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Motion in one demansion, photogate timer problem

  1. Dec 25, 2009 #1
    Please help me quick everyone, thank you in advance.
    1. The problem statement, all variables and given/known data
    http://imgkk.com/i/EasJ1d.jpg [Broken]
    a/We have two photogates placed as in the figure. We drop a marble through these at a negligble distance above the upper gate. The upper starts a timer as the ball passes through its beam and the second stop the timer.
    Prove that the experimental magnitude of freefall acceleration is given by
    [tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex]
    b/ For the setup and assume that gexp =9.81m/s2, what value of [tex]\Delta{t}[/tex] would you expect to measure.
    c/This time, the upper photogate is placed 0.50cm lower than the first time. But we still drop the marble from the same height as the first time.
    What value of gexp at this time will you determine?

    2. Relevant equations


    [tex]a = \frac{{\Delta v}}{{\Delta t}} = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}[/tex]

    3. The attempt at a solution

    Based on the problem, I think that [tex]\delta{t}[/tex] is the time when the marble passes through between two gates. But if it were, I couldn't get [tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex].
    If it were not, I could do like that
    We have [tex]{g_{\exp }} = \frac{{\Delta v}}{{\Delta t}}[/tex]
    Because v0=0,
    Hence,
    [tex]g_{exp}=2\Delta{y}/{(\Delta{t}})^2[/tex]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 25, 2009 #2

    Oddbio

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    Gold Member

    For part a, the [itex]g_{exp}[/itex] is your free fall acceleration.

    In order to prove it, start with these equations:
    [tex]a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}[/tex]
    as you have already written. But, also use this:
    [tex]v_{avg}=\frac{\Delta y}{\Delta t}[/tex]
    and rewrite it like this:
    [tex]\frac{v_{1}+v_{2}}{2}=\frac{\Delta y}{\Delta t}[/tex]

    In the second equation the Vavg is just the normal equation for "average" velocity, so you can rewrite it as the third equation, simply adding the two velocities (final and initial) and dividing by two (averaging them).

    So use the first and third equations, to solve for "a", but remember in this case "a" IS g_exp. A little algebra is required, you can eliminate v1 because it starts from rest (v1=0).

    Also you are correct, delta(t) is the time it takes to pass through the two gates. But I don't see why you say you couldn't solve for delta(t) from this:
    [tex]g_{exp}=2\Delta y/(\Delta t)^{2}[/tex]
    Because the only unknown in that equation is delta(t) which is what you want. The problem gives you g_exp, and delta(y) is just the distance between the two gates.

    For c, you will have to redetermine the equation for g_exp because you will now have an initial velocity. Originally the ball was assumed to have an initial velocity of zero, but now that the top gate is lower, the ball will obtain some downward motion by the time it reaches the upper gate.
     
  4. Dec 26, 2009 #3
    Thank you too much Oddbio.
    And I want too say more clearly about part c. That is the upper gate is placed accidentaly lower than before. So the problem ask us to calculate in that accident.
    (This is problem 113, chapter 2 - Motion in one dimension - Physics for Scientist and Engineers 6th edtion - Tipler Mosca - Univer of Washington)
    Here is my solution, is this correct, is there something need to be corrected?
    http://imgkk.com/i/5Be70O.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Dec 26, 2009 #4

    Oddbio

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    Gold Member

    Yes they all look correct to me. Good job :)
     
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