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Collision in the cm frame

  1. Aug 30, 2012 #1
    I have never understood why it's beneficial to do a collision in the cm frame, but then I did an example for myself and it was easier. I just don't understand why.

    I had two objects with masses m1 and m2 and velocities v1 and v2.
    I found the velocity of the centre of mass:

    vcm = (m1v1+m2v2)/(m1+m2)

    And found the velocities of the two objects relative to the cm.

    u1 = v1-vcm
    u2 = v2 - vcm

    And used conservation of momentum in the center of mass frame. And magically I got that:

    m1u1 + m2u2 = 0

    which I am guessing is the nice property of carrying out the math in the cm-frame. I just don't understand what leads to the above. Why should the equation:

    mu1 + mu2 = m1v1 + m2v2 - m1vcm - m2vcm = 0

    It is surely pretty obvious for you, so try to explain it for me :) And try to explain why it is obvious physically in terms of the center of mass being the point where only external forces work.
     
  2. jcsd
  3. Aug 30, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Did you plug in the expression for vcm and see what happens?

    Intuitively, the velocity of center of mass as measured in the center of mass frame must be zero. (Alternatively, the total momentum as measured in the center of mass frame must be zero.)
    Not sure what that means.
     
  4. Aug 30, 2012 #3
    The center of mass is defined specifically so that the entire momentum of the system be equal to that of an imaginary point of the total mass of the system moving with some constant velocity. Then the total momentum of the system with respect to that imaginary point has to zero.
     
  5. Aug 30, 2012 #4
    hmm yeh I didn't think probably. Since the total momentum is zero for a system moving with the center of mass it must be so. There is just a little thing that bothers me. On transforming to the center of mass you use the idea of galilean relativity. And as we can see the transformation makes sense in terms of the equations for conservation of momentum. Does that mean that galilean relativity is in some way incorporated into conservation of momentum in classical mechanics?
     
  6. Aug 31, 2012 #5
    Technically, the momentum is conserved because of Newton's third law and linearity of Newton's second law. When you have multiple particles, you can sum their equations of motion (second law). On one side of the resultant equation you will have the derivative of total momentum; on another, the sum of all forces, which the third law ensures must be zero (unless there are external forces). Because the derivative of total momentum is zero, the latter must be constant, i.e., conserved.

    This could be treated in another way. The laws do not depend on absolute positions. They all depend only on their relative differences. The system is said to have a translational symmetry, which means that if you take another frame of reference, shifted from the original one, its forces do not change. That's what then allows the second and the third laws to make the net force zero and produce conservation of total momentum.

    It can be generalized. Any invariance under a group of coordinate transformations implies an integral of motion, i.e., a conservation law, which is established in the Lagrangian formalism using Noether's theorem.
     
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