Push-Away Problems - Video Tutorials & Practice Problems

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1

concept

"Push-Away"Problems

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5m

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Hey guys. So in the last couple videos we sought to use conservation momentum solve problems where objects are interacting well, interactions between objects usually fall into two broad problem types. We've already seen the first one, which is called a collision anytime you have a collision. The basic setup of these problems, you have objects that are moving towards each other and they're going to collide. So we saw a couple of different outcomes of this and we'll talk about that in a later video. What I want to talk about in this video is the second type of problem, which I like to call a push away problem. So we're gonna see a couple of examples of this. Some really common examples include objects are thrown or like the recoil of a gun, which is actually the one we're going to solve down below. We'll also see things that are pushing off of each other, like ice skaters on a frozen lake or astronauts in space or something like that. You might even see some problems involving fireworks, explosions. But the general setup, the reason I like to call these push away problems is because objects are initially together and what happens is they're going to push away from each other and then afterwards are going to be moving in opposite directions. And really this is just because of action reaction. But what's fortunate for us is that regardless of the problem type, whether it's collisions or push away problems, we're gonna solve all problems by using conservation of momentum. Any problems involving interacting objects can be solved by using conservation momentum because momentum always has to be conserved. That's a sacred law of the universe. Whether it's collisions or push away, momentum has to stay the same from initial to final. So, let's take a look at our problem here. I'm gonna go ahead and skip this for just a second. We're gonna come back to it in just a second. All right, so, we have a four kg sniper rifle that shooting a five g bullets. So, we're gonna go ahead and draw our before and after. Basically, what's happening here is I've got this little gun like this, that's my little gun, and I've got the bullet that's inside. Now, this is before. But then afterwards, what happens is that the sniper rifle actually fires the bullets? Right, So, I've got my gun like this, and then I've got the bullet that's going away like this. So, I've got the mass of the gun which is four kg, and I've got the mass of the bullets that's inside. This is equal to 0.005. Now, what happens is we know that the initial velocity of the before a case before the bullet is actually fired out of the barrel is actually gonna be zero, right? The initial speed for both of these objects, the bullets sitting inside the gun and the gun is not moving anywhere. So both of these objects have an initial speed of zero afterwards. What though is I have the two finals equals to 600 m per second like this. What I want to do is I want to figure out what is V one final. What is the recoil speed of the gun? So that's the first step drawing the before and after. Now. We were at my my conservation of momentum. M1 V1 initial plus M two V two initial equals M one V one final plus M two V two final. All right. So, we're looking for here is we're looking for how do we solve for this V one final right here. So let's take a look at each of our terms, you know, the masses of all the objects. Now, I just need to know the speeds well, like we just said, the initial speeds for both of these objects, the bullet and the gun is just equal to zero. What does that mean? It just means that for both of the terms on the left side they're actually both going to equal zero. So I end up with zero momentum on the left side. In fact, this is going to happen in a lot of your push away problems and most of your push away problems objects are gonna be initially at rest, which just means that the total momentum of the of the system initial Is equal to zero. So that actually means that the conservation of momentum equation is going to simplify for us. Let's check out how, so we have zero equals M one V one final plus M two V two final. If you start out with zero momentum you have to end with zero momentum. So how do I solve for this? V one final? Basically I can just move this to the other side like this and what you'll end up with is negative M one V one final equals M two V two final. If you have zero momentum, initial and final and if you have these things that are moving in opposite directions, it means that the magnitude the momentum's have to be the same. They're just in opposite directions. That's what this negative sign means. So this is gonna be the general setup negative M1 view and final is negative em to the two final here. So we can solve for this V one final and we can say that V one final is equal to negative. Now we're just gonna plug in the numbers M two is equal to 0.5 Your V two final is 600 you're gonna divide the mass, which is four. If you go and work this out, you're gonna get negative 0.75 m per second. So as expected, what happens is that the gun has actually been recalled to the left. The bullet was traveling to the right at 600 m per second and so it has some momentum in this way. So the gun has to recall backwards and its speed is 0.75 m per second. Basically what happens is if the gun or if the bullet has gained 10 momentum to the rights, the gun has to recoil and gained 10 momentum to the left. I'm just using 10 as an example there. Right? So that's the speed. It's negative 0.75. All right. So that's it for the problem. I have one last sort of conceptual point to make which is that momentum is conserved. Like we had in this example here only if your system is isolated. So we've seen that word before isolated. Just means that all the forces that are acting in your system or in your problems are internal. So remember we define the system as the gun plus the bullets. And if you look at the forces, what's happening is that the gun exerts a force on the bullet to the right. That's what shoots it out of the barrel. But because of action reaction, the bullet has to exert a force backwards on the gun. Now, even though there's two forces here, if you define your bubble to be the gun and the bullets, then these forces are internal, which means that momentum is going to be conserved. If I had an external force, if I actually put my hand against the gun with a bullet inside and pushed it, that's an external force that's coming from outside of the system and therefore momentum is not going to be conserved. All right. So in your problems where you define your system as both of these interacting objects, these forces are usually gonna be internal, which means your momentum is going to be conserved. That's why we can use our conservation momentum equation. All right. So that's it for this one. Let me know if you guys have any questions.

