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Collisions in 2D

  1. Oct 12, 2004 #1
    okay I think my brain is fried cause I've never before had so many problems in one day anyways...

    1) Two billard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. A red one is moving upwards along the y axis at 2m/s, and a green one is moving horizontally (to the right) along the x axis at 3.7m/s. After the collision (elastic), the green ball is moving along the possitive y axis. What is the final direction of the red ball and what are their two speeds?

    now since I use conservation of momentum I am getting that the green ball would be moving up the y axis (90 degrees) @ 2m/s, and the red ball is moving along the x axis (0 degrees) @ 3.7m/s. I do not have this question in the answer key but I am more then 100% sure that this should be correct, am I missing something or is it really right?

    now 2nd one I don't really have a solution as it doesn't involve numbers and only algerbra and I hate algebra, so this one is where I really need some help if possible:

    2) An atomic nucleus of mass m travelling with speed v collides elastically with a target particle of mass 2m(initially at rest) and is scattered @ 90 degrees.
    a) At what angle does the target particle move after the collision
    b) What are the final speeds of the two particles
    c) What fraction of the initial KE is transfered to the target particle

    now for this one I am lost as to where to even start, I'm pretty sure I should be using conservation of momentum equationsalmost like in the previous one, except that the mass is different so that cannot be cancelled out, but the letters are messing with my head.
  2. jcsd
  3. Oct 12, 2004 #2
    Redo the first one using vectors (x and y components).
  4. Oct 17, 2004 #3
  5. Oct 30, 2004 #4
    To derive the solution from first principles is not trivial. But the result is that for equal masses, an elastic collision results in the masses moving with velocities at 90 degrees to each other (except for perfectly head-on collisions). That is all you have to know. (This is a very useful rule when playing pool. If you want to set up for the next shot you have to know the direction the cue ball will go). I think the answer is pretty obvious from that: You know the directions. You then set the speeds in order to conserve the x and y components of momentum, which is what you have concluded.

    In this case, the momentum of the system is [itex]mv_0[/itex] before and after the collision. So [itex]2mv_2 cos\theta = mv_0[/itex] where [itex]\theta[/itex] is the angle of the second (larger) mass from the original direction of [itex]m_1[/itex]. Also [itex]2mv_2 sin\theta = mv_1[/itex] .

    Also conservation of KE means that [itex]v_0^2 = v_1^2 + 2v_2^2[/itex].

    The rest is just math.
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