1. The problem statement, all variables and given/known data A mass m with velocity v0 makes an inelastic collision with second mass M that is suspended by a string of length L. The velocity v0 is perpendicular to the vertical string. After the collision the combined masses, m+M, rotate in a vertical plane around the point of suspension of the string. Find the minimum value v0 such that the string remains always under tension for a complete rotation. 2. Relevant equations ∑Fr=mv2/r T=1/2mv2 U=mgh 3. The attempt at a solution The condition for the an inflexible string to remain under tension is that it never goes slack, i.e. length=L for the entire rotation. Since the collision is inelastic, momentum is conserved while kinetic energy is not. So, mv0=(m+M)v' where v' is the magnitude of the x-directed velocity of the combined masses (m+M) after the collision. so v'=m/(m+M)v0. At the bottom of the vertical circle (In Polar Coordinates), letting m'=(m+M) ∑Fr= Tb-m'g=m'v'2/L (*) At the top, ∑Fr=Tt+m'g=m'v''2/L (**) Using Energy, 1/2m'v'2=1/2m'v''2+m'g(2L) (***) v'2=v''2+4gL v''2=v'2-4gL I am going to pause here to see if there is anything I am missing.