For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Linear Momentum

*Linear momentum*(→p) is the product of both velocity and mass:→p= m →v. - Notice that linear momentum uses
__velocity__, not speed. That means momentum is a__vector__quantity. - The unit for linear momentum is kg ·[m/s].
*Impulse*(→J) is a way to talk about changes in linear momentum:→J= →Ft. - The change in linear momentum is equal to the impulse and vice-versa-→J=∆→p.

### Linear Momentum

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Introduction to Linear Momentum 0:04
- Linear Momentum Overview
- Consider the Scenarios
- Linear Momentum 1:45
- Definition of Linear Momentum
- Impulse 3:10
- Impulse
- Relationship Between Impulse & Momentum 4:27
- Relationship Between Impulse & Momentum
- Why is It Linear Momentum? 6:55
- Why is It Linear Momentum?
- Example 1: Momentum of a Skateboard 8:25
- Example 2: Impulse and Final Velocity 8:57
- Example 3: Change in Linear Momentum and magnitude of the Impulse 13:53
- Example 4: A Ball of Putty 17:07

### High School Physics Online Course

I. Motion | ||
---|---|---|

Math Review | 16:49 | |

One Dimensional Kinematics | 26:02 | |

Multi-Dimensional Kinematics | 29:59 | |

Frames of Reference | 18:36 | |

Uniform Circular Motion | 16:34 | |

II. Force | ||

Newton's 1st Law | 12:37 | |

Newton's 2nd Law: Introduction | 27:05 | |

Newton's 2nd Law: Multiple Dimensions | 27:47 | |

Newton's 2nd Law: Advanced Examples | 42:05 | |

Newton's Third Law | 16:47 | |

Friction | 50:11 | |

Force & Uniform Circular Motion | 26:45 | |

III. Energy | ||

Work | 28:34 | |

Energy: Kinetic | 39:07 | |

Energy: Gravitational Potential | 28:10 | |

Energy: Elastic Potential | 44:16 | |

Power & Simple Machines | 28:54 | |

IV. Momentum | ||

Center of Mass | 36:55 | |

Linear Momentum | 22:50 | |

Collisions & Linear Momentum | 40:55 | |

V. Gravity | ||

Gravity & Orbits | 34:53 | |

VI. Waves | ||

Intro to Waves | 35:35 | |

Waves, Cont. | 52:57 | |

Sound | 36:24 | |

Light | 19:38 | |

VII. Thermodynamics | ||

Fluids | 42:52 | |

Intro to Temperature & Heat | 34:06 | |

Change Due to Heat | 44:03 | |

Thermodynamics | 27:30 | |

VIII. Electricity | ||

Electric Force & Charge | 41:35 | |

Electric Fields & Potential | 34:44 | |

Electric Current | 29:12 | |

Electric Circuits | 52:02 | |

IX. Magnetism | ||

Magnetism | 25:47 |

### Transcription: Linear Momentum

*Hi, welcome back to educator.com, today we are going to be talking about linear momentum.*0000

*From our work and energy, we already know that the mass and the speed of an object is able to determine its kinetic energy.*0006

*But, when we were dealing with energy, we only had speed as the way of determining energy.*0012

*There was not anything talking about direction.*0017

*Kinetic energy was great for telling u slots of stuff, but it did not tell us if we were going to the north, south, up or down.*0019

*To capture that, we are going to introduce a new idea: Linear Momentum, what your motion is along a line.*0026

*We want the linear momentum to talk about an object's motion in a given direction, just like energy gave us an idea of speed and mass for an object, linear momentum will give us something that tells us about the motion of an object.*0032

*Consider two scenarios: We got a box moving at the same speed but in opposite direction.*0045

*To capture the difference, we will not be able to just use the speed, because it is moving in the speed in both the cases.*0052

*But we will need to also capture its direction.*0057

*To do that, we need to use vector.*0060

*We are going to have to use v, not just 'v', the speed, we need to have its actual velocity.*0062

*Also, what if the boxes had different masses?*0071

*If we had two different boxes, that were both moving in the same speed, one of them was say 1 kg, and the other a 100 kg, we probably want to think of them as being different things.*0073

*It will take a whole lot more effort to stop a 100 kg box than the 1 kg box.*0083

*So, direction is part of it, but we are also going to have to take into account, the mass.*0088

*Clearly, it is going to a good idea to include mass in our idea of linear momentum.*0093

*So, we are going to have to be able to deal with the velocity vector, not just the speed, and also mass.*0097

*Put in that together, we get, linear momentum, and we define it as, mass×(velocity vector), m v.*0106

