Momentum Study Guide

Question

A .105kg hockey puck moving at 24 m/s bounces backwards off a 75 kg goalie at 10 m/s (who was initially at rest). What is the speed of the goalie after getting hit by the puck?

Answer 1

 

 The important thing to remember about momentum situations is that, within your system, momentum is conserved. This means that the total momentum before the "event" must equal the total momentum afterwards. The goalie's momentum is 0 beforehand (since they're at rest) so the total momentum is that of the puck before: .105 x 24 = 2.52 kg m/s. I'm going to call that initial puck direction "positive." The puck's momentum afterwards is -10 x .105 = -1.05 kg m/s. The total momentum needs to be 2.52, so the goalie's "after" momentum must be 2.52+1.05 = 3.57 kg m/s (3.57 +(-) 1.05 = 2.52). The goalie's speed is derived by dividing their momentum by their mass so, 3.57/75 = 0.05 m/s. Just in case the question really means velocity and not speed, I'd indicate a + direction or say that the goalie would be moving in the same direction that the puck was BEFORE the collision.