Question

Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 2.20 x 105 m/s. What is the distance of closest approach

Answer 1

r = 2.85 * 10^(-12) m

Step-by-step explanation:

The distance of closest approach is the minimum distance at which both protons can be. This is due to the repulsive force that acts on both protons.

We know that the total kinetic energy of both protons is equal to the potential energy due to one proton acting on another, hence:

½mv² + ½mv² = kq²/r

mv² = kq²/r

=> r = kq²/mv²

Given that:

k = Coulumbs constant

q = 1.6023 * 10^(-19) C

m = 1.673 * 10^(-27) kg

v = 2.20 * 10^5 m/s

Then, r is:

r = (9 * 10^9 * [1.6023 * 10^(-19)]²) / (1.673 * 10^(-27) * [2.2 * 10^5]²)

r = 2.85 * 10^(-12) m

This is the distance of closest approach.