Protons are projected with initial speed vi = 9.81 103 m/s into a region where a

Question

Protons are projected with initial speed vi = 9.81 103 m/s into a region where a uniform electric field E = (-720 j) N/C is present as shown in the figure below. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons cross the plane and enter the electric field.

a) Find the two projection angles that will result in a hit.
° (smaller angle)
° (larger angle)


(b) Find the total time of flight for each trajectory.
ns (for the smaller angle)
ns (for the larger angle)

Explanation / Answer

ux = vi cos theta

delta x = ux*t

delta x = vi^2 sin 2 theta/ (-a)

Sin 2 theta = -a*delta x/vi^2 = qE/m * delta x/ vi^2

Sin 2theta = 1.6*10^-19*720*1.27*10^-3/1.67*10^-27*9.81^2*10^6

Theta = 32.775 degrees (Smaller angle)

Theta = 90- 32.775 = 57.224 (Larger Angle)

B)

Time of flight = 2*vi*sin theta/g

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