1. The problem statement, all variables and given/known data An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of height 1 m. The bullet remains in the block, and after the impact the block lands 2 m from the bottom of the table. Determine the initial speed of the bullet. 2. Relevant equations Conservation of Momentum (for perfectly inelastic conditions) m1v1(initial) + m2v2 (initial) = (m1+m2)vf m1 = mass of bullet (.008 kg) v1 = velocity of bullet m2 = mass of block (.250 kg) v2 = velocity of block (0 m/s) vf = final velocity of block and bullet right after collision Conservation of Energy KE + PE (after collision) = KE + PE (on ground), rewritten as: 1/2mvi^2 + mgyi = 1/2mvf^2 + mgyf m = mass of bullet + block (.258 kg) vi = initial velocity of bullet + block after collision g = 9.8 m/s^2 yi = 1 m vf = 0 m/s yf = 0 m 3. The attempt at a solution I think I have this problem solved correctly, but wondered if someone could confirm that for me (those darn even-numbered problems at the back of the book without answers!). First, I use the conservation of momentum and put in my known values, ending up with: .008v1 = .258vf Next, I use the conservation of energy and put in the known values, ending up with 4.427 m/s for the initial velocity of the block and bullet, which si the final velocity in my first equation. I substitute 4.427 m/s into my first equation and get an answer for the initial speed of the bullet as 142.75 m/s.