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For more information, please see full course syllabus of AP Physics 1 & 2
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Lecture Comments (22)

1 answer

Last reply by: Professor Dan Fullerton
Mon Mar 7, 2016 6:10 AM

Post by john lee on March 6, 2016

I am a little confused. What's the relationship between work-energy theorem and momentum?

1 answer

Last reply by: Sally Acebo
Fri Dec 26, 2014 3:44 PM

Post by Sally Acebo on December 26, 2014

hey professor example 6,i used F=ma to get the a by dividing 6000/2000 and get an acceleration of 3, then used the kinematic eqn. to get the final velocity, why wasn't I able to get your answer (I got 9m/s) if the final velocity was all I was looking for?

1 answer

Last reply by: Professor Dan Fullerton
Thu Nov 6, 2014 8:04 PM

Post by Scott Beck on November 6, 2014

Hello! On the AP exam on example 8 would positive 2400 be acceptable because it asked for the magnitude? And I was told magnitude meant to put absolute value bars around the answer.

1 answer

Last reply by: Professor Dan Fullerton
Sat Aug 2, 2014 10:45 AM

Post by Jungle Jones on August 2, 2014

Why isn't the audio working?

4 answers

Last reply by: Professor Dan Fullerton
Tue Dec 17, 2013 7:52 AM

Post by Burhan Akram on December 15, 2013

at 9:59, you say, "something over something = 1 ". is that really a correct mathematical description?

1 answer

Last reply by: Professor Dan Fullerton
Fri Sep 13, 2013 11:28 AM

Post by Larry wang on September 13, 2013

Momentum (1:25) is a measure of how hard it is to stop a moving object. You also mentioned that both P and V are naturally in the same direction. Doesn't this two quantity somehow conflict with the definition of stopping a movable object?  

2 answers

Last reply by: Larry wang
Sat Sep 7, 2013 11:38 AM

Post by Larry wang on September 6, 2013

Hello once again, what's the reason behind adding the center of mass of point particles in this lecture series? In addition, knowing the equation of CM in both x and y direction  relate to momentum and impulse. Thank you.

3 answers

Last reply by: Professor Dan Fullerton
Fri May 10, 2013 2:00 PM

Post by Nawaphan Jedjomnongkit on May 10, 2013

From deriving impulse momentum theorem that we can relate J with F delta t but when we have to assume that m is constant , but what will happen in the case that momentum change due to change in mass for example the object broke into smaller pieces?

Impulse & Momentum

  • Momentum is a vector quantity describing how hard it is to stop a moving object. Momentum is equal to the product of an object's mass and its velocity.
  • The change in linear momentum is the product of the mass and change in the velocity of the center of mass.
  • Change in momentum is a vector known as impulse. The impulse vector is in the direction of the net force and occurs over a time interval.
  • Velocity of the center of mass cannot be changed by an interaction within the system.
  • Forces that systems exert on each other are due to interactions between objects in the system. If the interacting objects are part of the same system, there will be no change in the velocity of the center of mass of the system.

Impulse & 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
  • Objectives 0:06
  • Momentum 0:31
    • Example
    • Momentum measures How Hard It Is to Stop a Moving Object
    • Vector Quantity
  • Example 1: Comparing Momenta 1:48
  • Example 2: Calculating Momentum 3:08
  • Example 3: Changing Momentum 3:50
  • Impulse 5:02
    • Change In Momentum
  • Example 4: Impulse 5:26
  • Example 5: Impulse-Momentum 6:41
  • Deriving the Impulse-Momentum Theorem 9:04
  • Impulse-Momentum Theorem 12:02
  • Example 6: Impulse-Momentum Theorem 12:15
  • Non-Constant Forces 13:55
    • Impulse or Change in Momentum
    • Determine the Impulse by Calculating the Area of the Triangle Under the Curve
  • Center of Mass 14:56
    • Real Objects Are More Complex Than Theoretical Particles
    • Treat Entire Object as if Its Entire Mass Were Contained at the Object's Center of Mass
    • To Calculate the Center of Mass
  • Example 7: Force on a Moving Object 15:49
  • Example 8: Motorcycle Accident 17:49
  • Example 9: Auto Collision 19:32
  • Example 10: Center of Mass (1D) 21:29
  • Example 11: Center of Mass (2D) 23:28

