In this video, let's work on impulse and momentum relationship. This is simply the time integral of the F equals ma as opposed to the displacement integral that gives us the work in energy relationship. So to start with, let's start with the Newton's second law, F equals ma, and then acceleration is by definition is dbdt, and here we define the linear momentum as the mv here. So that's what we call the linear momentum noted as G. So that means the force is going to be the time change of the linear momentum. So if we integral Newton's Law left and right of all time delta t, what we can have is the change of linear momentum, which is delta mv, and that's the linear impulse and momentum relationship. So here, the time integral of the applied forces is what we call the linear impulse, and change for the mv is linear momentum change. So assuming that the delta is from state one to state two, then what we can have is mv1 plus linear impulse force Integral over time. We'll keep as mv2, the momentum at status two. So there is no force that exists, or if we integrate the force over really short amount of time, and if the force is finite, this could be approximated as zero. So in that case, we call that the linear momentum is conserved. So to solve the problem, generally you should define the coordinate, obtain the three body diagram and obtain the equations of motion, and then either tried to do integral over the displacement to obtain the work in energy relationship or to integrate over time to obtain the impulse momentum relationship. Let's work on the example. There is a box of m, the red one box m is initially at rest and it's been released. So it's going to be falling down, and then there's a cart which has straight part here and that the curry part from here. So that the blog was going to be straight fall down and then follow through the curve, and at some time it will separate from the card. So it's impending separations about to separate. So and then neglect the friction, friction between the box and the cart and also cart and the surface. So find the absolute velocity of the box and the cart when it's being asked, right it's impending separation. To do so, let's first start with defining the coordinate, and then once your obtained equations of motion in this problem, we are going to take the time integral to obtain the impulse momentum relation. However, you can also simply just plug in all the values to the formula, as for me as student, I would strongly recommend you to approach two methods, integrating the equations of motion over time, and also check your and double-check your answer by plugging in values into the formula. So where should I set the coordinate? I'm going to set the Cartesian coordinate, and it is reset along with the initial partition of the box, and I'm going to describe the motion of small mass and capital M, small mass is going to be falling down straight, and then during those time, the M will stays there and it is actually start coming into those curvy part. What will happened? Maybe cart will move to the left hand side, and since this small mass is or along the curvy side, it's part of circular motion. Whenever there's a circular motion, you can think of centripetal acceleration, and the centripetal forces. So in that case, what will be forced source for the centripetal force? Yeah, the normal force, and also since this capital M, the cart is going to obtaining the velocity, that means there is a force is applied, there is no external force here. What force would cause the capital M to move? These are the question that you could think about. Let's start with the free body diagram. I'm going to approach the separate masses like a cart and box separately, and how many contacts does this cart have? This one has a contact with the blog and contact with the surface. So the two contacts, so two horizontal and vertical forces per each contact and the gravity that will generate the acceleration, and put the small box, there's only one contact with the card, so there is a horizontal and vertical force and the gravity will generate the acceleration. Since there are the contact between the cart and the block. Those N2 and forces are actually action-reaction pair, same mechanism than different opposite direction. So having in considering only X direction, I'm going to have our X direction on most equations of motion for the capital M and small m, and if we sum them together, those normal forces being canceled out. So I can have is capital Ma capital M, and then small ma small m, and if there's no friction, this turns out to be zero. Now what if we handle the both cart and the box as a whole system altogether, then you should think of the equations of motion at the center of mass. Center of mass actually in this case will be located near to the center of mass of the cart, because it's cart mass is going to be a lot larger than the block mass, and while the blog is straight down, this is going to be slightly moves down to the motion of those small block, and when the small blog is on the curvy part of those cart, and this is going to be slowly moved to the right and slightly move to the down, and that's the actual dimension of the center of mass. So if we handle everything as a whole, the only contact is with the surface. So there are two forces, of horizontal and vertical, and the gravity that will generate the acceleration. So from those pictorial relationship, we can obtain the equations of motion of the whole system. Only external force X direction is going to be a friction, and that will generate the capital m and small m and a acceleration. Now if we compare both, equations of motion for a separate math will be, should be equivalent to the equations of motion for the whole system mass, So are you going to ask you, are these equivalent? Well, how you can relate the a? Is