https://youtu.be/fk2xtnJnNqQ
PHYS 1101: Lecture Eighteen, Part Two
Let me ease into our new material with, again this big picture of where this momentum fits in. It really is a completely new tool that we’re learning about. In other words the collection of problems that we’re going to be able to solve with encompass a whole new set, a new collection of leaves if you will. So we’re going to be learning, again, how to start from the basic equations that apply to all the momentum conservation problems, all of these problems, how do we customize that general equation and move logically to our specific solution.
As I alluded to above, there are two main tools that we’re going to learn, so in another way two types of problems in here. One type of problem, we’re just focused on a single object in an abrupt event, and that’s this equation where we need to think about the average force on this object, times the time interval of this collision, that product is equal to p final minus p initial. That’s one tool. If on the other hand you’re able to isolate, or realize that these two objects interacting aren’t in contact with anything outside, there’s no additional forces, then we can apply the more useful equation, which is the idea of total momentum conservation. The total momentum of both of these objects before can’t change, one object might pick up momentum, another one loses the same amount, but when you add them together, you got to get to the same. So if you mathematically set that constraint up at the beginning, it then tells you what, say, a final velocity has to be or what the mass of one object has to be.
Okay here on this next screen I’ve summarized these points that I’ve already made, but here they’re set a slightly different way, and this is a good place to summarize them. This notion of momentum tool, number one, if you ever realize in a problem that you have an abrupt event, something was thrown, something was caught, a collision happened or an explosion, immediately consider using what you’re about to learn about, these momentum tools. This is the case where you don’t care about the details during this abrupt event, you just want to know what the consequence is. If you want to focus on, you only care about one object involved in the event, see if you can’t apply this equation. We’re first going to start out with some examples and show you what this means and how to apply this to lots of general problems. Then if the focus is on usually just two objects, or both of these objects that are involved in the event, then you get to apply total momentum conservation. This one ends up being more powerful to use because this one completely bypasses the details of the collision, there’s no force in this equation anymore during this abrupt event, there’s no time duration of it. I get to skip over that detail entirely. It’s if I want to ask what the force is on this object during the collision, or the duration of the collision, that I’ll end up needing to use equation two as my starting point.
Let’s start with this, this first scenario, this basic equation that shows the consequence of a force in terms of changing the momentum. The first place to begin with that discussion is first carefully defining what we mean by momentum, as with all of these terms it’s mathematically very specific. The momentum of an object, small p, is given by, its equal to, the product of its inertia, the objects mass, times the velocity. The magnitude comes from the magnitude of the velocity, the speed times the mass, and the directions always the same as the velocity. Often in these problems the mass of the object doesn’t change, but the velocity will, and so we’re going to see the momentum change as a reflection of the velocity change. Okay, to appreciate the meaning of this connection between momentum being related to mass and velocity, I give you four scenarios here, one through four, I show you how the masses compare between them, and how the velocities compare, they’re all moving to the right so these also reflect just how the speeds compare. And I want you to choose the option here that represents the proper ranking of the momentum of these objects, and order them from largest to smallest. Here’s another question so you start to picture the connection here to a real physical event and this notion of momentum. Consider that you have an object, it’s moving to the right, it has an initial velocity in the positive x direction, positive to the right, and it’s plus 2 meters per second. At a later time, its velocity is plus 1 meter per second. How is the objects momentum changed, whats p final minus p initial. Has it increased, stayed the same or decreased?
Okay, to ease you into this first tool, let’s talk through it in the scenario of a tennis ball being hit by a racket. Our focus is on a single object, and the main tool equation that applies tells us what the consequences are of this force that’s on the object, times the duration of the collision, and the consequence of that being that the momentum changes. Whatever momentum I started with it has to change by a certain amount, I have to end up with the certain p final that reflects this force times time. So here I’ve got these labels to try to clearly indicate what you have to think about and plug into each of those terms. Some important points here, we’re going to be working with equations now where there can be a lot of different symbols, lot of different subscripts and it’s easy to get confused and to think I have a whole collection of different equations when, in reality it’s all the same equation just mathematically rearranged.
