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Conservation of Four-Momentum

  1. Oct 27, 2012 #1
    Hello. I've uploaded the question onto pictures and linked them. I hope this method of asking a question works.
    I'm trying to find the mass of the ball, but I get the square root of a negative number. Can someone take a look to see if I'm doing this problem correctly?

    IMAG0012.jpg
    IMAG0013.jpg
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 27, 2012 #2

    Simon Bridge

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    What immediately occurs to me:
    In ##p^0=E/c## is that E the rest-mass energy, total energy, or kinetic energy?
    (Closer look - you seem to have calculated gamma ... apologies: it is difficult for me to see your reasoning.)

    It can help see where you went wrong (or right!) if you leave the actual numbers out until the end.

    eg. In space-station coordinates: ##P_i = (M+m)\gamma_i(1,0,0,u)^t;\; P_f=M(1,0,0,0)^t+m\gamma_f(1,0,0,v)^t## (...or something) you have to find ##v## from the conservation laws and knowing ##M##, ##m##, ##\gamma_i##, and ##u##.

    The laws give you two equations.
    If you do this as a list, formally, you are less likely to get the numbers mixed up at the end.
     
    Last edited: Oct 27, 2012
  4. Oct 28, 2012 #3
    OH I see. The mass of the astronaut in the final momentum is not 100kg like I thought. Have to give it the Variable M. Ok This will be a piece of cake then. Thanks.
     
  5. Oct 28, 2012 #4

    Simon Bridge

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    No - I just used variables to keep track of what's what.
    Remember that the questions says to assume the astronaught's mass does not change.
    All the masses I used are rest-masses.

    What I am trying to get you to do is go back over the calculation more formally.
     
  6. Oct 28, 2012 #5
    No actually the question says it is the mass of the ball that does not change.
    Here is my answer. I'm about 90% sure of it. Can you please check it out and see if what I did is good?
    IMAG0016.jpg
     
  7. Oct 28, 2012 #6

    Simon Bridge

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    Oh so it does - I'm corrected. I find it ambiguous anyway ... why would it change?
    Why do you make people work so hard to read your working? It'll cost you marks.
     
  8. Oct 28, 2012 #7
    Is my handwriting/work too sloppy? I will try to make it much neater.
     
  9. Oct 28, 2012 #8

    Simon Bridge

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    It's the layout - your working is all over the place so I cannot easily follow your reasoning. Someone marking your work could easily think you've left something out of misunderstood something. At least there's boxes around your final answers but those are worth, at most, one mark each.

    You can reality check your results though ... should the speed of the ball be very high to cancel the momentum? Why would the astronaught's mass change (and is that a reasonable mass for an astronaught?)
     
  10. Oct 28, 2012 #9
    Ok here is what I've concluded about the velocity of the ball and the astronaut's mass. The velocity of the ball must be extremely high (close to c=1) to balance out the fact the the astronaut has velocity 0 in the final momentum in the x direction. Also the astronaut's mass in the beginning is 100kg whereas her final mass is 50.5 kg. This is because in the beginning the astronaut was moving with v=3/5 which is a big enough velocity to make her much more massive than when she is standing still. 50.5kg is about 110lbs which makes sense for a human being.

    ps. Is the following statement correct?
    As velocity increases approaching c=1, the more massive you get.
     
  11. Oct 28, 2012 #10

    Simon Bridge

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    So the masses quoted are not rest-masses, but kinetic energy contributions seen in the space-station frame? (Nobody literally gains mass by virtue of going fast.)

    This also suggests a very small astronaut doesn't it?

    I always thought the m in the 4-momentum was rest-mass though.
    If the initial mass is not rest mass them don't we put ##\gamma_i (M+m)=101kg##?
     
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