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MCATPhys

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Say you have two protons 10nm apart (at rest). If they are released, they naturally tend to accelerate away from each other. But what's the velocity after they are very far apart?

Basically, all the initial potential energy gets converted to kinetic energy. So I equal them to each other like this:

PE = qV = kq^2/.000000010 = 0.5mv^2

q is the charge of the proton, k is the constant, m is the mass (1.66*10^-27), v is the final velocity

When I solved it, the answer I get is around 5.3*10^3 m/s... but the right answer is supposed to be 3.8*10^3

Can someone please tell me what I am doing wrong... just the theory.

Basically, all the initial potential energy gets converted to kinetic energy. So I equal them to each other like this:

PE = qV = kq^2/.000000010 = 0.5mv^2

q is the charge of the proton, k is the constant, m is the mass (1.66*10^-27), v is the final velocity

When I solved it, the answer I get is around 5.3*10^3 m/s... but the right answer is supposed to be 3.8*10^3

Can someone please tell me what I am doing wrong... just the theory.

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