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Calculating how much kinetic energy is transfer?

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A neutron in a nuclear reactor makes an elastic head on collision with with the nucleus of a carbon atom initially at rest. (the mass of the carbon nucleus is 12 time the mass of the neutron)

    a) what fraction of the neutron's kinetic energy is transferred to the carbon nucleus?



    2. Relevant equations

    momentum P(after)=P(before)
    M(1a)V(1a)+M(2a)V(2a)=M(1b)V(1b)

    Kinetic Energy
    M(1a)V(1a)+M(2a)V(2a)=M(1b)V(1b) "I cancel the 1/2 already"


    3. The attempt at a solution
    I know that I can make the mass any number I want but for simplicity it would be
    M(1)=1
    M(2)=12

    I though I could calculate the speed of the second mass after the collision and then subtracted from the speed of the first mass and calculate the difference but I dont know the initial speed either can some show how to solve this problem? please
     
  2. jcsd
  3. Oct 20, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    You must pretend that you know the initial velocity. You could just use any number for that and you'll get the same answer regardless of its value, but it will be much more satisfying for you to use a variable. I used u, v and w for the three velocities to save writing. I solved the momentum equation for v, subbed into the energy equation (note that you need to square your velocities!) and got a quadratic in w that I could solve for the non-trivial case and get the ratio of w to u. That is all you need to find the relative kinetic energies you need.
     
  4. Oct 21, 2009 #3
    OK I did something like this.
    c= carbom
    n=nucleus

    Mn=1kg
    Mc=12kg
    Vin=U
    Vfn=V
    Vfc=W

    so
    MnV+McW=MnU "now solve for V"

    V=(MnU-McW)/Mn or V=U-12W "now subbed into the energy equation"

    MnV^2+McW^2=MnU^2 "going to make it easier by diving everything by Mn"
    V^2+12W^2=U^2
    (U-12W)^2+12W^2=U^2

    U^2-24WU+144W^2+12W^2=U^2 "I subtract U^2 from both sides "

    -24WU+156W^2=0

    12W(-2U+13W)=0 "divide by 12W"

    -2U+13W=0 "solve for W"

    13W=2U >>>>>> W=(2U)/13

    now I am stuck here becuase if I choose an arbitrary number for U let say 3

    W=6/13 >>>> W=.4615 but the answer says .284 ???????

    is something I did wrong or is there more steps I need to do ?????
     
  5. Oct 21, 2009 #4

    rl.bhat

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    Homework Helper

    KE of neutron = 1/2*m*U^2
    KE of carbon nucleus = 1/2*12m*[2U/13]^2
    Take the ratio.
     
  6. Oct 21, 2009 #5
    I got it thanks guys
     
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