Professor Anderson

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>> I had a question. >> Joseph did you have a question? >> Yeah, I had a question about one of the homework problems on where it was a spaceship and it got blasted apart in three different parts and how we set up the equation of the different integers. Because I think one of them is going backwards, so I wasn't sure if I was supposed to add or subtract it. >> Okay. I think that's an excellent question and a great place to start. So let's take a look at explosions, alright. We're talking about, of course, conservation of momentum. And what we discovered earlier is that in collisions, things are conserved, momentum is conserved. But an explosion is just sort of like a collision in reverse, right. So if I think about two cars coming together and crashing in a collision and sticking together, if I run that tape backwards, it looks like the cars separate and blow apart. So an explosion is really just like a collision in reverse. Let's talk about explosions with the idea of conservation of momentum. So let's say we do the following. Here's our rocket ship. And it's flying along in this direction, okay. And it's going at some initial speed VI. And we will say that it has mass M. Okay. And now there's an explosion, and the rocket ship blows apart. And it goes into three pieces. So here's one piece. That's the tail section. And we'll say that it goes back in its original direction. We'll draw the explosion here. And then we have a chunk of a rocket that's going to head off in this direction. And we have another chunk of the rocket that's going to head off in that direction. Okay. This is the explosion of the rocket. Now, we need to know a little bit about some of these pieces, okay. So let's call this piece M1, M2, and M3. And let's say that this is now V1, this is V2, and this is V3. And let's see if we can set up this equation that governs this explosion, conservation momentum equation in terms of all these parameters. Okay. So if this is where the explosion happens, it probably make sense to label this the X axis. We'll label that the Y axis. And then we need to define some angles here. Let's call this one theta 2. And we'll call this one theta 3. And this one is directly back along the negative X axis, okay. So this is what the problem looks like. And if you have numbers, you can worry about putting in the numbers at the end. Let's now analyze this from the point of view of conservation of momentum. So, in our before picture here, what can I say about the initial momentum of the system? Well we know that momentum is just mass times velocity. Everything is heading in the positive X direction. So this is just M times VI. Alright. In the final picture, over here, we have a couple different things, right. We have P final in the X, and we have P final in the Y. And what I know is that conservation and momentum tells me the following. P initial in the X is equal to P final in the X. P initial in the Y is equal to P final in the Y. Now what I wrote up here, is that P initial in the X or is that P initial in the Y? Joseph, what do you think? >> Is it both? >> Well, this thing was really heading horizontal, so it's just. >> Just X. >> X, right. P initial in the X. P initial in the Y is therefore what? >> Zero. >> Zero, right. This thing is not going up or down. It's going perfectly horizontal. Okay, so these are the initial conditions. And now down here we're going to write the final conditions. And then we'll worry about setting them equal and solving. Alright. What do I do? Well, P final in the X. It looks like I have one, two, three components that are going to have some contribution in the X direction. And we'll just go through them one at a time. So, M1V1 is this guy right here. But it's heading in the negative X direction. And so I put a negative sign in front of it, right. This is always a velocity, so the direction is negative. What about M2V2? M2V2 is the momentum of this particle, but it's not entirely in the X direction. There's some component in the X direction. So Joseph, is that component in the X direction, is that sine or cosine? >> Cosine. >> Cosine, right. This is my triangle right here. And so this side of the triangle, since this is the right angle, it's going to be cosine theta 2. And likewise for M3 we're going to have M3V3 cosine theta 3. Alright. That looks complete for the X. Now for the Y, what do we have? Well this one is perfectly horizontal we said. So there's no contribution from this one in the Y. But both of those have some contribution in the Y. And it looks like we're going to have M2V2, since we used cosine before, we're going to use sine, which is this side of the triangle. And that's going up, so we keep everything positive. And we have M3V3 going down, so we put a negative sign in front of it. So, these are the important conditions. There's our initial conditions. Here are the final conditions. And we need to satisfy conservation of momentum by setting them equal according to this. Okay. So, we're not going to bother doing all the math here. But this is how you set up the problem. And you would then go to this step right here to solve it. Alright, questions about that? Did that make sense, Joseph? >> Yeah. >> Okay. Everybody else okay with that one? Alright.