Transcription: Impulse & Momentum

Hi, everybody and welcome back to

This lesson is about impulse and momentum.0003

Our objectives are going to be to define and calculate the momentum of an object, to define and calculate the impulse applied to an object, and use impulse to solve a variety of problems.0006

We will also talk about how we interpret and use force versus time graphs and finish up by talking about how you find the center of mass of a system of point particles.0018

So with that, let us start by talking about momentum.0029

You probably use this term all the time.0032

A car speeds toward you out of control at a velocity of 60 miles per hour (mph), 27 meters per second (m/s).0035

Can you stop it by standing in front of it with your hand out? Probably not a good idea.0040

Momentum measures how hard it is to stop a moving object.0046

That car speeding toward you has too much momentum.0050

You are going to wind up in little leaky pieces somewhere on the pavement.0054

Momentum is a vector quantity and its symbol is p.0058

Don't ask me why -- Momentum, p, one of those physics things.0061

The units are kilogram meters per second (kg-m/s) which is equivalent to a Newton-second (N-s).0067

The formula for momentum, p = mass × an object's velocity.0072

So, since it is a vector, the momentum vector is in the same direction as the object's velocity.0079

That is probably obvious, but cannot really hurt to say it.0084

By the way, as we look at these units, a N-s -- remember a Newton is a kg-m/s2, so a N-s times a second is just a kg-m/s.0088

A little bit of dimensional analysis to reinforce that.0103

All right. Let us take a look at a quick example.0107

Two trains -- Big Red and Little Blue, have the same velocity.0110

Big Red, however, has twice the mass of Little Blue -- Compare their momentum.0114

To begin with, we are going to have to use our momentum formula, p = mass × velocity.0120

Let us let M equal the mass of Little Blue.0127

Having said that then, if we wanted to find out the momentum of Big Red, that is just going to be -- since it has twice the mass of Little Blue, it is going to be two times Little Blue's mass times whatever the velocity is.0137

They have the same velocity.0156

Compare that to the momentum of Little Blue, which is just equal to its mass times velocity.0158

The difference then, of course, is two times larger.0168

So, we would say that the momentum of Big Red is equal to twice the momentum of Little Blue.0172

Pretty straightforward. Let us take a look at another one.0183

The magnitude of the momentum of an object is 64 kg-m/s.0189

If the velocity is doubled, what happens to the magnitude of the momentum of the object?0193

If our initial momentum is 64 kg-m/s, we are going to multiply that by 2 because of the velocity and that is going to be equal to m × 2V.0199

The momentum then is going to be 128 kg-m/s -- double the velocity, double the momentum -- Answer 3.0216

Another example where we have a changing momentum.0230

We have this D3A bomber with a mass of 3600 kg and it departs from its aircraft carrier with a velocity of 85 m/s due east. What is its momentum?0233

Momentum is m × v which is 3600 kg × 85 m/s, so it is going to be 306,000 kg-m/s, and it is a vector -- needs a direction -- East.0245

Once the bomber drops its payload, though, its new mass is 3,000 kg and it attains a cruising speed of 120 m/s.0269

What is its new momentum?0277

Now, it's mass times velocity or 3,000 × 120 m/s or 360,000 kg-m/s, in again, still, East -- it is a direction.0280

As you observed in the previous problem, momentum can change.0302

A change in momentum is known as an impulse. It is a vector and it gets the symbol J.0306

So, J (impulse) is equal to a change in momentum, which is the final momentum minus the initial momentum.0312

So, let us take that bomber again which had a momentum of 360,000 kg-m/s East and it comes to a halt on the ground.0325

What impulse had to be applied?0333

Well, its initial momentum was 360,000 kg-m/s and its final momentum was 0.0336

The impulse then, which is change in momentum, which is final momentum minus initial momentum, will be 0 - 360,000 kg-m/s East.0349

Therefore, we could say that the impulse is -360,000 kg-m/s East which is equivalent to saying that the impulse was 360,000 kg-m/s West.0361

