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Modern Physics: Finding pressure using energy.

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A beam of pions ([tex]\pi[/tex]-mesons, with rest energy m[tex]\pi[/tex] c2=140 MeV) consisting of 1010 pions/cm^2s moves with speed such that the total energy of each pion is 10^5 MeV. What is the pressure the beam exerts on a surface that absorbs it and brings it to rest?

    Please note: For some reason my pi's are showing up strangly. The first one is supposed to say pi-mesons and the second one is supposed to be a subscript. Thanks.

    2. Relevant equations



    3. The attempt at a solution

    I am not sure where to start. Any help would be appreciated.
     
  2. jcsd
  3. Sep 16, 2010 #2

    diazona

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    Well, what do you know about pressure?
     
  4. Sep 16, 2010 #3
    Well I know that it is force per unit area.
     
  5. Sep 16, 2010 #4

    diazona

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    OK, now remember that pressure is always calculated over a surface. To find pressure, you pick a surface, find the force on the surface, and divide by the area of the surface. Can you do that?
     
  6. Sep 16, 2010 #5
    I see what you are saying, but I am still unsure how to proceed with tis problem. I'm having trouble finding the relationship between pressure and energy. My idea is that I have to find the Force, and for that I need mass and acceleration. I can find mass from the rest energy, but the acceleration is what is confusing me.
     
  7. Sep 16, 2010 #6

    diazona

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    No, actually you don't need mass and acceleration. There is another, more general form of Newton's law that you can use - do you know what it is?
     
  8. Sep 16, 2010 #7
    Is it F + U*dm/dt=m*dv/dt?
     
  9. Sep 16, 2010 #8

    diazona

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    No, I'm thinking of something else, more general than that.
     
  10. Sep 16, 2010 #9
    Hmmm I'm not sure then. That was the most general I could think of.
     
  11. Sep 16, 2010 #10

    diazona

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    Have a look at the Wikipedia page for Newton's second law. You'll find it there.
     
  12. Sep 16, 2010 #11
  13. Sep 16, 2010 #12

    diazona

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    Yep, that's the one.

    So, you need to pick a surface, find the force on that surface using F = dp/dt, and then divide the force by the area. Can you make any progress on that?

    By the way, in this case, since the force is constant over time, the force will be equal to its average. That means that you can use either Favg = Δp/Δt or F = dp/dt, whichever one makes more sense to you.
     
  14. Sep 16, 2010 #13
    Hmm ok I'll work on it tonight and let you know tomorrow afternoon.
     
  15. Sep 16, 2010 #14

    diazona

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    OK; I probably won't be able to respond at that time (traveling without internet access) but someone else can take over.
     
  16. Sep 17, 2010 #15
    Ok so I wasn't able to make much progress on this problem. Can anyone else who reads this help me? Thanks in advance.
     
  17. Sep 17, 2010 #16
    [tex]\Delta P =P(final)-P(initial)[/tex]
     
    Last edited: Sep 18, 2010
  18. Sep 17, 2010 #17
    Hmm yeah I understand that but I am having trouble getting to the point where I can use that equation.
     
  19. Sep 18, 2010 #18
    Consider only one particle moving to the wall.
    Can you get
    [tex]
    \Delta P
    [/tex]
    for this particle?
     
  20. Sep 18, 2010 #19
    Well I know P=gamma*m*v. I think I can find 'm' from the rest energy, and 'v' from finding first the Kinetic Energy, then solving for 'v.' Finally I think I can find gamma after I have all these other values. Is this correct?
     
  21. Sep 19, 2010 #20

    diazona

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    Seems reasonable to me. You can try it, and if you run into any problems you can't track down yourself, post your work here and we'll help you check it.
     
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