Hi, this is module 11 of Two Dimensional Dynamics. Today's learning outcomes are to derive the impulse momentum relationship, and to define impact. So let's start off with the momentum form of Euler' 1st law. Let's look at the definition of momentum for a system of particles. So I have a system of particles here, each particle has a mass m, and so this is m sub i, this is m sub j. Number of particles say N, capital N particles. Each one of these particles has a momentum, a linear momentum, and the momentum is defined as mass times velocity, and I show that there. So, the summation of the momentum for the system of particles is each one of these masses, times their velocities summed up over a total number of capital M particles. And, I can then take the position vector from some fixed origin in an inertial frame form this O to each of the particles. And I can take the dot, time derivative of that position vector, and it's the same as the velocity. So I've replaced it there. And then we saw in a previous module, that the summation of mR over a system of particles is equal to the total mass times a position vector from O, to the mass center by the definition of mass center. And so now we can extend that to, we can say we have an infinite number of particles, and we extend that to a continuously distributed mass. Now if I take the linear momentum again, linear momentum's given the symbol capital L. The mass is constant in this course, and so we have the derivative of the position vector from O, to the mass center is the velocity of C. And then I can take the, the derivative of that linear momentum for the body. It's equal to mdvcdt since mass again is constant. That becomes ma and we're back to the sum of the forces equals ma of the mass center, which is Euler's 1st law. And so what we find, is that the summation of the forces, on a continuously distributed mass is equal to the time derivative of the linear momentum, just the same as it was for a particle itself. So here's the momentum form of Euler's 1st law. And now let's come up with what we'll call the impulse-momentum relationship. And so we're going to integrate these momentum form of Euler's 1st law. And we have from some time, t1, to some time t2, the sum of the forces acting on the body, is equal to the change in the linear momentum. And so this left hand side, the intergal over some period of time, the force is just called an impulse. And the change in capital L is the change in momentum or in this case specifically the linear momentum. So, we now have this impulse-momentum relationship, a particular kind of impulse is called an impact. And an impact occurs when we have a collision of bodies over a very short period of time and so this delta t goes to 0. And then if the sum of the external forces in the direction of the momentum vanishes, if this left side vanishes, what we have is conservation of momentum. And typically in these types of problems where we have a very large impact force, we ignore relatively small forces, oftentimes ignoring gravity and other negligible forces. So let's look at an example. I have a block at rest on a smooth horizontal surface being struck by, so this is the block at rest here, being struck by an identical block sliding with a velocity, or a speed actually of v. And I want to find the velocity of the two blocks after the collision assuming that the blocks stick together. And so let's look at a little demo of this situation. So, I have a block that's sitting at rest. I have another block. We're going to assume this is a smooth surface, so it's friction-less. And as this block comes along with a velocity v, it sticks to the other block, and they go off together. And so we're going to neglect friction. We don't have any forces in the direction of motion, and so we can use conservation of momentum. And if I go back here now, by conservation of momentum, the total momentum before is equal to the total momentum afterward. The only momentum I had before the collision is the left block times the initial velocity V. The right block has no velocity. After the collisions just they both move together. The m left and the m right are going to have the velocity V after, which is what we're seeking to find for this problem. And so I know that V after, since the blocks are sliding together is going to be the block left and the block right's velocity. I can just do the algebra here. I've got M left times left divided by M left plus M right. We said that the blocks are identical, so they have the same mass. So V after becomes vm left over 2m left, or v over 2. And so, as these blocks slide together after they've impacted, they stick together. The original block has a velocity V. Both blocks sliding off together by the conservation of momentum have half that velocity. So, I would like you to know that if the blocks don't stick together, then conservation of momentum, linear momentum is not enough to find the subsequent velocities because I only have one equation. And we're going to have two unknowns for the velocity of each block and so we'll need another piece of information. So let's slide back over here and look at our situation. In the problem we just solved, the blocks go off together in the same velocity, but in reality in most cases, when the blocks hit each other, they bounce off and we have two different velocities. And so, we're going to need another piece of information and we'll pick that up at the next module.
