1. The problem statement, all variables and given/known data An alpha-particle with velocity 3.5 x 106 strikes a block of gold, atomic number 79 and mass number 197. Find the distance of the closest possible approach between the alpha particle and a gold nucelus, assuming Coulomb's law holds over such distances. Assume mp = mn = 1.67 x 10-27 2. Relevant equations p = mv KE = 1/2mv2 EPE = kq1q2/r2 3. The attempt at a solution Considering conversation of momentum and conservation of energy I worked out p = mv = (4 x 1.67 x 10-27)(3.5 x 106) = 2.3 x 10-20. The particle would repel the nucleus, at the closest approach they must have the same velocity. So p = mauv' + malpha v' v' = p/mau + mualpha v' = 2.3 x 10-20 / 201 x 1.67 x 10-27 v' = 6.4 x 104 So the change in KE in KEv minus KEv'. At this point I believe I've gone wrong as I've calculated the change in KE to be negligible yet the online question pp I'm using tells me to consider conversation of momentum (here). As I understand the remainder of the question is simply equate change KE to change in EPE, rearrange to find r and plug in values.