ShmoopTube

Where Monty Python meets your 10th grade teacher.

Search Thousands of Shmoop Videos


Physics Videos 34 videos

Physics: Isaac Newton
33 Views

Isaac Newton. Who was he? Why do we need to know about him? In a physics course, no less? Well, he's only the most famous physicist in history, and...

Physics: The Basics of Trigonometry
35 Views

What are the basics of trigonometry? And why are we learning about this in a physics course? Both good questions. In this video, you'll learn about...

Physics: Unit Analysis and Graphical Data Analysis
36 Views

It's time to make our liters and meters work together. Enough of the bickering, right? In this video, we'll do some unit analysis, covering SI Unit...

See All

Physics: Conservation of Momentum 72 Views


Share It!


Description:

The Law of Conservation of Momentum says that total momentum always stays the same in a closed system. In an open system, though, all hell usually breaks loose.

Language:
English Language
Subjects:

Transcript

00:00

Conservation of momentum. It's not just a good idea, it's the law. [mumbling]

00:32

All right, well it's important to have a healthy respect for the law. [cop in cop car]

00:36

No kneeling around here and don't seed when you drive, don't litter and don't

00:41

forget to tell your mom you love her. That's not a law, that's just you know being

00:45

nice. Alright but the truth is, there are some laws that you just can't break. Like

00:49

it's impossible, it can't be done. Try as hard as you want, it ain't gonna happen.

00:52

Those are the laws of physics, not the NFL. And if you try to break one of those [two people in lab]

00:58

physics laws, well you won't go to physics jail, or anything. But you might

01:01

not be too happy with the result. Try and break the law of gravity, for example.

01:05

I'll just watch from a safe distance. Well one of these laws of physics, is the

01:11

law of conservation of momentum. This law states, that in a closed system, the total

01:17

amount of momentum always stays the same. Now in the real world we don't run into

01:21

a lot of closed systems. For example when you roll a bowling ball down the lane,

01:25

the bowling balls momentum is transferred to the pins, or it's [bowling ball and bowling pins]

01:28

transferred to the gutter. Yeah it happens. But when the ball hits the pins,

01:32

pins don't just bounce around forever, maintaining their momentum, that gains

01:35

momentum is transferred to the side walls and the lane and whatever the

01:39

technical term is for the back of the lane. All of those surfaces deform

01:44

slightly, as the momentum of the pins, is transferred. So the momentum goes to

01:49

these other things, within the system. But the effect is so small, we can't see it.

01:54

Of course, physics tends to look at ideal systems. And as we know from the fact [cop in cop car with Ferrari speeding by]

01:59

that we still don't have our own Ferrari, well the real world is almost never

02:03

ideal. But there's an equation, that expresses the conservation of momentum

02:07

and it's this one right here. Well this is just fancy math way of saying, that

02:13

the sum of all momenta, for all objects within a closed system, before the

02:19

objects interact, equals the sum of momentum for the objects, after they [equation on note page]

02:25

interact. And fortunately we can write it in a much simpler format. Yah, there

02:30

we go. But really it doesn't matter, because when we start doing the math, for

02:34

actual objects, actually interacting with each other. We replace both sides, of

02:39

these equation,s with the numbers for the objects. So if we just so happened, to

02:44

have someone, oh I don't know, sitting on a skateboard and then throwing a

02:48

medicine ball to someone. Well our equation would look like this. This equation says

02:54

that the initial momentum of the skateboarder, that's the P sub s I, plus [equation]

02:59

the initial momentum of the medicine ball P sub B I, equals the final momentum

03:05

of the skateboarder, plus the final momentum of the ball.

