Jack (mass 55.0 kg) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 48.0 kg), who is initially at rest. After the collision, Jack is traveling at 5.00 m/s in a direction 34.0° north of east. What is Jill's velocity (magnitude and direction) after the collision? Ignore friction.

Video transcript

everyone welcome back in this problem. We have J. He's skating okay? He's on a frozen horizontal surface and he's going towards a crossroad. Um Kay is crossing the crossroad, she's gonna get shocked and stop. Okay, Upon spotting J. They're going to collide and we're told some information about J after the collision and we want to determine the magnitude and direction of Kay's velocity following the collision. Okay, We're told that we can ignore friction. Alright, so let's go ahead and take up into the right as positive. Okay, We're going to take up as north and to the right as east, and the first thing we want to do is just draw a diagram of what's going on in the system and fill in any of the information that we know. Alright, so, before the collisions, we have J Who is skating? We're told that Jay has a mass of 60 kg. Okay, we're gonna put a J here. So we know that this is Jay. He is skating or he has a massive 60 kg. Okay, And he's skating 10 m per second due north. Okay, So he's going straight north. We've chosen north to be up, so he's going straight up. Okay, that's the positive y direction. So his velocity Is going to be a positive 10 m/s. All right. Now, we have K as well, K is up here because they're going to collide. So this is K. She has a massive 51 kg. Were told here. Alright, and again, we're told that she stops when she spots him before they collide as if she stops, her initial velocity will be zero. She is not moving. Okay, So that's what's happening before the collision. Let's move to after the collision. So after the collision we have J. Mhm. They collide and we're told that J Is going 6.36.27 m/s and it's directed 28° east of north. Okay, so we have the east direction. We'll just do it in dotted line here then we have the north direction here. Okay. And he's going 28 degrees east of north. So you start at north and you move 28 degrees towards east. You're gonna be going in this direction here where this angle is 28 degrees and that's at a speed of 6.2 m per second. Alright, great. So we know what's going on with J. He's gonna have the same mass as well now with K. K with K. She's gonna be over here somewhere. She has the same mass as well. 51 kg, same masses before the collision, but we don't know anything about her velocity. Okay. And that's what we want to try to find what is. Kay's final velocity, the magnitude and direction of it. Alright, so let's work out jay's velocity. Okay, we're going to need to know the components, the X and Y, components of his velocity. So we know that we have D triangle given like this. Okay, so this hypotenuse is 6.2 m/s At an angle of 28°. Now along the top here this is going to be Jay's final x velocity. And along the bottom the left side here sorry, is going to be that's my philosophy and we noticed that these are going up and to the right. Okay, so they're both going to be positive. We just need to look at the number now. Alright, so if we have 28°, if we consider the sign Of 28° Hey, that's gonna give us the opposite side divided by the hypotenuse. So that's going to give us the velocity of J in the extraction final over 6.2. Okay, so we multiply 6.2 times sine of 28. When we get the velocity of J in the extraction final is going to be equal to 2.91 m/s. And similarly if we do this with coast coast of 28 degrees, we're going to have the adjacent side. So this time the Y direction of jay's final velocity divided by the hypotenuse. 6.2. And again 6.2 times coast of 28 degrees velocity of J in the Y direction final is going to be equal to five .474 m/s. Alright, great. Alright, so we have those, let's go ahead. We need to find caves velocity. We're told to ignore friction. So that means that we have no net external forces acting on the system. So let's use conservation of momentum. Now, in this case we have motion in both the X and Y directions. So we need to look at conservation of momentum in both the X and Y direction. So let's start with the X direction. Conservation of momentum tells us that the momentum in the system and the extraction actually is going to equal to the momentum in the system in the extraction after the collision. Well, what does our system consist of? We have K and J. Okay, so we need to consider the momentum from both K and J. So for each of these we're gonna have the momentum of J in the X direction, initially plus the momentum of K in the X direction initially can. On the right hand side we have the momentum of J in the X direction. After the collision. And the momentum of K in the X direction after the collision. Okay, Alright, great. Now let's recall what is momentum, momentum has given mass times velocity. So for each of these components, we're gonna have the corresponding mass times velocity. So we get the mass of J, velocity of J in the X direction. Initially the massive K velocity of K in the extraction initially. And on the right hand side, the massive J. The velocity of J in the extraction final final after the collision and the massive K velocity of K. In the extraction after the collision. All right now, before the collision, we know that K stops when she gets scared. Okay, So she stopped so her velocity is zero. So before the collision case velocity is zero. So she's gonna have zero momentum. That term is going to go to zero. We're thinking about J and we're talking about the extraction while jay is moving due north. Okay, so straight north, straight up. So he has no X motion. Okay? He's not moving left to right at all. So his X velocity is going to be zero. Initially he's on the left hand side, we're just gonna have zero On the right hand side. We're gonna have the massive J 60 kg the velocity of J in the X direction, final. Well, that's what we found over here. Okay, this 2.91 m per second. 