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Pendulum projectile momentum question

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A 0.010 kg pendulum bob is dropped from a height (h) above its equilibrium position. When the bob reaches its equilibrium position, the string breaks and the bob now acts as a projectile. After the string breaks the bob falls 1.5 m while moving 2.0 m horizontally. Calculate the height (h) from which the bob was released.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 4, 2012 #2

    tiny-tim

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    welcome to pf!

    hi jackneedshelp! welcome to pf! :wink:

    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Feb 4, 2012 #3
    I don't even know where to start! It should have something to do with momentum and impulse, but I can't even get started on it! Please help!
     
  5. Feb 4, 2012 #4

    tiny-tim

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    first find the speed of the bob (as a function of h) when the string breaks …

    show us what you get​
     
  6. Feb 4, 2012 #5
    I set potential energy equal to kinetic energy, and solved for v.
    v = sqrt(19.62h)
     
  7. Feb 4, 2012 #6

    tiny-tim

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    yup! :smile:

    now use that as the initial horizontal velocity of a projectile …

    write standard constant acceleration equations for x and y (separately)
     
  8. Feb 4, 2012 #7
    so I did the horizontal part, and got the horizontal acceleration to be 4.905h.
    how do I do the vertical part?
     
  9. Feb 4, 2012 #8

    tiny-tim

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    how can there be any horizontal acceleration? :confused:

    there's no horizontal force :redface:
     
  10. Feb 4, 2012 #9
    so I'm solving for time?
     
  11. Feb 4, 2012 #10
    or final velocity?
     
  12. Feb 4, 2012 #11

    tiny-tim

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    write out some equations!!
     
  13. Feb 4, 2012 #12
    d = 2.0m
    vi = sqrt(19.62h)
    vf = 0
    t = ?

    2 = (0+sqrt(19.62h))/2 *t
    t = 4 / (sqrt(19.62h))

    is that what I'm supposed to do?
     
  14. Feb 4, 2012 #13

    tiny-tim

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    something like that

    but vf isn't 0 (and you aren't told what it is)

    and that's the wrong equation anyway

    this time, first write out the equation you're using, in letters,

    then write it out with the numbers in​
     
  15. Feb 4, 2012 #14
    vf srd = vi srd + 2ad
    vf srd = (sqrt(19.62h))srd + 2(0)(2)
    vf = sqrt(19.62h)

    so vf = vi
     
  16. Feb 4, 2012 #15

    tiny-tim

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    well, yes, but that's a bit obvious

    horizontally, a = 0, so v is constant

    you need an equation with t in it ​
     
  17. Feb 4, 2012 #16
    d = vi t + (0.5)(a)(t)sqrd
    2 = sqrt(19.62h) t + 0
    t = 2 / sqrt(19.62h)
     
  18. Feb 4, 2012 #17

    tiny-tim

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    yes :smile:

    and now use that value of t in a similar equation for the y direction :wink:
     
  19. Feb 4, 2012 #18
    Ohhh kaaaaay! I get it! Thanks!
     
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