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Impulse momentum help

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data
    https://scontent-a-sjc.xx.fbcdn.net/hphotos-prn2/t1.0-9/10167935_1403417599934459_6123061969742894932_n.jpg


    2. Relevant equations



    3. The attempt at a solution
    I used ∫Fdt = m(vf-vo)

    and came up with -.822 m/s for the final velocity...

    I have been reworking this problem over and over and cannot come up with a different answer...

    Am i wrong in using the impulse momentum theorem?
     
  2. jcsd
  3. Apr 16, 2014 #2

    SammyS

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    You haven't given any details regarding how you came up with -.822 m/s for the final velocity.

    Why are you trying to find the final velocity anyway ?
     
  4. Apr 16, 2014 #3

    Andrew Mason

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    Your approach is correct. Can you show us how you integrated? What is the anti-derivative of sin(ωt)?

    AM
     
  5. Apr 16, 2014 #4
    sorry, should have done that from the get go, here is what i did: this is for when t = .55 seconds btw

    https://scontent-a-sjc.xx.fbcdn.net/hphotos-prn2/t1.0-9/10001366_1403433926599493_210320460460652531_n.jpg
     
    Last edited: Apr 16, 2014
  6. Apr 16, 2014 #5
    Im not trying to find the final velocity but the velocity at time = .55 seconds
     
  7. Apr 16, 2014 #6

    Andrew Mason

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    Your integral is correct. I can't tell from your answer how you got the 9e-5 value but it is not correct. The given answer is correct. Remember the argument for cos(ωt) is in radians, not degrees.

    According to your equation:

    vf = (1/m)∫Fdt + v0

    If you work that out you will get the answer that is given.

    AM
     
    Last edited: Apr 16, 2014
  8. Apr 16, 2014 #7
    Worked it out still got v(.55) = -.812 m/s

    the answer is suppose to be v(.55) = -.451 m/s
     
  9. Apr 16, 2014 #8

    SammyS

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    As AM said, ωt is in radians .

    Your result looks like your calculator is in degree mode.
     
  10. Apr 16, 2014 #9
    Hahaha thanks guys
     
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