1. The problem statement, all variables and given/known data A 2.3 kg wood block hangs from the bottom of a 1.3 kg, 1.3 m long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 35 degrees. 2. Relevant equations Conservation of Energy. Conservation of Momentum 3. The attempt at a solution (1/2)(m)(v^2) = mgh Drawing a triangle, h is 1.3 * cos 35 = 1.06m So (1/2)(v^2) = (9.8)(1.06) v = 4.568 m/s Conservation of momentum: (mB)(vB) = (mB + mP)(v) (.012)vB = (2.3+.012)(4.568) vB = 10.561 / .012 = 880.21 m/s Not sure where I went wrong.