Newton's Laws Legacy Problem #22 Guided Solution

In this problem, like any problem, you have to begin by getting a good visual picture of what's going on. And what's going on in this problem is a baseball is moving along and hits a catcher's mitt and it comes to a stop. Let's presume for a moment that the baseball is moving towards the left. The moment it touches the catcher's mitt, it begins to slow down. Not only that, it begins to push upon the catcher's mitt and push the mitt to the left. The catcher, whose hand is in the mitt, begins to resist this force and push back to the right up on the ball. And so there's a force on the mitt and a force on the ball and the force on the ball is what slows it down. The force on the mitt is what moves it backwards in the direction of the ball's motion. We're told that the mitt recoils a distance of .135 meters. What we're asked to do is to calculate the force of the ball of Brandon's hand up on the ball during this .135 meters of distance. Now that's a very short distance and if you can just picture a baseball being caught, the moment there's contact with the ball, the mitt begins to move backwards. It's true of any catching situation. What we're told is the ball moves, it hits the mitt moving at 38.2 meters per second and eventually stops. Now there's three pieces of kinematic information here. The original speed of the ball, 38.2. The O equals 38.2 meters per second. The final speed of the ball, the F equals zero meters per second. And the distance over which the ball moves when the force is applied to it, D equals .135 meters. The first question they ask me to calculate is the acceleration of the ball. So that's a kinematic question, not a Newton's laws question. And I can find a kinematic equation that has these four variables in it. I can find one that goes VF squared equals VO squared plus 2AD. And I can use it to calculate the acceleration of the ball. The VO, the VF squared is zero. The VO is 38.2, so I go zero equals 38.2 squared plus 2 times the unknown A times .135 meters. And I solve for my acceleration. Now I can answer the question of part B, which is to find the force acting up on the ball. Now if I think about the ball during the catching phase of its motion, when it's slowing down, there's at least two forces acting up on it. There's the F grab, which is always MG and always directed straight down. And then there's the applied force by the catcher on the ball. Now that force is going to be really large compared to the down force of gravity. And that F applied force on the ball by the catcher's mitt is going to be mostly horizontal. So if I think about what I currently know, I know the mass of the ball, 0.145 kilograms, and I've calculated the acceleration. If I go MA, it gives me the net force, and it tells me what the two forces add up to. Now of the two forces, the F applied is really, really big, and the F grab is about 1.5 newtons. So a simple way to get at the answer is just to say that the F applied plus the F grab is going to be pretty much equal to the F apply. And so the net force is going to be equal to F, the applied force. And so once you go MA to calculate the net force, just call that the applied force. The gravity force doesn't make much difference here.