Question

Q) A proton's speed as it passes point A is 3.00×10⁴ m/s . It follows the trajectory shown in the figure.
(Figure 1)

What is the proton's speed at point B? Express your answer with the appropriate units.

Figure 1 of 1 Br 30V 10 V -10 V

Answer 1

The kinetic energy at point A is:

KE = (1/2)*m*v^2 = (1/2)*1.673*10^-27*(3.00*10^4)^2 = 752.7*10^-21 J

It accelerates through a potential difference of 40 volts. (Acceleration is apparent, because it is a positive charge, moving away from a positive charge that creates the positive potential, toward a negative charge that creates the negative potential). It has a charge of +e

ΔKE = (30 - (-10))*1 = 40eV

40eV converts to 6.409*10^-18 J

Total KE at point B is:

KE = 752.7*10^-21 J + 6.409*10^-18 J = 7.162*10^-18 J

KE = (1/2)*m*v^2

v = sqrt( 2*KE/m) = sqrt( 2*7.162*10^-18/1.673*10^-27) = 92.5*10^3 m/s