Anderson Video - Momentum and Force

Professor Anderson
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>> All right, let's do it. Hello class, Professor Anderson here. Welcome back to another episode of physics with Dr. Anderson. I'd like to talk to you a little bit more about Newton's third law and particularly how it might apply to something called conservation of momentum. So remember what we said before was Newton's third law was the action/reaction law. In every action there is an associated reaction. The other way to think about that is no isolated force exists in the universe. So even with things like gravity, when gravity is pulling me down towards the earth, the gravity of me is also pulling back up on the earth. Everything comes in tandem and this is the idea behind Newton's third law. Equal and opposite forces. If I have tension in a rope, tension on one end of the rope, T pointing that way, there's tension on the other rope, T pointing that way. And we're going to see how this applies to a general conservation principle which is very powerful in physics which is called conservation of momentum. So before we get to that we need to talk a little bit about force and momentum. So we are familiar now of course with force. What do we know? We know that force equals mass times acceleration, Newton's second law. But from the kinematic equations we also know a little bit about acceleration. Acceleration is the derivative of velocity so I can just write it like that. Now if the mass doesn't change at all then it's just a constant number. I have a block of steel. It's one kilogram. That mass is basically there in the steel and if we don't take apart the steel the mass is going to be the same. So I can move that mass right inside the derivative. I can rank DDT of MV and this MV is a very special quantity -- that is what we are calling momentum. Momentum P is just equal to MV. So this is a new quantity that we are now going to talk about in physics. If F equals DPDT what can we say? Well, first off we really know that force is a -- Force is a vector. So any time we write force we should really write that. If you want the magnitude of F it's without the arrow on top. If you want the vector it's with the arrow on top. So if F equals DPDT then this thing, P, must be a vector. If we've got a vector on the left side we have to have a vector on the right side and so we can say that the momentum is also a vector and it's given by mass times velocity. Obviously mass is a scaler, it's a number. It's one kilogram, it's 10 kilograms. V is a velocity. It has direction associated with it. P equals MV and F is therefore DPDT. How does the momentum change as a function of time? That tells you about the forces that are acting on it and this should make sense. If I have an object that is stationary and I want to get it moving what do I have to do? I have to apply a force to it. I have to push on it. And if I push on it and I do that in some amount of time I'm going to pick up some velocity and therefore I have a change in my momentum. So this is how all of this stuff is tied together and let's take a look at going the other way now. Let's say we want to get momentum from force. Okay, what we had was force equals DPDT. Now let's just multiply both sides by DT -- And then let's integrate both sides. So the integral of FDT is equal to the integral of DP. The integral of DP is delta P -- and this is something very special which we are calling the impulse and we're going to use J as the symbol for impulse. What is impulse? It is the change in momentum. Just delta P. What is delta P? It's the integral of FDT. So impulse J is the integral of FDT. If I give you a function of F and a function of time then you just have to integrate that to get the impulse. Now a lot of times it simplifies so for instance let's say that F is a constant. If F is a constant let's look at this integral. J is the integral of FDT but we know if F is a constant it comes out of the integral. I can put it out in front. And if I integrate DT what do I get? I just get delta T. Some T initial to some T final and so in the case of a constant F, J is this, in general J is that. Another way to write this is J is equal to F bar, delta T where F bar is the average force. It's still a vector but we put the bar on the top to indicate average and that bar is going to double as both our average sign and our vector sign. Okay, this is impulse. Impulse is the application of a force for some amount of time and if I apply a force for some amount of time I can get things to move. I can make them accelerate. If they are accelerating for some amount of time they're going to pick up some velocity. Since P is just MV it's got to pick up some momentum. It has a change in momentum. So all this stuff is tied together through these relationships and now we can write out this delta P a little more carefully. All right, are we good? Did you guys fix the problems with how poor I am appearing? Are you going to make me better looking on the video? All right, good, thank you. Like I look up there and I go is that really how pale I am? Holy cow! >> Oh by the way that's a great shirt. It looks really good on camera. [ Laughter ] >> Is Sherry dissing me from the other room? >> No, I'm sorry. I just sound sarcastic whenever I talk. [ Laughter ] But seriously -- >> Okay, all right, good. I feel all rattled now you know? It's like I expect you guys to heckle me but when they start heckling me from the back room that's a whole different story. Okay, let's talk about something that's important to you guys which is of course homework. Let's look at a specific homework problem and let's look at the following problem. We're going to have an object that is moving to the right. And it has some initial speed -- let's say it is two meters per second, and now we're going to act on it with a force. We'll tell you what the mass is. The mass is let's say half a kilogram. And now we're going to act on it with a force that looks like the following. And we want to calculate what is the final speed of this object? And the force that