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Physics Problems: Using Linear Momentum & Impulse

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  • Type: Video Tutorial
  • Length: 13:20
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 144 MB
  • Posted: 07/01/2009

This lesson is part of the following series:

Physics (147 lessons, $198.00)
Physics: Momentum (8 lessons, $16.83)
Physics: Momentum and Its Conservation (5 lessons, $9.90)

This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.

Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.

Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and real-world applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for Non-Newtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.

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We've defined two new physical quantities, momentum and impulse. Momentum is defined by the equation p equals mv and impulse, I, is defined, in fact, as I is just p. The fact that p is equal to the integral of force times time is Newton's Second Law. You can prove that p is just integral of Fdt. It arises simply from integrating this equation. Force is change in momentum with time.
These are defined quantities, but they clearly have some physical meaning. Momentum is oomph. If a bowling ball that's rolling in a certain direction, you certainly have the sense that not only does it have energy, but it has momentum, it's carrying something in the direction of travel and that something is momentum. p, impulse, is also physically important. It tells you the change in momentum. From the beginning of the story to the end of the story, there's been some change in momentum, and that means both a change maybe in magnitude or in direction or both of velocity times mass. So these physical quantities that carry information, they're useful.
When you stare at the impulse equation, p is the integral of F times time, you can already see a case where impulse is practical in problem solving. It's really practical when forces are very strong and very short-lived. So Newton's Second Law, to apply it directly would be difficult. It would be hard to measure force; it might be complicated. What you care about in many problems is not the details of force as a function of time, what you care about is the result, the integral, p. Example: the classic tablecloth trick, where you snatch a tablecloth out from underneath some settings. What is that demonstrating? What's the physics: You could say it's demonstrating inertia, the objects, like the plates, are initially at rest and Newton's First Law says an object rest wants to remain at rest. And that's true, but if you think about this demo a little bit more carefully, there is a force. There's friction force between the tablecloth and the silverware and the plates and stuff. So how come, in fact, they are still sitting still? Look at the equation; what you care about, if you want the trick to work well, is that p should be small. You would like the final mass times velocity to be very small, because the initial mass times velocity is zero. They began sitting still. So what you want is for p to be small. How do you do that? Let's look at this equation in the average sense, where it's F [average ]times t. It's just easier to think about. This is what you care about and there is a frictional force. How do you do this trick? You make t small. You've got to snatch the tablecloth. If you just slowly pull the tablecloth, that trick's a disaster. Long t times some small friction force will build up and everything moves and there's no nice trick there. So even when the friction force is relatively large, you can still get away with the snatching of the tablecloth, just be sure you make t really small.
So that's a case where what you're interested in is p and you can make use of this connection between p and force times time, in order to think about how to get the physics to work the way you want it to. The fact that force is p divided by t when you divided by sides by t is also really useful in problems. Example: a car crash. It's a little bit of a morbid idea, but if you're in a car crash and you're moving at 20 meters per second east, and then, boom, you hit a brick wall. What's going on? The first thing you might think about now is change in momentum. So whose momentum are we going to worry about? It's a little bit of a tricky problem here, but what's most important, I think, is my head. My head is really what's doing most of the moving. It's my head that's going to hit the windshield, so let me focus my attention on the head as the object. So p[initial] is going to be mass of my head - I don't know what that is, something like 5 kilograms, and 20 meters per second, so I get 100 kilogram meters per second. And it's in the east direction. What's p[final] after the crash? It's zero. So p, p[final] minus p[initial] is going to be 100 kilogram meters per second and, if you take zero and you subtract a right arrow, you get the left direction. If you're in a car crash and you come to a halt, the direction of the impulse is opposite the direction that you were originally going. The forces on you are opposite the direction that you were originally going, so the signs make sense. And is this what you care about? I don't think so. p is not the physically important quantity. What's physically important is the force on your head, so you have to use p divided by t. I'm not quite sure what the time is in a car crash, but it's going to be something on the order of a hundredth of a second. So p divided by t, that's going to be the magnitude of the force, is going to be, if you divide 100 kilogram meters per second by .01 seconds, a hundredth of a second, you get about 10^4 newtons in this direction. And 10^4 newtons, how big is that? If you convert that to English system, British system of units, that's about a ton, about 2000 pounds of force. So that's what you feel is 2000 pounds of force on your head, and that's bad news. It's probably going to crack your skull. It's the force that's going to kill you in this situation.
