(adsbygoogle = window.adsbygoogle || []).push({}); 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?

Notation:

[tex]\begin{array}{l}

{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] = ? \\

\end{array}[/tex]

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:

[tex]\begin{array}{l}

{\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} \\

\end{array}[/tex]

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?

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

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