03:07

And since momentum equals mass, times velocity. We can swap those variables

03:12

into the equation, in place of P. So now we have this equation. Which says that,

03:22

the skateboarders mass, times his initial velocity, plus the ball's mass, times its

03:28

initial velocity, equals the skateboarders mass, times his final [formula]

03:33

velocity, plus the ball's mass, times its final velocity. Looking at this you can

03:38

see that if we have more than two objects, the equations can get a little

03:42

long-winded. Which is why we use that shorter

03:46

equation earlier. But let's go ahead and see how we would solve this equation. In

03:50

our scenario, we have teenagers performing physics in public. There's

03:54

nothing criminal in that, although it is a little weird. I've got my eye on you

03:57

citizens. At the start of this experiment, the skateboarder and the ball are not in [two people throwing ball to each other]

04:01

motion. Which means each initial velocity is a big fat zero. Let's put those values

04:07

into our equation. Before we go any further, we can solve the left side of

04:10

this thing right now. Whatever masses we're dealing with, are

04:12

irrelevant. Anything times zero in California, equals zero. So the initial

04:18

momentum of the skateboarder is zero, and the initial momentum of the medicine

04:20

ball zero and zero plus zero equals, seventy three billion 360. Oh wait hold

04:24

on, I spilled my coffee on my calculator earlier, it's on the fritz. [coffee on calculator]

04:27

All right zero, plus zero, is zero. Got it? So we got zilch. That's right, right hand

04:31

side will also have to evaluate to zero. That's only gonna happen if one of the

04:36

momentum on the right hand side is negative. Well remember momentum is a

04:40

vector quantity, so it has a magnitude and a direction.

04:44

Well the convention, to have any motion from left to right be a positive, value in any

04:48

motion from right to left be negative. Really doesn't matter though. You can set

04:52

whichever direction you want is positive and the other as negative. Just make sure

04:56

you're being consistent. The important thing to recognize here, is that we have [man throwing woman a ball while on skateboard]

04:59

two masses acting in opposite directions. One will have a negative velocity and

05:05

therefore have a negative momentum as well. So now we can set one velocity to

05:09

be negative in our equation, or just make the overall momentum be negative, like

05:13

this. And once we've done that we can see something interesting. With a bit of

05:18

algebra, we can add the negative momentum, to each side of our equation and find

05:22

that the final momentum for the skateboard, equals the final momentum for

05:24

the medicine ball. See how everything balances out all nice and pretty and [long equation]

05:29

laws of physics can be you know satisfying like that. Okay folks

05:34

experiment over. Let's keep moving, nothing else here to see. In that

05:38

skateboard experiment, we had two objects interacting and it was pretty friendly

05:41

interaction. But alot of times interactions between objects are gonna take the form

05:45

of, collisions. The collision can be a minor fender bender, or a five car pileup.[major car crash]

05:50

And collisions can be elastic, or inelastic. An elastic collision, occurs

05:57

when two objects collide and then bounce off each other going in different

06:02

directions. Think of playing pool. When the cue ball, hits the eight ball, the

06:06

balls don't just stick together, they go in different directions. Unless both

06:09

balls are covered in syrup, which is why waffle Wednesdays and pool tables are

06:14

not a good mix. An inelastic collision on the other hand, occurs when two objects

06:20

collide and then stick together. Say you're riding a bike, when a possum [man riding bike with possum on his back]

06:24

decides to hitch a ride. Well since conservation of momentum is

06:28

the law and you and your bike have now gained mass, in the form of a probably

06:33

dead marsupial. You'll therefore lose velocity. After all that's the only way

06:38

that momentum would stay the same, with an increase in mass. Just be careful that

06:43

this one looks kind of bitey, even if he is

06:45

potentially dead. Alright well because these problems can get confusing fast,

06:49

let's walk through the steps one by one, so we have a roadmap to solve them.

06:52

Step one, identify all the objects that interact with each other. We have two [chalk board step, and biker with possum on back]

06:57

objects that start off stuck together already, like the bike rider and his

07:01

bicycle. We can usually consider those as one object. Step two, if there are objects

07:07

moving in opposite directions, at any point, assign the velocity for one of the

07:11

directions a positive value and the velocity of the opposite direction a

07:15

negative value. Got that? Good. With our biker possum collision, all the velocity

07:21

was in the same direction. But typically we use right as the positive direction.

