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Anyone heard of this?

  1. Jan 9, 2007 #1
    so apparently, this problem usually would require, many hours, and quite a few pages to write a proof for this, but today, i was shown a way to solve it in only a few mintues with only a few steps...

    an object of mass m collides elastically with a second object of equal mass initially at rest. If the collision is a glancing collision, prove they separate at 90 degrees.

    where m1 = mass of object 1, m2 =mass of object 2, v = velocity of m1, x = velocity after collision of object 1, y = velocity of object after collision.



    so then

    1/2(m1)v² = 1/2(m1)x² + 1/2(m2)y²
    cancel out and simplify
    v² = x² + y²

    would that be completely valid?
  2. jcsd
  3. Jan 10, 2007 #2

    Andrew Mason

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    Almost. You have to add the condition that [itex]\vec v = \vec x + \vec y[/itex], which is a consequence of conservation of momentum. Since this is a triangle and v² = x² + y², as you have shown, then the angle between [itex] \vec x \text{ and } \vec y[/itex] must be 90 degrees.

    This is a very useful thing to know for pool players, who use this principle to determine where the cue ball will go.

  4. Jan 10, 2007 #3
    Consrvation of momentum states that:
    [tex]m_1\vec v=m_1\vec x + m_2\vec y[/tex]

    So why is that your condition is true?
  5. Jan 10, 2007 #4


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    Because we are given that m1= m2?
  6. Jan 10, 2007 #5
    Oh my bad... I am sorry, I missed that.
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