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What angle does it scatter at if the yellow ball is scattered

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    The white ball (2kg) in the figure has a speed of 1.74 m/s and the yellow ball (1kg) is at rest prior to an elastic glancing collision. After the collision the white ball has a speed of 1.37 m/s. what angle does it scatter at if the yellow ball is scattered at 280 degrees?


    2. Relevant equations

    mva=mvacos(@)+mvbcos(@)

    3. The attempt at a solution
    2(1.37)Cos(@)a+0.58Cos280
    2.74Cos@+.10
    =92 degrees

    I think I'm close, but not quite.
     
    Last edited: Nov 28, 2009
  2. jcsd
  3. Nov 28, 2009 #2
    Re: momemtum

    anyone please, a hint?
     
  4. Nov 28, 2009 #3

    Doc Al

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    Staff: Mentor

    Re: momemtum

    That's momentum conservation in one direction. What about the other? And what about the fact that the collision is elastic?

    I don't understand what you're doing here. I don't see the full equation being used. Where did you get '1.37' and '0.58'? Show all your steps.
     
  5. Nov 28, 2009 #4
    Re: momemtum

    1.37 is given as the speed of ball a after the collision.

    0.58 is what I had found the speed of ball b to be after collision, but I think it's not correct as I'm not sure how to really do this problem. I can't really understand what the author of the book is trying to say on a similar problem.
     
  6. Nov 28, 2009 #5

    Doc Al

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    Re: momemtum

    OK.
    How did you get this? (Hint: That's where the fact that the collision is elastic will come in handy.)
     
  7. Nov 28, 2009 #6
    Re: momemtum

    ok, so I think .58 was incorrect.

    mava+mbvb = mava'+mbvb'

    2(1.74)+0=2(1.37)+vb'
    .74m/s = vb'

    Right?
     
  8. Nov 28, 2009 #7
    Re: momemtum

    When I put this in I still don't get the right answer.

    2(1.37)Cos(@)+.74Cos280
    2.74Cos@+.13
    =92.72 degrees
     
  9. Nov 28, 2009 #8

    Doc Al

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    Re: momemtum

    No. That equation isn't valid. (Momentum is a vector--direction matters.)

    Instead, make use of the fact that the collision is elastic. What does that mean?
     
  10. Nov 28, 2009 #9
    Re: momemtum

    KE is also conserved
     
  11. Nov 28, 2009 #10

    Doc Al

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    Re: momemtum

    Right! Use that to determine the speed of the yellow ball after the collision.
     
  12. Nov 28, 2009 #11
    Re: momemtum

    3.0276 = 1.8769 + .5(v^2)
    (3.0276 - 1.8769)/(.5) = v^2
    Vb' = 1.51 m/s
     
  13. Nov 28, 2009 #12

    Doc Al

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    Re: momemtum

    Looks good. (I get 1.52, when I round off.)
     
  14. Nov 28, 2009 #13
    Re: momemtum

    0 = 2(1.37)Sin@+1.52Sin280
    0 = 2.74Sin@ - 1.50
    = 33 degree

    I think that's right.

    Thanks a million. Took me while to understand it, but I got it. Thanks.
     
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