Tow identical masses are released from rest in a smooth hemisherical bowl of radius R. One mass is on the top of the bowl a height R from the bottom and the other mass is on the bottom when they are released. There is no friction. If the masses stick together when they collide, how high above the bottom of the bowl wil the masses go after colliding? My work so far is: stage one: K1 + U1 = K2 + U2 0.5*mAVa2(squared) = mAgR Va2 = sqrt(2*g*R) Va2 the speed of mass A on the bottom stage two: K1 + U1 = K2 + U2 Mgy2 = 0.5*M*2*g*R where M = 2*m so y2 = R The right answer is y2=R/4 I can't see what I am doing wrong or not doing. Could someone please give me a hint to this problem?