Momentum, Collisions and Explosions Legacy Problem #30 Guided Solution

This is a very difficult problem, one that's going to require that you do a lot of thinking, visualizing, organizing of a solution, and practice the habits of an effective problem solver. What we're told is we have two students, a boy and a girl, at rest on the ice rink, and they push against each other such that the girl moves one direction, the boy moves the other direction. We're given their initial mass, and we're given the speed of the girl after the collision. We're told that when they finally separate from each other, the impulses cease. They're 1.4 meters apart. And we're asked to determine the distance of separation between them five seconds after they've slidden at their post-impulse speed. Now this is what I call a momentum-plus problem, in which you're doing a momentum analysis to determine something about the post-collision speed of the objects, and then you're doing another analysis in which you're using some other form of physics, kinematics, or Newton's laws, or a combination, in order to determine another quantity describing the motion after the collision, the impulse, or the explosion. So here, I'm going to begin by doing the momentum analysis of this momentum-plus problem. I'm going to look at the boy at rest and the girl at rest, and then I'm going to look at the fact that after they push on one another, the girl's moving, she has momentum in one direction, the boy therefore must have an equivalent amount of momentum, but in the opposite direction. And I can find the velocity of the boy afterwards using momentum conservation. I do that by saying that the momentum of the girl after the collision, the 48 kilograms times the 6.8 meters per second, is equal to the momentum of the boy, which would be 72 kilograms times v. So if 72v equals 48 times 6.8, I can solve that for v, and end up getting 4.533 meters per second. That's the momentum part of this momentum-plus problem. Now for the plus part, in which I'm asked to determine how far apart are the boy and the girl after 5.0 seconds. Well, they're on ice, and since there's no other information given about this, I'm just going to presume that that's a constant velocity motion once they're done pushing on one another. And because it is a constant velocity motion, I can find the distance the girl traveled by going speed times time. 6.8 times 5 seconds would give 34.0 meters, which is the distance that the girl travels in the 5 seconds after the impulse ceases. And for the boy, I can do a similar thing with the boy's velocity of 4.533 meters per second times 5 seconds. The boy would travel a distance of 22.6 repeating meters in the 5 seconds after impulse. Now I have to answer the question, perhaps, where the most thinking and visualizing comes in, is at this point. We've just calculated the distance the girl traveled and the distance the boy traveled, and they traveled away from each other, a net distance away from each other of 56.66 repeating meters. Now what I have to find is their separation distance. And to this 56.66 meters, I have to add the 1.4 meters, the original separation distance, when they, with outstretched arms, finish pushing off one another. And so when I take all these distances and add them together, I get 58.066 repeating meters, and I can round that to 58.1 meters.