**Concept:** Total Momentum of a System

Hey guys. So, now that you have a good idea of how momentum works, we're going to talk about conservation of momentum, which is one of the key aspects of momentum and the classic example is when you have two objects colliding whenever two objects colliding, we say that momentum of the entire system is conserved, let's check it out. So, as I said a key aspect of momentum is that when you have two or more objects interacting it could be a collision or some other form of two object interaction, the momentum of this system is conserved. Now, remember a system is just a group of objects, it's collection of objects, the total momentum of a system is the sum of the individual momenta, momenta is just the plural for momentum, it's not Momentum's, it's funny but that's what it is. So, if I have two objects, the momentum of the system, of the collection of the objects is just a momentum of one plus the moment or the other. So, momentum of a system is the sum of the individual momenta and if you have two objects, for two objects you would have something like p1 plus p2. Now, remember p equals m, v. So, we're going to take this one step further instead of P I'm going to write M, V. So, it's going to look like this m1 v1 plus m2 v2 and if you've gone to class and your professors talk about momentum already you will at the very least remember that he wrote this a lot, right? So, we're going to see a lot of m1 v1, m2 v2. Alright, let's do a very quick example here, I have two objects A and B, they're moving towards each other, it says that A moves to the right, if two objects are moving towards each other the one that's moving to the right is the one that's in the on the left side, okay? So, just by reading that and kind of think about it carefully, A is on the left it has a mass of 4 kilograms and it's moving to the right with 9 meters per second, B has a mass of 6 kilograms and it moves to the left since they're going towards each other with 6. Now, first things first here, velocities are vectors, they're going in opposite direction, they have to have opposite signs. So, this one is going to the right, I'm going to get positive. So, this has to be negative and obviously these are meter per second, calculate the total momentum of the system, we're just going to use this right here. So, p system is m1 v1 plus m2 v2, that's it. So, what I like to do is I look at the masses 4 and 6 and I just do this 4 parentheses 6 parentheses and then I slow down I try to figure out what number goes here and what number goes here. So, this is the velocity of the 4, the velocity for is plus 9 and this is the velocity of the 6, which is minus 6, when you do this you have 36 plus negative 36, which is 0, 0 kilograms meters per second. Now, I did this on purpose just to make the point that just because you have two objects moving it doesn't mean that their collective momenta is a nonzero number, if the momentum of this guy is 10 and this guy is 10 they add up to 0, okay? So, just in case you see something like this you're not weirded out, this is perfectly fine, let me do another quick example here and then we'll keep going. So, here it says, calculate the momentum of the system below if objects have mass 2 kilograms, so the objects are both 2 kilograms. Now, I want to remind you that a momentum is a vector, p is a vector because V is a vector. So, when I combine the momentum of these two guys, let's call this m1, let's call this m2, when I combine the momentum of these guys and I say p total or p system is p1 plus p2, this is not enough, this only works for one dimension but because these guys are going two dimensions the momentum can't be simply the addition, okay? Let me calculate the momentum real quick, momentum is p equals m, v, this guy is a mass of 2 and a velocity of 10. So, I can write that p 1 is 2 times 10, 20, let me put that in here, p 1 is 20, this mass is also 2, this velocity is 5. So, p2 equals 10, that's the momentum of each of them. Now, the sum of all momenta is not going to be 10 plus 20, that only works if they're pointing in the same direction. So, what we're going to have to do here is we're going to have to use vector addition to do this and the way that this works is that the total, the final answer p system will actually be, will actually be a vector at an angle as well, this is just vector addition, simple vector addition. So, to find the magnitude of a vector at an angle, we're always going to use the Pythagorean theorem. So, it's going to be px square plus py squared. So, what you're really doing first of all is finding the total px and the total py, okay? And since p system on a vector level is p1 plus p2 then all you do is you do px equals p1 x plus p2 x and py equals p1 y plus p2 Y. So, I'm going to have to find the four individual components of 1 and 2, once I have that I'm going to add them up and find my total px and my total py and those numbers will then go here, okay? So, to find these four guys here, 1, 2, 3, 4 I can just decompose here, okay? I'm going to remind you that px is simply p cosine of theta and py is simply p sine of theta. Now, this works as long as the angle is against the x-axis, right? So, this is a quick a quick reminder of how vectors work, this angle is against the x-axis but this angle here is against the y-axis. So, instead of using that angle, I'm going to use a new angle here, let me make it green and I'm going to use 53 and that's because 90 minus 37, which is the same right here equals 53, I have to get the complementary angle. Alright, so let's decompose this, I don't have tons of space here for the decomposition. So, what I'm going to do is I'm just going to tell you what these numbers end up being but you got to do this in the calculator yourself, right? So, if I decompose this it's going to look like this, this is going to be p2 in the y-axis and this is p2 in the x-axis. Now, p2 and the y axis is going to be 5 sine of 37, on the y axis is sine, okay? So, five sine of 37 is 3, actually I'm so sorry, it's p. So, it's 10 sine of, let me just do one of these, let me just do one of these, I'm going to write p2 y is p2 sine of theta 2. So, it's 10 sine of 37, which is 6, so this is actually 6 right here. Now this is going to be 10 cosine of 37, 10 the cosine of 37, which is 8. Now this is going to the left so it's negative 8. I'm going to do the same thing here, maybe you might want to pause the video and try to this yourself as well, see if you get these the vector decomposition here. So, we're looking for p1 x and all the way up over here, p1 Y, so this is the p, right? I want the p and not the V, p1 x is 20, cosine of 53, which is 12, positive 12, p1 Y is 20 sine of 53, which is positive 16. Now that I have all the four components I'm ready to plug them in here. So, p1 x is positive 12, p2 x is negative 8, this adds up to 4 and the Y's are 6 and 16. So, 16 is the first one and 6 is the other, this gives you 22, once you know this. once you know the PX total and the py total you can actually draw this out and it kind of looks like this, in the x-axis it's a 4. So, px is a 4, on the y axis it is a 22. So, that's what it looks like and this is your total p right here, which is what we want, if you see it forms a little triangle, which is why we're able to use the Pythagorean theorem. So, p is going to be the square root of 4 squared plus 22 squared and the final answer is 22.4 kilograms meters per second, cool? So, let me just separate that over there and that's how you would add momentum. So, it's a very straightforward problem, which is kind of annoying to get to decompose all the stuff but this is essentially a vector problem, cool? That's it for this one, let's go on to the next part.