2

Problem

Problem

An 80-kg astronaut is stranded floating in space is 30 m away from his spaceship. He wants to return to his spaceship in 20 s. How fast must he throw his 2-kg space hammer, directly away from the spaceship, to accomplish this?

A

40 m/s

B

120 m/s

C

240 m/s

D

60 m/s

3

example

Sliding Football Player

Video duration:

3m

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Hey guys, let's take a look this problem here. Hopefully get a chance to work it out on your own. Not here's a little bit help. So the idea here, we have a football player on frictionless ice who has a football. So they're both moving along together. He's got this football like this and they're both moving at two m per second afterwards. Them the player is actually going to throw the football ahead of about 15 m per second. So this is gonna be a push away problem. Both objects are together and now they're moving separately afterwards. Let's go ahead and draw our before and after. So this is our before, what is our after look like? Well, basically, now what happens is you have the football player like this and now he's throwing the football ahead of him And we actually can figure out what the speed is. So what I'm gonna do is I'm going to call this M1 which is 70 and I'm going to call this M2, which is 0.45 And we're told that the player is going to throw the ball with 15 m per second relative to the player. What does that mean? Well, this is gonna be my V. Two final here. What this means is that they are both initially were already moving at two m per second. Then the football player throws it an additional 15. So what actually happens is this is gonna be two plus 15 and that's going to equal 17. That's the new final velocity. We want to figure out what's the players velocity after that happens after this push away. All right. So let's go ahead and set up our momentum conservation equation. This is going to be M one V one initial plus M two V two initial equals M one V one final plus M two V two final. Okay, so basically I've got this 70 times something plus 0.45 times something equal, 70 times something plus 0.45 times something. All right, so, what what goes inside these parentheses? Remember initially they're actually both moving together. So these speeds are actually the same and they're both moving with initial two m per second. So that's what goes inside here. So that's two and two afterwards, we're looking for the final velocity of this V once. This is actually what goes inside here. That's my target variable. What about the football? Remember the football we just calculated is moving at 17 m per second forwards. That's what goes inside here. Now, we just go ahead and simplify the left and right sides. So that this becomes here is this becomes um let's see on the left side to get 140. And on the right side, I'm gonna get 70 times v one final plus. And then this is going to be 7.65. So all they do is just move and subtract the 7.65. I end up with 133.2 25 equals 70 times V one final and I can just go ahead and divide by the 70 once you get is you get V one final is equal to 1. m per second. So let's see basically what happens is that the football player is still moving to the rights just slightly slower than he was initially. Initially. Both objects were moving at two m per second. And then because the football is a lot lighter, the fact that it's going at 17 m per second now means there hasn't been a lot of recoil and the much heavier football player. So the football player has a final velocity of 1.9, so he's only slowed down a little bit, but he's given a lot of momentum to the football. All right, so that's the idea here, guys, let me know if you have any questions and I'll see the next one.

4

concept

Push-Away Problems with Energy

Video duration:

