1. The problem statement, all variables and given/known data I am going through past Olympiad qualifying exams as practice for this year's test. I came upon the following problem in the 2008 F=ma and continue to come up with the wrong answer. I keep coming up with C and that is not the answer provided in the answer key. 22. A bullet of mass m1 strikes a pendulum of mass m2 suspended from a pivot by a string of length L with a horizontal velocity v0. The collision is perfectly inelastic and the bullet sticks to the bob. Find the minimum velocity v0 such that the bob (with the bullet inside) completes a circular vertical loop. (a) 2√Lg (b) √5Lg (c) (m1 + m2)2√Lg/m1 (d) (m1 − m2)√Lg/m2 (e) (m1 + m2)√5Lg/m1 2. Relevant equations m1v1= (m1+m2)v2 1/2 mv ^2 = mgh 3. The attempt at a solution m1v1 = (m1+m2)v2 v2 = (m1v1)/(m1+m2) So the kinetic energy of the block should be: 1/2 (m1+m2)v2^2 Then I set the initial kinetic energy of the block equal to the potential energy it would have at height 2L. 1/2 (m1+m2)v2^2 = 2L(m1+m2)g Since v2 = (m1v1)/(m1+m2): 1/2(m1+m2)*((m1v1)/(m1+m2))^2 = 2L(m1+m2)g m1+m2 should cancel: ((m1v1)/(m1+m2))^2 = 4Lg (m1v1)^2 / (m1+m2) ^2 = 4Lg m1 ^2 * v1 ^ 2 = 4Lg / ((m1+m2) ^2) Take the Square Root of the whole thing: m1 * v1 = 2sqrt(Lg)/(m1+m2) SO my answer is (c) (m1 + m2)2√Lg/m1, which is apparently incorrect. If someone could help me correct this that would be great. Thanks.