To protect their young in the nest, peregrine falcons will fly in...

Question

To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one such episode, a 600-g falcon flying at 20.0 m/s hit a 1.50-kg raven flying at 9.0 m/s. The falcon hit the raven at right angles to its original path and bounced back at 5.0 m/s. (These figures were estimated by the author as he watched this attack occur in northern New Mexico.) (b) What was the raven’s speed right after the collision?

Accepted Answer

Hey everyone! Welcome back in this video. We have a game with two birds. Okay? We have a hawk that is diving and then a duck that is flying and they're going to collide at a right angle. Okay, what we want to know is what is the hawk speed immediately after the collision? Okay. What we want to do with a collision problem like this is go ahead and draw a sketch of what's going on before and after the collision, and we're going to take up into the right to be the positive directions. Alright? So before the collision we have two things in our system, we have our hawk over here, we're going to pretend that this box is a bird can label h and we have our duck down here, we're gonna label it. Alright, so we're told that our Hawk is 4.6 kg and that he's diving. Okay, So we're gonna take the hawk to be going down and its speed is 35.2 m per second. Okay, so that means that the velocity of the hawk initially while we took up to be positive, the hawk is going downwards. So the velocity is going to be 35.2 m per second negative. Now, for the duck, the duck we're told is 1.3 kg and they're going to collide at a right angle. So the duck is going to be going to the right and its speed Is 13. m/s. Okay, so the velocity of the duck initially Is going to be 30 or 13.4 m/s. Okay, all right, great. Now, after the collision, we're gonna have the same two things, we're gonna have our duck label it d it has the same mass as before. 1.3 kg. And we're told that it bounces back. So now it's going to be going to the left at six m per second and because it's going to the left and we've chosen right to be the positive direction, that velocity is going to be negative. So the velocity of the duck final is going to be negative six m per second. And we have the Hawk after the collision. Okay. We know it's gonna have the same mass 4. kg. We don't know anything about its velocity or its speed and that's what we want to find out. We want to find out the speed of the Hawk after the collision. Alright, so, we have a collision problem like this. We want to think about conservation of momentum. We know that momentum is conserved. Okay. And in this case, because we're looking in both the extraction and the Y direction, instead of having one conservation of momentum equation, we're gonna have to we're gonna have conservation of momentum in the X direction and then we're also gonna have conservation of momentum in the Y direction. Ok. So let's start with conservation of momentum in the Y direction. In this case. So, we know that the momentum in this system in the Y direction initially is going to be equal to the momentum system in the Y direction at the end. Okay, well what's in our system, we have our deck and we have our talk. The momentum in the system in the Y direction initially is going to be the momentum of the duck in the Y direction initially plus the momentum of the Hawk in the Y direction initially that's going to be equal to the momentum of the duck in the Y direction at the end, plus the momentum of the hawk in the Y direction at the end. Okay, Alright, now recall, momentum is going to be mass times velocity. So for each of these terms we're gonna put mass the corresponding mass times velocity. So the mass of the duck times the velocity of duck in the Y direction initially plus the mass of the hawk velocity of the hawk in the Y direction initially is going to equal the mass of the duck. The velocity of the duck in the Y direction finally, plus the mass of the hawk velocity of the hawk, the Y direction finally, and I just missed the Y direction. So just make sure that you're keeping your notation clear. Okay, so that you know what you're doing and you don't mix up any of this. The subscript. Okay, so when we're looking at this, we know that the duck is only moving right and left. Okay, so it's why velocity is actually gonna be zero both before and after the collision. So when we're talking about why momentum, the duck isn't gonna have any. Okay, so this term is going to go to zero and this term is going to go to zero again because those duck velocities in the Y direction are zero. So now we're just left with two terms were left with M. H. The mass of the hawk. The velocity of Hawk in the Y direction initially is going to equal the mass of the hawk in terms of velocity of the hawk in the Y direction. Finally. Okay, what we see here is that the mass will cancel because it's the same on both sides. And we're just left with the velocity of the Hawk in the Y direction at the end will be equal to the velocity of the Hawk in the Y direction initially. Okay, well, we know that value. We know that that's gonna be minus 35.2 m per second. Mhm. Alright, great. So we have