posted by maddie .
Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 1.5e6 m/s. What is the distance of closest approach?
PLEASE HELP! im very confused.
It will be at a value of the separation distance R such that the initial kinetic energy, 2*(1/2)M V^2, is equal to the potential energy k e^2/R.
Note that you have to add the kinetic energies of the two particles, but the potential energy only has to be counted for bringing one particle within a distance R of the other.
R = ke^2/[M V^2]
k is the Coulomb constant (which you may need to look up) and e is the proton charge (same as the electron charge, except for sign). M is the proton mass, and V = 1.5*10^6 m/s