If the bomber was going East, the impulse to stop it has to be applied in the opposite direction -- akes sense.0390

Let us take a look at combining impulse and momentum.0401

We have a 6 kg block sliding to the east across a horizontal frictionless surface with a momentum of 30 kg-m/s and it strikes an obstacle.0404

The obstacle exerts an impulse of 10N-s to the west on the block.0414

Find the speed of the block after the collision. Let us start with what we are given here.0418

We know the mass of our block is 6 kg and we know that it has a momentum of 30 kg-m/s0425

We know the impulse applied, J -- since it is in the opposite direction, it must be -10N-s and N-s and kg-m/s are the same units.0435

We are asked to find the final velocity of the object.0447

Let us start with the definition of impulse.0454

Impulse is change in momentum, which is the final value minus the initial value, but momentum is mass times velocity.0457

So, final momentum is mass times final velocity - initial momentum -- mass times initial velocity or V0.0467

This implies that impulse plus mV-initial must equal mass times velocity.0476

So if I want final velocity, that is just going to be impulse plus mV-initial divided by m, where if I substitute in my variables, velocity is equal to impulse (-10) + mass (6)...0486

It does not give us initial velocity. It has the initial momentum -- that is 30, so, 10 + 30 divided by our total mass (6)-- -10 + 30 = 20/6.0514

Or 3.33 m/s and in the positive direction which, throughout the problem, has been East.0531

That is combining impulse and momentum.0541

Now, let us talk about the concept of the impulse-momentum theorem.0545

That is a great tool that we can use for problem-solving here in physics.0549

We are going to derive it ourselves.0554

We are going to start with impulse being equal to change in momentum, but momentum is mass times velocity, so, impulse is change in mass times velocity.0555

But hopefully the mass of an object is not going to change.0567

If mass is constant, then we can write this as impulse is equal to mass times change in velocity.0570

Now we are going to pull one of those math tricks again.0579

Remember we can multiply anything by 1 and get the same value.0581

We are going to multiply this by 1, but we are going to write 1 a little bit differently again.0585

We are going to write 1 as δt/δt.0590

That red term there is equal to 1 -- something over something is 1, but if I multiply that, that allows me to make some transformations.0595

So then, I am just going to rewrite this and rearrange it a little bit to say that impulse is going to be equal to mass times δV over δt times δt.0604

All I did was I slid that δt over.0620

What that allows me to do now is, if I look at this -- this piece here -- δv over δt, the rate of change of velocity with respect to time -- that is the definition of acceleration (a).0623

So then I can write this as impulse is equal to mass times acceleration multiplied by δt.0645

But there is another transformation we can make here.0655

Now we have this ma -- Newton's Second Law, net force equals mass times acceleration, so I can now rewrite this again as J (impulse) equals mass times acceleration of force times your time interval.0660

So putting it altogether, impulse is a change in momentum, which is equal to a force applied over some time interval.0680

How do you apply an impulse? You apply a force for some amount of time.0693

How do you change an object's momentum? You apply a force for an amount of time.0697

It has some momentum, so if you want to change it, you have to apply a force.0702

The longer you apply the force, the greater the change in momentum.0706

The larger the force you apply, the larger the change in momentum.0710

That is what is known as the impulse-momentum theorem.0713

What it is saying again, when an unbalanced force acts on an object for a period of time, a change in momentum is produced and that is known as an impulse.0722

Here is our example problem -- A tow truck applies a force of 2,000N on a 2,000 kg car for a period of 3 s.0735

First off, what is the magnitude of the change in the car's momentum?0744

Well, let us start with number 1 here first.0748

Change in momentum, we know, is force times time.0751

We applied a force of 2,000N for a period of 3 s, therefore, the change in momentum is going to be 6,000N-s or 6,000 kg-m/s.0757

Now for 2 -- it says if the car starts at rest, what will be its speed after 3 s?0772

Well, we know δp is mass times change in velocity, which is mass times final velocity minus mass times initial velocity.0780

Therefore if we want the final velocity -- that is just going to be δp + mv initial/m -- just a little bit of algebra to rearrange to get v by itself.0792

