# Understanding elastic collisions of objects with the same velocities

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.

• Have you written out the energy conservation equation for both masses?
– user12029
Commented Oct 22, 2015 at 0:59
• @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
– Ovi
Commented Oct 22, 2015 at 1:15
• There are numerous issues with the given quotation. Are you sure you have it down correctly? Commented Oct 22, 2015 at 1:31
• @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?
– Ovi
Commented Oct 22, 2015 at 1:43
• 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.
– user12029
Commented Oct 22, 2015 at 2:03