Ch 09: Momentum & ImpulseWorksheetSee all chapters
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Concept #1: Push-Away Problems

Transcript

Hey guys. So, continuing here with conservation of momentum, I have a few extra things to talk about, we'll do a few more problems as well. So, when, remember that we only have two objects, two or more objects interacting, the momentum of the system is conserved, so the two main types of interactions of two or more objects is collision, when two things are going towards each other or push away when two things are going or pushing away from each other those are the two big categories. Now, the key idea here is that the momentum that is conserved is the total momentum of the system, the individual momentum of, the individual momenta are likely not conserved, okay? So, when two things collide their individual velocities will change. So, let's say goes like this and then after the collision goes like that the initial velocities will, the velocities will change of the individual object but the momentum of the system is conserved, okay? So, the individual will not be conserved because it's the total that is conserved. Now, this idea that in every interaction between two or more objects that momentum is conserved only applies for a system of two or more objects, if you have just one object the momentum is not necessarily conserved. So, in this case you have to look at the situation. So, I have a quick example here, two kilogram object is falling, ignore air resistance, is momentum conserved? The answer is no and I'll show you why. So, let me go over here, if you have a 2 kilogram object falling is momentum conserved? Well, momentum is m, V, the mass stays the same but as you fall your speed is changing. So, I'm going to put a little Delta here, your speed is increasing and your velocity is technically decreasing because it's going down, but because your velocity changes your momentum will change as well okay? So, for a single objects p changes, if V changes, okay? So, that's that idea, let's keep going here.

In many pushaway problems. So, we did a simple collision problem, I'm going to talk about pushaway problems, in many of those problems and most of those problems, in fact, the system is initially at rest. So, for example, you got to block you got two objects and then let's say, actually let's say you throw something, right? So, if you throw something you at rest and the object is initially at rest, when you throw it that way that object gains momentum and because of action-reaction the object pushes back on you and then you gain momentum back. So, that's the idea but, initially, initially, meaning before you threw it, before the event, everything is at rest, that's how it's going to be in most cases, what that means that the conservation of momentum equation is going to look a little simpler. So, let's do this example and I'll show you what I mean, it says here, find the recoil speed of a 4 kilogram rifle, if it shoots a bullet of this mass, with this speed. So, let me draw a little rifle here, I can't draw rifles but let's go with that, mass 1 is 4 kilograms, the bullet is going to come out this way, mass 2 is 0.005 kilograms and what else do we have? the muzzle speed is a speed with which the bullet leaves the gun. So, it's v2 final is 600 and the recoil speed is the v1 final and that's what we want to know, this is an interaction, it's a push away interaction between two objects. So, as always, we're going to use the conservation momentum equation, I'm going to write the full version here, m1 v1 initial plus m2 v2 initial equals m1 v1 final plus m2 v2 final m1 v1, m2 v2, m1 v1, m2 v2, so the thing here is initial means before the event, in this case before you took the shot, and final means after. So, imagine the gun before you shoot the gun has no velocity and the bullet is inside of the gun has no velocity either. So, both of these things are 0, what that means is that I can rearrange this equation and make it a little bit simpler. So, let's write this again, I'm going to move this guy here to the left so it becomes a negative. So, it's going to look, well, let's just do it like this m1, m2 v2 final equals the negative of m1 v1 final. So, you still start from the conservation of energy equation but it becomes simpler, when I said it's simplified is because instead of having four elements it just has two elements, instead of having eight variables it just has four variables, okay? So, that's the equation, that's a little bit simpler, the idea here, just one more time is that initially there's no momentum and momentum is conserved. So, if I give, if the object, that's going this way gains 10 of momentum this object has to gain 10 momentum this way, the momentum was 0 and now it's 10 plus minus 10 or the other way around and these two numbers cancel, you can't create momentum out of thin air. So, if you create positive momentum this way there has to be a negative momentum that pushes you back, cool? That's the idea. Now, we can just plug this in here, we're looking for v1 final. So, v1 final equals m2 v2 final minus, divided by negative m1, and the masses are 0.005 divided by 4 and all of that times the 600 and if you plug everything you get a negative, I got this here, negative 0.75 meters per second why did we get a negative? The reason why we get a negative is because the recoil speed is to the left, if you're shooting to the right that gun is going to recoil to the left so that makes sense, okay? That's it for this one very, very straightforward problem, again starts from 0 they push away, so the initial velocities are 0 but we to go with the momentum equation, it just simplifies a little bit. Alright, I've got a practice problem here for you guys to try, let's give that a shot.

