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Homework Help: Momentum and angles problem

  1. Dec 3, 2004 #1
    I had made an error on my previous post so i will take it back to the beginning. I can't get this one for some reason.

    First look at the diagram: http://bosco.iwarp.com/diagram.jpg

    OK here's the problem:

    Two identical balls (mass m) undergo a collision. Initially one ball is stationary, the other has kinetic energy of 8J. The collision is partially inelastic with 2 J energy converted to heat. What is the maximum deflection angle (alpha or beta) at which one of the balls is observed?

    I have come up with the following relationships using conservation of momentum and energy:

    1. Vo = V1 cos a + V2 cos B

    2. 0 = V1 sin a - V2 sin B

    and 12 = m (V1f^2 + V2f^2) which i believe can be written

    3. 0.75Vo = V1^2+V2^2

    by rearranging initial kinetic energy of ball one to solve for m.

    How do i go about comparing alpha and beta? Are my equations correct
    Help please! thanks
     
    Last edited: Dec 3, 2004
  2. jcsd
  3. Dec 3, 2004 #2

    Andrew Mason

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    1. and 2. are fine. 3. should have Vo^2:

    This is not a trivial problem to solve. Use 1 and 2. to find [itex]cos\beta \text{ in terms of } sin\alpha \text{ and } cos\alpha[/itex]

    Then look at the ranges of values that [itex]cos\beta[/itex] can have. There should be a [itex]\sqrt{.75}[/itex] term in there somewhere.

    AM
     
  4. Dec 4, 2004 #3
    Well i have attempted this another way and have got the following:

    [itex]2 = mv_{1}v_{2} (cos \alpha }+ \beta)[/itex]

    now i do not know how to compare [itex]\alpha[/itex] and [itex]\beta[/itex]

    any help
     
  5. Dec 4, 2004 #4
    ??????????
     
  6. Dec 4, 2004 #5

    Andrew Mason

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    It has to have [itex]\beta[/itex] in terms of [itex]\alpha[/itex] ONLY. You can do this with these equations: determine v2f in terms of v1f using 2. and substitute into 1. to find v1f in terms of [itex]\alpha \text{ and } \beta[/itex]. Substitute also into 3. to find v1f. [itex](v_0^2 = 16/m)[/itex]. Then combine the two to find [itex]beta[/itex] in terms of [itex]\alpha[/itex] (the m falls out).

    AM
     
  7. Dec 5, 2004 #6
    I dont know why this is not working out but after trying what you said i got:

    [tex] 3sin^2\beta (cos^2\alpha + 2cos\alpha + cos^2\beta) = 4(sin^2\beta+sin^2\alpha)[/itex]

    It should probably be simpler shouldnt it?

    I really want to get this one done but it just isnt working out
    Thanks for all your help on this one
     
    Last edited: Dec 5, 2004
  8. Dec 5, 2004 #7
    ???????????
     
  9. Dec 6, 2004 #8

    Andrew Mason

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    I told you it was non-trivial. Have a look at the solution for an elastic collision at: http://rustam.uwp.edu/201/L12/lec12_w.html (scroll down to the collision in 2 dimensions). For elastic collisions the two angles add to 90 degrees. The question is asking how the sum of the two angles in an inelastic collision where 25% of the energy is lost compares to 90 degrees.

    AM
     
    Last edited by a moderator: Apr 21, 2017
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