1. The problem statement, all variables and given/known data A proton and an alpha particle (q= +2e, m=4u) are fired directly at one another from far way, each with an initial velocity of 0.01c. What is their distance of closest approach, as measured between their centres? 2. Relevant equations mivi = mf vf and KEi +Ui = KEf + Uf U = (q1 x q2 ) / (4επr) KE= 0.5mv^2 Mass of proton: 1.67E-27 Charge of Proton: 1.692E-19 c = 3E8 3. The attempt at a solution I'm sorry this question has been asked before but I didn't understand the explanations in those threads. I tried using the conservation of momentum: (1.67E-27 * 0.01c) - (4 * 1.67E-27 * 0.01c) = -(1.67E-27 * velocity of proton) + (4* 1.67E-27 * velocity of alpha particle) ∴ -9000000= -velocity of proton + 4*velocity of alpha particle but then I don't understand how to find the velocities of the particles individually or how to get this equation into the KE equation. When I tried using the conservation of energy I got: ½*1.67*10^-27*(0.01c)^2 + ½*4*1.67*10^-27*(0.01c)^2 = (3*1.602*10^-19) /4επr ∴ r = 115008.95 which is so far off, as the answer in the back of the book says: 1.93x10^-14 Which equation am I using wrong and how can I fix it?