Practice: An 80-kg astronaut is floating in space is 30 m away from his spaceship. He wants to return to his spaceship in 10 s. With how much speed must he throw his 2-kg space hammer, directly away from the spaceship, to accomplish this?

Example #1: More Push-Away Problems

Transcript

Hey guys. So, I'm going to show another example of these pushaway problems but in this one we're not going to have numbers, we're going to solve this with letters, it's a literal problem. So, here we have two skaters of masses M and 1.3 M those are the masses that are at rest, and this is what I mean by literal, there's no numbers. So, instead of having something like 100 you're just going to have M, if you had 100, 1.3 M would have meant 130 but we don't have masses but that's fine, we can solve with this. So, we're just going to plug those in instead, when they push against each other for T seconds. So, they do this for an amounts of time delta t equals big T, the lighter moves back with the speed of V, let's draw this real quick. So, push against each other pretend these are people and they're pushing against each other. Now, the problem doesn't say who's left or, right? So, I'm going to put M here, 1.3 here, just doesn't matter the direction, the arrangement. Now, they're going to push and then this guy's going to go back this way with the speed of V. Now this is after the events, after they pushing each other so this is V, let's call it 1, v1 final, let's call this guy 2, this v1 final is big V and because it's to the left, I'm going to add a negative here. Now, I've written, I've drawn them differently, it would have been, if this, if M was here, it would be going to the right. So, it would've been positive v, the second mass v2 final has unknown velocity after the push and that's what we want to know, what is the speed of the heavier skater, we want to derive an expression, which means instead of a number, we're going to have some letters M then V's, okay?

The way, we're going to do, this is by using conservation of momentum since this is a push away problem. So, I'm going to write this equation here, real quick, m1 v1, m2 v2, initial m1 v1, m2 v2 final, next thing I'm going to do is replace the masses m1 and m2 with these masses here. So, big M, and I'm going to leave spaces for the velocities, put a little m here or big M rather. Alright, so now let's put the velocities, the initial velocity here is 0 for both of these guys, because initially they're just look at each other at rest. So, there's no initial velocity the final velocity however of the first one is right here, it goes to the left with negative V and the second one, we don't know. So, I'm going to put V2 final and that's actually, what we want to know, we want to calculate that or derive an expression for, okay? This whole thing here, becomes 0 and now we're going to move everything out of the way so that V2 final is by itself, it means that this goes to the other side as a positive and this goes to the other side dividing, cool? Notice that the masses cancelled and you're left with the final answer, which is that V2 final equals big V over 1.3, the answer was supposed to be, meters per second, the answer was supposed to be in terms of M and V, it doesn't mean that all the variables have to show up it just means that no variables other than these could show up. So, this should make sense because the heavier object, this is the heavier object, ended up being slower by the same factor. So, it's heavier by a factor of 1.3 and end up being slower by a factor of 1.3 and that should make sense. Alright, so Part B is a little bit more tricky, we want to know what is the average force. Now here, I hope the word average kind of gives you a hint that you should be using impulse. Remember, impulse J is force average, time delta t equals the change in momentum. Remember, change of momentum means p final minus p initial, in this particular problem there is the momentum initial because the two objects are moving, right? They're initially at rest. Now, the two of them are pushing each other with the same force, you can think of this as F1 and F2 but they're really the same magnitude. So, I can just think of it as F, okay? F. Now, they are pushing on each other and what makes this confusing is you have to look at a single object to be able to do this. So, what we're going to do is we're going to say, impulse on object one, on one, we're going to go with one because one has a simplest mass, the simpler mass, just M's that are 1.3M. So the impulse on 1 is the force on 1 delta t and it is the final momentum of 1 so the force on you gives the impulse of you, gives your impulse, okay? I'm looking for this, so the force on object 1, the time is big T, and this is going to be M since it's 1 it's going to be m1 v1 final. Now, let's replace these numbers F on 1, the mass of one is big M, the final velocity is negative big V and we divide everything by T, so the force applied on the first object is negative M, V over T. Now, that's the final answer negative is because it is to the left, we probably just want the magnitude. So, let's just say the magnitude of the force average of them on each other is M, V over T, okay? Either one is fine, this guy here gets pushed this way with a positive force, right? So, it'd be positive M, V over T, that's it, cool? So it's a little bit tricky, a little bit unusual but I wanted to do this one so you guys know how to do it. Alright, that's it.