07:27

If you're feeling wild and rebellious, well go ahead and flip it around. [rebellious people]

07:32

Alright step three, set our balanced momentum equation. It should look

07:36

something like in this one. With mass and initial velocity for each object on the

07:40

left side and mass and final velocity for each object on the right. If we have

07:44

more than two objects interacting with each other, then we add the extra objects

07:47

to both sides of the equation, there we go.

07:50

Step four, if two objects stick together after a collision, we know the final

07:54

velocity will be the same for both objects. That means we can simplify the

07:58

right side of our equation, by adding up the masses of each object and [equation]

08:01

multiplying that sum, by the final velocity. I'll save a little work that

08:04

way. Alright, step 5, plug in whatever values we've been

08:08

provided, or have found experimentally. Since we set up our equations in advance,

08:13

it should be pretty straightforward. Just make sure not to mix up your masses

08:16

and velocities. You don't want to multiply object ones mass and object

08:20

twos velocity. I can't write you a ticket for that, but I can give you a [cop talking]

08:24

stern look. So be careful. Alright step six, remember to make the velocities

08:29

negative for whichever velocity direction, we've designated as our

08:33

negative one. If you skip this step well your right side, won't balance with your

08:37

left side. Which means the equation will crash and burn.

08:40

Step 7, solve for our unknown variable. A lot to say on this one I trust that you

08:45

have the requisite math skills under your belt. Alright, step eight, celebrate

08:50

with a doughnut. Oh sure, cops and donuts are a cliche but [cops in donut shop]

08:53

tell me you don't like, a nice fresh glaze and a circle of fried dough.

08:57

Yeah that's what I thought. Alright one last thing before we look at some problems.

09:01

All the motion we've looked at so far has been along one dimension. I'm not

09:08

sure if you've looked around lately, but life is 3d. So although we're not going

09:12

to deal with it right now, in the future you're likely to run across problems

09:15

that deal with momenta, in two, three, or if supernatural forces are involved, five

09:21

or six dimensions. Who knows? Which means we might have to use trigonometry to

09:25

break the momentum down to XY and z components. We're just telling you now so[people freaking out in classroom]

09:29

you don't freak out, in some of the advanced physics class later. Alright

09:33

well we've got your attention. Let's talk about driving safely in dangerous

09:36

conditions. Say we've got an icy road, three cars are driving, with car a in

09:40

front, driving responsibly. Car B is making sure to keep an appropriate

09:44

amount of space between her car and car A. Meanwhile some jerk, in car C,

09:49

comes barreling in. Going way over the proper speed limit for the weather. Car

09:54

C collides with car B and they stick together going forward, until car B hits [three vehicle car crash]

09:58

car A. Now all three cars are conjoined at the bumper, headed in the same

10:03

direction and we've got a big mess to straighten out.

10:05

What's the conservation of momentum equation, for this scenario and we need

10:10

to break it down to masses and velocities. Okay, so we know the shorthand

10:14

equation, the sum of momentum before interaction, equals the sum of momenta

10:18

after the interaction. But we could also write it like this, with P sub a P sub B

10:23

and P sub C. But we need to break it down to masses and velocities. So let's work on the

10:28

left side first. Makes sense, since it's the, you know, before side. We'll go step [equation formula]

10:34

by step. We know our three objects are cars A, B

10:37

and C. So we'll start with M sub a times, V sub AI and we'll add, M sub B, times V

10:42

sub B I and last but not least we got the one that started it all, M sub C

10:47

times V sub C I, aka jerk. All right, how about the right side. We definitely need

10:54

to add up the masses of each car, but since they're all sliding along this icy

10:58

low friction road together. We'll only have one velocity to deal with. [cars conjoined in crash]

11:03

So the right side of the equation will be, M sub A, plus M sub B, plus M sub C and

11:09

all of that will be multiplied by V sub F. Alright now we don't have any values

11:14

here, we're just setting up the equation. Do we have any negative momentum to deal with? Well

11:18

no, this wasn't a head-on collision. All the cars we're moving in the same

11:22

direction, so we don't need to worry about anything negative there. Here's how

11:26

we want our equation to be, before we plug in any values, right there. Okay,

11:30

let's think about another scenario. And say we've got two ice skating kangaroos. [kangaroos ice skating]

11:34

Which may sound crazy to you, but I've seen some things in this line of work

11:38

man. Yeah, I've seen some things. Anywho, well they're working on their ice

11:43

dancing routine and at one part of the routine they come together on the ice,

11:46

standing motionless, facing each other. It's really a beautiful, marsupial moment.