So let me give you a couple of examples of that. All four of these that I’ve written down are all the same thing, for example, here I write down this main starting equation. The average force on this tennis ball, times the duration of the collision, causes the balls momentum to change. You’ll see this written as f Δt is equal to j, only because of the simple substitution that the definition of impulse is change in momentum, so this is just a substitution on the right. Let me highlight this for you because this is important. The definition of the impulse is the Δp, p final minus p initial. You’ll also maybe see it instead of p final minus p initial being written in terms of one simpler term, it might be useful for you to break it down, meaning substitute in for each of these what momentum is given by. Well it’s the mass times the velocity, this would be m v final, and this would be m v initial, so this is the same equation that we started out with. If your focus is on calculating the average force on the tennis ball or on this object, then it might be useful to divide both sides by Δt, and then that leaves you with this equation, it’s all equivalent, the average force is equal to Δp over Δt, the change in momentum divided by the change in time.
Quiz question four just has you go through an exercise of calculating p final minus p initial for a particular scenario, and we’re going to be, often in these problems, dealing with motion that’s only along a line, say two objects collide head on so they bounce of each other, or somebody throws a ball straight to the right or you hit a tennis ball straight horizontally. That means that momentum and velocity, they are vectors, but because we’ll only be doing the analysis, say horizontally, we can go back to those scalar comp1nts, or the scalar representations, meaning again, I just have to use the proper sine to mathematically capture the direction of this vector. So here I have a 2 kilogram object that’s moving to the right, with a velocity of plus 2, at a later time its velocity is minus 1, what’s the right calculation for Δp, p final minus p initial.
Here I have written our starting equation, again the focus is on a single object, let’s take just both sides of that equation and take our time and be sure we understand what has to be plugged into, or substituted in, to properly carry out this equation. Let’s start with the left side, what does that mean ff? That has to be, represent the force that’s on this object during the collision. Δt is the duration of the collision. What do I think about it is the left captures what’s happening somewhat during the collision, and the right hand side of the equation says, what happened during the collision has to equate to this overall consequence, or this change in momentum, of final after the collision minus before the collision. So let’s think about this in the context of this tennis ball. So this would be just before, the ball’s coming in, when it’s in contact with the racket, it’s experiencing this force. If you think about our criteria of forces, I have gravity down, but then the side of this tennis ball’s in contact with the racket, that’s a surface putting a normal force, or let’s just call it f, it’s just a push a contact force on the ball. While that ball’s in contact I’m going to continue to have this horizontal force, and then the ball in response to that force, is going to slow down, come to a stop, and turn around, and so after the collision, it leaves contact with the racket and it’s going to be headed back out towards the right.
Okay, comment on the magnitudes, a collision is usually a very catastrophic event, meaning the size of forces are big. Usually they’re much bigger than the size say, of the force of gravity, on this object. If you were to look at a plot, let’s just try to read this plot and see what useful information it’s telling us. The vertical axis here is telling us the size of this force that the racket’s putting on the ball, the bottom axis here is time. So this means that every time interval, I can read this graph, in other words, at this time I can, oh you know what, the force at that instant in time was this big. What this graph says is that, it looks like at times earlier than t 0, my force is 0. This would represent before the ball is even in contact with the racket, so it’s close to the racket but it’s not touching it yet. The instant the ball makes contact, it starts experiencing this push from the racket, that’s a positive force, as the ball continues to move into the racket, this force starts to grow, and it gets really big. As the ball starts to turn around, it starts to leave the racket, it’s still in contact with it though so there’s still a force, but that force is getting smaller. Beyond this time that’s labeled here t f, the ball has now left contact with the racket, it’s now bounced off and it’s no longer in contact and it’s headed to the right.
Okay the Δt that shows up in that equation on the left hand side represents the duration of the collision. T final minus t 0, as an aside for collision, they happen quick, you don’t even see them with your eye, you can’t watch that tennis ball respond as it hits this racket. This usually takes about a millisecond, that’s 001 seconds, high speed camera you can capture a couple of frames as this ball squishes and goes into the racket and comes back out, but you’re not going to see these details with your eye. Okay, the main equation on the left side if force times time, at every little snapshot if you were to go in here and calculate the force times the time, you’re going to get the area of all these little segments. When you do that for the whole duration of the collision, you end up getting the total area, when you add up all of this force times time, the left side of that equation equates to the area under this curve. Area meaning the height, times the base, but I’m restricted to this unusual shape, I need the area here, I’m going to have units of force times time. The area under that curve is the left side, that is then the impulse, that is the change in momentum.
