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(c)A second, identical toy car is introduced into the system. The second toy car is initially placed at rest at the lowest point on the track as shown in the figure...

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(c)A second, identical toy car is introduced into the system. The second toy car is initially placed at rest at the lowest point on the track as shown in the figure below_Car 1Car 2Car 1 is released from rest from position A and undergoes & perfectly elastic collision with car 2. A student measures the highest point in the trajectory of car 2 to be a distance of y above the starting position of car 2. How does the maximum height reached by car 2 (y) compare with the maximum height reached by

(c) A second, identical toy car is introduced into the system. The second toy car is initially placed at rest at the lowest point on the track as shown in the figure below_ Car 1 Car 2 Car 1 is released from rest from position A and undergoes & perfectly elastic collision with car 2. A student measures the highest point in the trajectory of car 2 to be a distance of y above the starting position of car 2. How does the maximum height reached by car 2 (y) compare with the maximum height reached by the system with only one car (y)? Ly>y Cy=y Ly< y; Justify your answer: ii) The two cars are modified such that they now stick together upon impact: The cars are returned to the same starting positons and car 1 is released from rest_ A student measures the horizontal distance that the cars travel in their trajectory to be x. How does this compare to the horizontal distance (x ) traveled by the car in the single car system? X>x X=r Cx<x Justify your answer



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A railroad flatbed car is set up for a physics demonstration. It is set to roll horizontally on its straight rails at a constant speed of $12.0 \mathrm{~m} / \mathrm{s} .$ On it is rigged a small launcher capable of launching a small lead ball vertically upward, relative to the bed of the car, at a speed of $25.0 \mathrm{~m} / \mathrm{s} .$ (a) Compare [by making two sketches] the description of the motion from the point of view of two different observers: one riding on the car and one at rest on the ground next to the tracks. (b) How long does it take the ball to return to its launch location? (c) Compare the ball's velocity at the top of its motion from the viewpoint of each of the two observers and explain any differences. (d) What is the launch angle (relative to the ground) and speed of the ball according to the ground observer? (e) How far down the rails has the car moved when the ball lands back at the launcher? Compare this distance to how far the ball has moved relative to the car.

And this example I'm just going to go over a quick quick little example of center of mass motion. Right? What we're going to be looking at here is we have two masses mhm And they are connected by a string which has a winch in it which brings them slowly together. So this could be at time one. This is what the system looks like at time to. It's going to look like this. Q. And what we want to do is just describe the emotions of the two masses. And we're going to do this for two situations. The first one I call that a We have mask one equals mass to And the 2nd 1 B. We have mass one equals dream math too. All right. So how do we do this? Um We know our equation for center of mass or uh the location of the center of mass. We know X center of mass equals the sum over all of our masses To end here in our example and equals two of mhm location of mass I times mass I divided by mass I All right. So let's write this out for our first case case A. We have that are less equals position of the first mess subsection B I plus position of the second nous over the thumbs of the masses. So we have M one plus M two. All right. And sent our masses are equal. We have and times one plus two over to em And just to make the math simpler I'm going to say that the center of mass. I'm going to call this the origin for this problem. So here we start at X equals zero. That's the mid point between the masses. Alright, so that equals zero. Mhm Which gives me X one plus X two equals zero. Once I cancel out the two masses which I can do there Uh zero is multiplied by anything is zero. Uh huh. Right, So we have X one equals negative X two. All right. So what this tells us is that as the objects are being pulled closer together you're going at the same rate. So they're moving they're approaching the center of mass are zero point here at the same rate and they're always at the same distance from the origin relative to each other. Right? And we do something similar for the case where And one mass is three times larger than the other on what that gives us here is Oh sorry, three x 1 equals negative X two. X one equals now at To over three. So, our masses are again approaching the center of mass points which is at the origin. They're both approaching it at the same. Great. But Our first mass X one is always going to be three times further away from it. Which makes sense. As X two is more massive object in the center of mass will be located closer to that mass rather than the first object. That's just simple center of mass motion

And this example I'm just going to go over a quick quick little example of center of mass emotion. Right. What we're going to be looking at here is we have two masses. Mhm And they are connected by a string which has a winch in it which brings them slowly together. So this could be at time one. This is what the system looks like at time to. It's going to look like this keeps you and what we want to do is just describe the emotions of the two masses and we're going to do this for two situations. The first one I'll call that a We have mask one equals mass to And the 2nd 1 B. We have mass one equals three. Math too. All right. So how do we do this? Um We know our equation for center of mass. Birthday the location of the center of mass. We know X center of mass equals the sum over all of our masses To end here in our example and equals two of mhm location of mass I times mass I divided by mass. I All right. So let's write this out for our first case case A. We have Okay, center less equals position of the first mess, obsession being I plus position of the second nous over the thumbs of the masses. So we have M one plus M two. All right. And since our masses are equal we have and chimes one plus two over to em. And just to make the math simpler I'm going to say that the center of mass. I'm going to call this the origin for this problem. So here we start at X equals zero. That's the mid point between the masses. Alright, so that equals zero mm Which gives me X one plus X two equals zero. Once I cancel out the two masses which I can do there and jeroboam Multiplied by anything is zero. Um All right, so we have X one equals negative X two. All right. So what this tells us is that as the objects are being pulled closer together they're going at the same rate. So they're moving they're approaching the center of mass or zero point here at the same rate and they're always at the same distance from the origin relative to each other. Right? And we do something similar for the case where um one mass is three times larger than the other on what that gives us here is Oh sorry, three x 1 equals negative X two X one equals negative X To over three. So our masses are again approaching the center of mass points which is at the origin. They're both approaching it at the same rate but Our first mass X one is always going to be three times further away from it. Which makes sense. As X two is more massive object in the center of mass will be located closer to that mass rather than the first object. That's just simple center of mass motion