6m

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Hey guys, on the last couple videos we saw how to use momentum conservation to solve these kinds of push away problems. We have objects that are pushing away from each other. But in some of these problems you're actually gonna be asked for energy before or after a push away events. For example, this example we're gonna work down here has these two boxes that are pushed up against this spring. And in part B we want to calculate how much energy is stored inside of that spring. So I'm gonna show you really quickly have to deal with these push away problems with energy. And the main idea here is they're actually going to use to conservation equations. We're going to have to use momentum. We're always going to use momentum conservation with these problems, but we're also going to have to go back and use energy conservation as well. Let's go ahead and take a look at our example here, we've got these two blocks, three and four kg and their push up against this light spring and you release it in the spring basically is going to fire both of them off in opposite directions. So the blocks are gonna get released. So we want to draw a diagram for before and after. This is the before and afterwards. What happens is you sort of have these boxes that start pushing away from each other, like this, this four kg box goes like this and the three kg box is now going to the left. We actually know what the final velocity of this three kg block is, it's launched at 10 m per second. So what happens is I'm gonna call this M one and M two, which means that the one final is equal to negative 10 because it's going to the right, the left and then V two final here is actually what we want to solve for in part a the recoil speed of the four kg block. So this is where you look for in part A. And basically we're just gonna use momentum conservation. We know how to solve these kinds of push away problems. So once we ever before and after we're just going to write our momentum conservation. So for part A We're going to do M1 V1 initial plus M two V two initial equals M one V one final plus M two V two final. Now remember what happened to these kinds of problems? Is that initially the system is at rest. So right after you release your hand, the box, the blocks are still actually at rest, which means that the the initial velocity is equal to zero. What that means is that basically both of our left terms are actually going to go away and we have zero initial momentum. So the system's initial momentum is equal to zero. And that means that our momentum conservation equation is going to simplify because we know that the final momentum has to be zero as well. So basically what happens is we're gonna have negative M1 V1 final equals M two V two final. If the right block gains 10 momentum this way, the left bloc has to gain negative 10 momentum that way. And so they have to cancel each other. So we can we can basically solve for this V two final here and we can start plugging in values. So I have negative and what I have here is the mass is three and the velocity is 10. But is it 10 or is it negative 10? Remember we have to input this negative sign when we plug this in because it's important to keep track of the science. So we have three times negative 10 divided by four, which you end up getting here is that V two final is equal to 7.5 m per second. So notice how we got a positive number. And that should make some sense if we actually got a negative number. So if we forgot one of these minus signs, we would have gotten a negative number and that would have meant that this block is actually going to the left and that doesn't make any sense. So it's always a good idea to keep track of your numbers and make sure that makes sense. So this is a positive answer. Positive 7.5. And that's just using momentum conservation. Let's take a look at Part B. Now, part B is asking just to figure out how much energy is stored inside the spring. So what's going on here is that you've had these blocks that are pushed up against this spring and there's some elastic potential energy that is stored inside of the spring because they're compressed once you release. What happens is that these two blocks start firing away from each other, right? They separate like this and the spring decompress is which means that the final velocity you start, the final potential energy is equal to zero. So what happens is there's some stored energy here and then this spring is no longer storing the energy. Where does that energy go? Well, basically what happens is that you have these two blocks that have now mass and speed and therefore there's some kinetic energy. This is K two final and K one final. So basically we're going to use energy conservation to figure out the energy instead. So you're going to use K. Initial plus you initial plus work done by non conservative K final plus you final. Now what we said here is that the initial velocity of the system is zero. So there's no kinetic energy. The initial energy that's stored here is going to be elastic potential energy. There's no gravitational because everything is sort of on the horizontal plane like this. What about work done by non conservative. Once you release your hands and the boxes start flying this way, there's no work done by you and there's no work done by friction as well. So we're on a smooth floor. And then finally what happens is we have some kinetic energy because the boxes are moving but we have no potential energy because this spring has decompressed. So what happens is we have you elastic is going to equal the kinetic energy. Now what happens is we have two blocks that are moving. So the kinetic energy is actually gonna be one half and one V. One final squared plus one half. Em to the two final squared, the kinetic energy is made up of two objects that are moving now. So we actually have all those numbers can go ahead and calculate this. So this you elastic here. The stored energy inside the spring is one half times four. I'm sorry, this is mass one. So this is gonna be three times negative 10 squared and it becomes positive once you square it plus one half times four times um 7. squared, which end up getting is 262.5 jewels. So that's the energy that was stored inside of the spring. Let's take a look now at part C. Which is the only want to figure out how much how much the spring was compressed before launching the blocks. So we we just calculated was we just calculated the elastic potential energy 262. jewels. What we want to do now is we want to calculate the compression distance which is really just gonna be X. Initial. So we can do is we can just relate the the the elastic potential energy to one half K. X. Squared. So we're looking for here is X. Initial. And we have The elastic potential energy and we also have the spring constant. It's 800 uh newtons per meter. So we can just go ahead and I'll just go through as quickly. This is two times you elastic divided by K. And we're gonna square root that that equals X. Initial. So when you plug this in, what you're gonna get is 525 divided by 800. And you're gonna get 0.81 m. That's final answer. So these are all your answers, right? This is a use, momentum and energy conservation to solve these kinds of problems that this one guys let me know if you have any questions.

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