Momentum, Collisions and Explosions Legacy Problem #25 Guided Solution

A good problem solver reads the problem carefully and develops a mental picture of what's going on. You'll notice that my effort to develop a mental picture involved drawing a diagram of the two objects involved in the collision. A good problem solver also identifies known and unknown information. The diagram is a great place to record this type of information. Finally, a good problem solver identifies the unknown quantity and then uses physics principles and conceptual reasoning skills to plot out a strategy as to how to get from the known information to the unknown information. This problem involves a collision between a cue ball and a thigh ball. We're told the pre-collision velocities of both objects. We're told that the cue ball is moving 82 centimeters per second, catches up and collides with the thigh ball, which is moving slower at 24 centimeters per second. Both objects are moving in the same direction before the collision. We're told that immediately after the collision that the thigh ball speeds up to 52 centimeters per second and we're asked to determine the post-collision speed of the cue ball. Now in this problem we have to assume that the mass of both objects is the same. We'll just call this mass m. They're about the same, so we're going to call the mass of that ball m. Now what I'm going to do is I'm going to use the principle that the total momentum of the system before the collision is equal to the total momentum of the system after the collision. If you were to add the individual momentums of cue ball and thigh ball before the collision, it should be equal to the individual momentums summed together after the collision. So for the cue ball, I'm going to call its momentum m times 82 centimeters per second. And for the thigh ball, I'll call its momentum m times 24 centimeters per second. Together this adds up to 106 centimeters per second times m. This is the pre-collision momentum of the system. It should be equal to the same number for the post-collision momentum of the system. So I'll call the post-collision momentum of the system m times v for the cue ball, where v is my unknown quantity. And for the thigh ball, I'll call it m times 52 centimeters per second. So I've written a statement that goes like this. 106 times m equals m times v plus m times 52. I can divide through each term of the equation by m, and the equation becomes 106 equals v plus 52. Subtracting 52 from each side gives me a final answer for v of 54 centimeters per second. This is the post-collision velocity of the cue ball.