**Concept:** Two Types of Conservation Problems

Hey guys. So, continually here you might remember I mentioned that collisions are the most typical problem or a type of conservation of momentum problem but there's really two broad categories, so it says, your interaction between two objects will fall into two broad categories one is collisions and this is very straightforward object collide when one goes against the other. Now, they don't have to necessarily go against each other, they don't necessarily have to be head-on, it could be that this guy just has a much faster velocity and that guy maybe isn't moving or maybe it's moving but slower. So, as long as they sort of go into each other in some way it's a collision, the other type doesn't really have a very obvious name. So, I kind of made this up, I call this a push away situation where instead of two objects coming together as they would in a collision, they're pushing away from each other, okay? So, there's some examples here, throwing or recoil. So, if I throw something I'm pushing the object that way but because of Newton's third law, action reaction, as I push the object the object pushing back on me. So, we're both basically pushing away on each other, every time you throw something you're pushing the object the object pushes you back, if you are on frictionless ice and you're throwing something that way you're going to move back, okay? Same thing with recoil, when you shoot a gun you're pushing the bullet that way therefore the gun gets pushed back this way, same thing with mutual pushing, if you have two, the classic examples two ice skaters face each other and they push away and they both move away, if you think about it recoil and throwing are both cases of mutual pushing, when you throw something you're pushing the object and the object is pushing against you, it's just that in the case of skaters they both are people, they're not objects, right? So, they're not innate objects and they're sort of applying force under on each other for your muscles and whatnot as opposed to in a box, if you throw the Box away it's pushing on you because of a reaction force, cool? So, it's a little bit different but you can think of it as being the same way and the last example is an explosion. So, you have an object and if you have a bomb it explodes and it breaks into two parts, one part is going to go this way and the other part is going to go this way or if one part goes that way the other part is going to go directly opposite this way that forms a line, and that's because when the bomb explodes these parts are pushing away on each other. So, all of these are really just different variations of mutual pushing problems where things are pushing on each other. So, typically in a lot of these problems the two pieces or the two things or two people are initially at rest, most of the time, and they will push away on each other here and then they end up going in opposite directions, okay? So, you can have collisions and you can have push away problems in, the nice thing is that they both work almost the same way and we're going to look at both of them, okay? So, in problems that say, where objects are interacting and usually you're just going to have two object interacting, you might get a bomb explosion problem that splits into three fragments or a three way car collision but a vast majority of the time you're only going to have two objects interacting. So, when we have that we're often going to use the conservation of momentum equation. So, whenever you think two objects interacting, collision or push away, which is sort of the opposite of a collision I want you to immediately think of conservation of momentum equation, okay? So, I'm going to show you what the conservation momentum equation is, conservation momentum is the momentum is conserved, doesn't change. So, p initial is the same as p final. Now, because we have two objects, because we have two objects, this is p initial of the system, p final of the system, right? So, this is actually p1 initial plus p2 initial and this is p1 final plus p2 final, it's not that that individual momenta are conserved it's the combination of the two that's conserved. Remember also that p equals m, v. So, my