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Elastic collision neutron problem

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A neutron has an elastic collision with a stationary deuteron. Mass of neutron is 1 amu, mass of deuteron is 2 amu. the neutron is deflected through 90 degree angle. Prove that the neutron loses 2/3 of it's initial kinetic energy to the deuteron.

    2. Relevant equations

    I'm guessing the conservation of momentum and kinetic energy equations


    3. The attempt at a solution

    I really have no idea where to start with this one. If the neutron is deflected through 90 degrees, does that mean the deuteron is too? I don't want the exact answer - just point me in the general direction of reasoning!!

    thanks for any help
     
  2. jcsd
  3. Sep 9, 2009 #2

    Hootenanny

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    Have you learnt about four-moment yet? If you haven't it doesn't matter, it would just makes things a little quicker.
     
  4. Sep 9, 2009 #3
    No - I have no idea what that is. This is a first year physics course if that helps?
     
  5. Sep 9, 2009 #4

    Hootenanny

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    Okay, no problem. In that case, I suggest you split the momentum into two components: (a) parallel to the initial velocity of the neutron, and (b) perpendicular to it. Can you write down two equations representing conservation of momentum for each component, (a) & (b)?

    P.S. You have done SR, right?
     
  6. Sep 9, 2009 #5
    Yeah - I've been playing around with those equations last night (cos theta for x component, sin theta for y component etc)....The neutron goes from having full x component momentum to full y component momentum etc

    But what about the deuteron? Does it go 90 degrees down if the neutron goes 90 degrees up? I'm guessing no because the law of conservation of linear momentum of the center of mass of the system says that the COM should carry on in the original direction right??

    dude - i guess i'm just totally lost on this one - it is frustrating me and I'm getting all the formula jumbled up the more worked up I get haha

    PS - sorry what does SR stand for?
     
  7. Sep 9, 2009 #6

    Hootenanny

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    Correct.
    That is what we're here to work out. It's very simply, start by writing down the basic equation for conservation of momentum.
    Special Relativity.
     
    Last edited: Sep 9, 2009
  8. Sep 9, 2009 #7
    ok man - i take back what i said earlier about not wanting the answer. If you know how to do it please could you show me. I have started with the basic equations like you said (although this is about the tenth time I'm doing that). I still don't get anywhere near the answer I'm looking for. I'm not seeing something. I have the equation for conservation of kinetic energy, the x component momentum and y component momentum equations, theta = 90, phi = unknown, mass1 = 1.0 amu and mass2 = 2.0 amu.

    I'm not going to write out the equations here cos it will be a mission - but due to the fact that cos90 = 0 and sin90 = 1 they reduce into really simple equations. But I still don't see how to get the 2/3 ratio?

    The only way I can sort of come right is if I use the one dimension elastic equation:

    v2final = [2(mass1) / mass1 + mass2 ] v1initial

    in that equation the 2/3 ratio works out.

    We have done SR but this is just straight forward stuff - no need to get involved in the relativistic equations here...

    thanks,
     
  9. Sep 9, 2009 #8

    kuruman

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    Can you show us the equations you wrote down? There should be three, two for momentum conservation in the x and y direction and one for energy conservation. If we see what you have done, we could provide guidance about where you went wrong or where to go from there.
     
  10. Sep 9, 2009 #9
    ok man - this is going to look ugly but here goes:

    x - component momentum conservation:

    mass1*v1initial = mass2*v2final * cos phi

    y - axis:

    mass1 * v1final = mass2 * v2final sin phi

    kinetic energy:

    (note sqr() means the square not the square root)

    1/2mass1 * sqr(v1initial) = [1/2mass1 * sqr(v1final)] + [1/2mass2 * sqr(v2final)]

    I've been trying to get the answer from these 3 eqs but I am missing something (probably something really dumb, I admit)

    thanks for any help!!
     
  11. Sep 9, 2009 #10

    kuruman

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    If you think that's ugly, it is going to look uglier before you are done. :eek:

    Your two momentum conservation equations are incorrect. After the collision, both masses are moving, so there should be two terms on the right hand side. That's probably what kept you from getting to the end.
     
  12. Sep 9, 2009 #11
    yeah - but the neutron is deflected 90 degrees - so that term becomes zero in one equation and 1 in the other, reducing the equations to the ones I have written, not so? Or am I still wrong?
     
  13. Sep 9, 2009 #12

    kuruman

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    I'm sorry, you are correct. I got confused a bit by your equations. To be 100% correct, the y-equation for momentum must have a negative sign in front of one of the terms. But that is not very important because the next step would be to get rid of the angles. What would happen if you squared the two momentum equations and then added them together?
     
  14. Sep 21, 2009 #13
    Let's say the neutron moves in the +y direction before the collision. It is moving at right angles, let's say along the negative x axis, to this afterward, so the deuteron has acquired a momentum equal to this along the positive x axis. since momentum is conserved. For the same reason, the dueteron's y momentum is the same as the neutrons prior to collision. The description above conserves all momentum. The answer desired should follow readily from this.
     
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