Mechanics - Collisions

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  • #1
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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...
 
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Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #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]?
 
  • #5
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Agree. You're right. Anyways, is this the way to proceed?
 
  • #6
CAF123
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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##.
 
  • #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?
 
  • #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.)
 
  • #9
vela
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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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