Two ice skaters, Daniel (mass 65.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of 53.1° from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink. (a) What are the magnitude and direction of Daniel's velocity after the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

Video transcript

Hey everyone, welcome back in this problem. We have science students who want to do an experiment. Okay, they're gonna be on the horizontal frictionless rink. We have a student, john is gonna be standing a student James is gonna skate and collide with johN okay. And we're told some information about their speed before and after the collision. And what we want to do is determine the change in kinetic energy of the system following the collision. Okay. Alright, so the first thing we want to do with a problem like this is go ahead and draw out the system, fill in any of the details we know. Okay, and then we can go from there. So in our diagram, we're going to take up into the right as the positive directions, K, Y and X direction, respectively. All right now, before the collision, We have our two people. Okay, so we have John who is 70 kg and he is standing still. Okay, so we're gonna John over here in red. Okay, this will be john He is 70 kg and he's standing still. And what that tells us is that his velocity we're going to call John Person one. Okay, so he has a one. The velocity of John initially is zero. All right now, let's get on with the other person. So we have James, K who is 53 kg will draw James in blue over here, this is James? Yes, 53 kg, Whoops! 50! 53 kg, Sorry. Um and he's skating in a straight line and he's going to collide with john So he's going towards john the velocity of 12.5 m per second. Okay, He's going towards john that's our positive extractions was velocity. It's positive. So the velocity and we'll call James the second person. So the velocity of James initially is 12. m per second. Alright, so this is what's going on before the collision. Now, let's look after the collision. Ok? After the collision while we have our two people again. So we have James and blue, He's still gonna be 53 kg. We're told that he's deflected 46.8 degrees from his original direction with a speed of 6.5 m per second. So, his original direction was straight out in the X direction. Okay, He's deflected 46.8 degrees. So his new direction is going to be up like this. Okay, where? This angle is 46.8 degrees and we're told that his speed is 6.5 m per second. Okay, so along this factor here, he's going 6.5 m per second. Alright. And we can also see if we were here, he will have some X velocity and some Y velocity. Okay, And we can work those out. We're not gonna do it right now, but we can work those out if we need to in the problem. Alright. And then we have john John is still going to be 70 kg. Okay. And we're not told anything about his final speed or velocity. We don't know. Alright, so what this problem is asking is for the change in kinetic energy of the system. Okay, We talk about change in kinetic energy. We're talking about delta K. And the change is just going to be the final kinetic energy minus the initial kinetic energy cannot. Alright, so this delta K. This is what we're looking for. Now. Let's go ahead and start with K. No, well, let's recall. Kinetic energy is one half M B squared. Okay. With our system, we have two people, we have johN and we have James. Okay, so they both can contribute to the kinetic energy. So we need to consider the kinetic energy from both of them. So the initial kinetic energy is going to be the initial kinetic energy of johN and the initial kinetic energy of James. Well, this is gonna be as I just said, 1/2 V Squared for each 1/2 m squared for each of them. And we're looking at initial kinetic energy. So we're talking about those initial speeds or initial velocities. Okay, well, we know JOHN is standing still initially so his initial velocity or his initial speed is zero. Okay, so this entire first term goes to zero. We're just left with this second term filling in the information. We know we have the mass of James which is 53 kg. Okay, times 12.5 m/s. His speed all squared. And this is going to give us an initial Kinetic energy of four, jewels. Alright, and how did we get from the units we were using to jules? Well here we have kilogram meter squared per second squared, OK, kilogram meter squared per second squared. We can write this as kilogram meter per second squared times meter. Okay, so we just pulled one of the meters out. Now kilogram meter per second squared recall. That's a newton. So this can be written as newton times meter And we know that a newton meter is a jewel. Okay, so we are in fact in jewels the unit for kinetic energy that we were wanting. Alright, so we found cannot What about K F K F. Same thing. We're gonna have the kinetic energy from john and we're going to have the kinetic energy from dreams recall once again, kinetic energy one half mv squared one half M. V. And in this case we're talking final speeds one half M the word. Alright, so we know the masses. We know the final speed of James but we don't know. Excuse me. The final speed of john. So what we need to do is find this quantity V1F in order to figure out our delta K. The change in kinetic energy of the system. Okay, and this is B one F. Here to see with the red. So this is the one F that we are looking for. Okay, how can we find that velocity of this guy? Well, let's recall our conservation of momentum. Okay, we are on a frictionless rink so we have no net external forces acting on the system. So we know that we will have conservation of momentum when we talk about conservation of momentum, we have conservation of momentum in the extraction and in the Y direction. Okay, so let's scroll down a bit. So we have some more room to work and we are going to start with our conservation of momentum. So conservation of momentum tells us that the momentum in the system in the X direction initially is going to equal the momentum in the system in the X direction. Finally. Well, what's in our system? We have James and we have john Okay, so we have to consider the momentum from each. So the momentum in the system initially is going to be the momentum of john initially in the extraction plus the momentum of James initially in the extraction. Okay. And then we're gonna have the momentum of john in the extraction after the collision. And same with James. Okay, recall momentum mass times velocity. So for each of these terms we have the corresponding mass times the corresponding velocity Massive to Velocity of two. It's not And we'll see here on the right hand side that we have this V one X F. Okay, well that's gonna be a component of this. V one F. Okay. We want to know the velocity of john after the collision. Well, this is going to be the X component of that