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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.

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  1. JamesJ
    • 4 years ago
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    Step 1: draw a diagram. I'd imagine the body initially at rest at the origin; the moving body approaching it along the negative x-axis. After the collision, by Conservation of Momentum, the x-components of the two bodies momentum must sum to the original momentum; and the y-components must sum to zero.

  2. anonymous
    • 4 years ago
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    I think more important here is the conservation of energy, since the conservation of momentum solely cannot really constrain the angle between the final vectors.

  3. JamesJ
    • 4 years ago
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    Yes, you'll need both. I was just giving a starting point.

  4. anonymous
    • 4 years ago
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    Actually, to proof the 90°, I think you need only the conservation of energy.

  5. JamesJ
    • 4 years ago
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    I'll be impressed if you can write down an argument with only the conservation of energy.

  6. anonymous
    • 4 years ago
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    I did this and I don't know what to do after. attached a file

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  7. anonymous
    • 4 years ago
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    What 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 \[ \vec v'_1 \perp \vec v'_2\]

  8. anonymous
    • 4 years ago
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    Okay, I see, you were right. Implicitly I was already using the momentum conservation. Sorry.

  9. JamesJ
    • 4 years ago
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    I was secretly hoping you were right. It's always good to be surprised. But in this case, I think we really do need both.

  10. anonymous
    • 4 years ago
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    I 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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