we are going to impart on it looks like this. Here is F sub X as a function of time and we will say it is it zero until this point and then it drops down to some negative value and then it goes back to zero. And we'll give you some numbers here. We'll say that this is negative three and this time is 0.5 seconds. Ball moving to the right, we're going to apply this force to it and we're going to see what the final speed is. So let's see if we can draw a picture of this thing. Here's our ball. It is moving to the right. We can call that the positive X direction. Initially it has some [inaudible] VI and later after we've applied this force it's going to have some VF. We don't know if it's to the right. We don't know if it's zero. We don't know if it's to the left, but let's just draw it to the right to start with and we'll see if we can figure something out. So, let me ask you guys a question. Will VF be less than VI, equal to VI or bigger than VI? Anybody have a thought on that? Somebody give me a thought on that. Bueller? Bueller? That was way before your time, right? >> No. >> No? Ferris Bueller's Day Off. You okay? What's your name over there? >> Cody. >> Cody. Can you hand the mic to Cody and let Cody and I have a chat now that we have Ferris Bueller in common. >> Oh, man. >> All right, so here's the question Cody. The ball is moving to the right at this speed. I'm going to apply this force to it. Is it going to slow down? Is it going to be equal to its initial speed or is it going to speed up? What do you think? >> Does it depend on what it's running into? >> Well, it's running into this force. That force is applied in this region here somewhere. So we don't know exactly what that region consists of. Maybe it's somebody's hand, maybe it's a baseball bat, maybe it hit some jello. >> I'd say less than. >> Okay, why would you say less than? >> Because if it's imparting a force then it's going to be impacted in some way. >> Okay, Cody likes that one. Anybody else agree with Cody? Okay. A lot of people agree with Cody. So why did you think it was less than? Say it again. >> Because something else is impacting -- >> Oops, I did the wrong one here, sorry. >> Yeah. >> Okay. Why did you think it's less than Cody? >> Because something is affecting the initial force. >> Okay. It was flying along. We applied some force and then it had some final speed but do we really know which direction the force was applied to the object here? Well, I think we do. Because on our graph here there was no force and then there was a negative force and then there was no force. So that negative means that F sub X is negative and that means that there is a force applied in that direction. It's against the motion in this case. And so there's no way this thing could speed up. The only option is to slow down or come to a stop or in fact turn around and go the other way. So this seems like a very good guess but it's a very good guess because of this fact that it's negative. If this was positive then the force would be the other direction and it would in fact speed up. Okay, you're swatting the ball faster instead of slowing it down with your hand. Okay, let's see if we can figure out what this number is now. Okay, what we said about impulse was the following. Impulse is delta P but delta P is just MV final minus MV initial and we also know that impulse is equal to the integral of FDT. In this case it's in the X direction and so it's the integral of F sub XDT. Now remember what our force diagram looked like. It was flat, drop down to a value and then came back up. That value was negative three and this time was 0.5 seconds. So if I'm doing an integral over this curve it's zero for the first part, then it has some value and then it's zero again, so the only region that I'm really worried about is the center region and in that region F is in fact constant. It has a constant value and so this comes out of the integral and we get F sub X times delta T. And now we should have just about all the information we need. Let's look at this. M times VF minus VI equals F sub X, delta T. This is of course a vector, F sub X. We'll put a vector sign on that. If you read the component there, F sub X, you don't always have to put the vector on top of it. Let's solve this for VF. VF minus VI is equal to what? Well I divide by the M, I get F sub X, delta T, divided by M and now I just have to add VI to the other side. We get FX delta T over M plus V sub I. And now we have all those numbers. Let's plug them in and see what we get. F sub X is negative three in that region. Presumably that would be Newtons. Delta T we said was half a second. M we said was a half kilogram, right? VI was -- ? What did we have for VI? >> Two meters. >> Two meters per second. And so look what we get. We get 0.5s cancel out. We get minus three plus two which is minus one meter per second. So in our initial picture we had this. VI -- moving along it goes through this force that is applying in this direction and what we drew was this, V sub F. By getting V sub F being a negative number, what does it mean this object is doing? >> Going the opposite way? >> Yeah. Is it Ben? >> Yep. >> Cody can you hand the mic to Ben? So Ben, if we get a minus one here for our initial picture what does it mean about the final velocity of the ball? >> It's negative. >> It's negative which means it's going which way? >> The opposite direction. >> The opposite, okay? So this ball is in fact going to go back the other way, going to go back in that direction because it's negative and it will have a speed of magnitude [inaudible]. So any time you draw a picture like that initially and you have the arrow going to the right that's totally fine. It just means that if you get a minus sign in your answer you had the arrow pointing the wrong way and the physics is going to tell you whether you had your picture right or wrong. You can draw it in general whichever way you want. Questions about that one? Obviously, if this answer was zero this thing would be at rest. If it was still a positive number it would be moving to the right. All right, good.
