For more information, please see full course syllabus of AP Physics 1 & 2
For more information, please see full course syllabus of AP Physics 1 & 2
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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
- Objectives
- Momentum
- Example 1: Comparing Momenta
- Example 2: Calculating Momentum
- Example 3: Changing Momentum
- Impulse
- Example 4: Impulse
- Example 5: Impulse-Momentum
- Deriving the Impulse-Momentum Theorem
- Impulse-Momentum Theorem
- Example 6: Impulse-Momentum Theorem
- Non-Constant Forces
- Impulse or Change in Momentum
- Determine the Impulse by Calculating the Area of the Triangle Under the Curve
- Center of Mass
- 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
- Example 8: Motorcycle Accident
- Example 9: Auto Collision
- Example 10: Center of Mass (1D)
- Example 11: Center of Mass (2D)
- 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
AP Physics 1 & 2 Exam Online Course
Transcription: Impulse & Momentum
Hi, everybody and welcome back to Educator.com.0000
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 Educator.com.1561
Make it a great day, everyone.1565
1 answer
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
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
Sat Aug 2, 2014 10:45 AM
Post by Jungle Jones on August 2, 2014
Why isn't the audio working?
4 answers
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
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
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?