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

`rarr v = 7.45 xx 10^(5) m//s ` u =0

By the law of cons. Of energy `(1)/( 2) mv^(2) + 0 = ( Ke^(2))/( r_(0))+ ( 1)/( 2) mv_(1)^(2) + ( 1)/( 2) mv_(1)^(2)` ....(1)
By the cons. of momentum `mv+ 0 = mv_(1) + mv_(1) `
`:. v_(1) = (v )/(2)`
From equation `(1) (1)/(2) mv ^(2) = (ke^(2))/( r_(0)) + ((v)/(2))^(2)`
`r_(0) = ( 4)/( mv^(2)) xx ke^(2) = ( 4 xx ( 9 xx 10^(9)) ( 1.6 xx 10^(-19))^(2))/( ( 1.66 xx10^(-27) ) ( 7.45 xx 10^(5))^(2)),`
`r_(0) = 1.0 xx 10^(-12) m`