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Pendulum problem i really need help

  1. Apr 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A small (65g) steel ball bearing is attached to a 50 cm long string as shown (link below) The ball bearing is lifted (with string stretched) to an angle of 55 degrees and released from rest.

    a) when the small bearing reaches its lowest point it collides with another steel ball bearing of larger mass (150g) on a string of the same length. if the collision would have been elastic, to what maximum angle would each bearing reach after the collision.

    b) if the small ball bearing actually bounces back to a maximum angle of 15 degrees( collision was NOT elastic), to what maximum angle will the larger ball bearing reach?

    c) if the bearings behaved as stated in part (b) how much energy is lost during the collision.

    http://s615.photobucket.com/albums/tt233/gaby596/
    click on the only image there called pendulum problem thanks!



    2. Relevant equations
    this are the ones i think should be used
    v1=m1-m2/m1+m2*v initial
    v2=2m1Vi/(m1+m2)
    and the equations for finding delta KE and Delta PE



    3. The attempt at a solution
    ive tried it several different ways but i keep getting the wrong answer ive seen the right answers from a answer sheet but they show no explanation on how to work it out
     
  2. jcsd
  3. Apr 19, 2009 #2

    rl.bhat

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    The ball bearing is lifted (with string stretched) to an angle of 55 degrees and released from rest.
    In this position what is the height of the ball from the lowest point of the pendulum?
    Using kinematic equation find its velocity.
     
  4. Apr 19, 2009 #3
    The ball bearing is lifted (with string stretched) to an angle of 55 degrees and released from rest.

    ok so i did this, to find the change in height
    i cant figure out this equation stuff on this site but i found the change in height to be 21 cm is that right?
    so do i use this equation?
    Delta KE + Delta PE=0

    1/2m(v^2-Vi^2)+mgdeltah=0
     
    Last edited: Apr 19, 2009
  5. Apr 19, 2009 #4

    rl.bhat

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    Yes. You are in right track. Proceed.
     
  6. Apr 19, 2009 #5
    ok so for that formula i just made im confused is the initial velocity zero?
     
  7. Apr 19, 2009 #6

    rl.bhat

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    Yes. The initial velocity is zero.
     
  8. Apr 19, 2009 #7
    im i supposed to convert grams into kilograms and cm into meters?
     
  9. Apr 19, 2009 #8

    rl.bhat

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  10. Apr 19, 2009 #9
    since initial is zero can i do this?
    Delta PE= KE
    (.065)(9.8)(0.21)=1/2(.065)V^2
    .13377=.0325(V^2)
    .13377/.0325=V^2
    4.116=v^2
    V=2.02879m/s

    is that right?
     
  11. Apr 19, 2009 #10
    so if im right can i use this equations to find the before and after now right?
    v1=m1-m2/m1+m2*v initial (for before)
    v2=2m1Vi/(m1+m2) (for after)
     
  12. Apr 19, 2009 #11

    rl.bhat

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    Yes. It is right.
     
  13. Apr 19, 2009 #12
    yay :)

    ok so V1=(.065-.15)(2.02879m/s)/(.065+.15) for before the collision and
    V2=2(.065)(2.02879m/s)/(.065+.15)

    V before= -.80208 m/s
    v after= 1.25 m/s

    ?
     
  14. Apr 19, 2009 #13
    for part b i need to solve for delta h again since it changed from 55 degrees to 15 degree?
    so do the same thing again that i did for the first one
    l= lenght of string
    Delta h=l-lcos
    Delta h=50cm-50cmcos15degrees
    ?
     
  15. Apr 19, 2009 #14

    rl.bhat

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    Here V1 is the velocity of .065 kg mass after collision and V2 is that of 0,15 kg.
    With these velocities find the heights attained by them using V^2 = 2gh formula.
     
  16. Apr 19, 2009 #15
    ? im lost now, so are those formula right?
     
  17. Apr 19, 2009 #16

    rl.bhat

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    With these velocities find the heights attained by them using V^2 = 2gh formula.
     
  18. Apr 19, 2009 #17
    V^2 = 2gh formula im not familiar with this formula
    can you show me V1 because i got an extremely small number and i don't know if i did it right.

    h= -.032823 and h=.079719 for V2

    ?
     
    Last edited: Apr 19, 2009
  19. Apr 19, 2009 #18

    rl.bhat

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    Yes. for V1 it is correct. Using Δh formula find the angle. Repeat the same thing for V2.
     
  20. Apr 19, 2009 #19
    We are still in part a i thought we were answering the other ones never mind ha ha so we are answering if the collision would have been elastic, to what maximum angle would each bearing reach after the collision. But i still have no clue how to solve for the angle, :(
     
    Last edited: Apr 19, 2009
  21. Apr 19, 2009 #20
    So im i using this formula?
    Delta h=l-lcos

    im not sure how to arrange this in terms of cos, can you help me?
     
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