For this problem, there are two covers. The car is moving along at a constant velocity. Carby starts at rest and accelerates to meet car A that distance deep at a constant acceleration across the entire time. First we have conceptual questions for a car B's final velocity greater than cars A due to be starting at zero, needing to catch our traveling faster for me, the average velocity of Karbi relative to car A can be equal to or greater depending on the situation for C the distance they meet at. Then using X equals Vienna at times T plus one half A t squared. The acceleration of B can be found Were given that x equals 460 m T equals 210 seconds. As noted above, we lose came back equation of X equals V, not T plus one half a t squared for car A The acceleration is zero as is a constant velocity, So then x equals v. eight times T plug in the numbers, we get 460 equals v. A. times to 10, Which makes via equal to 2.19 m/s. For Car B. We use the same formula And plug in the numbers. We get 160 equals zero because it starts at rest plus one half times acceleration will be Times to 10 squared, Rearranging. We get the acceleration of the equals 460 times two Divided by 210 squared. So the acceleration equals 0.21 m per second squared, notify the velocity of carby. We use the evacuation of v equals v not plus 80 plug in the numbers we get V equals zero plus 0.21 times to 10. So the final velocity of car b equals 4.41 m/s. Yeah, verify this. With the conceptual question of we get the average velocity of Carby should be the final plus V initial divided by two, which is 4.41 plus zero, divided by two, which equals 2.205, which is slightly greater Than the velocity of car one At 2.19 m/s.

Okay for this problem. We're gonna take a look. We're gonna analyze the height of gravel coming off the back of a truck right here. So, for part, a weird givens and information about this piece of gravel. Namely, that the velocity is 30 miles per hour. And the book was nice enough to tell us. That's also 44 feet per second is what we're using. The formula, our angle is gonna be pi over four. And the distance that it travels is 40 feet right here. So we can plug in all that information into the formula up there. And that gives us that the height is going to be 13.553 feet. Okay, for part B. Now, what we're gonna do is we want to set Why, equal to zero. So that's gonna be the height. Um, that's gonna be when the gravel hits the ground again. Um, So I went ahead and rewrote this formula right here without the tangent and the C camp. So using our basic trig functions right here. So we want zero to be equal. This X sign Alfa over co sign Alfa minus 16 x squared over the velocity squared Times one over the co sign squared Alfa. Now, immediately, you can probably notice that we can multiply the entire equation by co sign Alfa. And that will get rid of one of those, um, and Weaken Divide both sides of the equation by an ex as well. And that'll get rid of this X and this X right here. And then I'm gonna go ahead and just subtract this piece or add this piece to both sides. So we come up with the equation 16 x over v times one over co sign Alfa has to equal just sign Alfa. So I'm gonna multiply both sides of this equation by a V cosign Alfa because an Alfa over 16 and we'll do it to the side. Vico, sign Alfa over 16. So that's going to get rid of this and this and this, um and so that tells us that X is going to equal um v over 16. And I apologize. These were all squared the V squared Sze I'm I lost that along the way. So it's V squared over 16 times sine alfa co sign Alfa, which I'm gonna rewrite that as v squared over 32 times to sign Alfa Co sign Alfa and then we can go ahead and use the formula given in the problem, the trigger identity. So that's V squared over 32 times. Sign to Alfa so we can see the identity that they wanted. I'm from the problem. So for this next problem, we're gonna use our formula. Um and they give us some initial data once again. So they go ahead and tell us that the angle it's going to get thrown at is pi over three this time and they tell us that the velocity is 44 feet per second again, and we just want to know on how far the gravel's gonna travel. So we plug in those numbers and we're gonna come up with the distance traveled is about 52.395 feet right there. Now, in Part D. We want to find d x d Alfa, which simply says, um, how for each radiant of angle we increase. How much is the distance going to change right here. So we'll just use some simple calculus right here, And this is assuming we have a constant velocity so that would turn into V squared over 32 times the co sign of Two Alfa. And then we multiply that by two by the chain rule. So it comes out to be V squared over 16 co sign to Alfa. Um, and that is the maximum distance that it could fly for. That's part D and then in Part E. We want to find out, um, for a car traveling 60 miles per hour and for 60 miles per hour, that's double the speed we've been working with previously. So that's 88 feet per second's. We want to know how far that Stone is going to travel at maximum. So what we're really looking for is when is D. X D Alfa equal to zero that will find our critical points where we can get a maximum at So again, the formula was V squared over 16 times co sign to Alfa, so I know V squared over 16 isn't zero. So we're just really trying to find out where that piece is equal. That Alfa there have been that's equal to zero, and that tells us that Alfa is pi over. Four will do that doing. We double it. We have pie over two, and the coastline of that would be zero. We only really need one critical point even though this is periodic, because it's gonna throw it off up to 90 degrees. So we only need to consider upto that. And then we're just gonna calculate why when? Um, Alfa is pi over four. So using our formula right here. Sorry. Not are why we're looking for our X right here. So we get X is equal to 88 squared over 32 times. Be sign of to Alfa. Now, We actually didn't need to know that. Um, Alfa is pi over four. We only needed to know that the co sign of two Alfa has to equal zero. Because if co sign of to Alfie equals zero, then sign of two. Alfa has to equal one, which means the maximum distance is 88 squared over 32 that comes out to be 242 feet. Is that the maximum distance that a piece of gravel could be thrown if you're going 60 miles per hour


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