Okay but rather than having to know all these specific details during this collision, if we say to ourselves, well you know what, we just care about the end result, what is the impulse, what is the change in momentum, we can get the same area by determining, or estimating just an average force, an average force over the same time interval, that product will give you the same area. So let me put an arrow here and say we’re going to use f av, then f av times the duration of the collision equals the same area. So that also is going to give me the impulse or give me the number for p final minus p initial. Maybe you can see by eye here that certainly the area, if I compare the pink to what I marked here in purple, certainly the area where they overlap it’s going to be equivalent. The purple rectangle here I pick up some more area here and here, and that ends up filling in, or compensating, for all this area up there that I missed. The purple hashed area is the same as the pink area, okay. So I need the average force during that collision. That average force is always going to be less than the peak force, but it’ll still be a big number, and it’s usually perfectly reasonably to just, in your mind, think about the average force or use that as a representation of what’s happening during the collision if you want. So that’s why we use f av.
Here are two quiz questions for you to get you to think about that a little bit more. I’ve got two scenarios drawn, scenario one and two. Both are showing you the force on an object during a collision versus time. It’s the same kind of graph we were looking at above. The difference is that in collision one, the duration is shorter than collision two, and also the peak force is larger in the first collision than for the second collision. So you can think of these as being perfect triangles, I didn’t draw them very carefully but they’re supposed to be exact triangles, so you should be able to calculate the areas of triangles, remember the area of a triangle is one half base times height. Consider both of these questions, question five and six. Which is going to have the larger average force, and which represents the larger impulse on the tennis balls, so these are two different collisions of a racket with a tennis ball. And remember the impulse is going to be the product of a rectangle of f av time Δt, where the product of this has to be the same area as the triangles, the area below the real force versus time curves.
Okay well how about the right side of our basic starting equation, I have it drawn here again for us our basic equation, what about the right side. What goes in for p final and p initial. Well as I say, in this class we’re going to focus on simple one dimensional collisions so that motion, all the action’s going to be happening either horizontal or vertical. Here’s a horizontal example, a ball comes in initially heading to the right, initial velocity to the right. Collides with a wall, the force that the exerts on the ball during the collision is going to be to the left, and then out comes the ball as a consequence headed to the left, everything is horizontal. These are vector quantities, so I need a sine convention, I’m going to call positive to the right, so now these vectors that I’m working I can simplify it, and just use the sine to represent the direction. So my initial momentum I would call positive, the final is negative, and the force, also a vector, gets a negative sine to represent that it’s pointing to the left in the negative direction. So we’re back to vectors, back to looking at the one dimensional part of the vectors, and back to using a sine to represent direction, not value, not big versus small, purely the sine, the direction that these vectors points, represented by the sine.
Here’s a couple of quiz questions to give you some practice calculating, p f minus p 0, you can pause the movie here and read those. Again we’re going to define the positive axis to the right, perfectly fine let’s go with our standard convention. Alright let’s do an example. This first tool, meaning we’re working with the equation f av times Δt is equal to p final minus p initial, and this is going to be a one d vector equation. So I am going to have to put the proper sines in for our vectors, my force and my momentum’s to use this equation properly. I’ve got two scenarios to compare here, I’m dropping a small apple, it hits the ground with a speed of ten meters per second, it doesn’t bounce back up. I’m going to compare two cases, in case A the ground is really hard, the collision takes 1 millisecond. Case B, the ground has some give to it, maybe this is carpet, the collision takes a lot longer because it’s a softer surface, it takes 100 milliseconds, it’s a hundred times larger. How does the average force of the ground on the apple compare? It’s that average force that you’re most concerned about because that’s what ends up doing the damage to your apple, and of course your intuition you’d much rather accidentally drop your apple on the carpet then you would on the concrete in the grocery store.
