Intro to Conservation of Momentum - Video Tutorials & Practice Problems

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1

concept

Total Momentum of a System of Objects

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3m

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Hey guys. So now that we have a basic understanding of momentum in the next couple of videos will search to see what happens when objects are interacting with each other or that you're going to see often is the word collide. And the key idea of momentum is that when you have two or more objects that are interacting with each other, the momentum of the system is going to be conserved. We're gonna talk about this a lot more detail in the next couple videos. For now. I just want to focus on the words system and in this video want to show you how to calculate the total momentum when you have a system of multiple objects. So let's go and check this out. Remember that? The idea behind the system is that it's just a collection of objects. It's just one or two or even three objects. However many you define in your problem. So the idea here is that I have these two objects right? You've got objects A and B. Just to get to the problem here, I've got the massive A. Is four and the mass of B is five. So they both have mass and they also both have speeds, object A moves to the right with 12 m per second. That's 12 object moves to the left with nines, This is nine right here. So the idea here is that both these objects have mass and speed. So what I can do is I can say this one has momentum P. A. This one has momentum P. B. They both have momentum, but in these kinds of problems I don't really care about the individual momentum's of each object. What I care about is the momentum of the entire system as a whole. And the momentum of the system is just gonna be the sum really just the vector sum to be more specific of each object, individual minds up. So the equation for this is that peace system is going to be the sum of all your momentum's. But most of the time almost always your problems are going to come down to just two objects. So it's gonna be P one plus P. Two. These are arrows. So what I can do is I can rewrite this because I know P is equal to M. V. So if P equals M. V then I could rewrite this as M one V. One plus M two V. To remember these are arrows. So this is how you calculate the momentum of a system of objects. Were gonna be writing this a lot in the next couple videos. So let's go ahead and get some practice. Alright, so like we said we want to calculate the total momentum of the system. This is peace system right here and to do this, we're just gonna add together P. A. Plus PB. Remember these are vectors right here. So what I can do is I can rewrite this and say M. A times V A plus MB times VB. So now what I have to do is look at the problem. So I have one velocity that points to the rights and have another velocity that points to the left. Whenever I had these arrows have pointed different directions, I have to pick a direction of positive. That's going to be super important in these problems. So I want to pick the right direction to be positive, which means that this velocity is positive and this velocity is negative. That's really important here. So now what happens is that r. P system is extremely the massive A. Which is four times the speed of A. Which is 12 and it's positive Mass B. Is five. This is gonna be plus five and then VB is gonna be negative nine. Super important that you keep track of the negatives and positives in these kinds of problems. So what you get here is you end up getting 48 plus negative 45 you get a net peace system of three kg meters per second. So that's the answer. So the idea here is that even though you have some objects that are moving to the right into the left As a whole, your system has three kgm per second of momentum to the rights. All right. So that's it for this one. Guys, let me know if you have any questions.

2

Problem

Problem

Object A moves at 10 m/s at 53° and Object B moves at 5 m/s at –37° as shown below. Calculate the magnitude of the system's total momentum if both objects have a mass of 2kg.

