Having trouble with a Force problem

[Blue]

Two protons are initially held at rest 2.5*10^-10 meters apart. (Both would have positive charges, I would assume equal to the elementary charge, 1.60*10^-19 coulombs.)

If one of the protons is released, what is its speed when it is 8.0*10^-10 meters from the fixed proton?

(+) <-- 2.5*10^-10 --> (+)

initial (one proton is fixed)

(+) <----- 8.0*10^-10 ----> (+)

final

v_f = ?

Remember F = (kQ₁Q₂) / r²

where F = force, k = Coulomb's constant (9.00*10⁹Nm²/C²), Q₁ = charge of the first object, Q₂ = charge of the second object, r = distance between the two objects (in meters)

E_p = (kQ₁Q₂) / r

where E_p = Potential Energy

E = F/Q

E = kQ / r²

I'm stuck

Edited May 12, 2014 by Blue

[Brotoro]

Find the difference in potential energy. That is where the kinetic energy of the proton comes from. Find what velocity from the kinetic energy equation.

[LLlAMnYP]

E_k=kQ²(1/r₁-1/r₂)=m_pv²/2.

v=(2kQ²/m_p(1/r₁-1/r₂))^1/2=2.75*10⁴m/s

That's if the absolutely fictitious idea of having a "fixed" proton holds.

[YNM]

Use the fact that mechanic energy never changes. So

Em1 = Em2 // Em1 is while at initial position, Em2 is at end position

(k.Q1.Q2)/r1 = (k.Q1.Q2)/r2 + (1/2).m.v^2 // Em2 = Ep2 + Ek ; r1 is initial distance and r2 is end distance

The rest goes as the post above. Mind you that this doesn't take gravity into account (a good thing to be written down, if it's not really stated there).