A positicely charged particle starts at rest 25 cm from a second positicely char

Question

A positicely charged particle starts at rest 25 cm from a second positicely charged particle which is held stationary throughout the experiment. The ffirst particle is released and accelerates directly away from the second particle. When the first particle has moved 25cm, it has reached a velocity of 10 m/s. What is the maximum velocity that the first particle will reach? The answer is 14m/s but I am not sure how to work it....
I really need someone to help me understand how to work this question.

Explanation / Answer

After travelling change in kinetic energy =change in potential energy

change in potential energy = (kq^2/0.25 - kq^2/0.5) = kq^2/0.5

change in kinetic energy = 1/2mv^2 = 1/2m(10)^2 = 50m

50m = kq^2/0.5 ......(1)

As particle reaches infinity

Change in potential energy = (kq/0.25 - 0) = kq/0.25

This is equal to the change in kinetic energy = 1/2m(v1)^2

1/2m(v1)^2 = kq/0.25........(2)

Using equation (1) and(2)

(v1)^2 = 200

v1 = 14.1 m/s

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