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COM/COE Problem

  1. Dec 20, 2004 #1
    I'm going on 10 years since I got my (physics) degree, but I haven't done much work besides conceptual stuff in that time and it shows. :mad: Anyway, I decided to crack open my Modern Physics book from school and start working my way through it. The first few problems were OK, but I just can't quite seem to get a handle on how to work through this one.

    Here it is:

    A beryillium atom (m[tex]\approx[/tex]8.00u) is moving in the positive x direction with a kinetic energy of 60.0 keV. It splits into 2 Helium atoms (m[tex]\approx[/tex]4.00u) with the release of 92.0 keV of Energy. Particle 1 is found to move at 30.0o to the x-axis ([tex]\theta_{1}[/tex]). Using Conservation of Energy and Momentum (i.e., no coordinate transformations), find v1, v2, and [tex]\theta_{2}[/tex].

    Conversion factor: [tex]1u = 9.3150 x 10^5 \frac{keV}{c^2}[/tex]

    I think I've got the COE part set up correctly. Given that [tex]m_{1}=m_{2}\equiv m[/tex]

    [tex]60.0 keV + 92.0 keV = \frac{1}{2}m(v^2_{1} + v^2_{2})[/tex]

    By substituting known values, I get [tex]v^2_{1}+v^2_{2}=8.16x10^{-5} c^2[/tex]

    The COM portion seems to be throwing me for a loop somehow. No matter how I set them up, when I substitute back into the COE equation, I end up with a really nasty looking quadratic equation. I know that the angle given [tex]\theta_{1}[/tex] and the relationship it implies for v1 to v1x and v1y must be the key, but I just can't seem to make it fit.

    Thanks for any help you can offer,

  2. jcsd
  3. Dec 21, 2004 #2


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    For CoM :

    [tex](2m)v = mv_1 cos30 + mv_2 cos \theta _2[/tex]


    [tex]mv_1 sin30 = mv_2 sin \theta _2 [/tex]

    In both equations, you can cancel off m.

    Hope that helps. But I still suspect you'll wind up with a quadratic that you must solve.
  4. Dec 21, 2004 #3

    Andrew Mason

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    Translate into the frame of reference of the original beryillium nucleus. The quadratic equation will be easier. The velocities of the two He nuclei have to be equal and opposite in that frame. So you have:

    [tex]\frac{1}{2}m(v'^2 + v'^2) = mv'^2 = E = 92Kev[/tex]
    (1)[tex]v' = \sqrt{E/m}[/tex]

    (2)[tex]v'sin\theta' = v_1sin30 = v_1/2[/tex]
    (3)[tex]v'^2 = (v_1cos30 - v_0)^2 + (v_1sin30)^2[/tex]
    (4)[tex]sin\theta_2 = v_1sin30/v_2 =v_1/2v_2[/tex]
    (5)[tex]v_2^2+v_1^2 = 2(E + E_0)/m[/tex]

    From (3):
    [tex]v'^2 = (v_1cos30 - v_0)^2 + (v_1sin30)^2[/tex]

    [tex]v'^2 = v_1^2cos^230 + v_0^2 - 2v_1cos30v_0 + v_1^2sin^230[/tex]

    [tex]v'^2 - v_0^2 = v_1^2(cos^230 + sin^230) - 2v_1cos30v_0[/tex]

    This is a quadratic equation of form
    [itex]v_1^2 + av_1 + b = 0[/itex] where [itex] a = -2cos30v_0 \text{ and } b = - (v'^2 - v_0^2)[/itex]

    Solve for v_1. [itex]\theta'[/itex] follows from (2). Getting v_2 and its direction follows using (5) and (4),

    Last edited: Dec 21, 2004
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