>> All right, let's do it. Hello class, Professor Anderson here. Welcome back to another episode of physics with Dr. Anderson. I'd like to talk to you a little bit more about Newton's third law and particularly how it might apply to something called conservation of momentum. So remember what we said before was Newton's third law was the action/reaction law. In every action there is an associated reaction. The other way to think about that is no isolated force exists in the universe. So even with things like gravity, when gravity is pulling me down towards the earth, the gravity of me is also pulling back up on the earth. Everything comes in tandem and this is the idea behind Newton's third law. Equal and opposite forces. If I have tension in a rope, tension on one end of the rope, T pointing that way, there's tension on the other rope, T pointing that way. And we're going to see how this applies to a general conservation principle which is very powerful in physics which is called conservation of momentum. So before we get to that we need to talk a little bit about force and momentum. So we are familiar now of course with force. What do we know? We know that force equals mass times acceleration, Newton's second law. But from the kinematic equations we also know a little bit about acceleration. Acceleration is the derivative of velocity so I can just write it like that. Now if the mass doesn't change at all then it's just a constant number. I have a block of steel. It's one kilogram. That mass is basically there in the steel and if we don't take apart the steel the mass is going to be the same. So I can move that mass right inside the derivative. I can rank DDT of MV and this MV is a very special quantity -- that is what we are calling momentum. Momentum P is just equal to MV. So this is a new quantity that we are now going to talk about in physics. If F equals DPDT what can we say? Well, first off we really know that force is a -- Force is a vector. So any time we write force we should really write that. If you want the magnitude of F it's without the arrow on top. If you want the vector it's with the arrow on top. So if F equals DPDT then this thing, P, must be a vector. If we've got a vector on the left side we have to have a vector on the right side and so we can say that the momentum is also a vector and it's given by mass times velocity. Obviously mass is a scaler, it's a number. It's one kilogram, it's 10 kilograms. V is a velocity. It has direction associated with it. P equals MV and F is therefore DPDT. How does the momentum change as a function of time? That tells you about the forces that are acting on it and this should make sense. If I have an object that is stationary and I want to get it moving what do I have to do? I have to apply a force to it. I have to push on it. And if I push on it and I do that in some amount of time I'm going to pick up some velocity and therefore I have a change in my momentum. So this is how all of this stuff is tied together and let's take a look at going the other way now. Let's say we want to get momentum from force. Okay, what we had was force equals DPDT. Now let's just multiply both sides by DT -- And then let's integrate both sides. So the integral of FDT is equal to the integral of DP. The integral of DP is delta P -- and this is something very special which we are calling the impulse and we're going to use J as the symbol for impulse. What is impulse? It is the change in momentum. Just delta P. What is delta P? It's the integral of FDT. So impulse J is the integral of FDT. If I give you a function of F and a function of time then you just have to integrate that to get the impulse. Now a lot of times it simplifies so for instance let's say that F is a constant. If F is a constant let's look at this integral. J is the integral of FDT but we know if F is a constant it comes out of the integral. I can put it out in front. And if I integrate DT what do I get? I just get delta T. Some T initial to some T final and so in the case of a constant F, J is this, in general J is that. Another way to write this is J is equal to F bar, delta T where F bar is the average force. It's still a vector but we put the bar on the top to indicate average and that bar is going to double as both our average sign and our vector sign. Okay, this is impulse. Impulse is the application of a force for some amount of time and if I apply a force for some amount of time I can get things to move. I can make them accelerate. If they are accelerating for some amount of time they're going to pick up some velocity. Since P is just MV it's got to pick up some momentum. It has a change in momentum. So all this stuff is tied together through these relationships and now we can write out this delta P a little more carefully. All right, are we good? Did you guys fix the problems with how poor I am appearing? Are you going to make me better looking on the video? All right, good, thank you. Like I look up there and I go is that really how pale I am? Holy cow! >> Oh by the way that's a great shirt. It looks really good on camera. [ Laughter ] >> Is Sherry dissing me from the other room? >> No, I'm sorry. I just sound sarcastic whenever I talk. [ Laughter ] But seriously -- >> Okay, all right, good. I feel all rattled now you know? It's like I expect you guys to heckle me but when they start heckling me from the back room that's a whole different story. Okay, let's talk about something that's important to you guys which is of course homework. Let's look at a specific homework problem and let's look at the following problem. We're going to have an object that is moving to the right. And it has some initial speed -- let's say it is two meters per second, and now we're going to act on it with a force. We'll tell you what the mass is. The mass is let's say half a kilogram. And now we're going to act on it with a force that looks like the following. And we want to calculate what is the final speed of this object? And the force that we are going to impart on it looks like this. Here is