2.91 m/s. The massive K is 51 kg in the velocity of K in the X direction. Final. Well, that's what we want to try to find. Okay, we want to find k's velocity. The magnitude and direction of our final philosophy. That means we need to find the velocity and the extraction and Y direction. Ok, so, we need to find this term here. V K X. F. Okay, working this out, we're gonna move this term over. Okay, we have kilograms times meters per second here when we divide by the 51 kg a unit of kilogram will cancel. We'll be left with the unit of meters per second, which is exactly what we want for velocity. Okay, so I'm going to scroll so we have a little bit more room to work here. All right, so we're gonna end up with the velocity of K in the extraction Finally, after the collision is going to be equal to minus 3. 2 3, 5 m/s. And that's that green value, the X component of her velocity we were looking for. Alright, so we've done the X component. Now let's go ahead and do the Y component like the X direction. We also have momentum in the Y direction. So the momentum of the system in the Y direction initially is going to equal the momentum of the system in the Y direction After the collision. Just like before we have K. And J in our system. So we need to take the some of their momentum's the momentum of J in the Y direction, initially plus the momentum of J. In the wider whoops, momentum of K. Sorry, in the Y direction initially is going to equal the momentum of J in the Y direction. After the collision plus the momentum of J in the Y direction. After the collision. And that second one should be K. I did it twice there. So the momentum of K in the Y direction. Finally, Sorry for that. All right. Now again momentum mass times velocity. Okay, so for each of these terms corresponding mass times corresponding velocity, corresponding mass corresponding velocity right hand side. The mass of J. The velocity of J in the Y direction after the collision. Plus the massive K. The velocity of K in the Y direction after the collision. Alright, now, just like in the extraction, K is not moving at all. She stopped so her y velocity will also be zero. And so this term on the left hand side will also go to zero. Okay, filling into the details for the rest of the terms we have the massive J is kg. His initial y velocity while his entire velocity is due north. So his entire velocity is a y velocity. It's gonna be 10 m/s. On the right hand side again. The massive J 60 kg. Okay, his final y velocity. Well, that's what we found over here. We'll draw it in purple. Those final wind velocity was 5.474 m/s. So 5.474 m/s. The massive k is 51 kg in her final Y velocity. Well, we don't know. Okay, that's that second component that we're looking for. We found the X component. Now, we're looking for this Y component. All right, let's give ourselves some more room to write. Okey Dokey on the left hand side. We're going to get 600 kgm/s. On the right hand side. We are going to get In 28. Plus 51 kg times the k Y. And the units here on the 328 should also be kilogram meters per second because we have kilograms times meters per second. All right. All right, so we can move that over to the other side. We're gonna get 271. kg meters per second. On the right hand side. We have 51 kg times V K Y f. Alright. And dividing by 51 kg. The unit of kg will cancel. We're going to be left with the VKYF is equal to 5. 244 m per second. The unit that we were looking for for velocity, the units work out. Alright, so that's the Y component of case. Final velocity. Okay, so now we have the X component and the Y component. Let's use those to find the direction and the magnitude. Alright, so let's go ahead and draw a little diagram here. So we know that case why velocity is positive. We've taken up to be positive. So she's going to be moving up Okay with the speed of 5.3244. We also know that her ex velocity is negative. Okay, we've chosen right to be the positive extraction. So that means she's going to be moving left. Okay. And that speed there is 4, 235 m/s. And so her total velocity will be adding these vectors kind of tail. So it will be this high pot noose here. Okay, this is her total velocity V k f that we are looking for. Alright. And we have some angle here, Theta. Okay, so let's start with the magnitude of the velocity. Okay, so V K F. If we're looking just at the magnitude of her velocity squared, we can use pythagorean theorem. So we're gonna have three 0. squared plus 5.3244 squared. Hey, V k f squared adding those up. We're gonna have 40.1 and taking the square root we get plus or minus 6.33 m per second. And again, we're just worried about magnitude. So if we're just worried about magnitude, we can just take the absolute value or take the positive and we're gonna get 6.33 m per second. Okay, so that's the magnitude of her velocity. That's great. We figured out one thing that we were looking for now we need to do the direction. Okay, so let's figure out what this value of data is. Okay, well, we know that tan of theta is going to be the opposite over the adjacent side. So we're gonna have 10 data is equal to 3.4235 divided by 5.3244. Well, that means that theta is going to be equal to the inverse tangent Of that value. 3.4235 divided by 5.3244. And working that out, we're gonna get 32.7°. So we have 32.7°. Is our fate a value. Okay, now we need to interpret that. So let's go back to our results. Hey, we know we're moving 6.32 m per second. Okay, So we're either option C. Or option E. And our directions are given in terms of west of north. Okay, Well if we're looking at our theta value, we're starting with the North and we're moving towards west. So our theta is going to be 32.7 degrees west of North. Okay, that's gonna be answer. E. Thanks everyone for watching. See you in the next video.