye

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  12. shaanMay 9, 2011 at 1:52 PM

    P+ ki KE= Ke^2/(2 root2*a)

    find e+ ki KE by conserving initial energy..

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  13. ShivankerMay 9, 2011 at 3:26 PM

    hi amritpal..
    i never said i "frame my own questions"!!
    this one is from INPhO-09.. well, does that really matter??

    @Shaan: 5 packets of potential energy are used up to bring the positrons into motion, while only the last one provides energy to the protons.. that's the reason behind the word "packets".. ;)

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  14. shaanMay 9, 2011 at 7:38 PM

    nice )

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  15. AnujMay 14, 2011 at 4:48 AM

    hi all..
    regarding my comment, actually i'd seen this question during a test (and solved it too :)), so i thought that it might have be plain..

    i guess that question had given more hints..

    ReplyDelete

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