Okay we’re going to be using our main equation, for sure to use it properly, I’m going to have to draw some pictures so I have a nice scenario of p final p initial, the direction of the force, so I can start properly assigning the sines to these terms, and deciding what I know, what I don’t know, and what variable I need to go after. Let’s draw case A on the left and case B on the right. So the difference here was the duration of the collision is really quick when I dropped it on the concrete, a millisecond, that’s 0.001 seconds, versus the carpet, 100 milliseconds is 0.1 seconds. Okay let me draw a picture, this is going to be one d motion, up down, I’m going to go with the sine convention, let me draw it here between these two, because I’ll use the same for both, with plus y being up. Here’s my ground, here’s my ground, what you want to draw for these, you’re focused on what does the scene look like just before and just after the collision, so on the left I’m going to draw my apple just before it hits the ground, after the collision, here it is. Do the same thing on the right. I started out with a velocity heading down, my momentum then has to point in that same direction, there’s p zero, tells me that the apple doesn’t bounce back, so p final has to be zero for both of these cases, and in fact both cases have the same initial momentum. I’m coming into this collision from above, the balls falling down, and I have a speed of 10 meters per second.
Often, between these two quick sketches, just before and just after I often just sketch a little free body diagram in between to just remind myself the directions of forces. Well there is a force of gravity always on the apple, I’m going to draw that just tiny, because it’s going to be really small compared to this collision force. What’s the collision force? When that ball makes contact with the ground, surfaces push, they support the load, it’s not just an apple sitting on the ground anymore, my normal force isn’t just balancing f, that force is big, it has to decelerate this apple and bring it to a stop, big force. Force, I’m going to call this, instead of calling it a normal force maybe I’ll start being more specific and indicate what agent is causing that force. This is the force that the ground puts on the apple, I’m going to draw that nice and big. What’s our list of knowns? I know the mass of the apple is the same, 0.1 kilogram, I know that my initial speed is 10 meters per second, I’m going to note here that that means my initial vector, scalar comp1nt for the momentum because it’s down, is going to be minus 0.1 kilograms the mass times the velocity. The velocity is negative, that’s why this ends up being overall negative. So I have p zero is equal to minus 1 kilogram meter per second for both objects.
What is different? I’m going to jot that down up above. The time of the collision, the duration, is significantly different, hundred times different. Δt is 0.001 on the left, and 0.1 on the right. Now what do I want, well as always, I’m after a variable that’s either directly in my starting equation or I know is going to show up in my starting equation. I’m asked to compare the average force of the ground on the apple. Well this equation tells me that the average force on this object during the collision, times the time is equal to p final minus p initial, is this the force that I’m after? The force on the object during the collision, is that the force that the ground is going to put on the apple, well during the collision I have the force of gravity and the force of the ground, technically the left hand side gives me the effect of all of that, this is all of the forces on this apple during the collision. I’m going to use that to represent just the force of the ground because it’s so much bigger, you’ll see, than the force of the apple, or the gravity on the apple. So what do I want? In my starting equation, I’m going to go after f av, let’s take this starting equation and just rearrange it, and solve for f av. Here it is, I’m going to divide both sides by Δt, just so I’m looking at an equation that says the variable that I want on the left is equal to the consequence, the mathematical implications on the right. Let me write it freshly here, f av, the thing I want to solve for, is p final minus p initial divided by Δt.
Okay that’s the same starting equation I’m going to use for both cases, what do I plug in? Well what is p final and what is p initial. I did work out what p initial is and I put that in my list of known’s, well now I realize I could have concluded something earlier about what p final is as well. I did here up in my drawing but I didn’t put it in my list of known’s, I’m going to go ahead and do that. P final is zero for both cases, p final is zero. Let’s plug in numbers, I’m ready now, I’ve got p initial, I know what the time is for the two scenarios, case one I’ve got f average is equal to the negative of p zero, which is minus 1 kilogram meter per second, and then I have to divide by this Δt, 0.001 seconds. For case B, f av is again negative minus 1, but this time I divide by only 0.1 seconds. Doing the simplification and the math on the left, I end up with an average force on the apple on the collision that’s 1000 newtons. Notice the negative and the negative cancel to leave me with a positive force, that would mean a force that is pointing up, is that consistent with our picture? Yes, the ground has to put a large upward force to decelerate quickly and bring it to a stop, the acceleration has to point up for this apple during the collision. Same scenario here on the right.
