A little question causing big problems

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SUMMARY

The discussion focuses on calculating the distance of closest approach between a proton and an alpha particle, each with an initial speed of 0.141c. The conservation of energy principle is applied, where the initial kinetic energies of both particles are equated to the potential energy at the closest approach. The equation used is Enet1 = Enet2, leading to Kp + Ka = Kq1q2/r. The user expresses uncertainty regarding the assumption that the alpha particle comes to a complete stop at closest approach, indicating a need for further analysis of the system's dynamics.

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Homework Statement



A proton and an alpha particle (q = +2.00e, m = 4.00u ) are fired directly toward each other from far away, each with an initial speed of 0.141c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)


Homework Equations



Enet1 = Enet2
Kp + Ka = Kq1q2/r

Then solve for "r".


The Attempt at a Solution


I would figure that you could use conservation of energy in the sense that the energy of the system initially is the kinetic energies of the two particles combined (Enet1 = Kp + Ka). At the point of closest approach, their speeds should be zero, and hence Enet2 = Uelec = Kq1q2/r. From here it should be straightforward:

Enet1 = Enet2
Kp + Ka = Kq1q2/r

Then solve for "r".

However, this is incorrect. Perhaps my assumption that the alpha particle (4 times the mass, 2 times the charge) stops completely is wrong. At this point, I really have no idea.
 
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what about W= intergral of F(x) dx
 

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