Question

55. A proton moving at 3.0 x 10t m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 NiC is acting down, how long does it take the proton to return to the horizontal plane? (Hint: Ignore gravity. mproton 1.67 x 1027 kg, gproton 1.6x 1019 c) a. 7.8 x 10-7 s b. 1.7x 10 6s c.3.9x10 6s d. 7.8 x 10° s

Answer 1

First we need to form an equation of motion for the proton.

Newton's 2nd Law applies in this case:

ΣF=ma => Fel.= ma (weight ignored, as we're advised to ignore gravity)

So EQ=ma => a=EQ/m => a=(400x1.6x10^-19)/(1.67x10^-27) => a=3.83...x10^10 m/s^2

This acceleration is directed vertically downards (it has the same direction with Fel., and hence with E)

What we really care about is the vertical motion of the proton, as we're asked to find the time for it to return to the horizontal plane, i.e. the time to return to its original vertical position.

Vertically, the motion is described by uniformly decelerating motion:

y = y0 + (uy)t - (1/2)at^2.......(*)

From the data, uy, the initial vertical velocity component is found to be

(uy)=[3x10^4]xcos60˚=1.5x10^4 m/s

y0 = 0, for obvious (? allow me to say so) reasons.

a=3.83...x10^10 m/s^2

From (*): y = (1.5x10^4)t - (1.916...x10^10)t^2

When the proton returns to the horizontal plane, y = 0, so the time at which this occurs is given by:

0 = (1.5x10^4)t - (1.916...x10^10)t^2

Solve this by taking t as a common factor.

This gives t=0 as the first root (rejected; this is the time of projection).

The remaining root yields t=(7.82x10^-7)s, or (7.8x10^-7)s to 2s.f.; the correct answer would be a.