Two protons approach each other from a great distance. What speed must each have if they are to reach a minimum separation of 10^{15} m?
The before and after states are shown below. To just barely reach separation d the particles will be at rest in the final state. To be at rest in the final state, they must have equal and opposite velocities to begin with. (Use momentum conservation to prove this, as we did in the class example.)
Then:

Before 
After 
Kinetic energy 
2(mv^{2}/2) 
0 
Potential energy 
0 
ke^{2}/d 
Total energy 
mv^{2} 
ke^{2}/d 
So, setting total energy before equal to total energy after:
This speed is about 1/30 of the speed of light.