velocity. So this is what we're looking for from this equation. So filling in some of the details that we know, Well, we know that johN is not moving before the collision. Okay, So john's initial speed is going to be zero, so V one X nine is zero. So this term is going to go to zero. All right now, when we're looking at this, we're gonna need to break down um James's final velocity and I mentioned that earlier that we may have to do that, but we didn't need it yet. So now we need those components, the V two X. F. And the V two Y F. So let's go ahead and find those so we can write that V two Y F. Okay, looking at our triangle, we have sine, the sine of the angle is going to be the opposite side divided by the hypotenuse. Okay. And then we can multiply so V2 YF is going to be signed. The angle were given 46.8° times the Hippopotamus, m/s. And this is going to give us a wide velocity or wind speed of 4.738 meters per second. Okay. And similarly in the extraction. Okay, for the X direction we're going to use coast because we're gonna have the adjacent side divided by the hypotenuse. So we're gonna have coasts of 46.8 degrees times 6.5 m per second. This will give us an X velocity of 4.45 m per second. Okay, So now we've broken up that velocity into its components and we can use them in our conservation of momentum equations. Alright, so initially the mass of James is 53 kg. Okay, in his initial speed. Well, let's just go up and check our diagram. Okay, he's initially going 12.5 m per second directly to the right. Okay, so his initial velocity is all in the extraction. And so when we have our equation here, we're gonna have 53 times 12.5 m/s. His initial velocity it's all in the extraction. Okay? We have the mass of John 70 kg times his the x component of his velocity after the collision, which we're looking for. Okay, the mass of James again, 53 kg. And his X velocity finally, which we've just found it is 4.45 m per second. Okay, so working this out on the left hand side, we're gonna have 662.5 kg meters per second on the right hand side, we have 70 kg times the velocity of john in the X direction after the collision and 235.85. And again, our unit here is kilogram meter per second can be multiplied kilogram meter per second. Alright, moving this over to the left hand side, 426. kg meters per second is equal to kg times V one X f K. And when we divide the unit of kilogram will cancel, we're gonna be left with the velocity of jOHN in the extraction after the collision is 6.095 meters per second. Okay, so our units look good that meters per second is a unit of velocity that we would have wanted. So that's great. Alright, so this is the X component that we were looking for. Now, we need to find the Y component. Now remember we looked at momentum conservation of momentum in the X direction. Okay, we can do the exact same thing in the Y direction. So let's move down once more. Give ourselves a bit more room to do conservation of momentum in the Y direction. So we're gonna have the same thing momentum of the system in the Y direction initially is equal to momentum of the system in the Y direction. After the collision, our system is made up of jOHN and James. So we need to consider the momentum from each. So we have the momentum of jOHn in the Y direction initially plus the momentum of James in the Y direction initially is equal to the momentum of john in the Y direction. After the collision plus the momentum of James in the Y direction. After the collision, momentum is mass times velocity case over each of these. Mass times corresponding velocity, mass times corresponding velocity. And on the right hand side, M one V one, Y F, M two V two, Y F. Okay, now, filling in what we know again, john is standing still initially, so his velocity is zero in both the X and Y direction. Ok, so this first term is going to go to zero and when we think about James, James has an initial velocity but remember he's moving directly to the right, okay, he's moving in the X direction only he's not moving in the Y direction at all, He doesn't have a Y velocity. So the second term is also going to go to zero. Okay, so on the left hand side we're just gonna have zero. On the right hand side, we're gonna have the massive john 70 kg times V one Y F. Okay, that's the value we're looking for. That's the Y component of the velocity we're looking for. Okay, the mass of James, which is 53 kg and the Y component of his final velocity and if we go back up we found that it's 4.738 m per second. Okay, So here Times 4. m/s. Alright, so, working this out multiplying and moving this to the left hand side, we're gonna have 70 kg times V one Y F is equal to minus 251. kg meters per second. Okay, dividing by 70 kg. A unit of kilogram will cancel. And we will be left with The one YF. Is equal to 3.587 meters per second. Alright, that's great. So we now are the X component of our velocity. We know the Y component of our velocity. So we can find the magnitude of the velocity that we need. Okay, So when we're talking about magnitude of velocity, but we know in the X direction we have a positive velocity. So we're going to the right, okay, And in the Y direction we have a negative velocity. So we're going down. Okay. To the right is 6.95, and on the left is 3.587. Okay. What we want to know is this here the magnitude of that total velocity? Okay, so we can call this V one F. This vector here and V one F squared. We can just use pythagorean theorem 6.95 squared Plus 3.587 Squared. Oops, this is going to equal 50.015594. We take the square root, we're going to get the positive and negative root. In this case we only want the positive we just want to know the magnitude. Okay, so taking the magnitude of the square root. We at 7.072 m/s. Okay. And that's the value we needed above. Okay, so we're gonna take this 7.072 m/s. We're gonna go back to what we were doing now remember we're trying to find delta K. That total change in kinetic energy. We found the initial kinetic energy. But in order to find the final kinetic energy, we needed that value of V one F. Now that we have that. Let's go ahead. So we have one half the mass of John is 70. Okay. V1 Act that we just found 7. squared. And that's meters per second. Sorry? Plus one half the mass of James, 53 kg times his final speed. 6.5 m/s. All squared. Alright, now we work this out. This is going to give us a final kinetic energy of 2800 and 70 jewels. Okay, so now we have K. F. And K. Not, we can get back to our delta K. So delta K, which is what we are looking for is equal to K. Final minus K. Initial Is equal to 2870 jewels -4141 Jules. This gives us a final Result of - Jules. And if we write this in scientific notation, we know we get 1.27 times 10 to the three jewels. Okay. That's going to correspond with answer d. Thanks everyone for watching. See you in the next video.