Protons are projected with an initial speed v0 = 9,850 m/s into a region where a

Question

Protons are projected with an initial speed v0 = 9,850 m/s into a region where a uniform electric field of magnitude E = 320 N/C is present (see figure below). The protons are to hit a target that lies a horizontal distance of 1.12 mm from the point where the protons are launched.

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

(b) Find the total duration of flight for each of the two trajectories. (Enter the smaller duration of flight first.)

Explanation / Answer

v0 = 9850 m/s ; E = 320 N/C ; R = 1.12 mm

a)The force experienced by the protons in electric field will be:

F = q E ; also F = m a

equating the above two to obtain acceleration.

ma = q E

a = q E/m

a = 1.6 x 10^-19 x 320 / 1.67 x 10^-27 = 3.07 x 10^10 m/s^2

We see that the proton is to follow a projectile motion, with range given by

Range = R = v0^2 sin(2theta)/a

sin(2theta) = R a/v0^2

sin(2theta) = 1.12 x 10^-3 x 3.07 x 10^10/(9850)^2 = 0.354

2theta = sin-1(0.354) = 20.75

theta = 10.38 deg

theta2 = 79.63 deg

b)for a projectile

t = 2 v0 sin(theta)/a

t1 = 2 x 9850 x sin10.38/3.07 x 10^10 = 1.16 x 10^-7 s

t2 = 2 x 9850 x sin79.63/3.07 x 10^10 = 6.31 x 10^-7 s

Hence, t = 1.17 x 10^-7 ; t2 = 6.31 x 10^-7 s

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