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Kinematics motion

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A puck moving at 32m/s slams into a wall of snow 34cm thick. it emerges moving at 18m/s. Assuming constant acceleration, find
    a) time taken for puck to spend in the snow (ans:0.014s got it)
    b)the thickness of a snow wall that would stop the puck entirely

    2. Relevant equations



    3. The attempt at a solution

    This should be a simple question but the kinematics equations I utilized does not yield me the answer.
     
  2. jcsd
  3. Dec 6, 2013 #2

    Doc Al

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    Show what you did. Did you find the acceleration as it travels through the snow?
     
  4. Dec 6, 2013 #3
    from part (a), t = 0.014s

    a = dv/dt = [vf-vi]/0.014s
    a = 1000m/s^-2

    vf^2 - vi^2 = 2a(xf-xi)
    0-(32m/s^-1)^2 = 2(1000)(xf-xi)
    xf-xi = -0.512m

    answer states 0.512m
    Could someone explain to me the negative in front of my answer?


    EDIT: Also, it would be good too if an explanation as to why acceleration is required for me to arrive at the answer?
     
    Last edited: Dec 6, 2013
  5. Dec 6, 2013 #4
    Alright something is wrong. Vf ought to be 0ms^-1.
    I'm pretty lost.
     
  6. Dec 6, 2013 #5

    haruspex

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    Which direction are you taking as positive for x? Is the acceleration +ve or -ve in that direction?

    The kinematics equations you are using are only valid if acceleration is constant.
     
  7. Dec 6, 2013 #6
    I think the answer was a fluke.

    in arriving at 0.512m, my vf was 18ms^-1.
    Shouldn't vf = 0ms^-1 due to the fact that it crash into a snow wall?

    vf = 0ms^-1
    vi = 32ms^-1
    a = [0ms^-1 - 32ms^-1]/0.014s = -2286ms^-2(does it make sense to use the previous time,t = 0.014s, from part (a)?)

    vf^2 - vi^2 = 2a(xf-i)
    -(32ms^-1)^2 = 2(-2285)(xf-xi)
    xf-xi = 0.22m

    snow wall must be at least 0.22m thick in order for the puck to stop upon collision. However, this does not coincide with the answer 0.51m on my textbook.
     
    Last edited: Dec 6, 2013
  8. Dec 6, 2013 #7

    CWatters

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    The question asks for the thickness of snow (s) needed to stop the puck. So I would use one of the suvat equations that begins s=....

    http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

    You know..

    vi = 32m/s
    Vf= 0 (because it the problem asks for the thickness of snow needed to stop the puck)

    Do you know the time t? No. The snow walls is thicker so t will change.

    What about the acceleration (a)? You can assume that the acceleration is the same as the first part of the question, however you got that slightly wrong. The acceleration is not 1000m/s but it's hard to give you a hint without making it too easy. Does the puck go faster or slower after hitting the snow? Haruspex also gave you a hint which you appear to have missed..

     
    Last edited: Dec 6, 2013
  9. Dec 6, 2013 #8

    Doc Al

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    OK, close enough. Note that since it slows down, the acceleration should be negative.

    Sure. You took the direction of motion to be the positive direction, which would make the acceleration negative. Using a negative value for acceleration will give you a positive distance.


    That's one parameter that is assumed to be constant for any thickness of snow.
     
  10. Dec 6, 2013 #9

    haruspex

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    Not at all. Your answer was almost right - just one detail that CWatters, Doc Al and I have all tried to draw your attention to.
    This is part b), right? In your earlier post, you correctly used vf=0 in part b). You seem to be confusing yourself.
    No, it makes sense to do what you did originally: use the acceleration from part a). (But you first have to get that acceleration right.)
     
  11. Dec 11, 2013 #10
    The only confusion that persists is setting the equation a = [vf-vi]/t where
    vf = 18ms^-1
    I would have thought that upon collision, the vf of the puck would have been at 0ms^-1.
    But were I to set vf = 0ms^-1, I would have gotten a = -2285.7ms^-2 instead of -1000ms^-2.
    Inserting a=-2285.7 into vf^2 - vi^2 = 2a(xf-xi) would yield a different answer.
     
    Last edited: Dec 11, 2013
  12. Dec 11, 2013 #11

    haruspex

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    Sorry, but I'm not following you.
    In part a), you use the given distance and vf = 18 m/s to find the acceleration.
    In part b), you use the acceleration from part a) and vf=0 to find the distance.
    What's the problem?
     
  13. Dec 12, 2013 #12

    Doc Al

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    When the puck first collides with the snow it doesn't stop dead it just begins to slow down (with some acceleration). You are told in part a that after going through a certain distance of snow it only slows down to 18 m/s.
     
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