Question

Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 1.50 x 105 m/s. What is the distance of closest approach?

Answer 1

To solve this problem we will start from the conservation of energy. Here we will have that the kinetic energy must be equivalent to the potential energy. That is

`KE = (1)/(2) mv^2`

Here,

m = mass

v = Velocity

`PE = (kq^2)/(r)`

`k = 8.99*10^9 N\codt m^2\cdot C^(-2) \rightarrow` Coulomb constant

`q = 1.602 x 10^(-19) Coulombs \rightarrow` Charge Proton

r = Distance

Equation both expression we have that

`2((1)/(2)) mv^2 = (kq^2)/(r)`

The kinetic energy is multiplied by two, by the existence of the two protons.

`r = (kq^2)/(mv^2)`

Replacing the values we have that

`r = (( 8.99*10^9)(1.602 x 10^(-19))^2)/((1.67*10^(-27))(1.5*10^5)^2)`

`r = 6.1402*10^(-12)m`

Therefore the distance of closest approach is
`6.1402*10^(-12)m`