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Problem: A proton follows the path shown in (Figure 1). Its initial speed is v0 = 2.4x106 m/s .What is the proton's speed as it passes through point P?

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Law of conservation of energy:

Ki+Ui+Wnc=Kf+Uf

Kinetic energy, K:

K=12mv2

Potential energy, U:

U=-kq1q2r

Wnc = 0

The law of conservation of energy is now written as:

12mv02-kq1er0=12mvf2-kq1erf12mvf2=kq1e(1rf-1r0)+12mv02

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Problem Details

A proton follows the path shown in (Figure 1). Its initial speed is v0 = 2.4x106 m/s .

What is the proton's speed as it passes through point P?

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