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. (b) What are the magnitude and direction of the net momentum?

Answer

(a) $P_{x}=p_{A x}+p_{B x}=m_{A} v_{A x}+m_{B} v_{B x}=(2500 \mathrm{kg})(-14.0 \mathrm{m} / \mathrm{s})+0=-3.50 \times 10^{4} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$
\[
P_{y}=p_{A y}+p_{B y}=m_{A} v_{A y}+m_{B} v_{B y}=(1500 \mathrm{kg})(+23.0 \mathrm{m} / \mathrm{s})=+3.45 \times 10^{4} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}
\]
(b) $P=\sqrt{P_{x}^{2}+P_{y}^{2}}=4.91 \times 10^{4} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .$ From Figure $8.4, \tan \theta=\left|\frac{P_{x}}{P_{y}}\right|=\frac{3.50 \times 10^{4} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}}{3.45 \times 10^{4} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}}$ and $\theta=45.4^{\circ}$

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Physics 101 Mechanics

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Chapter 8

Momentum, Impulse, and Collisions

Section 1

Momentum and Impulse

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Video Transcript

once again welcome to a new set of chapter problem. This time we're going to be dealing with momentum and motion problems involving mass and velocity. So in this chapter problem, we're given two two vehicles, you know, like a pickup truck, which we're going to call an Esso has a muscle, um, two thousand five hundred kilograms. That's what we're given. And then also, we're being told that the pick up truck is traveling east west. So if you think about directions, your East West direction looks like this. So this is this's not self. This's no, this is so and then East West. So the pick up truck, two thousand five hundred kilograms, uh, is traveling westward. And then also there's another sedan, which happens to be one thousand five hundred kilograms, and I'm going to call that Andy and the moss off the sedan is one thousand five hundred kilograms, and it happens to travel south to north. So the sedan is struggling upwards south to not assume also that the positive ex direction is that way, and then the positive wide direction eyes upwards. So those are the assumptions that we're given, um, the pickup truck him and is troubling out of the lost City, thanks to the that happens to be fourteen point zero meters per second and then their sedan is traveling at a velocity that happens to be twenty three point zero meters per second. So that's the information we were given in terms ofthe diagramming. You can assume that the pickup truck the sedan is traveling, not self. So if you had a next axes, positive eggs, axes and positive Y axes. The sedan is traveling, not salt, so it's going upwards. If this is the middle, it's going to be going up once like that. And that's a That's the sedan, which I'm calling A and then also the The pickup truck is traveling east the westward God way so you can see eventually. He's a potential of having them a resultant results and momentum. Remember, this is our momentum problem. So we're dealing with masses and also of the lost ism, a different vehicles. So that's what we're given in part A. We want to find the X and Y components off the net momentum, so we want to find the X and white component off the net momentum, which we're going to call p. And then that's just but a in part be. Want to find the money to and direction ugly, too, and direction off the net momentum. That's a goal and target. Want to find the magnitude and direction of the metal men. So we're going to start with party and, ah, I will deal with the pickup truck and then we'LL deal with the the sedan. So for the pick up truck, we have to but two types off momentum. So we're going to call P X. That's the momentum of the pickup truck in the ex direction. S O P a X, you know, wanna call the P X because we have has the happens to be the pickup truck. Okay, so is the pick up drunk? How could a pickup truck So in this problem in the X direction, you know, P X member positive axes towards the right and then negative axes towards the left and the pickup truck. As you can see, it's moving towards the left that way. And so we have to find that momentum. If it's moving towards the left, it's going to be negative. Leg momentum. Um, some negative empty times the velocity of a oh which is obviously in the ex you could call it, you know, exhale. Just a If plug in the numbers we get, we'LL get negative. The mosses The Mars happens to be twenty five hundred because our masters twenty five hundred kilograms times the velocity which is fourteen point zero meters a second. And so we have a momentum off leg thirty five thousand kilogram meters per second. And that's that's the well momentum in the X direction for for the pick up truck. I'm just going to call this negative kilogram meters per second like that, that's in the ex direction. The next thing we want to find is the momentum for the pickup truck in the white direction. Remember the velocity? It's just m e times d y. That's the velocity in the UAE. But this time it's going to be we don't have a sign. We don't have a sign because the velocity in the UAE happens to be zero. So this is zero mirrors for second, so if you plug in the numbers, we get twenty five thousand kilograms times zero meters per second and so momentum in the white direction for the pick up truck is zero kilogram meters per second. The next part of the problem is too. I'm going to