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Homework Help: Postion vs time graph without time or initial speed?

  1. Sep 24, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider the system shown to the right. Particles 1 and 2 are fixed in place while particle three is placed at the location shown and released.


    1. Let’s look at the mechanics of q3, which has a mass of 3.00 X 10-5 kg.

      • What will be the net force on q3?

      • What will be q3’s acceleration?

      • Can you use the acceleration equation [x(t)=1/2at^2+v0t+x0] to predict where q3 will be after 0.50 seconds? Explain.
    1. Project: You al make a spreadsheet that graphs q3’s position vs. time. Here are the requirements:

      • Spreadsheet accurately models the movement of particle 3.

      • You can adjust the magnitude and sign of each of the charges.

      • You can adjust the initial position and mass of q3 (including to the left or right of both charges or right in the middle).
    2. Relevant equations
    How would i make the graph without time or any intial or final speed? do you assume that initial speed is 0?

    3. The attempt at a solution
    i solved for the fnet on 3 through coulomb's law and then plugged the net force in to the equation F/M=a to find the acceleration. but i was not really sure where to go from there? I am not really wondering about the spread sheet as to what equations would i use to create a x VS t graph?
  2. jcsd
  3. Sep 24, 2016 #2

    Simon Bridge

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    Homework Helper

    Welcome to PF;
    You sketch it ... however, in this case the initial speed is implied in the problem statement.
    I never "assume". That would make a <donkey> out of u and me ;)

    You look in the problem statement for clues - you should never assume anything without a good reason to do so.

    In this case the problem statement says that the particle is "placed" and then "released".
    In English, this sort of construct is usually intended to imply that it was initially held stationary (if it was not held, then how can it be "released"?). If it was intended to say that it had some initial velocity the they would say the object was "fired" or "thrown" or some other word would be used that implies movement. If you look through all other problems involving motion that you have done, you should see this is almost always the case.
    (Though the statement I quoted could be a translation - in which case: look to the linguistic conventions for the original language.)

    The force depends on position. You have the initial position and velocity - and you know how Coulombs law applies to each point on the diagram.
    The next equation you need is Newton's laws of motion ... and the suvat equations.
    The assignment is to do the problem numerically using the spreadsheet ... you have probably had some example from classwork to help you.
    Part of the assignment is to figure out how to use these so we are limited in how much to help you ... show us how you are thin king and we can nudge you when you get confused.
  4. Sep 25, 2016 #3
    I know how to get on a spreadsheet i would think but I cant really put it on a spreadsheet if i dont know how to do it by hand. I take AP physics E&M which i am fairly good at but i always hated mechanics so that is why I might sound like i am a little more lost than i should be. I apologize. at my school juniors take pre-calc I take advanced pre-calc, since the school year is still young I dont really know that much calculus so maybe my derivatives were wrong nevertheless this is what I was thinking.

    1. q1*q3*k/(r^2)= F1_3(this is the electric force between 1 and 3)

    2. q2*q3*k/r^2)= F1_2(force between particle 1 and 2)

    3. from there i can find the net force between the two charges with f1_3+f1_2=fnet

    4. with the net force i would then find the acceleration using a=fnet/m (with m being the mass. (now everything above i was okay with doing but now heres where i get confused)

    5. take the acceleration and find velocity. v=-(at) being time intervals (i got that equation aver deriving it from the equation in step 6 and the intial equation was x(.05)=at+v

    6. take that velocity and previous acceleration and find new position: x=1/2at^2+v*t+x

    7. the value for x becomes the new position for particle 3 and now i go back to to the top in order to calculate electric force then acceleration etc rinse and repeat.

    the equations that i used derived them from suvat equation x=1/2at^2+v*t+x to find the values i needed. My position value for .00-.05 value didn't make sense because it equated to about 2000 meters I am not understanding where i went wrong. my process went like this:

    F1-3=-899 X componet
    F2_3= -4495 X componet
    Fnet3- 5394 N
    A=F/m or -5394/3E-5= -1.798E8
    from here this what i did since the problem wants a x vs t graph i decided to do the t in intervals
    x(.05)=at+v (here is where i get my velocity which is a whopping 8990000.0
    x(.05)=.5*-1.798E8*.05^2+ 8990000.0 *.05+.01
    this value of x equates to 224750 but how can the particle move to such a high value i know that something is off i just cant find it.
    Note: forgot to say i used the distance from 1-3 .01 meters and the distance from 2-3 .004 meters

    but if that value was to make sense i would take that x and put that as the new position of q3 and then do the distances between the charges again find fnet again new acceleration etc. and then repeat the process of equations till I have done as many time intervals as necessary.
    Last edited: Sep 25, 2016
  5. Sep 25, 2016 #4

    Simon Bridge

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    It helps to define a coordinate system, and use consistent notation. Here, let me help:

    ie. x is the distance between q3 qnd q1. The d=14cm is the distance between the fixed charged.
    Now you can talk about the distances properly (your 1 and 2 use the variable "r" to refer to different distances).
    ##F=kq_3(1/x^2 - 1/(d-x)^2)## ... and x is a function of time x(t).

    You are going to divide time into discrete "snapshots" some small time-step ##\delta t## apart.
    So define: ##t_n=t_0+n\delta t##, start the motion so that ##t_0=0##

    It follows that: ##x_n=x(n\delta t), x_0=x(0)=10\text{cm}## ... see how that works?
    Also: ##v_n=v(n\delta t), v_0=0,\; a_n=a(n\delta t)\cdots## ... so what is ##a_0##?

    Using this notation you should be able to relate ##a_1## to ##a_0## and, more generally, ##a_{n+1}## to the values before it. Same with the other variables.
    You have basically done that in your list above, but it helps to have a notation that makes it clear where you are going.

    In your spreadsheet... you need to work out how to generate the values.
    How that happens depends on the spreadsheet ...
    In general you will want the columb letter to correspond to a physical quantity, and the row number corresponding to the value of "n" in the above notation.
    So if A is time, then A0 may be ##t_0## (though that leaves no room for headings and titles).
    You can find out how to write equations into cells in terms of other cell-identifiers and make that general by copy-down or something.
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