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Billiard ball physics homework

  1. Oct 17, 2004 #1
    Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. One is moving upward along the y axis at 3 m/s, and the other is moving to the right along the x axis with speed 4 m/s. After the collision (assumed elastic), the second ball is moving along the positive y axis.

    (a) What is the final direction of the first ball?

    (b) What are their two speeds?

    OK, I use two different for each x and y component:
    x-direction: m*4=m*v1'*cos(theta)
    y-direction: m*3=m*v2'+m*v1'*sin(theta)

    Now I can cancel the masses, but still, I have two equations but 3 unknowns...
  2. jcsd
  3. Oct 17, 2004 #2

    Doc Al

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    Staff: Mentor

    There is a third equation you can use: Realize that the collision is elastic.
  4. Oct 20, 2004 #3
    No matter what I do, I can't get the answer.(yes, I used conservation of energy as my third equation) Can I assume that ball a will move along the positive x-axis after collision?
  5. Oct 21, 2004 #4

    Doc Al

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    Staff: Mentor

    So you should have 3 equations with 3 unknowns.
    Of course not! :smile:

    Here's a trick to simply the two momentum equations and eliminate theta: Rewrite your x-direction equation to isolate v1cos(theta) and your y-direction equation to isolate v1sin(theta). Square both sides of each equation and add them. Use a simple trig identity. Now combine with energy conservation to solve for V1 and V2. (Then go back and solve for theta.)
  6. Oct 21, 2004 #5
    In an elastic collision the the relative speed before the collision is equal to the opposite of the negative relative speed after the collision v1-v2=-(v1-v2)
    I'm not sure how this applies in 2 dimensions though, could someone explain?
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