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Nanoscale electronics and impulse force

  1. Oct 13, 2013 #1
    In nanoscale electronics, electrons can be treated like billiard balls. The figure shows a simple device currently under study in which an electron elastically collides with a rigid wall (a ballistic electron transistor). The green bars represent electrodes that can apply a vertical force of 8.90·10-13 N to the electrons. If an electron initially has velocity components vx = 1.30·105 m/s and vy = 0 and the wall is at 45.0°, the deflection angle θD is 90.0°. How long does the vertical force from the electrodes need to be applied to obtain a deflection angle of 136.°?



    v=√(vx^2+vy^2)

    Δp=J(impulse force)=F(average)*Δt

    We had a similar problem in class where a baseball was pitched over home plate at 5 below the horizontal with a velocity of 40.23m/s and than hit back at 35° above the horizontal with a velocity of 49.17 m/s. the baseball weighed 0.145 kg and the bat made contact for1.2 ms. We solved this one by
    Δvx=(49.17 m/s)(cos35°)-(40.23 m/s)(cos185°)=80.35 m/s
    Δvy=(49.17 m/s)(sin 35°)-(40.23 m/s)(sin185°)=31.71 m/s
    √((80.35 m/s)^2+(31.71 m/s)^2)=86.38 m/s
    Δp=mΔv=(0.145 kg)(86.38 m/s)=12.5 kg m/s
    and the we used impulse formula to solve for average force:
    F(average)=Δp/Δt=(12.5 kg m/s)/0.0012s=10.4 kN

    I understand that we had to break up the velocity vectors in order to solve for the change in velocity vectors, because that change will give us change in momentum. However I am really hung up on this problem when it hits a 45° wall. Would the the force applied by the electrodes be the impulse force or average force? If anyone could help me through this that would be great! Thanks!
     
  2. jcsd
  3. Oct 13, 2013 #2
    Here is a picture as well.
     

    Attached Files:

  4. Oct 13, 2013 #3
    Does anyone have any idea?
     
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