Questions what is the distance of closest approach when b120

Questions: what is the distance of closest approach when b=1.20*10^-13 m and when b=1.00*10^-14 m? (when b=1.00*10^-12 m, the distance of closest approach is 1.01*10^-12)

Solution

potential energy: PE = K*q1*q2/r0
kinetic energy: KE = 0.5*m*v0^2

Total energy: U(r0) = K*q1*q2/r0 + 0.5*m*v0^2

from conservation f momentum:

v1*m*b = v0*m*r0

         v0 = v1*[b/r0]

thus, the total energy:

         U(r0) = K*q1*q2/r0 + 0.5*m*v1^2*b^2/r0^2

this must equal the initial kinetic energy U(1) (15 MeV)

U1 = K*q1*q2/r0 + 0.5*m*v1^2*b^2/r0^2

but, 0.5*m*v1^2 = U1

so,

   U1 = K*q1*q2/r0 + U1*b^2/r0^2

U1*r0^2 - (K*q1*q2)*r0 - U1*b^2 = 0

r0^2 - (K*q1*q2/U1)*r0 - b^2 = 0

here, U1 = 10.5 MeV = 1.68*10^-12 J

solve: k*q1*q2 = 3.789*10^-26
solve: k*q1*q2/U1 = 2.255*10^-14

r0^2 - 2.255*10^-14*r0 - [1.20*10^-13 ]^2 = 0

r0 = 1.318*10^-13 m

----------------------------------------------------------------------------------------------------

when b = 1.0*10^-14

r0^2 - 2.255*10^-14*r0 - [1.0*10^-14 ]^2 = 0

r0 = 2.63*10^-14 m