SOLVED 1. The problem statement, all variables and given/known data A 120g granite cube slides down a 40 degree frictionless ramp. At the bottom, just as it exits onto a horizontal table, it collides with a 225g steel cube at rest. How high above the table should the granite cube be released to give the steel cube a speed of 185cm/s? 2. Relevant equations Ug=mgh K=0.5mv2 P=mv 3. The attempt at a solution I first calculated the momentum of the steel block (block 2): P=mv P2=(.225)(1.85) P2=0.41625kg*m/s Then, using conservation of momentum I deduced that the momentum of the granite block (block 1) before the collision must be equal to the momentum of the second block after the collision (P1=0.41625kg*m/s). Using this I can find the velocity of block 1 just before the collision: P=mv 0.41625=(0.120)v v=3.46875m/s Now I can find the kinetic energy of block 1 right before it hit block 2: K=0.5mv2 K=0.5(0.120)(3.46875)2 K=0.722J Since energy is conserved, Ug at the top of the ramp is equal to K at the bottom of the ramp (Ug=0.722J). We can now find the height of the block: Ug=mgh 0.722=(0.120)(9.8)h 0.722=1.176h h=0.6139m=61.39cm Where did I go wrong?