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Momentum 2

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data

    Similar to the other one...
    On billiard ball is shot east at 2.6m/s. A second, identical billiard ball is shot west at .8m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90 degrees and sending it north at 1.58m/s. What are the speed and direction of the first ball after the collision?

    2. Relevant equations



    3. The attempt at a solution
    I tried using (P1x)f + (P2x)f = (P1x)i + (P2x)i ----> (m1xv1x)f + (m2xv2x)f = (m1xv1x)i + (m2xv2x)i. I tried solving for (v1x)f. I assumed since the billiard balls are identical that the mass cancels out.
     
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  3. Oct 20, 2007 #2

    learningphysics

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    yes, the masses cancel. Looks like you have the right idea. can you show your calculations?
     
  4. Oct 20, 2007 #3
    V1f = V1i + V2i - V2f
    = 2.6 + .8 - 1.58 = 1.82
     
  5. Oct 20, 2007 #4

    learningphysics

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    but 1.58 is not in the x-direction... what is the Vxf of the second ball?
     
  6. Oct 20, 2007 #5
    I didn't realize that. The Vxf of the 2nd ball is zero since it isn't moving in the x direction anymore.
     
  7. Oct 20, 2007 #6

    learningphysics

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    exactly.
     
  8. Oct 20, 2007 #7
    Ok, so I tried V1f = 2.6 + .8 = 3.4, but that wasn't right.
     
  9. Oct 20, 2007 #8

    learningphysics

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    Oh, sorry I didn't notice... you should have -0.8, not 0.8.

    use east positive. west negative. north positive. south negative.
     
  10. Oct 20, 2007 #9
    Alright, this time I tried V1f = 2.6 - .8 = 1.8, but it again wasn't right.
     
  11. Oct 20, 2007 #10

    learningphysics

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    that's only in the east west direction... you also need to consider the north south direction.

    I thought the question was asking for the components separately. I just noticed now it only asks for speed. I'm sorry... wish I realized before you submitted again...

    get the north south component, same way you got the east/west component... then get the speed using pythagorean theorem.
     
  12. Oct 20, 2007 #11
    Blah....freaking components. Ok, for the north/south component, V1f = V1i + V2i - V2f -----> V1f = 0 + 0 -1.58. There's only one problem with using the pythagorean theroem. With the last problem, I figured out the angle really easily since both components were the same. I don't know the angle with this one.
     
  13. Oct 20, 2007 #12

    learningphysics

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    you can get the angle using arctan(vy/vx). be careful before you submit though... how do they ask for the angle? Do they want the angle from the east axis etc..?
     
  14. Oct 20, 2007 #13
    They ask for degrees in the south east direction.
     
  15. Oct 20, 2007 #14
    So the angle would be -41.276 degrees, but it would be positive since they want it in the SE direction, right?
     
  16. Oct 20, 2007 #15

    learningphysics

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    yeah, but it is East 41.276 degrees South...

    Do they just say SE... because it could be east of south or south of east... that makes a difference...
     
  17. Oct 20, 2007 #16
    It says south of east
     
  18. Oct 20, 2007 #17

    learningphysics

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    Cool. Then 41.276 is right.
     
  19. Oct 20, 2007 #18
    Ok so the angle was right, but I don't know the velocity....and I'm out of tries. I tried the -1.58 = vcos(theta).
     
  20. Oct 20, 2007 #19

    learningphysics

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    The 1.58 is the sin. so vsin(41.276) = 1.58. v = 2.395m/s.

    But it's best to just use the pythagorean theorem when you need the speed... you don't need to worry about the angles. sqrt(1.58^2 + (1.8)^2) = 2.395m/s
     
  21. Oct 20, 2007 #20
    Alright.
     
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