Electrostatic question. HELP IS REALLY REALLY NEEDED!

[dorjiako]

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Two protons (mass = 1.66 x 10^-27kg) initially are at rest at a distance of 10nm from each other. They are released and accelerate away from each other. How fast are they both going after they are very far apart. K = 9 x 10NM^2/c^2.
How did they got the answer 3.8x10m/s? Any input will be appreciated.

 

[dorjiako]

Thanks all.

 

[docntrainin]

Thanks all.

Do you mind posting the solution?

 

[wish2bMD]

I think you would use conservation of energy here. U=kq1q2/r and KE=1/2 (mv^2)
set them equal to each other and calculate the answer. I'm not positive since I didn't actually do the calculation.

 

[dorjiako]

Do you mind posting the solution?

Einitial = Efinal. potential potential energy(Kqq/r) = 2(1/2 mv^2)
K.E is multiplied by 2 because both particle were moving.
(2.304 x 10^-20) = (1.6 x 10^-27)(v(final)^2 -v(initial)
(vfinal ^2 -(0))= 14400000
vfinal = 3794
Therefore V = 3.8 x 10^3 m/s.