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Scattering angle in the CM frame

  1. Sep 23, 2007 #1
    [tex]\alpha[/tex] particle of mass m is scattered by a nucleus of mass M. [tex]\Theta[/tex] is the scattering angle of the [tex]\alpha[/tex] particle in the LAB reference frame, and [tex]\theta[/tex] is the scattering angle in the CM frame.

    What is the relation between [tex]\Theta[/tex] and [tex] \theta [/tex] using conservation of energy and momentum?
    I am suing v1 as velocity of m and v2 as velocity of M

    [tex]v_{CM} = \frac{m v_1}{m+M}[/tex]

    in the CM frame (denoted by v'):
    [tex]v'_2 = v_1 - v_{CM} = \frac{M v_1}{m+M}[/tex]
    [tex]v'_2 = v_{CM}=\frac{mv_1}{m+M}[/tex]

    for elastic scattering:

    [tex]v cos \theta -v_{CM} = v'_1 cos\Theta [/tex]
    [tex]v cos \theta = v'_1 cos\Theta +v_{CM} [/tex]

    we also know:
    [tex]v \sin \theta = v'_1 \sin \Theta [/tex]

    dividing the two expressions, we get:

    [tex]\tan \theta = \frac{sin\Theta}{\cos \Theta + \frac{m}{M}}[/tex]

    i was wondering this was the correct logic to solving this problem? I didnt use conservation of energy, and was wondering if there was something that i missed.
  2. jcsd
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