A

$21.6\operatorname{kg}\cdot\frac{m}{s}$

B

$22.4\operatorname{kg}\cdot\frac{m}{s}$

C

$29.7\operatorname{kg}\cdot\frac{m}{s}$

D

$9.17\operatorname{kg}\cdot\frac{m}{s}$

3

concept

Conservation Of Momentum

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5m

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Hey guys. So in the last couple videos we were taking a look at how we calculate the total momentum when we have a system of objects. And the reason why that was important was because when objects are interacting, it's not the individual momentum of each object that we're looking at, but rather the total momentum of the system that were actually concerned with. And we said that that total momentum of the system P system is going to be conserved. So in this video I'm gonna show you what the conservation of momentum equation looks like and how we use it to solve problems. So let's check this out what is conserved. Actually mean we've seen this before, we talked about energy conserved. Just means that whatever you start off with is whatever you have to end up with in the case of energy, it was the initial and E equals E. Final most of the time. So this is the same idea with momentum. What happens is the momentum here? This piece system has to be the same value from initial to final. So we say here is that peace system, initial is equal to p system final. So whatever you have multiple objects that are moving around like this and you calculate the total momentum for that system. That number has to remain the same from initial to final. So we can say here is that for two objects, remember that piece system is just P one plus P two. So we can rewrite this and say that P one initial plus P two initial equals P one final plus P two final. So it gets conserved. Is not the individual momentum's of each object but rather the combination or the some of the momentum's that has to remain the same from the left and the right side of the equation. So now what we can do is we can write this reason, Rewrite this using P equals np and say that this is M one V one initial plus MTV two initial equals M one V one final plus M two V two final. So this right here is known as the conservation of momentum equation. We're going to use it every time we have multiple objects that are interacting with each other. This is one of the sacred laws of the universe that cannot be violated. Whatever momentum you start off with has to be the momentum that you end up with the left and the right side. I have to be equal to each other. That's really all there is to it. Let's go ahead and take a look at an example here. So we have two balls are rolling towards each other. And basically what happens is you have ballet that's moving to the rights be is moving to the left and they're gonna collide and then something's gonna happen afterwards. So in these problems, the first thing you're gonna want to do is draw a diagram for the before and the after. So here's what's happening before the collision. We have these two balls are rolling towards each other. So this is a three kg ball, it's moving to the right with seven. And then we have a four kg ball that's moving to the left with five. So immediately what I can do here is because I have two different directions, I'm gonna choose the direction of positive. So my right direction is going to be is gonna be positive, which means that this is plus seven and this is negative five. Anytime you're writing velocities in your problems, always make sure to keep track of the science. So now what happens is these things are going to collide, right? They're gonna hit each other and we're told here, is that after the collision, but be right, this four kg ball is actually now moving to the right, It's moving at two m per second to the right like this. So now what happens is this is too because it points to the right, it's positive. And I want to figure out what happens to the three kg ball. I want to figure out the magnitude and the direction of ball is velocity. So this is the magnitude and direction right here. So, we're gonna figure out basically what is V. A. Final here. So how do we do that? Well, once we figured out our sort of diagrams for before and after, these are really helpful at figuring out like what exactly is going on in the problem. Now, we're gonna write our conservation of momentum equation. So this is gonna be M one V one initial plus M two V two initial equals M one V one final plus M two V two final. So we're gonna write this a lot in our problems. This is going to become kind of tedious at some point. So one thing I like to do is actually just already replace all the values for the mass that we know we know we're dealing with a three and four kg ball. So this is gonna be three times some velocity plus four times some velocity equals three plus four. And they were really really just have to figure out what is going on what values we plug inside each other of the parentheses. And to do that we just look at the diagram right? So we have this three kg ball that's initially moving at seven. So this is what goes into the initial four ball A. And then the four is going at at negative five because it's going to the left so we're gonna plug it in with the correct sign. Now what happens is this three kg ball, we're trying to figure out the final velocity, so that's actually what goes inside here and this is really our target variable, this is V. A. Final and then this four here is going to the right with two. So this is gonna be plus two. So really we just have one value that's unknown. We're just gonna go ahead and solve for this right? So this is gonna be 21 plus negative 20 equals three of a final plus eight. Now when you simplify this, what happens is this is gonna be one kg meter per seconds. So this is gonna be you're just gonna one on the left side and on the right side you're gonna get three V. A final plus eight. Now remember what we said? Is that momentum conservation means whatever you start off with is whatever you end up with on the left. The initial momentum of the system here, once you've added this all up together is actually just equal to one kg meters per second. So that's actually what we have to get on the right side. So all we have to do is just solve for this missing variable right here. So we're gonna do is you're gonna move this over to the other side. This is gonna be one minus eight equals three V. A. Final and this is gonna be negative 7/3 equals V. A. Final. And what you get is you're gonna get negative 2.33 m per second. So the magnitude here is 2.33 m/s. But what is this negative sign that we got here? Well, hopefully you guys realize that because we've chosen the direction of positive to be to the rights. This negative sign here really just means that this philosophy actually points to the left. So what happens is this velocity here is going to be 2.33m/s, except this ball is actually gonna be going to the left like this. Uh sorry, I meant to say to the left. So what happens here is that you have these two balls that are colliding with each other and then they're actually gonna rebound and go backwards. The initial momentum here is equal to the final momentum here. So that's how you solve these kinds of problems using conservation of momentum. Let's go and take a look at some of the problems

4

Problem

Problem

On a frictionless air hockey table, puck A of mass 0.250 kg moves to the right and collides with puck B of mass 0.38kg, which is initially at rest. After the collision, puck A is moving the left at 0.12 m/s and puck B moves to the right at 0.65 m/s. What was the initial velocity of puck A before the collision?

A

0.73 m/s

B

0.57 m/s

C

0.87 m/s

D

1.1 m/s

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