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A proton’s speed as it passes point A is 3.00 × 10^{4} 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.

The kinetic energy at point B 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 at a potential difference 40 volts. It is evident that it is a positively charged, which is moving away from the positive charge that creates positive potential to a negative charge that generates negative potential. It has a charge at +e

DKE = (30-(-10))*1 =40eV

40eV is equivalent to 6.409*10-18 j

The total KE at point A is:

KE = 752.7*10-21 + 6.409*10-18 JE = 7.162*10-18 JE

KE = (1/2).*m*v2.

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