tough problem--conservation of momentum? suppose you have a dumbbell standing up vertically against a frictionless wall |O || || || |O_____________ (given, m, L (length of rod)) the weights are of course the same mass. the bottom weight of the dumbbell starts sliding out so the top mass moves vertically, while the bottom moves horizontally. i am asked to find the velocity when the weights are both moving with the same speed. i'm not really clear as to how to approach this problem.. it seems if i apply conservation of momentum, mv1 = mv2; v1 = v2 which is obviously not true. so does this mean momentum is not conserved? if i take the system to be the earth and the dumbbell, the external forces would sum to zero, wouldn't they? i also figured that x^2 + y^2 = L^2; 2x * dx/dt + 2y * dy/dt = 0 but this doesn't seem to be enough. it seems as if the speeds should be the same since it is a rigid body. could anyone explain this to me? or give a hint or two? any help would be appreciated! thanks!