this value. Okay, we know the velocity in the Y direction. Ok, and again, we're trying to find the total speed of that hawk. So we need to know the speed. It's going in the Y direction and the extraction. So right here we found what it's doing in the Y direction. Now let's switch over and look at the X direction. So same with the momentum in the Y direction, the momentum in the X direction is going to be conserved. And we can write that the momentum in the system in the X direction initially is going to equal the momentum in the system, the X direction finally. Okay, alright, and again, that's going to consist of the momentum of the duck and the momentum of the hawk. Okay. And similarly after the momentum of the duck in the X direction at the end, plus the momentum of the hawk in the X direction at the end. Okay, just like in the Y direction, the momentum is going to be given by mass times velocity. So for each of these terms we get the corresponding mass times velocity a mass of the hawk velocity of the hawk in the extraction initially mass of the duck, velocity, the duck and the extraction finally mass of the This one should be hawk massive the hawk velocity of the hawk in the extraction finally. Okay, all right now one of these terms, we can see right off the bat is going to go away. Okay, so the hawk initially only has a wide velocity, it's going straight down so it's not moving to the right or the left. So this term with the velocity of the hawk initially in the extraction is going to go to zero. Okay. And then plugging in the other values we know. So the mass of the duck is 1.3 kg. The velocity of the duck in the extraction initially is 13.4 m/s. And that's going to equal the mass of the duck again. 1.3 kg times the velocity of the duck in the X direction. After the collision minus six m per second. Okay? We're talking about momentum. The direction matters okay. In the momentum equation we have a velocity we know in velocity, we need the direction and the magnitude and so we want to include the negative here to indicate the direction. Okay? And then for the Hawk we know the mass of the Hawk after the collision is going to be 4. kg and we wanna know what its speed in the X direction or what its velocity in the X direction is going to be after the collision. Alright, so working through this on the left hand side, we're going to have 17.42 and our units are going to be kilogram meters per second. Okay? So we have kilogram we're multiplying by meters per second. So we get kilogram meters per second on the right hand side. Similarly, we get minus 7.8 kg meters per second. Okay? And then we have 4.6 kg times that VHX. F. Alright. So moving this guy over to the left 25.22 kg meters per second. It's going to be equal to 4.6 kg. V h X f. Okay. And then to isolate, we're gonna divide the unit of kilogram is gonna cancel from both sides, We're gonna be left with V. H X. F. Equal to 5. m/s. Okay, so this is our other green value. Alright, so what we've done so far is we found the velocity of the hawks in the extra er sorry, in the X direction, yep. At the end after the collision, and we found the velocity of the hawk in the Y direction after the collision. Okay, so the question is asking what is its speed? Okay, so we need to now combine the extraction in the Y direction to get its speed. Okay, so let's draw a little diagram down here. Okay, now we're gonna we know that the hawk is going to be moving to the right, okay, so the X direction velocity we found was positive, so we know that that's gonna be moving to the right, okay, so we're moving to the right And the magnitude is going to be 5.48 and you can see I just left some extra digits here, just so we don't get any round off air and we also know that our hawk is going to be moving downwards right? So the y direction velocity we found was negative that indicates downwards, so this is going to be 35.2 m per second. Um Okay, the magnitude of that. And what we want to find is this here. What is the hypotenuse? That's going to be the speed of that hawk. Finally. Alright, so we have a right angle triangle. We know two sides. We can go ahead and use pythagorean theorem. So we know V H F squared. It's going to be equal to V H X F squared plus V H Y F squared. Ok. And you'll notice I didn't use arrows on here because we're talking just speed. We don't need to worry about the direction here. Okay, so plugging in our values. Okay, on the right hand side we're gonna have 5.4826087 all squared And then 35.2 squared. So V H F squared is going to be equal to 1269.099. Okay. Taking the square root, We're gonna be left with positive or negative. 0.6 24 m per second. Okay? So remember when we take a square root, we get the positive and the negative answer, we just need to interpret which one we want. So in this case we're looking for speed. So all we care about is a magnitude. So we'll just take the positive route. Okay, so VHF We're gonna take the positive, so we're gonna have 35. m/s. So that is our answer. That is the speed of the hawk after the collision thats gonna correspond to answer. C. Alright, thanks everyone for watching. I hope that helps you in the next video.

Watch next