last step here is going to be to replace every one of these guys with m, v. So, instead of p1 initial it's going to be m1, v1 initial and instead of p2 initial it's going to be m2, v2 initial and the other side it's the same thing except I got finals instead m1, v1 final plus m2, v2 final. So, m1, v1, m2, v2, m1, v1, m2, v2, this is the conservation momentum equation, which is the most important equation in this chapter and whenever you have two objects interacting, that's what we're going to use. Alright, so that's the conservation momentum equation, you got to know it, let's do a quick example here, we have two balls moving towards each other, ball A moves with 7 to the right, again, if they're moving towards each other, if you're moving and the ball is moving to the right that ball is coming from the left. So, A 3 kilograms moving with 7 to the right and B 4 kilograms is moving with 5, it doesn't say that it's moving with 5 to the left but you're supposed to figure that out, you can figure that out by knowing that they're moving towards each other, if one is going one way then the other has to be going the other way because they're going opposite directions they have different signs, please don't forget the signs, a lot of people make that very silly mistake, very simple mistake, as soon as I write the 7.5 I immediately put the signs to make sure I don't forget, okay? So, these balls are going to collide, it says, after the collision, so this is before the collision. So, I'm going to put a little hash tag before here, it's an Instagram picture, after the collision p moves with 2 to the right. So, after the collision B is going to move with 2 to the leg therefore it's plus 2, okay? We don't know what's gonna happen within, that's exactly what we want, we will know the magnitude direction of the velocity of A after the collision, after the collision means V final and we're talking about A. So, it's V, A final, what is V, A final. So, do we have two objects interacting? Yes, this is a collision. So, what we do? we immediately jump into the conservation of momentum equation, right up here, let's do that real quick, I'm going to write the long version of this equation first m1, v1 m2, v2, m1, v1 final, m2, v2 final. Now, if you do this a lot it's going to get kind of tiring to write this stuff over and over. So, what I actually do is, I don't write the full version of the equation and sometimes professors want you to start with the full version of the equation and then plug in numbers. So, then you have to do it that way but if you don't have to do it that way, if you can start plugging the numbers right away, what I like to do is I look at the masses 3 and 4 and instead of doing this I just go 3, that's mass one, space 4, 3, 4. So, I save some time by not writing this entire equation I just do 3, 4, 3, 4 and I put spaces for the velocity. So, I now have to figure out, which numbers go here, then I got to look at the picture. So, next to the 3, this is the 3 right here, the initial velocity of 3 is 7, that's a plus 7, the initial velocity of the 4 is a minus 5, the initial, the final velocity right here, of A, I don't have, that's what we're looking for, V, A final, and the final velocity of V is plus 2. So, if you look at this equation I got a bunch of numbers one unkown, I can solve this, we're done, so this is going to 21 plus negative 20 equals 3 v, AF plus 8, right? So, this is 1, this whole thing here's a 1 and then to solve for V, A, I'm going to send the 8 to the other side to go as a negative 3 v, AF and I'm going to skip one step here, this is just algebra, actually I won't skip a step. So, 1 minus 8 is just negative 7 then I have to divide by 3 then the final answer is negative 2.33 meters per second. Now, if you look at the question, the question is asking for the magnitude and the direction and what do you think the direction is? Hopefully you said left, it's just to the left and we know that because of the negative sign, we said that going to the right was positive therefore going to the left is negative, okay? So, this ball going to go this way, which kind of makes sense, if they're going this way and then this goes this way you would expect that A would go to the left.

