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Position of electron motion 2D

  1. May 21, 2016 #1
    1. The problem statement, all variables and given/known data

    consider an electron traveling away from the origin along the x axis in the xy plane with initial velocity vi= viî. As it passes through the region x = 0 to x = d, the electron experiences acceleration a = axî + ayj, where ax and ay are constants. For the case vi = 1.80 ✕ 10^7 m/s, ax = 8.00✕ 10^14 m/s2, and ay = 1.60 ✕ 10^15 m/s2, determine the following, at x = d = 0.0100 m. (a) the position of the electron, (b) the velocity of the electron, (c) the speed of the electron, and (d) the direction of travel of the electron (i.e., the angle between its velocity and the x axis)
    2. Relevant equations
    Xf = Xi + Vt + 1/2at2
    Vf2= vi2 + 2ax
    Vf= Vi + at

    3. The attempt at a solution
    Vf2 = vi2 +2ax
    Vf = sqrt((1.80x10^7)^2 +2(8.00x10^14)(0.01))
    Vf = 18×10^6


    Time taken to reach this velocity:
    Vf = vi + at
    t = 225s

    Therefore y position after 225s
    Yf = yi + viyt +1/2at2
    yf = 0 + 1/2(1.60×10^5)(225)^2
    Yf = 4.05× 10^9

    rf = (0.01î + 4.05x10^9j)m
    Magnitude = 4.05×10^9m

    (b) vfx = 18x10^6î
    Vfy= viy +at
    Vfy = 0 + (1.60×10^5)(225)
    Vfy = 36×10^6j
    Vf = (18× 10^6)î + (36x10^6)j
    Magnitude = 40.25 ×10^6
    Direction tan@ = 36x10^6/18×10^6
    @= 63.43 NE

    Are these answers right or did I make a mistake
     
  2. jcsd
  3. May 21, 2016 #2

    Simon Bridge

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    Without checking your arithmetic, it looks OK.
    I think you just need to tidy up your notation and add comments to show your reasoning.
    That would also help you (and others) troubleshoot your own work.

    ie.
    The initial velocity is: ##\vec v(0)= v_i\hat\imath##
    Acceleration is: ##\vec a = a_x\hat\imath + a_y\hat\jmath##
    Initial and final displacements: ##x_i=x(0)=0,\; x_f=x(T)=d,\; y_i=y(0)=0##
    ... ie. ##x=d## when ##t=T##.

    Position would be: ##\vec r(t) = x(t)\hat\imath + y(t)\hat\jmath##
    Also velocity is: ##\vec v(t) = v_x(t)\hat\imath + v_y(t)\hat\jmath##

    You need to find the final position: ##\vec r(T)=\vec r_f=d\hat\imath + y(T)\hat\jmath## and the final velocity, ##\vec v_f=\vec v(T)##

    See how to keep the variables clear? See how I've distinguished the vectors from their magnitudes?
    Also watch how clear my reasoning is when I add comments to show what everything is.

    Use suvat equations.

    Velocity is the easiest - so do that first:
    x-component of final velocity is given by:##v_x^2(T)=v_i^2 + a_xd##
    y-component of final velocity is given by: ##v_y^2(T)=a_yy_f##
    ... that what you had in mind?

    ##y_f## is given by: ##y_f = \frac{1}{2}a_yT^2## and ##v_x(T)=v_i+a_xT##
    So solve the second for ##T## and substitute into the first one. Which I think is what you did.
    Gets you something like ##y_f=\frac{1}{2a_x}\Delta v_x^2##

    That was the hard work.
    From there you can pretty much just read off the answers.
    (a) final position: ##\vec r_f = d\hat\imath + y_f\hat\jmath##
    (b) the velocity: ##\vec v_f = \sqrt{(v_i^2 + a_xd)}\hat\imath + \sqrt{a_yd}\hat\jmath##
    From the velocity, you get (c) the speed: ##v_f = |\vec v_f| = \sqrt{v_i^2 + a_xd + a_yy_f}##
    ...and also (d) the angle to the x-axis: ##\theta = \tan^{-1}\big( (v_i^2 + a_xd)/(a_yy_f) \big)##

    ... and we are done.
    If I've made a mistake in the above, it will probably be easy to find ;)
     
  4. May 21, 2016 #3
    Ok thank you I will setup my solutions like this from now on
     
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