Two vehicles are approaching an intersection. One is a 2500-kg pi...

Question

Two vehicles are approaching an intersection. One is a 2500-kg pickup traveling at 14.0 m/s from east to west (the -x-direction), and the other is a 1500-kg sedan going from south to north (the +y-direction) at 23.0 m/s. (a) Find the x- and y-components of the net momentum of this system.

Accepted Answer

Hey everyone welcome back in this problem. We have an SUv and a truck approaching a crossroads, Okay. We're given some information about the Suv in the truck and we are asked to determine the X and Y components of the total momentum of the system. Okay? And were told to take the plus Y and plus extractions as north and east. Alright, so the first thing we want to do is just to draw out our system and get an idea of what's going on. Okay, So we're told to take north and east as a positive Hawaiian X directions, respectively. Okay, so let's take north to be up. Okay, so this is gonna be north, it's like east to be to the right. Okay. And so these are positive directions. Alright, so when drawing our diagram, we have two things, we have the Suv Suv here call it s We're told it has a massive 1760 kg based at 1760 kg and we're told that it's headed from north to south at a speed of 18.6 m per second. Okay, so from north to south means that it's going from the top to the bottom, so it's traveling down. Okay? We've chosen up to be our positive y direction. So down means that the velocity of the SVV is going to be negative 18.6 m/s. So we have our suv there and then we also have a truck. We have a truck here, we will call it T for truck And it has a mass of 2500 kg and we're told that it's moving from west to east, so it's moving to the east, so it's moving to the right, the right is our positive X direction. So the velocity will be positive. We're told the speed is 11.5 m per second until the velocity of the truck is going to be 11.5 m per second. Alright, So we want to determine the X and Y components of the total momentum. Let's start with X. So, if we have the momentum in the X direction, well, we have two things in our system. Okay, we're looking for the momentum of the system. In the extraction, we have to consider everything in our system which in this case is the Suv and the truck. So that's going to be equal to the momentum of the SUv. And the extraction plus the momentum of the truck in the extraction. Let's recall what is momentum while momentum is mass times velocity. So, for each of these we have mass times velocity. So for the suv, the mass of the Suv times the velocity of the Suv in the extraction plus the mass of the truck, times the velocity of the truck in the extraction. Alright, so starting with the Suv. Well, the Suv is moving from north to south, it's moving down, it has no X component of its motion, it's not moving right to left at all. So its velocity in the X direction is just going to be zero. So this term is going to go to zero and we're only left with the momentum from the truck. Well, what is the mass of the truck? It's 2500 kg. And the velocity of the truck and the extraction, the truck is traveling completely to the right, so its total momentum is in the extraction. So the velocity will be 11.5 m/s. Working this out. This is going to give us 2800, sorry, 28,750 kilogram meters per second. Okay, and just writing this in our scientific notation, we're gonna get 2.88 times 10 to the four kg meters per second. Okay, so this is the momentum in the X direction. 2.8 times 10 to the four kg per meter second. Okay, Alright, so, we know we're looking at either answer either A or B. Now let's go ahead and do the same for the Y direction. Ok. The momentum in the Y direction. Again, the total momentum of the system. In the Y direction, we need to consider the suv and the truck. So we have the momentum from the suv in the Y direction, plus the momentum from the truck in the Y direction. Okay, momentum recall again, mass times velocity and same for the truck. The mass of the truck times the velocity of the truck in the Y direction. Alright, so in the UAE direction we had no velocity for the suv total velocity for the truck. Sorry, In the extraction, in the Y direction we have the opposite. So the suv is traveling up and down, that has a wide velocity and its velocity is completely in the Y direction. The truck, the truck is moving to the right, it's not moving up and down at all, its velocity in the Y direction is zero. Okay, so the second term is going to go to zero And we're left with the or the SVV. So the mass of the SVV 1760 kg times of velocity in the Y direction. Now This momentum equation recall momentum, the direction matters just like when we're talking about velocity. Okay, so this v in our equation is the velocity. So we need to include the negative to indicate the direction -18.6 m/s. And working this out, we're gonna minus 32,736. Our unit is kilogram meter per second. And just like we did for the first writing this in scientific notation, we get minus 3.27 times 10 to the four kg meter per second. That's our momentum in the y direction of that system. Okay, and so our answer is going to be a the momentum in the X direction is going to be 2.8 times 10 to the four kg meter per second. and in the Y direction minus 3.27 times 10 to the four kg meters per second. Thanks everyone for watching. I hope this video helped see you in the next one.

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Two vehicles are approaching an intersection. One is a 2500-kg pickup traveling at 14.0 m/s from east to west (the -x-direction), and the other is a 1500-kg sedan going from south to north (the +y-direction) at 23.0 m/s. (a) Find the x- and y-components of the net momentum of this system.

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