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anonymous
 4 years ago
Prove that in the elastic collision of two objects of identical mass, with one being a target initially at rest, the angle between their final velocity vectors is always 90 degrees.
anonymous
 4 years ago
Prove that in the elastic collision of two objects of identical mass, with one being a target initially at rest, the angle between their final velocity vectors is always 90 degrees.

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Step 1: draw a diagram. I'd imagine the body initially at rest at the origin; the moving body approaching it along the negative xaxis. After the collision, by Conservation of Momentum, the xcomponents of the two bodies momentum must sum to the original momentum; and the ycomponents must sum to zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think more important here is the conservation of energy, since the conservation of momentum solely cannot really constrain the angle between the final vectors.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, you'll need both. I was just giving a starting point.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, to proof the 90°, I think you need only the conservation of energy.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I'll be impressed if you can write down an argument with only the conservation of energy.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did this and I don't know what to do after. attached a file

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What about \[ \frac{1}{2} m \vec v_1^2 = \frac{1}{2} m \vec v'_1^2 + \frac{1}{2} m \vec v'_2^2 \] \[\Rightarrow \vec v_1^2 = \vec v'_1^2 + \vec v'_2^2 \] Now think about the vectorial sum and the Pythagorean theorem and it's clear that this holds only if \[ \vec v'_1 \perp \vec v'_2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, I see, you were right. Implicitly I was already using the momentum conservation. Sorry.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I was secretly hoping you were right. It's always good to be surprised. But in this case, I think we really do need both.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't even need to use any of the trig stuff to prove? So your equation is basically saying that the lines are in a pythygorean relation. The resultant is the initial momentum. dw:1328308175582:dw
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