A proton moves with a speed of `7.45xx10^(5) m//s` dirctly towards a free proton originally at rest. Find the distance of closest approach for the two protons.
Given : `(1)/(4pi in_(0))=9xx10^(9)(N-m^(2))/C^(2), m_(p)=1.67xx10^(-27) kg` and `e=1.6xx10^(-19) C`

Explanation

As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic repulsion other will also start moving and so the velocity of first particle will decrease while of other will increase and at closect approach both will move with same velocity. So if v is the common velocity of each particle at closest approach, by 'conservation of momentum'.
`m u+mv+mv rArr v=1/2u`
And by conservation of energy `1/2 m u^(2)=1/2 mv^(2)+1/2 mv^(2)+1/(4pi epsilon_(0))e^(2)/(r)`
So, `r=(4e^(2))/(4pi epsilon_(0)m u^(2))[as v=u/2]`
And hence substituting the given data, `r=9xx10^(9)xx(4xx(1.6xx10^(-19))^(2))/(1.67xx10^(-27)xx(7.45xx10^(5))^(2))=10^(-12) m`