F sub X as a function of time and we will say it is it zero until this point and then it drops down to some negative value and then it goes back to zero. And we'll give you some numbers here. We'll say that this is negative three and this time is 0.5 seconds. Ball moving to the right, we're going to apply this force to it and we're going to see what the final speed is. So let's see if we can draw a picture of this thing. Here's our ball. It is moving to the right. We can call that the positive X direction. Initially it has some [inaudible] VI and later after we've applied this force it's going to have some VF. We don't know if it's to the right. We don't know if it's zero. We don't know if it's to the left, but let's just draw it to the right to start with and we'll see if we can figure something out. So, let me ask you guys a question. Will VF be less than VI, equal to VI or bigger than VI? Anybody have a thought on that? Somebody give me a thought on that. Bueller? Bueller? That was way before your time, right? >> No. >> No? Ferris Bueller's Day Off. You okay? What's your name over there? >> Cody. >> Cody. Can you hand the mic to Cody and let Cody and I have a chat now that we have Ferris Bueller in common. >> Oh, man. >> All right, so here's the question Cody. The ball is moving to the right at this speed. I'm going to apply this force to it. Is it going to slow down? Is it going to be equal to its initial speed or is it going to speed up? What do you think? >> Does it depend on what it's running into? >> Well, it's running into this force. That force is applied in this region here somewhere. So we don't know exactly what that region consists of. Maybe it's somebody's hand, maybe it's a baseball bat, maybe it hit some jello. >> I'd say less than. >> Okay, why would you say less than? >> Because if it's imparting a force then it's going to be impacted in some way. >> Okay, Cody likes that one. Anybody else agree with Cody? Okay. A lot of people agree with Cody. So why did you think it was less than? Say it again. >> Because something else is impacting -- >> Oops, I did the wrong one here, sorry. >> Yeah. >> Okay. Why did you think it's less than Cody? >> Because something is affecting the initial force. >> Okay. It was flying along. We applied some force and then it had some final speed but do we really know which direction the force was applied to the object here? Well, I think we do. Because on our graph here there was no force and then there was a negative force and then there was no force. So that negative means that F sub X is negative and that means that there is a force applied in that direction. It's against the motion in this case. And so there's no way this thing could speed up. The only option is to slow down or come to a stop or in fact turn around and go the other way. So this seems like a very good guess but it's a very good guess because of this fact that it's negative. If this was positive then the force would be the other direction and it would in fact speed up. Okay, you're swatting the ball faster instead of slowing it down with your hand. Okay, let's see if we can figure out what this number is now. Okay, what we said about impulse was the following. Impulse is delta P but delta P is just MV final minus MV initial and we also know that impulse is equal to the integral of FDT. In this case it's in the X direction and so it's the integral of F sub XDT. Now remember what our force diagram looked like. It was flat, drop down to a value and then came back up. That value was negative three and this time was 0.5 seconds. So if I'm doing an integral over this curve it's zero for the first part, then it has some value and then it's zero again, so the only region that I'm really worried about is the center region and in that region F is in fact constant. It has a constant value and so this comes out of the integral and we get F sub X times delta T. And now we should have just about all the information we need. Let's look at this. M times VF minus VI equals F sub X, delta T. This is of course a vector, F sub X. We'll put a vector sign on that. If you read the component there, F sub X, you don't always have to put the vector on top of it. Let's solve this for VF. VF minus VI is equal to what? Well I divide by the M, I get F sub X, delta T, divided by M and now I just have to add VI to the other side. We get FX delta T over M plus V sub I. And now we have all those numbers. Let's plug them in and see what we get. F sub X is negative three in that region. Presumably that would be Newtons. Delta T we said was half a second. M we said was a half kilogram, right? VI was -- ? What did we have for VI? >> Two meters. >> Two meters per second. And so look what we get. We get 0.5s cancel out. We get minus three plus two which is minus one meter per second. So in our initial picture we had this. VI -- moving along it goes through this force that is applying in this direction and what we drew was this, V sub F. By getting V sub F being a negative number, what does it mean this object is doing? >> Going the opposite way? >> Yeah. Is it Ben? >> Yep. >> Cody can you hand the mic to Ben? So Ben, if we get a minus one here for our initial picture what does it mean about the final velocity of the ball? >> It's negative. >> It's negative which means it's going which way? >> The opposite direction. >> The opposite, okay? So this ball is in fact going to go back the other way, going to go back in that direction because it's negative and it will have a speed of magnitude [inaudible]. So any time you draw a picture like that initially and you have the arrow going to the right that's totally fine. It just means that if you get a minus sign in your answer you had the arrow pointing the wrong way and the physics is going to tell you whether you had your picture right or wrong. You can draw it in general whichever way you want. Questions about that one? Obviously, if this answer was zero this thing would be at rest. If it was still a positive number it would be moving to the right. All right, good.