Momentum, Collisions and Explosions Legacy Problem #29 Guided Solution

This is a very, very difficult problem, one that's going to require that you be organized, be a good thinker, use your understanding of a wealth of physics concepts, and most importantly, employ the habits of an effective problem solver. This is a type of problem which I call a momentum plus problem. A momentum plus problem is a problem that involves a momentum analysis plus some other form of analysis. In this problem, we have a baseball that's moving at high speed, and it hits a box on a table that's originally at rest. The collision of the ball into the box causes the ball and box together to move with a post-collision velocity. We're asked to determine the distance that the ball and box move across the tabletop after the collision. My solution will begin by the construction of a diagram representing the collision between the ball and the box. I draw a diagram, and on the diagram, I record known information. First of all, I know that the ball is moving. Its pre-collision velocity is 54.1 meters per second. Its mass is 315 grams, which is not a great unit for mass, so I'm going to convert it to 0.315 kilograms to make it consistent with the mass of the box. The box is at rest. Its velocity is 0 meters per second, and its mass is 3.54 kilograms. After the collision, we now have both objects moving and moving with the same speed. The mass of the objects combined is 3.855 kilograms. I determine that by summing the masses of the two individual objects. This box and ball are moving with some velocity, which we'll call v, and I can solve for v by using momentum conservation. That momentum conservation equation would be written something like this. 0.315 kilograms times 54.1 meters per second, that's momentum of the ball, plus 3.54 kilograms times 0 meters per second. The box doesn't have any momentum before the collision, is equal to 3.855 kilograms times v. That's the momentum of both box and ball after the collision. I can take this equation and solve for v. 17.0415 is what the left side evaluates to, and that's equal to 3.855 v. Dividing through by my 3.855 gives me the velocity of the box and ball after the collision. It comes out to be 4.4206 meters per second. Now I know the post-collision velocity of the box, and what I have to find is the distance it slides across the tabletop. Finding this stopping distance demands that I use Newton's laws and kinematics. Using Newton's laws, I'm going to try to determine the acceleration from the coefficient of friction. So, I construct a free body diagram representing the forces acting upon ball and box. There is of course a downforce called f-graph. Its value would be m times g, and there is an upforce, f-normal. This value would be equal to the downforce value, so we'll just call it m times g. Finally, there is a friction force acting off to the left on the forward-moving box, and that's going to be mu times f-normal. Now this friction force, mu times f-normal, is just going to be equal to mu times mg. And the mu-mg for friction force is the net force. So I can say A is equal to mu-mg over m. That's the A left net over m equation. Now if I cancel my m's, I end up with A equal mu-g, and I can calculate the acceleration from the coefficient of friction. It comes out to be 6.9972 meters per second squared. Now let's look at this box and ball embedded within it. It begins with a velocity of 4.4206 meters per second, and then begins to accelerate at 6.9972 meters per second squared in the negative direction. Its final velocity is 0 meters per second, and we wish to calculate the distance it slides. I can pick a kinematic equation out capable of solving for d through knowledge of Vf, Vo, and A. That equation is Vf squared equal Vo squared plus 2Ad. I can substitute 0 in for Vf, 4.4206 for Vo, and negative 6.9972 in for A, and I can solve for d. Doing good algebra gives me an answer of 1.3964 meters, and I'll round that to 1.40 meters.