1. The problem statement, all variables and given/known data A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation), and the two are dropped simultaneously from height h = 1.8m. (Assume the radius of each ball is negligible relative to h.) a. If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball? b. What height does the small ball then reach? 2. Relevant equations Ug=mgh K=1/2 mv2 p=mv 3. The attempt at a solution Define time t=0 to be before the balls are dropped. Time t=1 is just before the large ball collides with the ground. Time t=2 is just after the large ball collides with the small ball. K1 = Ug,0 = Mgh + mgh To find momentum at t=1, velocity is needed. Velocity will be the same for both balls, since they are dropped form the same height. Considering only the larger ball: K1 = Ug,0 1/2 Mv12 = Mgh v1 = √(2gh) V1 will also be the same for the smaller ball. When the collision occurs, the momentum immediately before and after the collision (t1 and t2, respectively) should stay constant. p1 = v1 (M + m) = √(2gh) * (M + m) = p2 After the collision, the velocity of the large ball is 0, so p2 = mv2 To find v2, the kinetic energy equations are needed again. All the collisions are elastic, so the kinetic energy will be the same at t1 and t2. K1 = Mgh + mgh = (M + m)gh K2 = 1/2 mv22 K1 = K2 (M + m)gh = 1/2 mv22 v2 = √(2gh(M + m) / m) So after some substitutions, it looks like p1 = p2 (M + m)v1 = mv2 (M + m) * √(2gh) = m√(2gh(M + m) / m) Everything is known except for m, so it seems like it should be able to be solved at this point. However, when I try to simplify this by squaring each side, I end up with M=0 or -m=M. Neither of those can be right, so I think I must have made a mistake in the physics part. I don't know where though. Can anyone spot the error?