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Elastic collisions between proton and helium nucleus

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data


    In an elastic collision between a proton and a helium nucleus at rest, the proton was scattered through an angle of 45 degrees. What proportion of its initial energy did it lose? what was the recoil angle of the helium nucleus?
    2. Relevant equations

    tan[tex]\theta[/tex]1=(sin[tex]\Psi[/tex]))/((cos([tex]\Psi[/tex])+[tex]\gamma[/tex])

    [tex]\theta[/tex]2=1/2(pi-[tex]\Psi[/tex])

    tan[tex]\theta[/tex]=(([tex]\gamma[/tex])+1)/([tex]\gamma[/tex]-1)*cot(.5*[tex]\gamma[/tex])

    E2/E0=4[tex]\gamma[/tex](sin(1/2*[tex]\gamma[/tex])^2/([tex]\gamma[/tex]+1)^2

    phi is the scaterttering angle in the ZM frame and gamma=m2/m1, the mass ratio of the two particles. [tex]\theta[/tex]2 is the recoil angle and [tex]\theta[/tex]1 is the scatter angle.

    3. The attempt at a solution

    mhelium=4*mproton , therefore gamma=1/4
    [tex]\theta[/tex]1=45 degrees

    tan(45 degrees)=sin(phi)/(cos(phi)+.25)
    1=sin(phi)/(cos(phi)+.25)==> cos(phi)+.25=sin (phi)

    not sure how I can determine phi with sin(phi)-cos*(phi)=.25 ; I know I need phi to determine the recoil angle.

    not sure how to determined how much initial energy was lost but I know it probably will have to apply this elastic formulae:

    E2/E0=4[tex]\gamma[/tex](sin(1/2*[tex]\gamma[/tex])^2/([tex]\gamma[/tex]+1)^2

    Does E0 represent the initial energy?
     
  2. jcsd
  3. Jan 22, 2009 #2

    tiny-tim

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    Hi pentazoid! :smile:

    (have a phi: φ :wink:)

    Hint: sin(φ - 45º) = sinφcos45º - cosφsin45º :wink:
     
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