do that in the next page. No, dunce. That's the bead on Derby is the sedan. This's this dad. So we'LL get the, um the momentum in the ex direction for this had done, and that happens to be the ex direction. Remember, the sedan is moving upwards. If you go, if you go back, you can see this down right here. It's moving upward. So we have the mass off the sedan times the velocity off the sedan in the X, which we're going to call X B. But this is zero meters second, because if you go back and check, you see that the sedan is moving in the white direction. But it's not moving in the ex direction, so we don't have any effective lost in that direction. The mass off the sedan happens to be a thousand five hundred kilograms, so we're gonna plug in those numbers one thousand five hundred kilograms times the last time you two zero meters per second. That's going to give us zero kill graham meters per second in the wide direction the wind direction. Sudan is moving upwards, so going to get a positive momentum of the wybie. The muscle. The sedan is one thousand five hundred kilograms. The velocity of the wind directions positive is the positive elastic. And if you go back, you see that that lost his twenty three point zero meters per second. So we have twenty three point zero meters per second and if you plug in the numbers in the equation, you end up getting No. Thirty, four hundred thirty four thousand Sorry, five hundred kilogram meters a second. The final requirement off the problem was, that's important. It was to find X and Y components off the net momentum. So the net momentum were going to call that p net off in the and so peanut off and be the fast one. Be the net momentum in the ex direction. So the net momentum in the ex direction that happens to be going to call it p ex navy. And that is simply the sum of the momentum for in and the momentum for being the ex for a the momentum Dex direction As you Khun call happens to be negative thirty five thousand kilogram meters for second, so negative thirty five thousand kilogram meters a second and then for being zero killed one meters a second. And so in the X direction. The resulting momentum for NBC's thirty five thousand kilogram meters a second in the Why direction, the momentum for and B that's the net Momentum is P. Why it plus B. Why d p y if you go back, you see it zero kilogram meters per second, zero kilogram meters of the second and then pee Wybie. This is it. It's thirty four thousand five hundred kilogram meters per second. So the momentum, the net momentum in the white direction is positive. Positive thirty four thousand five hundred kilogram meters for second, so we would have done with part. But we still have to complete. But B. But B is just telling us the uh so so you know we have two way. Have to backtrack a little bit. This is not the next moment, and this is the component. This's the component, not the net momentum. The net momentum wass the question in part B. What's the magnitude and direction off the net momentum So this is the momentum in the X the net momentum in the X on the net momentum in the Y on then. But be but be off the problem. We want to find the magnitude and direction itude and direction Off the net. Comment on Remember calling the nettle mentum P If you go back, you know, just calling it a p a b. Okay, so he may be on the formula. For that is simply you take this square off the total momentum for a they squared and then you take the square of the total momentum V squared and then two times momentum off eight times. Momentum will be co sign off. Ada data happens to be if the if the if the sedan is moving our points and then the pick up truck is moving towards the right means that the angle between them is nineties off. Data is ninety. Okay, so then if you plug in the numbers for the net momentum for they would get thirty five thousand kilogram Amigas the second we want to square that and then for being get well, actually, remember, it was it wasn't just thirty five thousand was negative thirty five thousand. So we have to recall that but eventually want to square that you see that it's going to be the negative will disappear. So this is kilogram meters per second. That's the squirrel. The momentum will be then n plus to P a T B Co sign ninety. Okay, remember, cause I ninety is no his ear co sign of ninety zero. And so if you sum this up, we end up getting forty nine thousand. You get forty nine thousand one hundred forty nine thousand one hundred forty five Just have some of that will be good back a little bit Could we just go back a bit? See, this is forty nine thousand one hundred forty five point one nine kilograms meters per second in terms ofthe direction. Remember the sedan if or Kobach sedan is going upwards and then the pick up truck is where it was left that Sudan is going south. No pick up truck is going east west and they travel forty five degree resultant with respect to the not access or the y axis. So we'Ll say that the the direction off the pickup truck well rather off the net moment. I'm not the peak of drug. No, we want to be careful with that. The direction off the net momentum off the net. Mentum is not west, not West. And that that happens to be forty five degrees. That's a problem for you. I hope you enjoyed it. Feel free to send questions or comments my way on. Make sure you have a wonderful day. Okay, Thanks. Bye.

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