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Nanoscale electronics and impulse force

  • Thread starter xdctassonx
  • Start date
  • #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!
 

Answers and Replies

  • #2
Here is a picture as well.
 

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  • #3
Does anyone have any idea?
 

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