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Comparison of velocity at the end of distance

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Carts A and B have equal masses and travel equal distances D on side-by-side straight frictionless tracks while a constant force F acts on A and a constant force 2F acts on B. Both carts start from rest. The velocities vA and vB of the bodies at the end of distance D are related by

    a.vB = vA.

    b.vB = vA.

    c.vB = 2 vA.

    d.vB = 4 vA.

    e.vA = 2vB.

    2. Relevant equations
    F = ma
    x = vot + 0.5 at^2
    Vf = Vo + at

    3. The attempt at a solution
    Answer is B but I have no clue why
    First, I just picked a random number to make it easier. I picked acceleration of 2m/s^2 for A and 4m/s^2 for B (since it has twice more force than A)
    and I picked x as 1m
    Then I'll get
    1 = 0 + 0.5 * 2 * t^2, t = 1 for A
    1 = 0 + 0.5 * 4 * t^2, t = 0.707 for B
    If I plug this in to Vf = V0 + at
    Vf = 2 for A
    Vf = 2.83 for B...
    To me vB = vA doesn't make sense.. Can anyone help me
     
  2. jcsd
  3. Mar 23, 2015 #2

    TSny

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    Was there any information given regarding the directions of the forces?
     
  4. Mar 23, 2015 #3

    PeroK

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    Answers a. and b. are the same. Are you sure b. isn't something else?

    Your calculations are correct. Can you see a way to generalise them to get the answer?
     
  5. Mar 23, 2015 #4
    That is exactly what question stated. I copied and pasted it.
     
  6. Mar 23, 2015 #5
    Probably the answer is wrong haha I'm just using random questions from the internet to practice so.
     
  7. Mar 23, 2015 #6

    PeroK

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    It would be good practice to get the answer for this without multiple choice.

    You could generalise the calculations you've already done. Or you could consider energy.
     
  8. Mar 23, 2015 #7
    Is the answer V2 = sqrt 2 * V1 ???
     
  9. Mar 23, 2015 #8

    PeroK

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    It is!
     
  10. Mar 23, 2015 #9
    Thank you!! :)
     
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