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I am having trouble understanding the following statement from my book:

The law of cosines tells us that if the sides of a triangle obey the Pythagorean formula, they must form a right triangle. This fact explains why the final velocities of two equal mass bodies that undergo a perfectly elastic two-dimensional collision must be orthogonal.

This doesn't really make sense to me; if there are two pool balls traveling with the same velocity along paths with an angle of intersection of 5 degrees, it doesn't seem like the balls should go at a 90 degree angle from each other once they collide. I also don't see how you can derive this fact from the Pythagorean theorem.

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    $\begingroup$ Have you written out the energy conservation equation for both masses? $\endgroup$
    – user12029
    Commented Oct 22, 2015 at 0:59
  • $\begingroup$ @NeuroFuzzy Well it would be $\dfrac 12 m v_{Ai}^2 + \dfrac 12 m v_{Bi}^2 = \dfrac 12 m v_{Af}^2 + \dfrac 12 m v_{Bf}^2$ which would simplify to $v_{Ai}^2 + v_{Bi}^2 = v_{Af}^2 + v_{Bf}^2$, but that doesn't quite look like the pythagorean theorem, unless one of the initial velocities were zero. I thought about that, but in the quote I wrote the book doesn't restrict one of the initial velocities to being zero $\endgroup$
    – Ovi
    Commented Oct 22, 2015 at 1:15
  • $\begingroup$ There are numerous issues with the given quotation. Are you sure you have it down correctly? $\endgroup$
    – DilithiumMatrix
    Commented Oct 22, 2015 at 1:31
  • $\begingroup$ @DilithiumMatrix Yes I double checked. The quote is actually from a worksheet which was quoting the textbook, so maybe the worksheet got it wrong (sadly I don't have access to the textbook). Could you please tell me what is wrong with the quotation? $\endgroup$
    – Ovi
    Commented Oct 22, 2015 at 1:43
  • $\begingroup$ Oh! I misread it and misread your question. @Ovi the author of that quote surely meant that you're in a frame where one particle is at rest! That is the usual result. $\endgroup$
    – user12029
    Commented Oct 22, 2015 at 2:03

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I think, what the given quote is trying to say is that in a perfectly elastic collision in which the velocities are initially perpendicular* then the final velocities will also be perpendicular.

In any case, this is really the only valid interpretation of this quote. So don't worry too much about it... it's very poorly written, and very understandable that you were confused!

*I guess that's what's meant by 'two-dimensional' collision... although very unclearly.

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