Example #2: More Push-Away Problems

Transcript

Hey guys. So, I'm going to show another example of one of these pushaway problems. Here we're going to have three objects check out. So, we have a 60 kilogram boy that is on a 2 kilogram skate holding a 5 kilogram ball. So, there's three things here and let me draw that real quick, let's got a dude here and he's holding the ball, the guy is 62 kilograms, the skate is 2 kilograms and the ball is 5 kilograms everyone is moving to the right, actually it doesn't says it's to the right but I'm just going to make it up here, that's what you would do, with initial velocity of 6, so the ball, the dude and the skates and that's because everyone's together, you're on top of the skate so the skate is moving with you and you hold the ball so the ball is moving with you, okay? Now, then he throws the ball forward with 10 meters per second, that's the ball's final velocity. So, here's a ball and I'm going to say V final for the ball is 10, he throws it forward the direction is moving so it's also positive, and what we want to know is what is his speed relative to the floor after he throws the ball. So, I want to know what is V final of the boy. Now here, we have three objects. So, you could, one way you could do this, is you could write the momentum equation for three objects, m1 v1 plus m2 v2 plus m3 v3 equals m1 v1, m2 v2, m3 v3, you don't actually have to do that, it's a little bit too much work, what you could do is because the boy never really separates from the skate and we know this because it doesn't say that he falls see you'll assume he stays on, we can treat the two of them as one single thing, one object, one body, okay? Of mass 62 kilograms and that's what we're going to do and so that's one thing that guy plus skate and the other thing is the ball, cool? So, then these two final velocities are really the same alright. Now, let's let's start it. So, m1 v1 plus m2 v2 equals m1 v1 plus m2 v2 and I'm going to plug in the masses, they are 62 and 5, 62 and 5, then we're going to put the initial velocities here, the initial velocities are 6 and 6, right? Everyone starts at 6 and after that the throw the ball, the ball is moving with 10. So, here is the ball plus 10, this is what we're looking for, which is the final velocity of the guy, which happens to be the v velocity skate because they're moving together, I got all the numbers so this is just a matter of plugging stuff in, really easy question, I just wanted to go over this ideal having three things together and merging two of them, let's wrap it up here. So, this is going to be, this whole thing here is 4 and 2, this is 50, which I can send over to the other side equals 62 v final, v final therefore is 352 divided by 62, which is 5.67 meters per second, I got a positive, which means it's going to the right, okay? So, the final velocity here is 5.67, if you think about it, the ball is moving with 6 and now it's moving with 10. So, you're going this way and you are throwing a ball that way? Well, what happens? We got a little bit of a recoil effect by throwing the ball that way, the ball action-reaction pushed against you and you come back a little bit, but you're moving with 6. So, what that little pushback does, it slows you down a little bit and actually it only slows you down with tiny bit, like 20, like a few percentage points off your speed and that's because of the mass difference, this is 62 and the ball is only 5, so the ball doesn't make that much of an impact, if you're throwing a heavier ball you would slow down more. Alright, so that's it for this one, hopefully that makes sense, let me know.