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>> I had a question. >> Joseph did you have a question? >> Yeah, I had a question about one of the homework problems on where it was a spaceship and it got blasted apart in three different parts and how we set up the equation of the different integers. Because I think one of them is going backwards, so I wasn't sure if I was supposed to add or subtract it. >> Okay. I think that's an excellent question and a great place to start. So let's take a look at explosions, alright. We're talking about, of course, conservation of momentum. And what we discovered earlier is that in collisions, things are conserved, momentum is conserved. But an explosion is just sort of like a collision in reverse, right. So if I think about two cars coming together and crashing in a collision and sticking together, if I run that tape backwards, it looks like the cars separate and blow apart. So an explosion is really just like a collision in reverse. Let's talk about explosions with the idea of conservation of momentum. So let's say we do the following. Here's our rocket ship. And it's flying along in this direction, okay. And it's going at some initial speed VI. And we will say that it has mass M. Okay. And now there's an explosion, and the rocket ship blows apart. And it goes into three pieces. So here's one piece. That's the tail section. And we'll say that it goes back in its original direction. We'll draw the explosion here. And then we have a chunk of a rocket that's going to head off in this direction. And we have another chunk of the rocket that's going to head off in that direction. Okay. This is the explosion of the rocket. Now, we need to know a little bit about some of these pieces, okay. So let's call this piece M1, M2, and M3. And let's say that this is now V1, this is V2, and this is V3. And let's see if we can set up this equation that governs this explosion, conservation momentum equation in terms of all these parameters. Okay. So if this is where the explosion happens, it probably make sense to label this the X axis. We'll label that the Y axis. And then we need to define some angles here. Let's call this one theta 2. And we'll call this one theta 3. And this one is directly back along the negative X axis, okay. So this is what the problem looks like. And if you have numbers, you can worry about putting in the numbers at the end. Let's now analyze this from the point of view of conservation of momentum. So, in our before picture here, what can I say about the initial momentum of the system? Well we know that momentum is just mass times velocity. Everything is heading in the positive X direction. So this is just M times VI. Alright. In the final picture, over here, we have a couple different things, right. We have P final in the X, and we have P final in the Y. And what I know is that conservation and momentum tells me the following. P initial in the X is equal to P final in the X. P initial in the Y is equal to P final in the Y. Now what I wrote up here, is that P initial in the X or is that P initial in the Y? Joseph, what do you think? >> Is it both? >> Well, this thing was really heading horizontal, so it's just. >> Just X. >> X, right. P initial in the X. P initial in the Y is therefore what? >> Zero. >> Zero, right. This thing is not going up or down. It's going perfectly horizontal. Okay, so these are the initial conditions. And now down here we're going to write the final conditions. And then we'll worry about setting them equal and solving. Alright. What do I do? Well, P final in the X. It looks like I have one, two, three components that are going to have some contribution in the X direction. And we'll just go through them one at a time. So, M1V1 is this guy right here. But it's heading in the negative X direction. And so I put a negative sign in front of it, right. This is always a velocity, so the direction is negative. What about M2V2? M2V2 is the momentum of this particle, but it's not entirely in the X direction. There's some component in the X direction. So Joseph, is that component in the X direction, is that sine or cosine? >> Cosine. >> Cosine, right. This is my triangle right here. And so this side of the triangle, since this is the right angle, it's going to be cosine theta 2. And likewise for M3 we're going to have M3V3 cosine theta 3. Alright. That looks complete for the X. Now for the Y, what do we have? Well this one is perfectly horizontal we said. So there's no contribution from this one in the Y. But both of those have some contribution in the Y. And it looks like we're going to have M2V2, since we used cosine before, we're going to use sine, which is this side of the triangle. And that's going up, so we keep everything positive. And we have M3V3 going down, so we put a negative sign in front of it. So, these are the important conditions. There's our initial conditions. Here are the final conditions. And we need to satisfy conservation of momentum by setting them equal according to this. Okay. So, we're not going to bother doing all the math here. But this is how you set up the problem. And you would then go to this step right here to solve it. Alright, questions about that? Did that make sense, Joseph? >> Yeah. >> Okay. Everybody else okay with that one? Alright.