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Projectile Motion in an Electric field

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Protons are projected with an initial speed of v0 = 10^5 m/s into a region in which a uniform electric field E = 800N/C [down] is present. The protons are to hit a target that lies a horizontal distance of 10.0 mm from the point at which the protons are launched. a) find the two projection angles theta that will result in a hit b) what is the total duration of flight for each of the two trajectories?

    2. Relevant equations
    (1) x = v0 cos (theta) t

    (2) y = v0 sin (theta) t - 1/2at^2


    3. The attempt at a solution

    x = v0 cos(theta) t y = v0 sin(theta) t - 1/2 a t^2

    when the proton hits its mark at 10 mm, the y displacement will be zero

    1/2*a*t = v0 sin(theta) ===> t = (2*v0 sin(theta))/a

    plugging t into the formula for the x displacement gives

    x = v0 cos(theta)*(2*v0 sin(theta))/a

    => x = [(v0)^2 * sin(2theta)]/a
    => theta = 0.5 *arcsin( (a*x)/(v0)^2)
    => theta = 2.2 degrees
    t = (2*v0 sin(theta))/a
    => t = 10^-7 s

    Now I don't really know how I am supposed to get the second angle
  2. jcsd
  3. Feb 11, 2016 #2

    Ken G

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    Gold Member

    2*cos(theta)*sin(theta) = sin(2*theta). So if 2*theta can be above pi/2, there's a second solution, since the inverse sine function is double valued.
  4. Feb 11, 2016 #3
    I don't really know what you mean by "if 2*theta can be above pi/2"
  5. Feb 11, 2016 #4
    Try using the mass and charge of a proton in your equations. You need to use
    [tex] qE = F [/tex]
    Then break it up to
    [tex] qE = ma [/tex]
    [tex] \frac{qE}m = a [/tex]
    First solve for the time of the vertical component.
    [tex] y = v_0t + \frac{1}{2}at^2 [/tex]
    [tex] 0 = v_0sin\theta t + \frac{1}{2}at^2 [/tex]
    [tex] -v_0sin\theta t = \frac{1}{2}at^2 [/tex]
    [tex] t = \frac{-v_0sin\theta}{\frac{1}{2}a} [/tex]
    Then substitute time into horizontal component.
    [tex] x = v_0t [/tex]
    [tex] x = \frac{v_0cos\theta (-v_0sin\theta)}{\frac{1}{2}a} [/tex]
    [tex] x = \frac{v_0cos\theta (2v_0sin\theta)}{-a} [/tex]
    [tex] x = \frac{v_0^2 sin2\theta}{-a} [/tex]
    [tex] \frac{-ax}{v_0^2} = sin2\theta [/tex]
    [tex] \frac{qEx}{mv_0^2} = 2sin\theta [/tex]
    Solve for angle which will be your first angle, and then subtract it from 90 to get your other angle. Then I believe you can get time no problem.
  6. Feb 11, 2016 #5
    Why does subtracting the original angle from 90 degrees give me the other correct angle?
  7. Feb 11, 2016 #6
    It surrounds the optimal angle of 45 degrees. They will hit the same spot, 10mm away. One will hit the spot because it'll be low to the ground and gravity will accelerate it downwards fast, while the other one will hit the spot because of the short wavelength of the parabola which equals a short distance across the x axis even though the amplitude/height is larger.
  8. Feb 11, 2016 #7
    I suppose that makes sense intuitively, but how can I prove that the other angle must be 90 minus the original angle in a mathematical way?
  9. Feb 11, 2016 #8
    It's pretty much just like taking the vertical component and rotating the page around to make it the horizontal component, or in other words from sine to cosine.
  10. Feb 11, 2016 #9

    Ken G

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    Gold Member

    Your original solution seemed fine, but you did not solve for theta. To do so, use the identity I gave. Then solve for theta using an inverse sine function. But the inverse sine function has two angles that will give the same result, one theta is below pi/4 (which is 45 degrees if you are not using radians), the other above pi/4.
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