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 a team of researchers studying collisions. K and we're on a frozen pond, which we're told is frictionless. We have two volunteers. Joy and Glad. Okay, Joy, we're told is standing still and Glad is going to slide towards her. Okay, They're going to collide and after the collision were told some information about glides movement and we want to find out what is the magnitude and direction of Joy's velocity following the collision. Okay, Alright, When we have a collision problem like this, the first thing we want to do is go ahead and draw out the system and fill in any of the information we know. Okay, so let's take up and to the right to be positive in our diagram. Now, before the collision, our system consists of two things Okay, we are glad and we have Joy. So glad will draw in red G for glad Here in red, we're told that she has a mass of 52 kg and that she is moving 11.5 m per second towards Okay, so we're going to take her movement to be to the right and her motion is 11.5 m per second. Okay, so her velocity is 11.5 m per second. She's going to the right. That's her initial speed and now we have Joy. Can you hear them? Blue J for joy? And joy has a massive kilograms And we're told that she is standing still. Okay, so her initial speed is going to be zero not moving. Alright, that's what's happening before the collision. What about after the collision? Well, again, we still have our two people, we have Glad She is still 52 kg. We're told that she is moving. Let me just move this 52 kg over to this side, so we have some more room. Okay, she is going seven m per second and that's directed 42.4 degrees from our initial direction. So our initial direction was straight out like this and so her new direction is going to be like this, where this angle is 42.4 degrees. Okay, Alright, and a speed seven m/s. And then we have Joy. Joy is still 62 kg and we don't know anything about her final velocity. That's what we want to find out the magnitude and direction of that final velocity. Okay, alright, now, before we get started let's just look come back to Glad here, after the collision we have her the magnitude and direction for velocity. Let's just break that down into the X and Y components. Okay, because we're gonna need them. So let's call this the velocity of Glad in the X direction, final. Okay, we have the velocity of Glad in the wide action final as well. Alright, well, how do we find these? Well, we know that the Y direction is related through sign. Okay, we're in the opposite directions were related through sign so we know that V G Y F is going to be equal to sign 42.4° Times 7m/s. Okay, It's going to be equal to 4. m/s. And similarly in the X direction V G X F instead of being related through sign. Now on the adjacent sides who are related to coast coast of 42. degrees times seven m per second. And this is going to give us an X. Component of that velocity of 5.1692 m per second. Okay, so now we have that information and we can use that in our calculations going forward. Alright, so we have a collision and let's use conservation of momentum. Okay, we know that we are on a frictionless pond so we have no net external forces acting on the system. And so we do have conservation of momentum. In this case we have a motion in the X and Y directions. So we're gonna have to consider the conservation of momentum in both the X and Y directions. Okay, so let's start with the extraction conservation of momentum tells us that the momentum in the system initially in the X direction is going to equal the momentum in the system in the extraction after the collision. Okay, what is our system consist of flow? We have two people glad enjoy. Okay, they both contribute to the momentum. So we need to consider the sum of their momentum's. So the momentum of the system and the extraction initially is the momentum of Glad the extraction initially plus the momentum of joy in the extraction initially and same thing after the collision we have the momentum of Glad after the collision in the extraction plus the momentum of joy. After the collision in the extraction recall what is momentum mass times velocity? So for each of these we get mass times the corresponding velocity M J V J X not and on the right hand side M G V G X F plus M JV J X F. And now let's recall, we're trying to find this velocity J F. The final velocity of Joy magnitude direction. So we need to find the X component. Okay, if we find the X component and the Y component we can find that total velocity. Okay, so we're looking for this green VJ Xf here let's fill in the information that we know. Okay, alright, so we're going to scroll down Now the massive glad we know is kg. Mhm And we know their initial speed in the X direction while Glad is moving completely to the right to her. Total velocity in the X direction initially Is 11.5 m/s. Okay, 11.5 m/s. Now the mass of joy, we know the velocity well Joy is standing still. Okay, so initially Joyce standing still, her velocity is zero and so this term goes to zero. Okay, on the right we get the massive glad again which is 52 kg when we get her ex Velocity at the end, which we