The math, or the numbers for case B, end up giving you an average force on the apple that’s only 10 newtons, that’s a hundred times smaller. If I were to go up here and draw my force picture, the force of the ground is a hundred times smaller here than it would be here. I haven’t begun to draw that in proportion so it looks right. This is why you want to drop your apple on a carpet. If you could stretch out the duration of this collision, in order to bring about the same momentum change I could use a smaller force to do the job, because that force acts over a longer time, it could change the momentum more gently than a force that can only act over a millisecond, it has to be big to bring about the same momentum change.
Question nine on the quiz asks you to think more carefully about that. This is the same type of problem as dropping an apple on the ground, but it’s in a very different context, in this case we have a stunt person that’s jumping from a building and lands on an air filled bag. Both cases the incoming velocity is going to be the same, if he uses the bag or doesn’t. The final momentum is going to be zero whether he uses the bag or doesn’t. Why does using the bag lead to no injuries, or prevent him from injuring himself. I have a collection of questions that follow, wanting you to think about the vector nature of some of these quantities, and let me walk you through what I mean by that. We’re now going to be thinking about the direction implications when I write these equations down.
Let me first on the left show you the mathematics and equating directions, and then we’re going to look at a picture on the right and see physically what the connection is. Here’s our starting equation, f av is equal to change in momentum. The direction of the force has to be the same direction as Δp, not the same direction as p zero or p final, but the difference between these two vectors. This we know is called the impulse, the change in momentum, so let me note here that any vector equation like that the direction on each side has to come from the direction of the vector on each side. So let me point out the impulse, the change in momentum has to point in the same direction as f av, and saying that again a different way, f av times Δt, if you think in terms of p final times p initial, it means the combination, the vector difference between these two, is the same as f av. Let’s bootstrap our way to that understanding by looking at the scenario on the right.
Everybody by now I think is comfortable with picturing a velocity vector, literally points in the direction the object is headed. Tennis comes in, collides, after the collision it’s heading back out to the right. Momentum equals m times v, these directions have to be the same, so if I were to sketch the momentum vectors, they would have to point in the same direction as the velocity. There’s p final, let’s calculate the impulse, which is p final minus p initial. Let’s do that vector difference, here is p final, made a copy of it there and brought it down, there’s the vector, p final, and now I need to add to it the negative of the vector p zero. Here’s p zero, I need to add the negative of that, that means I need to change the direction this arrow points, points to the right then. So to p final I need to add the negative of p zero, my result of this vector operation then goes to the initial tail to the final tip. This is the vector j, this is the vector Δp, the change in momentum. It’s a bit like the Δv exercises we were doing at a turning point where the object literally turns around. When you do the change, when you have a complete turnaround like this, you end up with a big vector that points to the inside of that change. There’s my j, my impulse, the change in momentum direction. If the direction of j points to the right, then that has to be the direction of the force. The net effect of all forces during this collision then also has to be to the right.
Quiz question ten, I’m going to have you follow through thinking about the implications of that. So above we identified the direction of the force that was on the tennis ball during the collision. So think about that force, go back and look at that picture, what’s the third law pair to that force. Let me say it again clearly, what we identified above was the force of the racket on the tennis ball, what’s the third law pair to that. Once you’ve thought about what that third law pair is, what is the direction of that force. We have five choices there. Here’s the implication that I want you think through, theses are the logical steps that’s I’m leading you to. The duration of that collision that was experienced by the racket is the same duration experience by the ball, as they, when they interact with that contact point they both feel that interaction, experience it for the same time. They both are feeling this equal and opposite third law pair during the collision. How does the magnitude of the impulse that the racket experiences compare to the impulse magnitude that the ball experiences. What’s the connection? You’ve thought through the forces and how they compare on the ball and the racket, remember that the impulse is the average force times the duration. So how do the j’s compare for the ball, compared to what the impulse that the racket must feel.
Question 13, think back and go back to the example, the apple falling, that we did. What’s the direction of the impulse for the apple in that example?