Δp -- we just determined up here, was 6,000N-s plus mass times initial velocity.0806

Well, if it started at rest, initial velocity is 0 divided by its mass (2,000 kg), therefore it is going to have a final velocity of 3 m/s.0816

But then, what do you do if there is a non-constant force?0832

In that case, we can draw a graph of the force versus time.0836

The area under the force versus time curve is equivalent to the impulse or the change in momentum.0840

The area under a force versus time curve is the impulse or change in momentum.0846

Let us determine the impulse applied by calculating the area of the triangle under the curve here for a force that ramps up for 5 s and then back down for 5 s.0853

Well, the impulse is going to be the area.0862

It is the area of a triangle, so that is one-half base times height.0865

One-half the base here is 10 s and our height gets up to 5N.0872

So, 1/2 × 10 × 5, I come up with an impulse of 25N-s.0881

Another way we can use graphs to help us solve problems.0890

Let us talk for a minute about center of mass while we are here.0895

Typically, real objects are more complex than these particles, these ideal objects we have been dealing with so far.0899

However, what is really nice in physics is we can treat the entire object as if it had all its mass concentrated at a specific point that we know is the object's center of mass.0906

To calculate the center of mass, for the x coordinate, all you do is you add up all the different individual masses times their x position and divide by the total mass of the object or the system.0917

The y center of mass works the same way -- add up all the individual masses times their y coordinates and divide by the total mass of that system.0930

That will give you the x and y coordinates or if you want to go extending to three dimensions, you can go to z as well for the center of mass of an object.0940

Let us finish up with a couple of more example problems.0949

We have a 2 kg body initially traveling at a velocity of 40 m/s East.0953

If a constant force of 10N due East is applied to the body for 5 s, the final speed of the body is what?0958

Let us go to our impulse-momentum theorem.0966

Impulse is change in momentum or force times time, which implies then that force times time equals mass times the change in velocity, the change in momentum.0969

Or it implies that change in velocity then will be force times time divided by the mass.0983

Well, δV -- that is V - V-initial, so, V - V-initial equals force times time divided by the mass.0994

Or, to get the final velocity by itself, that is just going to be equal to the initial velocity plus force times time divided by mass.1004

We are going to come back to that, so I am going to put a little dot there for a second.1015

Let us keep going to solve our problem -- V-initial (40) plus our force (10N) times our time (5 s) over the mass of our object (2 kg) and I come up with 65 m/s -- Answer 3.1019

But while we are looking at this, let us take a look at that formula for a second -- Force divided by mass -- Newton's Second Law, remember?1042

Force divided by mass is acceleration.1051

V = V-initial plus force divided by mass(at).1054

That is one of our kinematic equations.This all inter-relates.1062

Let us take a look as we analyze a motorcycle accident.1070

A motorcycle being driven on a dirt path hits a rock. How sad.1073

Its 60 kg cyclist is projected over the handlebars at 20 m/s into a haystack.1077

Don't worry, it is a nice, soft happy little haystack.1083

If the cyclist is brought to rest in half a second, find the magnitude of the average force exerted on the cyclist by the haystack.1086

Again, impulse is change in momentum which is force times time.1094

If we are looking for force then, force times time equals mδV -- or change in momentum just rewritten -- or that's M x V - V-initial.1102

Therefore, our force is equal to the mass times final velocity minus initial velocity all over our time interval.1116

So, 60 kg (cyclist), V-final (0) - initial (20 m/s) in the time interval of that 1/2 second means that there must have been a force on our cyclist of -2400N.1126

We have a negative sign again. What does that mean?1145

The force applied was in the opposite direction of what we called positive -- initially the cyclist's velocity.1148

So that is the opposite direction of the cyclist's velocity, which is what you would expect if you are going to bring the cyclist to rest.1156

Great. How about an automobile collision?1169

In an automobile collision, a 44 kg passenger moving at 15 m/s is brought to rest -- V is 0 -- by an airbag during a 0.1 s time interval.1175

That is kind of the point of airbags -- not only do they spread out where the force is applied to decrease the pressure at any given point, but they increase the time interval in which that force is applied.1187

