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Mechanics - Collisions

  1. Jan 14, 2015 #1
    A particle with mass m is at rest. A second one, identical to the last one hits the first one. Show that in the case of a perfectly elastic collision (Q=0) the directions of the two particles make a right angle.

    You can't assume that both final velocities will be equal. Here's what I've got from using: 1. the conservation of linear momentum. 2. The conservation of kinetic energy (Q=0). 3. Scalar product between the initial velocities.

    v=u1cosα + u2 cosβ
    u1 sinα = u2 sinβ
    v2=u12 + u22
    cos(α+β)=(u12 cosα cosβ + u22 sinα sinβ) / (u1 u2)

    And hell, I really can't solve this system...
    Last edited: Jan 14, 2015
  2. jcsd
  3. Jan 14, 2015 #2

    Stephen Tashi

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    You haven't defined the variables you used. You didn't say how you established the coordinate system. If you are going to portray a collision in 3-dimensions as 2-dimensional problem, you should say why this is possible.
  4. Jan 14, 2015 #3
    OK. That's my bad, I'm sorry. We're in 2 dimensions.
    v represents the initial velocity of the particle. (The one hitting).
    u represents the final velocity.
    α, β represent the angles made with the direction of the particle 1 (The one hitting)
    - α for the one that was moving previously and β for the one that was at rest.
    They way I got to that system is explained up there.
  5. Jan 14, 2015 #4

    Stephen Tashi

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    Why does your expression derrived from the inner product [itex] (u1 \cos(\alpha), u1 \sin(\alpha)) \cdot (u2\cos(\beta), u2\sin(\beta)) [/itex] have the factors [itex] u_1^2 , u_2^2 [/itex]? Why not the factor [itex] u_1u_2 [/itex]?
  6. Jan 14, 2015 #5
    Agree. You're right. Anyways, is this the way to proceed?
  7. Jan 14, 2015 #6


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    Hi, manipulate your equations from conservation of linear of momentum and kinetic energy (three equations) to eliminate all instances of ##v, u_1## and ##u_2##.
  8. Jan 14, 2015 #7

    Stephen Tashi

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    The simplistic way to look at it is that two momentum vectors add to be the vector velocity V and that conservation of energy implies that the lengh of the hypotenuse V squared is the sum of the squares of the lengths of the other two sides. Hence it is a right triangle. Can we put that into algebra?
  9. Jan 14, 2015 #8

    Stephen Tashi

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    Try the law of cosines for triangles.

    (By the way, the angle between the two velocity vectors wouldn't be sum of angles [itex] \alpha, \beta [/itex] , it would be a difference of them. The the addition law for [itex] cos(A +B) [/itex] should be consistent with the signs of the cosine terms in the inner product.)
  10. Jan 15, 2015 #9


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    Another approach is to work with the momentum vectors without breaking them into components. Conservation of momentum gives you
    $$\vec{p}_1 + \vec{p}_2 = \vec{p}'_1 + \vec{p}'_2$$ where ##\vec{p}_1 = 0##. Square both sides and use the fact that the kinetic energy can be written as ##\frac{\vec{p}^2}{2m}##.
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