never mind I got the solution:
KE(B) + PE(B) = KE(A) + PE(A)
(1/2)mvB^2 + qV(B) = (1/2)mvA^2 + qV(A)\
vB^2 = vA^2 + (2q/m)[V(A) - V(B)]
Answer: 68739.85m/s
I should have known it was a simple conservation of energy problem
A proton's speed as it passes point A is 53,000 m/s. It follows the trajectory shown in the figure below, in which V1 = 15 V and V3 = 5 V. What is the proton's speed at point B?
Figure: http://www.webassign.net/knight/p29-44alt.gif
Homework Statement
Mass of proton: 1.67e-27
Charge of...