*Notice that, since this is a vector quantity that we are dealing with, it is going to be, how much we are moving in the x coordinate, how much we are moving in the y coordinate, if you are also moving in the z coordinate, it is going to be, mv _{x} + mv_{y} + mv_{z}.*0116

*We are going to break it up as a vector, and m will just scale the vector.*0135

*These two characteristics, they define, what we are going to create as momentum.*0138

*Momentum, p = mv.*0142

*Why do we use a p?*0146

*I honestly do not have a good answer, I wish I did, there are possibilities, it has a Latin root, but I was not able to figure it out, sometimes there are mysteries in the world.*0148

*An important thing to notice here is that, this p is not just a scalar quantity, it is not just a single number, it is a vector.*0166

*If v comes in (x,y,z), our p is also going to have to come in (x,y,z).*0174

*Units of linear momentum are, m v, so kg×(m/s).*0183

*We are going to consider a new idea: Impulse.*0191

*What if we want to talk about how much an object's momentum changes, that is important.*0193

*If we got a box moving along, and if we put a force on the box, we are going to change the momentum of the box, because we will change the speed that it is moving at.*0198

*So we are going to define the idea of impulse.*0205

*What really changes the velocity of the thing?*0207

*It is just going to be the fore involved, but not just the force.*0212

*The same force is going to be very different if you put a 100 N on an object for 0 s, 1 s, 10 s, 100 s, totally different things are going to happen depending on the amount of time that the force is acting on it.*0217

*The objects mass remains constant, pretty reasonable.*0227

*The object's velocity, and thus its momentum, is going to change based on the force applied, and how the long that force lasts.*0232

*We define impulse as the letter 'j' (no particularly good reason here, just making sure we are using letters that have not already been taken by somebody else), j= force×time = Ft, just like before, j is a vector because force is a vector.*0238

*Makes sense, because we are talking about change in a vector quantity.*0255

*Note that impulse is a vector, and its units are going to be, Ft, so N s.*0259

*At this point, we have created some definitions, and we can see that linear momentum and impulse are connected because we wanted impulse to represent a way of shifting around momentum.*0268

*You put force into an object for a certain object of time, it is going to change the momentum that the object has, because we will be changing the speed that it is changing at.*0279

*But, what is the precise mathematical relationship?*0286

*Let us figure it out.*0288

*We look more closely at the formula for impulse.*0289

* j = Ft, if we expand that out, we can get this.*0292

* F = ma, so, j = mat = mΔv, (since 'a' is how much your velocity is changing with time).*0299

*Since mass is not changing, we can pull that change outside, and we get, Δv, because we do not have to worry about, since velocity is the only thing that can change, we are assuming that the mass is constant, so, mΔv is the same thing as, Δ(mv), because mass is just a constant, and velocity is the variable, at this point.*0328

*Remember, we defined, p = mv, so it has to be the case that, Δ(mv) is the same as, Δp.*0350

*So, in the end, j = Δp, so impulse is simply the change in the momentum.*0362

*Note that impulse and momentum have the same units, 'N s' is the same thing as 'kg m/s', because N comes from, if F = ma is kg m/s/s, so, N is kg m/s/s, and multiply with s, so kg m/s, is what we had for momentum.*0371

*It makes a lot of sense, our units wind up working out, so, j = Δp.*0404

*In this section, we have talked about what linear momentum is, but why have we talked about 'linear' momentum, when we have not heard about any other kinds of momentum, why is it called linear momentum if it is the only momentum that we are concerned with?*0417

*The thing that is going on, is there are other kinds of momentum, there is angular or rotational momentum.*0438

*Spinning objects, objects that are spinning, if you take a wheel and you spin it really fast, it will keep spinning, right?*0445

*It has a momentum, it is not moving anywhere, it is just sitting there in space and spinning, but it takes effort to start it spinning, and it takes effort to stop it spinning, so there are torques involved, we have not talked about rotational mechanics.*0452

*The entire thing in Physics you cannot talk about, but we just do not have quite enough math to really feel comfortable handling it, we are almost there, this is definitely close to being within our grasp, but a little too much math for us to tackle there, so we are holding off on it, that is why we have not talked about angular momentum, which is also similar to rotational momentum.*0465

*We have been talking about linear momentum, because we want to make sure that this is kept clear, as this is linear momentum as opposed to this other kind of momentum, but often when we are talking about linear momentum, we will also just refer to it simply as momentum, because that is the thing that is more common, but it is important to keep in mind that there are other kinds of momentum.*0485

*Just because we are talking about one of them, does not mean that there is nothing else out there.*0501

*Let us start with our examples.*0506

*A skate board of mass 4 kg is rolling along at 10 m/s.*0508

*What is its momentum? This one is pretty easy.*0512

*What is the basic definition for momentum? p = mv = 4 kg × 10 m/s = 40 kg m/s. (since one dimension, velocity becomes a single number).*0513

