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At the interscetion of Texas Avenue and University Drive, a yellow subcompact car with mass 950 $\mathrm{kg}$ traveling east on University collides with a red pickup truck with mass 1900 $\mathrm{kg}$ that is traveling north on Texas and ran a red light (Fig. 8.37$) .$ The two vehicles sticktogether as a result of the collsion, and the wreckage slides at 16.0 $\mathrm{m} / \mathrm{s}$ in the direction $24,0^{\circ}$ east of north. Calculate the speed of each vehicle before the collision. The collision occurs during a heavy rainstorm; you can ignore friction forces between the vehicles and the wet road.

$v_{A 1}=19.52 \mathrm{m} / \mathrm{s}$$v_{B 1}=21.93 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

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in this question. One car with a mass off 950 kg was moving to the east or in this case, to the right, and another car with a mass off 1000 and 900 kg was moving to the north, the velocity of both vehicles, Waas announced. Then they collide somewhere here and after that collision. Both vehicles travel together with velocity off 16 m per second in the direction making 24 degrees east off the north. Then we have to deter mine. What was the initial velocity off both cars? For that we have to use the law off. Conservation of momentum. That law tells us that the net momentum is conserved in every situation. The reform. It tells us that the net momentum before the collision is equals to the net momentum after the collision, but then noticed that in this question we are working with two different directions. One direction which is the vertical one and another direction, which is the horizontal one. So we must be careful and remember that this is a Victoria law. That is the momentum is a vector and it can point in any direction. Okay, that law must hold in both directions separately, that is, this is a factor law. So it must be true for the X axis, which is the horizontal axis that I'm choosing like this, and for the Y axis, which is the vertical axis that I'm choosing like this. So that law can be written as follows for the X axis. It's true that the net momentum in the X direction before the collision is equals to the net momentum. The X direction after the collision, the Y direction. It's true that the net momentum the y direction before the collision. Is it close to the net momentum in the Y direction after the collision? So using these equations, we will be able to dr Mind what was the velocities off course? Number one and two. So it goes as follows. Let us work before with the X direction. So in the X direction, the net momentum before the collision was composed on Lee by the momentum off car number one because it's the only car that has some velocity in that direction. So the net momentum before the collision is given by the mass off car number one. I am one times the velocity of car number one, which is the one and then after the collision. What happens is that both cars are moving in this direction with velocity off 16 m per second. So as you can see, this velocity has a component in the X direction. That is, we can draw the velocity like that and by doing that you can. They compose this velocity in its two components and then it's very easy to see that it points both in the Y direction on in the X direction. So the mo mentum after the collision is given by the some off the masses. So the total mass m one blood and true thanks the velocity after the collision. Let me call this big veep in the X direction. Of course, then the philosophy off car one before the collision is equals to the mass off car number one, plus the mass off car number two times the velocity after the collision, divided by the mass off part number one, and this can be written as follows. View one is equals to one plus m two, divided by m one times the velocity after the collision. X component. Now let us plug in the values in this equation. V one is equals to one plus the mass of car number 2 1000 and 900 divided by the mass of car number 1 950 times the X component off the velocity after the collision. That is the X. How can they calculate Vieques? Given V, it's not difficult to do. You just have to notice that we can make a triangle out off the velocity and its components. That triangle is just like this. This site is the Y component. Decide the X component on this side, the world velocity. Then what we know is that this angle in between V on the Y is 24 degrees, then the angle between V and V y, which is this one, is 24 degrees. Using that angle, you can calculate the value off Vieques in terms off V. You just have to use the sign off 24 because in this context, the sign off 24 is equals to the opposite side off the triangle. So VX divided by the hypotenuse, which is V the reform. You can see that Vieques is equals to V times this sign off 24 degrees and you know that is equals to 16 m per second. So Vieques is 16 times the sign off 24 degrees. The reform plugging this result in the equation for you one we get that you want is equals to this factor times 16 times this sign off 24 degrees. And these results in a velocity view one that is approximately 19 0.5 m per second On. By doing that, we solved the first beat off this question, which was to Dr Mind What is the Velocity V one? Now we have two different mind. What is the velocity V two? For that? We all need some space, but we will be doing almost the same thing. So I'm just raising this calculation so that I can have some more space to show you how to do that. Okay, now working. If the momentum conservation in the Y direction we have the following the net momentum in the Y direction before the collision is given on Lee by the momentum off car number two because it's the only car that has some velocity pointing in the Y direction. So before the collision, the momentum in the Y direction is given by the mass off car number two, which is m two times its velocity before the collision, which is V two. Then after the collision, the net momentum in the Y direction is given by the mass off both cars because they're moving together times the white component off that final velocity. Okay, now solving this equation for v choo we got the following Chu is equals to m one plus m two times v y. And this is divided by m two. Then we can write this equation as m one divided by m two plus one times V y now plugging the virus we got the following v Chu is equals to m 1 950 divided by M. Chu, which is 1000 and 900 plus one times the white component off the final velocity. How can we evaluate the white component off the final velocity? Well, to evaluate the X component, we use a design to evaluate the white component. We will use the curl sign. So in this context, the curl sign off 24 degrees is given by the badges and site off the triangle, so v y divided by the hypotheses, which is V. Therefore, you can see that the Y is equal to the cold. Sign off 24 degrees times V and you know that V is 16 so V y is 16 times the cold sign off 24 degrees. Therefore, if you chew is equals to this factor times 16 times the co sign off 24 degrees on these results in the velocity V two, that is approximately 21.9 m per second. So this was the velocity off car number two. So car number one was moving east with velocity off 19.5 m per second on car number two was moving north. We philosophy off 21.9 m per second and this is the answer to this question.

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