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Problem Involving Momentum & Collision

  1. Jun 10, 2008 #1
    Here's the problem :

    A ball of mass m = 2.2 kg moving with a speed of v0 = 20.1 m/s strikes an identical ball which was initially at rest. After the collision, the incoming ball (ball 1) goes off at q1 = 24° relative to its original direction and the struck ball (ball 2) moves off at q2 = 34° as shown in the diagram.

    I can't post the diagram, but ball 1 shoots off 24 degrees above the x-axis and ball 2 shoots 34 degress below the x-axis.

    --------------------------------------------------------------------------------
    a) What is the magnitude of the initial momentum of the two ball system?
    pi = 44.22 kg m/s

    --------------------------------------------------------------------------------
    b) What is the ratio of the balls' final speeds?
    |v1|/|v2| = 1.37

    --------------------------------------------------------------------------------
    c) What is the final speed of the incoming ball?
    |v1| = m/s

    Okay, I have another problem that I'm stuck on! I was able to figure out part A & B (answers above), but I can't seem to get part C.

    I tried setting the horizontal momentum of the two objects after the collision equal to the initial horizontal momentum (44.22) and then solving for one variable.

    44.22 = m1v1 + m2v2 or 44.22 = m (v1+v2)

    Then I thought I would have to use the ratio of v1/v2 in order to solve for a variable. So I did the following:

    v1/v2 = 1.37 ... v2 = v1/1.37

    I then plugged that into the momentum equation above and solved for the horiztonal velocity and got 11.619m/s for v1.

    Then I took that number and divided it by COS(24) to get the Vf since the ball shoots off at an angle. I got 12.72m/s, but it isn't right. Where have I gone wrong? :(
     
  2. jcsd
  3. Jun 10, 2008 #2

    G01

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    OK. I think you may have made a calculation mistake while finding v_1. I am getting v_1=8.375m/s.

    Check your calculations, again.
     
  4. Jun 10, 2008 #3
    I'm still getting the same answer as I have before.

    And I tried plugging your value in to solve for V1final and it was still wrong. :-\
     
  5. Jun 10, 2008 #4

    tiny-tim

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    Hi Pat!

    (A very well laid-out answer, by the way … I wish other people were as clear! :rolleyes:)

    Hint: You must take horizontal components of momentum! :smile:
     
  6. Jun 10, 2008 #5
    Isn't that what I was doing though? I thought v1 and v2 that I was solving for in the equation you quoted were the two horizontal component?
     
  7. Jun 10, 2008 #6

    G01

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    TT I think he means that v_1=horizontal component of momentum of the first ball.

    Then hes saying:

    [tex]v_1=v_{f1}\cos24^o[/tex]

    [tex]v_{f1}=\frac{v_1}{\cos24^o}[/tex]

    where v_f1 = final velocity magnitude.

    Am I correct, Pat?
     
  8. Jun 10, 2008 #7
    Yeah, I think so but I thought that's what I've been doing? I mentioned that I found v1 and then divided it by Cos24, but I must be getting the wrong value for v1.
     
  9. Jun 10, 2008 #8

    tiny-tim

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    v = v1cos24º + v2cos34º :smile:
     
  10. Jun 10, 2008 #9
    Now I'm a bit more confused sorry! lol

    I don't know V do I? I thought you had to add the two horizontal components of momentum and set them equal to the initial momentum of the first ball? I wouldn't know V in your equation would I?
     
  11. Jun 10, 2008 #10

    tiny-tim

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    v = 20.1​
     
  12. Jun 10, 2008 #11
    I finally got it! Thanks!

    I have to ask though...do the horizontal components always equal the sum of the initial horizontal compents? I just want to understand what I did :)

    And thanks again!
     
  13. Jun 10, 2008 #12

    tiny-tim

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    Yup! :biggrin:

    Good ol' Newton's second law, and conservation of momentum, are vector equations, so they're valid on all the components in any one direction. :smile:
     
  14. Jun 10, 2008 #13
    Oh okay I understand now! Thank you so much :)
     
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