*Same skateboard, m = 4 kg, is rolling along with an initial velocity of 10 m/s, just like before.*0538

*A force of F = -6 N is applied to it, for t = 6 s.*0547

*At the beginning of this problem, we got some skateboard rolling along on the ground, and it is moving this way.*0551

*However, as time moves on, there winds up being a force applied to it in this direction, so later on, this skateboard is going to be rolling along with a much smaller velocity vector.*0558

*It is still going to be moving forward, potentially, depends on how long that force is actually, may be that force is going to push it so hard that it winds up going in the other direction, we are going to have to do some math to figure it out.*0569

*But, the force is acting on it in the direction opposite of current travel, we are travelling in the positive direction (right), and now force is going to wind up in the negative direction.*0583

*That is the importance of using vectors, we know how positive and negative direction, even if we are still on one dimension.*0596

*So, What is the impulse vector?*0602

*Impulse, j = Ft = -6×6 = -36 N s, so what is the final velocity that it is going to have?*0604

*In this case, we know that change in momentum is equal to the impulse.*0629

*We already figured out what the initial momentum is.*0635

*In the last problem, it wound up being, 4×10 = 40 kg m/s, so the final one is going to be, p_{i} + j = p_{f}, since p_{f} - p_{i} = j.*0638

* p_{i} = 40, and the change is -36, so in the end we get, 4 kg m/s = p_{f}, is the final momentum.*0670

*Final momentum does not quite tells us the final velocity.*0684

*But we can figure that out pretty easily from there.*0687

* p_{f} = mv_{f}, 4 = 4v_{f}, so, v_{f} = 1 m/s, is the final velocity.*0688

*It is still moving in the positive direction.*0709

*It would be possible to figure this out without using momentum and impulse, but momentum and impulse may dissolve pretty simple things that we have to do, very direct, multiplying and then adding, and then doing some really simple algebra.*0711

*but, we could go back to doing this with Newton's second law.*0725

*If we want to do this in Newton's second law, we have got, F = ma, -6 = 4a, a = -1.5 m/s/s.*0729

*What does the change in velocity wind up being?*0753

*Δv = -1.5×t = -1.5×6 = -9 m/s, and so, if you started with initial velocity of 10 m/s, then, v _{f} = 10 + (-9) = +1 m/s.*0755

*So, if we wanted, we could do this in terms of basic fundamental, Newton's second law, but in this case it is pretty easy.*0780

*Remember, the way momentum works, the way we have defined it, is really just sort of jumping off the point of using Newton's second law.*0789

*That is how we got impulse.*0796

*Impulse was based around the fact that, the reason why impulse is equal to the change in momentum is because we used F = ma, at what point we got, Ft, so in the end, they are deeply interconnected.*0797

*So, we can decide to go with Newton's second law, but in lots of problems, it is going to wind up being the case that it is actually a little bit easier the way of linear momentum, especially when we are dealing with momentum problems.*0812

*In the next lesson, we are going to wind up seeing why it is really useful to have momentum when we get to the conservation of momentum, and that is why this stuff really matters.*0821

*Example 3: A ball of mass m = 0.5 kg is moving horizontally with v_{i} = 10 m/s.*0835

*It bounces off a wall, after which it moves with v_{f} = -7 m/s.*0842

*What is the change in linear momentum, what is the magnitude of the impulse?*0848

*We got this ball, moving along the positive direction, and it hits the wall, and afterwards it changes, rebounds, and it is moving in the negative direction.*0851

*What is the initial momentum? m v = 0.5×10 = 5 kg m/s.*0862

*What is the final momentum? 0.5×(-7) = -3.5 kg m/s.*0885

*So, the change is, final - initial, Δp = -3.5 - 5, seems a bit weird, but makes sense, we started off in the positive direction, and ended up going in the negative direction, so the entire change has got to be one of negative momentum occurring.*0898

*So, -8.5 kg m/s is the change in the linear momentum.*0937

*The magnitude of impulse, remember, the magnitude is the size of something, so the size of this, j = Δp = magnitude(-8.5 kg m/s) = 8.5 kg m/s.*0945

*In the end, when we are dealing with magnitude, it does not care about direction, it does not care about positive or negative, it just cares what [unclear] like the thing we are dealing with.*0972

*In this case, we had -8.5 as the change in momentum, but the magnitude of the change in momentum was just the total moment, 8.5.*0983

*This is sort of similar to what we saw in energy before, instead of using velocity, it was the length of the velocity vector that we cared about, its speed.*0993

*It did not matter if it was pointing flat, it was pointing straight up , pointing at a 45 degree angle, all that mattered was what the total length was.*1001