Why is an airbag a nice thing in a car crash? Well, the airbag cannot change p. You're stuck. You had some initial momentum, you're going to end up with zero momentum. p is just some fixed number. The airbag increases t. You see, what's the relevant time here? In the original scenario, with no seat belt, you're just flying forward. While you're flying forward, there are no big forces on you. In fact, your head is just an object in motion remaining in motion. The car has come to a halt and you're catching up to it. It's that brief instant in time, when there's big forces acting, that we had to put into this equation to find the force. And with the airbag, instead of that brief period of time of head against glass, it's the much longer period of time of head against airbag. So you've increased t, maybe by a factor of 10 or so. You've decreased the force by a factor of 10 of so, going from 2000 pounds to 200 pounds. I can survive 200 pounds of force. The airbag has another virtue, it spreads the force out instead of just being on one little spot. That's a separate piece of physics; also helps make you safer in the crash.
Let me show you a picture here that demonstrates what we've just been talking about. Here is the picture of force versus time for when you're just crashing into the windshield. It's a short amount of time, big force, and p is the area. This is the scenario with the airbag; it's spread out over a much longer time, the average force is much smaller. The area under the curves is the same. In both cases, the area under force versus time is the impulse, p, and that's a fixed common number. These have the same area, but clearly you would rather be in this situation, because the maximum force is much smaller. The average force is also smaller, because the time is longer. So the buzzword here is this is called a hard collision and this is called a soft collision. It's just a qualitative word, it doesn't have a technical definition. Hard collisions are just short and strong, soft collisions are spread out over time and not so intense.
Let's do another example, because impulse and momentum are vector quantities and it's worthwhile remembering that. There are a lot of problems where it matters. Consider tossing a tennis ball at a brick wall. So here's your tennis ball and you launch it with some speed v at some angle , maybe it's 30 degrees or something. I give you the speed and the angle, and then let's suppose we watch and this is a fairly ideal bounce, and so it comes off with the same speed v. Kinetic energy is conserved in this collision and comes out at 30 degrees again. So I would like to understand what's going on, I would like to be able to talk about these complicated forces over here that are very difficult to measure. Let's think about impulse. What's the impulse? Impulse is equal to the integral of force times time. I'm kind of stuck. I cannot calculate the impulse, because I really don't know what the forces are yet and I really don't know what the time is. I suppose I could measure them, but it'd be tough. These are, in principle, vector quantities and that's really important in this problem, because we're clearly in two dimensions. Remember impulse is also p. That I can work out. I know p[final], it's just mv[final] and p[initial]. It's a vector equation; let's look at the components.
We want to look at the x components and the y components separately. So I've got to look at components, so here's v, this is the initial. Here is v[initial] in the x direction, it's to the right, and here is v[initial] in the y direction, and that's up. You might draw this arrow over here, but it's the same to slide it over and it just fits in the picture better. How about v[final]? v[final] has a v[final] x component, which is leftwards now, and it's got v[final] upwards, v[final] y. So those are the components and they're easy to work out. The ix, for instance, is v sin and this is v cos . And since v is the same, that simplifies the story a little.
Let's look at the impulse in the y direction, momentum[final] minus momentum[initial] in the y direction, so that's m times this minus m times this. These are clearly equal, because it's the same angles and the same speeds, zero. No momentum transfer, no change in momentum in the y direction. Does that make sense? I think so. It was traveling upwards, it's still traveling upwards, no change. It teaches me something. What can I say about the force? The integral of force times time in the y direction is zero. Probably what that means is there was no force in the y direction at all. It was frictionless. Remember in the y direction that would be friction forces. I suppose it could be negative for a while and positive for a while, but most likely this is telling me that there is no friction and so I've learned something from thinking about impulse.
How about in the x direction? I've got v[initial] x and v[final] x. And again, the lengths of these two arrows are the same, so you think mv[final] minus mv[initial], same lengths, zero. Oops, that's wrong. It's an easy mistake to make. This is where the vector nature really comes into play. Look at the components in the plus x direction. v[final] is negative, so if you take a negative number and you subtract a positive number, you've got negative minus negative. You get a big negative number. Even though they have the same lengths, they're not canceling, in fact, they're kind of adding up in the vector sense. You get a big negative number and that makes sense. The change in momentum in the x direction, you were going this way and now you're going that way, and the change is to the left. Does it make sense about the force? Force has also got to be in the same direction as p and, sure, the wall is pushing you to the left. You were heading towards it, it whacks you to the left and off you go. If you prefer to think of this visually, here was p[initial] and here was p[final]. They had the same lengths, so you're tempted to say the difference is zero, but no, the difference of this arrow and that arrow - they're tail to tail, so it's just the arrow that goes from here to here. It's a big left p. p[initial] plus the change gives you p[final].
So momentum is a defined concept and so is impulse. They have some physical content and you can use them to solve problems. They're especially useful, as we have just seen, in problems involving collisions and crashes, where you've got big forces for short amounts of time. Momentum and impulse are useful quantities. We're going to see that momentum is useful in lots of other contexts, as well, and that's where we're headed.
Momentum
Momentum and Its Conservation
Solving Problems Using Linear Momentum and Impulse Page [3 of 3]

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