capital A, capital M acceleration, same as ax here that I defined it as acceleration of the center of mass. Now, as I said, since there are relative motion between this current and the block, those center of mass acceleration will be neither of the capital mass M or acceleration of small mass m. What will happen, there's a friction occurs between the cart and the surface. Well, as I said, the friction force is a reaction to some action force, right? What will be the action force here when the block slides down? When the block slides down, there is no external force, there is no action force to the system horizontal, right? So there's no friction force, so friction force is initially zero. When it comes to the curvy part between the motion of the box and the curvy structure will give back and force the force effect that will actually generate the horizontal forces. So there will be an action force that will lead to the generation of the friction force. So the friction force between the cart and the surface will increase from zero to some value and direct it to the right. Now, so coming back to the original question, we are supposed to find out the velocity at exit, like an impending separation. So here we have an acceleration, so we have to do the time integral to obtain the velocity. So if we do the time integral both in right, left hand side is a zero and right hand side, we will have V capital mass and v small mass hence this relationship. So the ratios are inversely proportional to the rate for the mass. So block will be faster than the heavier cart. Now, instead of just integrating the equations of motion, we can also obtain the relationship by simply plugging the values to the formula. So let's work on the easier way. So G_1, initially there's nothing, it's a zero, and there is no horizontal linear impulse because there's no friction. The final linear momentum will be capital Mv_M, and small mv_m. If you equate them, you will have 0 equals to the Mv, and mv. So those are equivalents. So either you could integrate over the time or if you just plug into the formula, you will obtain the equivalent impulse-momentum relationship. Now, what I have is a combined velocity formula, capital Mv_M and small mv_m, and I'm supposed to find out small vm, capital VM separately. I have two unknowns, one equation, so definitely I need more information. What should I do? Even though you just supply for the time integral, there's another one that you can do which is displacement integral, so work energy relationship. So instead of integrating the equations of motion, let's take a simpler way, just plug in the value to the formula. Initially, there is no kinetic energy, it's rest, initially. Then let's set potential energy reference as zero for the initial position. Now, there is no work done except the work done by the friction but there is no friction here, so this is going to be zero. The final state, there is a block has its own speed and the cart has its own velocity as well. Also, there are height changes of the small block m, from l plus r, right? The cart center of mass stays there. So if you equate them, the initial kinetic and potential energy and work done is going to be kinetic potential energy. Some at later status, you will be able to find out the relationship between vm and vM. If you combined with the relationship, what you just have obtained for the impulse-momentum relationship, you can solve specify what are the velocity for the cart and the block. Let's solve another one. There is a block of mass m on the inclined. The typical problem while we worked on in the Chapter 3, the first part of the Chapter 3. Now, its initially has a velocity to the left hand side. I mean, downward velocity be v_0. There is a static and kinetic friction coefficient between the block and the incline, and there's external force is applied with the time trajectory shown as a graph. Find the velocity at time t. Seems like initially it will be v_0 and without external force P, it'll just getting larger and larger because the gravity applied. If the P is applied backward, we'll see what will happen. So define the coordinate, and draw the free body diagram, and we're going to the point where we're going to do the time integral of the equations of motion to obtain the impulse-momentum relationship and double check our answer by plugging in values to the formula. So where are you going to set the coordinate? I'm going to set the coordinate cartesian along the incline. I'm going to draw the free body diagram or two contact, one with the surface and one with external force. So there are friction force, dormant force, and gravity, and the external force will generate the acceleration. So what I can have is x directional equations of motion like this and y directional equations of motion like that. So what is the direction of friction? Why do I set the friction to the right-hand side? The case if you cannot specify the direction of friction, you just assume one direction and just solve the problem. Once you obtain the final answer, check if your initial assumption is correct, is valid. Here, since it's initially having a downward speed v_0, that's why I set the friction to the right-hand side. To formulate the friction is mu_n, should I use mu_s or mu_k? Since this is moving, there is a relative displacement between the block and the surface. So I'm going to use the kinetic friction coefficient. So to obtain the velocity, I'm going to take the time integral of the acceleration over a of t. This is given, time function is given, and mg and the friction is constant value. So I will be able to obtain the v of t as a function of t by just taking the integral of all the external forces in the x direction over time. So this is it for simple concept review for the impulse and momentum relationship. Practice with the example.
