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Problem 6 Medium Difficulty

Two vehicles are approaching an intersection. One is a 2500 -kg pickup traveling at 14.0 $\mathrm{m} / \mathrm{s}$ from east to west (the $-x$ -direction), and the other is a $1500-\mathrm{kg}$ sedan going from south to north (the 1 y-direction at 23.0 $\mathrm{m} / \mathrm{s} )$ (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) $+3.45 \times 10^{4} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$
(b) $4.91 \times 10^{4} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$

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Video Transcript

In this question, a traveling pickup is moving to the left with a velocity off 14 m per second. Let have a mass off 2000 and 500 kg at the same time as sedan is traveling upwards to the north with a velocity off 23 m per second and the mass off 1000 and 500 kg. In the first item off this question, we have to find the X and Y components off the net momentum. For that. You have to be quite careful now because we are dealing with a problem that happens in two dimensions. That is, we have things moving in the Y direction and we have things moving in the X direction. So we evaluate the net. Momenta must follows. So the net momentum peer not has two components. It has an X component and it also has and why component and the X component off unit is composed on Lee by the momentum off the peak up because only the peak up is moving in the X direction. Actually, the minus X direction. This is because the problem gives us a reference frame, which is this one everything that is moving up is going to the positive direction and everything that is moving to the right is going to the positive direction. So knowing that we're now in conditions off evaluating the components off the net momentum, the X component is given by the momentum off the peak up on Lee. So we have a mass off 2000 and 500 kg times a velocity off minus 14 m per second. We have the minus sign because the velocity points the left toe. The negative direction off our exacts is then for the white component we have the following. The white component is given by the mass off the Sedin 1000 and 500 kg times the velocity off the season, which is 23 m per second. It's positive because it's moving in a positive Y direction Now. By performing this calculations, we get the following the X component off the net momentum is given by minus 35,000 kg times meters per second and at the same time, the white component off the net momentum is given by 34,000 and 500 kg times meters per second. On this WAAS, the first item off this question in the second item, we have to evaluate the magnitude and the direction off the net momentum. For that. It's very, very helpful toe actually draw the components off the momentum. So the white component off the momentum is a vertical arrow that points of ports, and it has a length magnitude that is given by 34,000 and 500. Now we can draw the X component off the momentum. The X component points to the left, so it's L Arrow that points to the left, and it has a length that is 35,000 kg meters per second. Now remember that we're dealing with vectors so we can move vectors around and this will not change anything. So what I'm doing is the following I You put the beginning off each vector together just like this. As you can see, there is an angle in between them, and that angle is a 90 degree angle. The reform. We can complete the triangle by doing something like that. The hypotheses off the triangle will be the magnitude off the net momentum so we can evaluate the magnitude of the net momentum by using the Pythagorean theorem. Let me call this pew net then. According to the Pythagorean theorem, pure net will be given by the square root off 34 500 squared plus 35,000 squared on these results in a magnitude for the net momentum, there is approximately 49,000 and 100 kg meters per second. Now we have to determine what is the direction off that net momentum. For that, we perform a slightly different calculation which goes as follows. So one of the components, the white component, points upwards. Then we have the X component pointing to left. So let us draw it like that. Then the direction off that net momentum is given by that red arrow that goes from the beginning to the end off the trajectory formed by both components. So we did not protect you here, go up here and turn to the left and ends up here. So this is the actual net momentum factor. As you can see, it forms an angle with the X axis that I'm drawing in green here on that angle, I'm calling Tita, and Tita is right here. Now how can we evaluate Tita? Well, it's not difficult again. You can use the triangle, So let me draw the white component here again. So here we have the white component. Let me draw the X component here. So as you can see, this triangle that is formed here by the components together with the net momentum is the same triangle as this one. But now we have to locate. Where is this, Tita? In this triangle. And as you can see, the Tita is the angle in between the X component on the net. Momentum. So here we have the X component. And here we have the net momentum. The reform. This is the angle, Tita. In between the X component and the net momentum. Now, how can we evaluate Tita? Well, it's very easy. We know that the tangent off that angle Tita is given by the opposite side off the triangle. 34,000 and 500 divided by the edges inside off the triangle, which is 35,000. Now we can simplify the zeros just like that. And then we invert that equation. Soul Tita is given by the inverse tangent sometimes known as the Arc Tangent off 345 divided by 350. On the resulting angle, Tita is approximately 44 point 59 degrees. So the answer is the following The magnitude is 49,000 and 100 under direction is 44.59 degrees to the north, off the west.

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