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Time It Takes

  1. Jan 26, 2009 #1
    You will have to use your imagination: 3 tracks are constructed as such... path 1 (P1) is similar to a "down-facing parabola," path 2 (P2) is straight, and path 3 (P3) is similar to an "up-facing parabola." P1 is "uphill," P2 is "through that hill," and P3 "under the hill, beneath P2, and exactly 'oposite' of P1."

    Okay! .... A mass m starting at point A (foot of the hill) is projected with the same initial horizontal velocity v along each of the three tracks (negligible friction) sufficient in each case to allow the mass to reach the end of the track at point B (oposite foot of the hill). The masses remain in contact with the tracks throughout their motions. The displacement A-B is the same in each case, and the total path length of P1 and P3 are the equal. If t1, t2, and t3 are the total travel times between A and B for P1, P2, and P3 respetively, what is the relation among these times?



    I said intuitively that: t2<t1=t3. correct?
     
  2. jcsd
  3. Jan 26, 2009 #2

    AEM

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    Gold Member

    What a nice little problem! Keeping in mind that the initial velocity is horizontal... Consider parabola (1) Will a horizontal velocity enable the mass to slide up over the hill??? (If so, will it for ALL parabolas?) Consider parabola (2) The horizontal velocity will enable the mass to slide down the track. With no friction, what does conservation of energy tell you about the velocity of the mass when it reached point B? Finally, what is the relationship between the lengths of the straight line track and the parabolic track.

    (As you may deduce from my comments, I have some qualms about the formulation of the problem, but it's still a nice problem. Substitute a cycloid for the parabola and tweak it a little and you have a classic physics problem...)
     
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