Edit to the title: CHALLENGE problem on electric potential energy* 1. The problem statement, all variables and given/known data A proton and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with an initial speed of 0.010c. what is their distance of closes approach, as measured between their centers? 2. Relevant equations u = proton mass = 1.67262158 × 10-27 kilograms e = elementary charge = 1.60217646 × 10-19 coulombs Kinetic energy Ke= 1/2 mv^2 U = Kq1q2/r K = 9.0exp9 deltaKe + deltaU = 0 3. The attempt at a solution (deltaKe1 + deltaU1) + (deltaKe2 + deltaU2) = 0 where 1 represents the u mass proton and 2 represents the alpha particle. ((0-.5u(.01c)^2) + (Ke*2e/r)) + ((0-2u(0.01c)^2) + (Ke^2e/r)) = 0 solve for r =2.45 *10^-14 the answer in the textbook is 1.93 * 10 ^-14 Iv done this over and over and over and am getting this 2.45 answer. please help!!!!! it is very frustrating! thanks!!!