A bullet with mass m and velocity v, penetrates a mass M suspended on a piece of string lenght R. The bullet loses half of its velocity in the impact and carries on in a straight line. What must the minimum magnitude of v be so that the mass would spin a full circle. p = mv, 1/2mv^2, mgh, 3. The attempt at a solution Ok so im not sure whether this is a moment of kinetic energy problem but im leaning towards the Ke option. Im assuming that the mass needs only to get to the top of the circle as a result of the collision where all of its kinetic energy has turned into potential energy which then allows it to fall back down thus making a full revolution. Bullet's initial Ke = 1/2mv^2 Bullet's final Ke = 1/2m(v/2)^2 = 1/2m(v^2/4) = (mv^2)/8 Ke gained by mass M = (mv^2)/8 Energy needed for mass M to get to the top of the circular path = Mgh = Mg2R Based on the assumption that work done in raising a mass to a certain height is independent of the chosen path. Thus (mv^2)/8 = Mg2R and solving for v we get v = SQRT(16MgR/m) A swing and a miss?