If you have an impulse, you are coming to rest at some point -- no matter what, that impulse is going to be applied, so you would rather have it applied over a longer period of time so that you have a smaller force.1199

Before the days of airbags, you would hit the windshield -- smack -- really short time, really big force. Game over.1210

I am trying to avoid those game overs, those really big forces over short time intervals.1218

So increase the time interval, lower the force -- greater survival rates.1222

So we are going to find the average force exerted on the passenger during that time.1227

Impulse is change in momentum, which is force times time.1231

Therefore, force equals change in momentum divided by time which is the final momentum minus the initial momentum divided by time.1239

Final momentum (0) came to rest minus the initial momentum (44) mass times velocity (15 m/s) in 0.1 s means that the force exerted on the passenger during that time is -6600 N.1253

If there was not an airbag, imagine that was 0.01 s.1274

Now we are talking 66,000N exerted on the passenger. Yay, airbags!1278

All right. Let us take a look at a couple of center of mass problems.1286

Find the center of mass of an object modeled as two separate masses on the x axis.1291

The first mass is 2 kg in an x coordinate of 2 and the second mass is 6 kg in an x coordinate of 8.1295

Just by looking at this, we should be able to guesstimate where this is going to be.1303

First off, it is going to be somewhere between those two objects.1307

Since the object over here on the right, the 6 kg mass, is bigger, I would anticipate that we are going to be somewhere closer to the 6 kg mass than they are the 2 kg mass when we find the center of mass.1311

So let us go back to our formula for center of mass.1324

In the x direction, x center of mass is m1x1 plus m2x2 plus for however many masses we have divided by all of the masses (m1 + m2 + ....).1328

In this case, mass 1 is 2 kg, so that is 2 and its x position is 2 -- m2 is 6 kg and its x position is 8.1344

We are going to divide that by the sum of the masses, 2 kg + 6 kg.1354

So I get 4 + 48/8 -- that is 52/8 which is going to be -- 6.5 should be the position on the scale of our center of mass.1361

1, 2, 3, 4, 5, 6 -- So we could draw it in here, right about there.1383

We could replace these two masses and treat it as if it is one 8 kg mass centered at a point that has an x value of 6.5.1391

The center of mass in the x direction.1403

Let us finish off by doing one that is a two-dimensional center of mass.1406

Find the coordinates of the center of mass for this system where we have three masses.1411

So we are going to have an x and a y coordinate now.1415

And right away, let us look at it again and go, "You know, chances are, in terms of the x position, we're going to be between this mass and this mass and in the y position, we're going to be between that one and that one."1419

So we are probably looking for a center of mass somewhere in that area.1429

That will help us see if we get this right or not.1433

The formula for the x center of mass again is m1x1 plus m2x2 and so on divided by the sum of all the individual masses.1436

In the y direction, the y center of mass works the same way -- m1y1 plus m2y2 plus however many more divided by all of the masses.1449

So let us figure out exactly where the center of mass should lie.1465

Start with the x coordinate. That is going to be...1469

Well, we have got 3 kg in x coordinate of 1 + 4 kg in x coordinate of 5 + 1 kg in x coordinate of 7 divided by the total mass, 3 + 4 + 1 = 8.1472

So that is going to be 3 + 20 + 7 -- that is going to be 30/8 or 3.75 for the x coordinate.1490

For the y coordinate -- now we are going to take our 3 kg × the y value (2) + 4 kg × the y coordinate (3) + 1 kg × the y coordinate (1) divided by the the total mass is 8.1500

So we have 6 + 12 = 18 -- 19/8 or about 2.38.1520

So our center of mass is going to be located at an x coordinate of 3.75 and our y coordinate of 2.38.1526

Let us see where that is -- see how that lines up -- 3.75, 2.38 -- centered somewhere in there.1538

We could treat this entire system as if we had an 8 kg mass centered at that point -- 3.75 in the x, 2.38 in the y.1545

All right. Hopefully, that gets you a good introduction to impulse-momentum and center of mass.1557

Thank you so much for your time and for watching

Make it a great day, everyone.1565