just calculated to be 5.1692 meters per second. Okay, we have the mass of joy which is 62 kg. And then this VJXF that we're looking for. So moving this quantity to the left hand side, it's simplifying everything On the left hand side we get 329.2016 in our unit is kilogram meter per second. Okay, on the right hand side we have 62 kg times. V J X F. Okay, dividing by 62 kg. A unit of kilogram will cancel. We're gonna be left with V j X F is equal to 5.3097 m per second. Okay, and so this is the X component of the velocity. We want to find the magnitude and direction of the total velocity. So now we need to do the same for the Y direction. Okay, so we consider conservation of momentum in the Y direction we get the same the momentum in the system in the Y direction, initially is equal to the momentum in the system in the Y direction after the collision. Okay, for each of these we have joy and Glad. So we get the momentum from both. With the momentum from Glad in the Y direction initially plus the momentum of Joy in the Y direction initially is equal to the momentum of Glad in the Y direction after the collision plus the momentum of joy in the Y direction. After the collision momentum mass times velocity. Okay, so for each of these terms mass times corresponding velocity. So we get the mass of joy velocity of joy in the Y direction initially plus a massive sorry Glad now massive joy velocity of joy in the Y direction initially OK, is equal to the mass of Glad velocity of Glad in the Y direction after the collision plus the mass of joy, the velocity joy in the Y direction after the collision. Okay, and this B J Y F. Well, that's the Y component of the velocity we're looking for. Okay, we found VJ xf Now we're looking for V J Y f. Let's substitute in the values we know Well, when we're talking about Glad, her initial velocity in the Y direction while she was moving completely to the right, okay, she didn't know why Velocity because she's not moving up and down at all. Okay, and so her why velocity is zero. So this term goes to zero. We know that joy is standing still, so her initial velocity is zero. So on the left hand side we're just gonna get zero on the right hand side, we have the massive glad, which is 52 kg times the velocity of Glad in the Y direction after the collision this was calculated okay with our little triangle. When we draw a diagram. So we have 4.72 m per second. The mass of Joy 62 kg. And this VJYF. That we're looking for. Alright, so working this all out, we're gonna multiply simplify and then move over and what we're gonna have, you're gonna have 62. Let me just move down a little bit. So we have a little bit more room to work 62 kg times. V J. Y. F is equal to minus 245. kg meters per second. Okay, When we divide the unit of kilogram will cancel. We're gonna be left with V J Y F. Is equal to negative 3. 5871 m per second. Okay, so this is the Y component of the velocity we're looking for and notice that it's negative. Okay, when we do momentum calculations, we include velocity K. The direction matters. And so the negative indicates that we're going in the opposite direction to our positive Y direction. And so are wide velocity is moving down. Okay, Alright, so we found the X and Y components of the velocity. What we want to find is the magnitude and direction of that total velocity. Okay, And so here is Joy. Okay and let's just draw her in blue so we remember that she was our blue person in our original diagram. Okay. She is moving. She has some velocity in the X direction 5.3097 m per second. And in the negative y direction 3.95871 meters per second. Okay, now the high partners here, Well, that's gonna be the J f k. That vector there. So we want to find the magnitude of the vector and we know that this data is going to be here as well. Alright, so the magnitude of the hypotenuse? Well, we can just use pythagorean theorem. So we have B J f squared is equal to 5. squared. Ok. And we have meters per second Um for each of these 3. squared. BJF squared is equal to 0.8643 m squared per second squared. Okay, taking the square root V j f Is equal to plus or -6.62 meters per second. And here we want the magnitude. So we just take the magnitude. We don't need to worry about the negative route. And we get 6.62 m per second. Okay, so this is the magnitude of the velocity. Now we just need to find the direction feta. Okay, and when we're thinking about direction data, what we can do Is relate are two sides through Tan. Okay. And so we know that theta will be the inverse tangent of the chu 3. 5871 divided by 5.3097. Okay. Now we're thinking about taking The inverse tangent of these. Think about what quadrant we're in. Okay. If we were to draw the axes were in quadrant four. Okay. And so tan is going to be negative here. Yes. When we take the inverse tan, the result is going to be negative 36.7 degrees. Okay, so we're going down 36.7 degrees. And so our final answer going back up to our options, we have the final speed. The final velocity sorry, of joy is going to be 6.62 m/s With the direction of negative 36.7° from the initial direction of Glad. Okay, thanks everyone for watching See you in the next video.