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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.
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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