An atom of mass m moving in the x direction with speed v collides elastically with an atom of mass 3m at rest. After the collision the first atom moves in the y direction. Find the direction of motion of the second atom and speed of both atoms (in terms of v) after the collision? Ans: mass m moves at v/sqrt(2), mass 3m moves at v/sqrt(6) in the direction theta=-35.3 degrees. Momentum before = momentum after MV1i + MV2i = MV1f + MV2f x-direction MV1i = MV1fx + MV2fx y-direction 0 = MV1fy + MV2fy this translates into mv = mV1fx + 3mV2fx ....................(i) 0 = mV1fy + 3mV2fy ....................(ii) mv^2 = m(0 + V1fy^2) + 3m(V2fx^2 + V2fy^2) ........(iii) where : M, V1i, V2i, V1f, V2f, V1fx, V2fx, V1fy, V2fy , etc are variables m,v are values x = x-component , y = y-component, f = final , i = initial The question is are these equations right, as I cannot get the above answers Thanks and regards.