*That is what we are seeing here, when we ask what the magnitude of the impulse, we are asking for what is the length of that thing.*1010

*In multiple dimensions, we take, sqrt(x ^{2}+y^{2}+z^{2}), because that is how we take the magnitude of a vector.*1017

*Last example: A ball of putty with a mass of 1 kg is about to fall on your head.*1027

*The velocity initially is -5 m/s, makes sense since it is falling down.*1032

*Which one is going to hurt worse, it lands and sticks on your head, so it lands and sticks, so the final velocity is zero, or, it lands and bounces off your head, with a final velocity of 4 m/s.*1037

*Let us call these cases, A and B.*1048

*In case A, it hits your head, and it sticks in place.*1052

*In case B, it bounces off your head.*1059

*Which one of these is going to hurt worse?*1062

*Normally we think of things being bouncy as good, they are easy right?*1065

*A super ball being bounced on your head, and bouncing off your head, it would not hurt much.*1069

*But what is really going on, which one is taking more force, the super ball that hits your head and just sort of sits there, or the ball that hits your head and ricochets right off?*1075

*We can break this down with Physics and Math.*1083

*In both cases, what is really going to be defined as hurting?*1086

*The way that we define as 'hurt' is probably the amount of force that it exerts on you.*1092

*So, you would rather have 2 N of force shoved on you than a billion newtons of force being shoved on you, a billion newtons force applied on you, your body is going to look like a pancake.*1097

*But 2 N of force, a small amount of force is going to hurt less than large amounts of force.*1108

*What we are looking for is, which one of these cases is going to produce less force on impact.*1115

*In both of these cases, it is going to be important to know how long the balls contacting your head, we will be able to figure out what average force is going to be.*1121

*Impact time for both of these cases will be 0.25 s, so let us figure out how much force is involved.*1128

*To do that, we need to figure out what the impulse in both cases are.*1135

*To do that, we need to know what the initial velocities are, what the initial momentum is, and what the final momentum is.*1139

*From there, we will be able to figure out the impulse, and from impulse, we will be able to figure out pretty easily what the force is.*1145

*In both them, we need to know what is their initial momentum.*1150

*Initial momentum = m× v_{i} = 1×(-5) = -5 kg m/s.*1154

*Case A, it lands and sticks on you head, so its v _{f} = 0.*1168

*If that is the case, what is the final momentum?*1174

*Still mass of 'm', but now it sticks, so it now has no velocity, so it has got a momentum of zero in the end, 0 kg m/s.*1177

*Compare that to B, where we have got a final momentum = 1×4 = 4 kg m/s.*1189

*It might make good sense at this point to go, 'Okay, the one with less momentum is case B, because we go from 5 to 4'.*1207

*But that is not the case.*1215

*The whole case is, we go from -5 to 4.*1217

*So which is a bigger change, going from -5 to 0, or -5 to 4?*1220

*We check that out, we get, Δp = (final) - (initial) = +5 kg m/s, is in A.*1224

*In B, Δp = 4 - (-5) = 9 kg m/s.*1240

*So, there is more change in momentum in case B than in case A.*1254

*So it is going to make more sense for B to hurt, because that force, to change that momentum, has to come from somewhere.*1257

*We have got a constant time, so it has got to be, the amount of force involved is going to have to be more in case B.*1264

*We will finish this out, at this point we can see that the answer is going to be case B.*1272

*So, j = Ft, we also know that j = Δp, so for case A, we got, 5 = F×0.25 s, F = 20 N.*1276

*So, 20 N is how much force you wind up undergoing for A.*1301

*Now, how much force does you wind up undergoing in B?*1306

* j =Ft = Δp, so, 9 = F×0.25, so, F = 36 N, in part B.*1310

*So, part B winds up putting more force on your head, in both cases, they are pretty small, so worst case you are going to have a little bit of headache, but 36 N is more force you have to suffer than 20 N.*1328

*So, it is actually better to have an object that lands on your head and just sticks there, something that goes 'splat!', than something that goes, 'boing!', because 'boing!', that force to make it bounce off is going to have to come from somewhere.*1340

*It is coming from your head, that is going to make it hurt more.*1351

*It is better to have it land and splat, it has less force, because it has to change its momentum if it is going to be able to bounce off your head.*1354

*I hope this lesson made sense, I hope you got a good understanding of linear momentum, because the next thing that is coming is conservation of momentum, and that is where the real point of momentum is.*1361

*Alright, good day!*1368

2 answers

Last reply by: Enoch Lee

Wed Jan 6, 2016 5:25 AM

Post by Farzana Meem on June 16, 2015

Can you explain how an airbag "cushions" the blow for a passenger in a car collision using impulse and momentum.