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Kinetic energy related

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    A stationary radioactive nucleus of mass M ejects an α-particle of mass 'm' at a speed of 2.0 x 10^7 ms^-1 . Given M=55m, calculate the kinetic energy of the α-particle as a percentage of the final total kinetic energy.


    2. Relevant equations

    K.E=1/2 x M x V^2 {half m[v(squared)]}
    Where m= mass,
    v= velocity

    3. The attempt at a solution
    Dont know where to start at this to be honest :(

    Thanks for any help given
     
  2. jcsd
  3. Sep 18, 2010 #2

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    Hello Psycotic.Goth,

    Welcome to Physics Forums!
    Since you know the velocity of the alpha particle, you should be able to calculate its kinetic energy.

    Using conservation of momentum, you can find the velocity of the resulting nucleus (Hint: don't forget, since it has ejected mass m, the resulting nucleus' mass is less than it was originally). And if you know its velocity, you can find its kinetic energy.

    The final total kinetic energy is the sum of both.
     
  4. Sep 18, 2010 #3
    Im not sure how to calculate the mass of the alpha particle
     
  5. Sep 18, 2010 #4

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    Just leave everything in terms of the variable m. Set up your equations, and see if anything cancels out. :wink:
     
  6. Sep 21, 2010 #5
    Hi again - thanks that did help - Well, i've done the first part of getting the velocity using the conservation of momentum formula - I got it out to be 363636.3636 - Now im stuck at finding the mass 'm' :((
     
  7. Sep 21, 2010 #6

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    Ummmm ...:uhh:... Please show your work, on this one. But let me give you a hint. Let's call [itex] v_{\alpha} [/itex] the velocity of the alpha particle, and [itex] v_n [/itex] the velocity of the resulting nucleus, then noting that mass of the nucleus = (M-m) = (55m - m) = 54m, then conservation of momentum says,

    [tex] v_{\alpha} m = 54v_n m [/tex]

    Try taking it from there. :wink:
    Look how the 'm's can cancel out in the conservation of momentum equation. See what happens when you set up your ratio of alpha particle's kinetic energy per total kinetic energy. Will the 'm's cancel out once again?
     
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