Example #3: Walking on a Canoe

Transcript

Hey guys. So, in this video I want to talk about a very particular kind of conservation momentum problem, which is, when you have someone walking inside of a canoe, that's on the lake. Now, like I said it's very particular. So, you might want to check if your professor even cares about this kind of problem so the basic idea is that, if someone's walking on a canoe that's not moving and you're walking you're stepping on the canoe and the way you walk is by pushing yourself forward. So, what's going to happen is as you walk forward the canoe is going to start to move back relative to the water. So, if you look at the water, if you look from far it's going to look like the canoe is going to the left meanwhile you're going to the right, these are conservation of momentum problems because you have two objects that are interacting with each other, that they're mutually pushing each other so we use conservation of momentum to solve this, let's check it out. Alright, so you have a 50 kilogram girl on a 90 kilogram canoe, let me draw this real quick, I got a canoe here and got a girl here and put a little hair, girl 50 kilograms, canoe 90 kilograms and it rests on calm water, okay? So, got some water here, like a kid, alright so the girl then walks three meters to the right of the canoe with a constant 2 meters per second squared. So, she's going to walk this way with 2 meters per second and she's going to walk a distance of Delta x equals 3 meters, I'm going to squeeze it in there, okay? Now, the first question is what is the magnitude of the direction, of the velocity that the canoe will attain. Remember, I mentioned that if you're walking this way the canoe starts walking that way. Now, how do you walk faster, we walk faster by running, and the way you do that is by pushing harder against the floor by putting more force into your legs push harder or if you do that guess what, the canoes also going to move faster in the opposite direction. So, to figure out that velocity, we're going to use the conservation momentum equation. So, m1v1 m2v2. Alright, so m1 v1, m2 v2, m1 v1, m2 v2 final, let's say the girl is 1 and the canoe is 2 you see that you can barely see that, m1 equals 50 alright, so the masses are 50 and 90, 50 and 90, initially their velocities are 0 because the canoe is not moving and the girl is not moving, this is right before the girl starts moving, then the girl starts moving with 2, I drew it where the girls go into the right. So, this has to be positive 2 as a result the canoe is going to move backwards. So, it's going to be a negative number, v2 final is what we're looking for, okay? This is all 0, 0, this is 100, which I can take it to the other side. So, v2 final is 1.11 negative 1.11, the girl moves to the right with 2 the canoe moves to the left with 1.11 the velocity canoe is lower than the girl because it canoe is heavier. Remember, heavier usually correlates with slower. Alright, so Part B. So, that's part A and we use conservation of momentum just like, we've used with other types of conservation with the problems.

The question that for Part B now is, how much does the canoe move relative to the water, so the girl moved 3 inside of the canoe, how much does the canoe move to the left? Well, to find this we have to first, so this is going to be a constant velocity problem, this is constant velocity. So, we're going to use V average or V constant equals Delta x over delta t and what we can do here is because the time that the girl is moving to the right is the same time that the canoe is moving to the left I can write the equation twice here and combine, let me show you, so the idea is that v1, which is the girl equals Delta x of the girl divided by time and v2 of the canoe is Delta x2 divided by time. Now, these times are the same, right? These times are the same, and what I'm looking for is the Delta x of the canoe, I have both velocities, I have Delta x1. So, what you can do is you can use this information to find t and then plug t here and solve for Delta x, okay? That's the way to do it, you have to find out first how long does it take for the girl to walk 3 with 2 meters per second and then you can figure out how long look, how much the canoe moved in that time period, okay? So, this velocity here is true this Delta x is 3, the time therefore is, I got to move some stuff up here to T equals 3 therefore T equals 3 over 2, t equals 1.5 seconds, this is for the first equation. Now, for the second equation, I have v2, which is the velocity of the canoe, which is negative 1.11 equals Delta x 2, that's what I want to know designed by time, I just got time 1.5, therefore Delta x 2 is negative 1.11 times 1.5 and if you multiply this you get negative 1.67 meters, okay? So, as the girl moves 3 meters this way the canoe moves 1.67 meters to the left, it should make sense that it moves less to the left because it is slower, okay? Less because it's slower. Alright. Part C is probably the trickiest part. Now, it asks, how much does the girl move relative to the water, so this is sort of a very strict follower, it's kind of simple but nonetheless still a relative motion problem relative velocity problem, so the idea is that the girl walks 3 inside of the canoe but as the girls walks inside the canoe the canoe is moving to the left, so the girl doesn't you doesn't really walk 3 relative to the water. So, if you draw this it's probably the easiest way to do it so the girl is going to walk 3 meters this way but while the girl is walking 3 meters the canoe also walk this way 1.6t7 meters to the left. So, let me put a little negative here, so the combined effect of the girl moving 3 while the canoes moving 1.67 is that the girl doesn't actually move relative to the water the entire 3, the Delta x of the girl relative to the water, is she goes 3 to the right but then the canoe takes her back 1.67, okay? Let me disappear here, which means she, relative to the water she only walked 1.33 meters, okay? Now, if you see a question like this you're not necessarily going to get these three parts but I wanted to show you sort of the different possibilities. So if you get something weird like this you're ready ,it's a different type of question from the more traditional ones but it's pretty easy, if you've seen it once at least. Alright, that's sort of what I wanted to do with this question. Alright, so hopefully this makes sense, let me know if you have any questions.