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Giancoli, 3rd Ed, pr. 55: Ellastic collision. Solved, but is there easier way?

  1. Aug 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A neutron collides elastically with an at-rest helium-nucleus. Masses are approximately "m" and "4m", respectively. Helium nucleus rebounds at 45-degree angle

    What angle does the neutron rebound at?

    What are speeds of two particles after collision?

    {v_{4m}} = \left[ {{\rm{pre - collision He - nucleus velocity}}} \right] = 0 \\
    {v_m} = \left[ {{\rm{pre - collision He - nucleus velocity}}} \right] = 6.5 \times {10^5}{\textstyle{m \over s}} \\
    {{v'}_{4m}} = \left[ {{\rm{post - collision He - nucleus velocity}}} \right] = ? \\
    {{v'}_m} = \left[ {{\rm{post - collision He - nucleus velocity}}} \right] = ? \\

    2. Relevant equations
    Kinetic energy conservation, simplified, and with appropreiate zero of initial He-nucleus velocity put in, and with factor of (1/2)*m cancelled out:
    [tex]{v_m}^2 = 4{v'_{4m}}^2 + {v'_m}^2[/tex]

    x-component momentum conservation with factor of "m" struck out:
    [tex]6.5 \times {10^5}{\textstyle{m \over s}} = {{v'}_m}\cos \theta + 4{{v'}_{4m}}\cos 45[/tex]

    y-component momentum conservation with factor of "m" struck out:
    [tex]\sin \theta = - \sin 45{\textstyle{{{{v'}_{4m}}} \over {{{v'}_m}}}}[/tex]

    3. The attempt at a solution

    Isolate [tex]{{{v'}_m}}[/tex] and put next to trig functions and square both sides in preparation to add them together to get a [tex]{\sin ^2} + {\cos ^2}[/tex] simplification:

    {\left( {{{v'}_m}\cos \theta } \right)^2} = {\left( {6.5 \times {{10}^5}{\textstyle{m \over s}} - 2\sqrt 2 {{v'}_{4m}}} \right)^2} \\
    {\left( {{{v'}_m}\sin \theta } \right)^2} = {\left( { - {{v'}_{4m}}\sin 45} \right)^2} \\

    Adding these equations and using the abovementioned [tex]{\sin ^2} + {\cos ^2}[/tex] leads to:
    [tex]{({{v'}_m})^2} = {6.5^2} \times {10^{10}} + 8{({{v'}_{4m}})^2} - 2 \cdot 6.5 \times {10^5} \cdot 2\sqrt 2 {{v'}_{4m}} + {\textstyle{1 \over 2}}{({{v'}_{4m}})^2}[/tex]

    Then: use the kinetic-energy balance mentioned way earlier to eliminate an unknown.

    THE QUESTION: This leads to the quadratic formula. I am studying to take the Physics GRE. Is there a way to avoid the quadratic formula? Some sort of physical-intuition-shortcut? I tried visiting the CM frame briefly, but that makes things a bit messy....but perhaps I didn't look hard enough?
  2. jcsd
  3. Aug 11, 2010 #2


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    There's a mistake in your momentum y equation. You dropped the factor of four that comes from the mass of the helium nucleus.

    Your approach is fine. The quadratic you get is factorable if you do the algebra correctly, with one root being v'4m=0. It might help not to plug in the value for vm so early.
  4. Aug 11, 2010 #3
    I mean yeah, unless you know scattering formulas off the top of your head you will have to derive the result just like you have here. Though fix your error as stated above.
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