# physics

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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?

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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

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