1. The problem statement, all variables and given/known data A 0.5 g pellet is fired horizontally at two foam squares resting on a frictionless tabletop. The pellet rips through the first block (mass = 1 kg) and embeds itself in the second block (mass = 0.5 kg). As the bullet interacts with the blocks, it causes them to move with speeds of speed = 0.43 m/s and speed = 1.1 m/s, respectively. What is the speed of the pellet as it emerges from the first block? 2. Relevant equations P=mv; conservation of momentum 3. The attempt at a solution We began by calculating the momentum of the system in its final state, by adding the mass of the second block to the mass of the pellet, then multiplying by the final velocity of the second block (with the pellet embedded): P3 = (0.0005kg + 0.5kg)(1.1m/s) = 0.55055 kg m/s We then reasoned that, by conservation of momentum, the momentum of the pellet just prior to entering the second block would be equal to the momentum of the block+pellet after the pellet had embedded in the second block: 0.55055kg m/s = (0.0005kg) v , where v is the entry velocity of the pellet into the second block. Solving for v: v = (0.55055kg m/s) / 0.0005kg = 1101.1 m/s Assuming that the difference between the velocity of the pellet as it exits the first block, and its velocity as it enters the second block, is negligible, the final answer would therefore be 1101.1 m/s. However, the solution from the instructor took a different approach, and got a different answer. She states: "based on conservation of momentum, the second block and pellet should be equal to the combination of the first block and the pellet leaving the block." Based on this reasoning, the solution given by the instructor is: 0.55055kg m/s = 0.0005v + 0.43 v = 241 m/s What was wrong with our reasoning?