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Conservation of momentum

  1. Oct 22, 2005 #1
    A He4 nucleus, with a mass of 4 amu moving with speed v breaks up into a neutron 1amu and a He3 nucleus 3amu . If the neutron moves in a direction perpendicular to the direction of motion of the original He4 nucleus with speed 3v, what is the speed of the He3 nucleus?

    if the neutron moves down. The nucleus will be pushed upward with 1/3v. the nuetron also has the original 1v in the horizontal direction. so should i just use pythagorean theorem?: so i get sqrt(10)v/3?
     
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  3. Oct 23, 2005 #2

    Päällikkö

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    Incorrect:
    Incorrect.
     
  4. Oct 23, 2005 #3
    which of those quotes are incorrect? should the horizontal velocity of the nucleus increase as well? im so confused.
     
  5. Oct 23, 2005 #4

    Päällikkö

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    "the nuetron also has the original 1v in the horizontal direction"
    "The nucleus will be pushed upward with 1/3v"

    The above are incorrect. Horizontal momentum must be conserved, and the only thing moving horizontally is the nucleus. So yes, its velocity in x direction must increase.
     
    Last edited: Oct 23, 2005
  6. Oct 23, 2005 #5
    so the nucleus will be pushed +v/3 in the x direction so the total x direction is 4v/3.
    In the y direction the velocity will be sqrt(3v^2 - v^2) for the neutron. and 1/3 that for the nucleus?
     
  7. Oct 23, 2005 #6

    Päällikkö

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    Yep.

    In y-direction it is given in the problem that the neutron has 3v. And yes, 1/3 (although I'm not sure how you got the ratio, did you use the conservation of momentum?) of the neutron's velocity in y direction is the nucleus' velocity in y direction.
     
  8. Oct 23, 2005 #7
    ahhh, now i get it, the velocity of the neutron is only in the y direction. (yes i used conservation of momentum) so the total speed of the neucleus is sqrt(v^2 + (4v/3)^2) which is 5v/3! thanks a lot
     
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