Assume: This collision takes place in outer space where there is no gravitational field and no friction. A small piece of putty of mass 31 g and negligible size has a speed of 2.5 m/s. It makes a collision with a rod of length 3 cm and mass 83 g (initially at rest) such that the putty hits the very end of the rod. The putty sticks to the end of the rod and spins around after the collision. After the collisions the center-of-mass has a linear velocity V and an angular velocity ω about the center-of-mass “+ cm”. What is the velocity V of the center-of-mass of the system after the collision? Answer in units of m/s. Icm = 1/12 MR^2 Ipm = MR^2 The Attempt at a Solution First I said that that the moment of inertia for the system is I=1/12(M+m)R^2 ("M" being the mass of the rod and "m" being the mass of the putty) Then using conservation of angular momentum i got: mR^2w = (R^2/12)(M+m)w substituting the respect v's (and Rs) in for w i got: mv1R = (R/12)(M+m)v2 then v2 = 12mv1/(M+m) plugging in my numbers i got [12(31)(2.5)]/[(31+83)] = 8.16 m/s