Now, that doesn't always happen, that's not always the case. So, you can't really assume that, for example, if you have a volleyball and a ping-pong ball and they're going with the same speed towards each other they're not going to bounce and then go back this way, the volleyball is much heavier so it has much more momentum. So, it's probably going to happen is that to the collision, well, actually what's definitely going to happen is that after the collision they're just both going to move that way, the volley ball might slow down by a tiny, tiny bit but they will keep going the same direction, here they had similar speeds and they have similar masses, in fact, the momentum of this guy here is plus 21 and the momentum of this guy here is minus 20. So, one wasn't much more overpowered than the other and in this set case it makes sense that if one goes to the right the other one goes to the left, cool? That's it for this one, let me know if you have any questions.

Adam ( m_{A} = 80 *Kg*) and Barbara (m_{B} = 60 *Kg*) are initially together at rest on an ice rink. When they push each other part, Adam moves with velocity ** v_{A}** = (3 m/s)î - (9 m/s)jˆ. Assuming that there is no friction from the ice, what is Adam’s velocity?

[a] (-4 m/s)î + (12 m/s)jˆ

[b] (4 m/s)î - (12 m/s)jˆ

[c] (-9/4 m/s)î + (24/7 m/s)jˆ

[d] (3 m/s)î - (9 m/s)jˆ

[e] (4 m/s)î + (12 m/s)jˆ

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Margie (of mass 41 kg) and Bill (of mass 65 kg), both with brand new roller blades, are at rest facing each other in the parking lot. They push off each other and move in opposite directions, with Margie moving at a constant speed of 18 ft/s. At what speed is Bill moving?

1. 12.5735

2. 8.4

3. 9.0

4. 13.871

5. 7.61905

6. 8.65574

7. 11.9118

8. 11.3538

9. 10.2857

10. 9.53333

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A man whose mass is 85 kg and a woman whose mass is 50 kg sit at opposite ends of a canoe 5 m long, whose mass is 45 kg. Assume the man is seated at x = 0 and the boat extends along the positive x axis with the woman at the other end. Suppose that the man moves quickly to the center of the canoe and sits down there, while the woman moves quickly 1/3 the length of the canoe towards the man. How far does the canoe move in the water? Assume force of friction between water and the canoe is negligible.

1. 0.972222

2. 0.864865

3. 0.72

4. 1.18056

5. 0.903226

6. 3.29

7. 1.59375

8. 1.13636

9. 0.944444

10. 1.35484

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Two blocks are at rest on a horizontal frictionless surface with a compressed spring of negligible mass between them. Block A has mass 2.00 kg and block B has mass 5.00 kg. The blocks are released from rest and move off in opposite directions, leaving the spring behind. If block B has speed 0.800 m/s after it leaves the spring, what is the speed of block A after it leaves the spring? (Block A is the less massive block.)

(a) 0.63 m/s

(b) 0.95 m/s

(c) 1.00 m/s

(d) 1.26 m/s

(e) 1.50 m/s

(f) 2.00 m/s

(g) none of the above answers

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Blocks *A* and *B* are initially at rest on a horizontal frictionless surface with a spring of negligible mass compressed between them. Block *A* has mass 5.0 kg and block *B* has mass 20.0 kg. The spring is released and the blocks move off in opposite directions. After the blocks, have moved away from the spring.

A) the magnitude of the momentum of block *A* is the same as the magnitude of the momentum of block *B*

B) the magnitude of the momentum of block *A* is less than the magnitude of the momentum of block *B*

C) the magnitude of the momentum of block *A* is greater than the magnitude of the momentum of block *B*

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Boxes *A* and *B* are at rest on a horizontal frictionless surface with a compressed spring of negligible mass between them. Box a has mass 2.0 kg and box *B* has 4.0 kg. When the spring is released the two boxes move off in opposite direction and the spring is left behind. After the boxes have moved away from the spring,

A) the magnitude of the momentum of *A* is less than the magnitude of the momentum of *B*

B) the magnitude of the momentum of *A* is greater than the magnitude of the momentum of *B*

C) the magnitude of the momentum of *A* is equals than the magnitude of the momentum of *B*

D) the kinetic energy of *A* equals the kinetic energy of *B*

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