1. A proton an an alpha-particle are projected with a velocity v=(Ke^2/mL)^0.5 each, when they are far away from each other. If their initial velocity vectors are anti parallel to each other, with a distance "L" between them, find the distance of closest approach.(mass of proton = m, charge of proton= +e, mass of alpha-particle= 4m, charge of alpha particle= +2e) Ans: L[5+(89)^0.5]/8 2. F=kqQ/r^2, U=kqQ/r 3. Conservation of momentum and conservation of energy are the two prime theorems that are applicable in this case, as there is no external agent doing work, but they still are insufficient to eliminate the number of variables. As both the particles experience force (changing with time), towards each other they deflect from each other and at some point they are closest to each other... this is all i could sort out... Plz help!!!!!