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Perfectly elastic collision; truck with mass of 5000 kilograms slams What was the ~ecd the truck before the collisian?stationary car 0OO kilograms The two vehicles ...

Question

Perfectly elastic collision; truck with mass of 5000 kilograms slams What was the ~ecd the truck before the collisian?stationary car 0OO kilograms The two vehicles stick together after the collision and are moving together with a speed of 10 m/sSelect one: 10Ms12 ms50 MsJ8Mis

perfectly elastic collision; truck with mass of 5000 kilograms slams What was the ~ecd the truck before the collisian? stationary car 0OO kilograms The two vehicles stick together after the collision and are moving together with a speed of 10 m/s Select one: 10Ms 12 ms 50 Ms J8Mis



Answers

A truck collides with a car, and during the collision, the net force on each vehicle is essentially the force exerted by the other. Suppose the mass of the car is $550 \mathrm{kg}$, the mass of the truck is $2200 \mathrm{kg}$, and the magnitude of the truck's acceleration is $10 \mathrm{m} / \mathrm{s}^{2}$. Find the magnitude of the car's acceleration.

Mhm. This lesson is going to look at contributing the momentum in two dimensions. All right. So what we're looking at is a hard crash. We have a car initially moving south mass of the car equals 1070 kg. and philosophy with our equals 1.4 five m/s. Right? And we have this colliding with a truck that has a mass of 3420 kg moving at a speed of 9.2 m/s in the West Direction. Right. And we have these two vehicles colliding. What we want to find is the velocity of the wreckage. One of the two cars stick together to beagle stick together. Sorry, We want to find their final velocity, Right? So we know from conservation of momentum that the initial jerry, P. I. Equals the final and we know P initial equals mass of our cars to kill them. one times negative. 1.45 m second Y direction. Uh huh. Massive trucks and series or wanting kilograms In the negative x direction. At a speed of 9.2 m second. Negative I'm putting our values in that gives us initial momentum equals. My name is 3.1 5 10 2 negative. Work. Very positive work. Mm Newton seconds in the X direction minus 1.55 times. Hang with 10 to the three seconds in the Y direction. And we know that our P initially guild RP final the P final equals mm car plus. And we're up times the velocity of that system. Mhm. So we end up with it's the final equals see I over and he puts and teeth. So in the previous step we have the numerical equation for P. I. And when we plug those values in, we get for the final equals negative seven m per second in the X direction plus negative 0.34 5m/s in the white directions. And this makes sense each vehicle was initially traveling in the negative direction, and the initial magnitude of the yeah, sorry, magnitude of the momentum was much greater in the Y direction. It makes sense that it would be much greater in this final is up as well.

So in this problem first we have to determine the velocity of the combination after circulation. So the equation for conservation of movement um is M. One V. One. Not that I M two. V. Two not is a girl too and one plus M two V. This equation one um there be one north is the initial velocity of the car and we do notice the final velocity of the car and V. Is the um we is the combined velocity of the car and the truck. So now rearranging this equation for the we will have and one we won not plus M. Two, we do not upon and one plus M. Two. The separation to us. Now, that's what the values and one is 1200 K G. B. One noticed 25 m per second and to 1500 kg and we do not. Just several meters per second upon anyone is 1200 kg plus 1500. Did you? So he is 11 m per second. Therefore, velocity of the combination after the collision is 11 mito per second, do they? Right now? In the second we have to determine the velocity of the combination. Um So from equation to we have V is equal to and one we were not bless them too. We do not upon and one plus M. Two. So let's put the values and one is 1200 kg. We won notice 25 m per second. It was empty is 1500 kg. Was he to not his 20? Meet the 1st 2nd upon 1200 kg verse 1500 kg. So basically 22 m per second. Therefore, the velocity of the combination after the collision, moving to the right with the speed of 20 m per second is 22 m per second to be all right. And in the third we have to determine the velocity of the combination after the collision. If the truck is moving to the left with the speed of 20 m for a second, so we have B is equal to and when we will not bless them too, we do not afford and when I'm too, I went to us and to So we have and one is 200 kg, 25 m per second, and to 1500 kg vitter is minus 20 m per second upon 1200 plus 1500 so there is a zero m per second. Therefore, the velocity of the combination of 30 collision is zero m dir per second, which reflects that the combination, that the combination will be in arrest after the collision.

In this question were wanting to find the total momentum of the car on the truck before and after the car. Israel rear ended, so the car is traveling along at 14 meters per second when a truck RIA ends it and that is where it collides into the back of it. Now we're told in the question. Sorry. The trick for Clank looks like it's floating. It should be on the same plane as the girl. We're told in the question that the truck is 2000 kilograms. Andi that it is traveling at 25 meters per second. So this is the instant just before the collision occurs. The truck, It's super close to the car. They haven't quite collided yet. So we've got the truck going at 25 meters per second, the car going at 14 meters per second. After this collision, we will have the truck on the car touching each other on moving along at the same speed. Remember to out all your zeros in, as I almost forgot one there. So this question is asking us for the total momentum before and after the collision. Now, in this case, we have conservation of linear momentum, and that is because there is no external force acting on the truck or the car throughout the collisions. That means the total momentum before is equal to the total momentum afterwards. That means we only need to consider one of the cases to find the total momentum as it is the same. So we can ignore finding the total momentum in the afterwards case because we don't know the speed of the joint, um, truck and car combination. So let's look at the total momentum of the truck and the car before just the second before they collect with each other. So let's remind ourselves, the equation for momentum is mass times velocity. So let's begin with the truck. The mass of the truck is 2000 and its velocity is 25 meters per second. Let's add that to the momentum of the car, which is it's mass. 1200 times by its velocity, which is 40 so 2000 times by 25 is equal to 50,000 on day, 1200 times 14 is equal to 16,800. So if we add these together we get 66,000 800 being the total momentum of the car and truck prior to the collision, which is the same as the total momentum of the current truck after the collision and I've just added the units on their kilograms meters per second, make sure we're always adding our units on.

For two identical number cars have a linear collision like this. Okay, um, And you in the same direction they were? You okay, So these guys, these guys have this collision with each other, right? And it's perfectly elastic. And also momentum is conserved. Right. Okay, So I'm just gonna use, um, letters here, and we call this velocity A And this would be this one See, in this one D right. So if energy is conserved, we know that 1/2 and a squared plus 1/2 and b squared right is gonna equal 1/2 I m c squared, plus 1/2 m d score Nolan doing saying that the total kinetic energy adds up to the same thing. Right. Okay, well, this boils down to a squared plus B squared, Michael C squared plus B squared. Right. But if we look at momentum, we know that I am a plus. M b equals, um c plus m d. And that boils down to a plus. B equals C plus D. Because we just divide that out. Right? So then the question is, you know what actually obeys this on? And the answer is that recruiters, guests by inspection. OK, that if a is 6.0 plus 5.6, well, guess what, 5.6 plus 6.0 that works, right? And if we put him in a square, that also will work, right? And so the answer is that for this guy is that they just trade velocities, that this one's going 5.6 zero meters per second And there there is no other solution that will work right, 6.0 meters per cent per second, right? There's nothing. There's nothing else we can put in these numbers that will make this work right. And clearly this does work if we just have them trade velocities. So that's that's the trick, right? And of course, you know this If you've ever played pool and you hit a ball dead on, you know, spin and you get it straight like that, right? Um, the way both stops in the other bowl moves in the direction that you had it. Okay, this gets a lot harder. Thes masses are not the same


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