11:49

Then they push off, going in opposite directions. You can hear the music

11:53

crescendoing, kangaroo one has a mass of 102 kilograms,

11:56

slides away with the velocity of 2.1 meters a second. Kangaroo two who we're

12:01

gonna call kanga two. Because well it just sounds cooler, has a mass of 109 [2 kangaroos ice skating]

12:04

kilograms. What's kanga two's velocity as he gracefully

12:07

slides away? Well let's walk through this thing step by step. First we identify our

12:12

objects. Easy enough, we've got kanga 1 and kanga two step 2. If we have

12:16

motion in opposite directions, we set one is positive, the other is negative. In

12:20

this case kanga one's velocity has already said is positive 2.1

12:24

meters a second. So kanga 2 will be negative velocity and momentum. Next we

12:28

set up our before-and-after equation. Sum of moment of each kangaroo, before they

12:33

push off each other, equals the sum of momenta, after they push off each other. [kangaroo formula]

12:37

All right, step four says, if two objects stick together after their interaction,

12:42

we add up their masses and multiply that by the final velocity. But in this case

12:47

we have two animals that start off together, then move apart. So we can

12:51

add up their masses and multiply that, by the initial velocity. Okay, so now we've

12:56

got our equation set up. Step five says, we can start putting in the numbers. All [equation written out]

13:01

right on the left side, we've got one hundred two kilograms, plus hundred nine

13:04

kilograms and an initial velocity of nada.

13:07

On the right we've got a mass of 102 kilograms, times velocity of 2.1 meters a

13:11

second, plus a mass of 109 kilograms, times velocity were solving for. What

13:15

is that number? All right, right away we can see the left side is going to equal

13:19

zero and we can do that math on kanga one, to find a momentum of two hundred

13:23

fourteen point two kilogram meters per second. Then we can subtract kanga two's

13:28

momentum, the mass, times the unknown velocity there, from both sides of the [equation worked out]

13:32

equation. Leaving us with negative 109 kilograms, times V sub 2f, equaling two

13:38

hundred fourteen point two kilogram meters per second. And now people all we

13:42

have to do is, divide each side by negative 109 kilograms, to solve for the

13:46

missing velocity. And Kenga two's velocity equals about negative 2.0 kilogram

13:51

meters a second. Which makes sense kanga two has a smidge more mass, than kanga one

13:56

so his velocity should be a smidge less. Truth is you can break most laws. I might [cop in cop car]

14:02

not see you speed on the highway and if you're super careful, jaywalking isn't

14:06

gonna be a big deal most of the time. And your mom might say you get away with

14:09

murder, but that's just a figure of speech. Please don't take her literally.

14:13

But yeah you're not breaking the law of conservation of momentum. Don't even

14:16

bother trying. All right if the Isaac Newton and Rene Descartes

14:20

say it's law, well you better take their word for it. [court room with people in it]

Related Videos

Jane Eyre Summary
123033 Views

When you're about to marry the love of your life, not many things could stop you. However, finding out that your future hubby is keeping his crazy...

What is Shmoop?
91411 Views

Here at Shmoop, we work for kids, not just the bottom line. Founded by David Siminoff and his wife Ellen Siminoff, Shmoop was originally conceived...

ACT Math 4.5 Elementary Algebra
492 Views

ACT Math: Elementary Algebra Drill 4, Problem 5. What is the solution to the problem shown?

AP English Literature and Composition 1.1 Passage Drill 1
1039 Views

AP® English Literature and Composition Passage Drill 1, Problem 1. Which literary device is used in lines 31 to 37?

AP English Literature and Composition 1.1 Passage Drill 2
683 Views

AP® English Literature and Composition Passage Drill 2, Problem 1. What claim does Bacon make that contradicts the maxim "Whatsoever is delig...