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Two football players collided on a field and neither of themgot pushed back. If we know that player A has a weight of 1500 Nand was moving at 2 m/s and player B has...

Question

Two football players collided on a field and neither of themgot pushed back. If we know that player A has a weight of 1500 Nand was moving at 2 m/s and player B has a mass of 150 kg, thenwhat was player B’s velocity?30.56 m/s3.11 m/s2.03 m/s0.54 m/s

Two football players collided on a field and neither of them got pushed back. If we know that player A has a weight of 1500 N and was moving at 2 m/s and player B has a mass of 150 kg, then what was player B’s velocity? 30.56 m/s 3.11 m/s 2.03 m/s 0.54 m/s



Answers

Two football players collide head-on in midair while trying to catch a thrown football. The first player is $95.0 \mathrm{kg}$ and has an initial velocity of $6.00 \mathrm{m} / \mathrm{s},$ while the second player is $115 \mathrm{kg}$ and has an initial velocity of $-3.50 \mathrm{m} / \mathrm{s}$. What is their velocity just after impact if they cling together?

Okay, so here we have two players there, two players, and they have a mass off 51 95 Casey and 100 15 kids. You. Right? So as the glyphs satisfied that first and one is a call to 95 busy on and to mass off the 2nd 4 wheeler, is he called to under on 15 cases. Okay, so now what happens? He's the equal. I'd read an in silver city off six meters per second, huh? Minus 3.5 million for second. That means the negative terms the guests, that they are in opposite directions now. They came to Iraq. So the initial moment, um is someone you one plus and to you too Now, finally they came to the right. So combined musk is and one percenter let us consider their abilities their combined velocities. After they spray the B To calculate this V began opinion Expression on one. You want bless to you too a poem and one less than now. Just substitute the values off m one. And you are you Do you get this? V is equal to zero point 79 meters per second with turns out to be positive, right? So it is the same direction as on that off. Must everyone the boat, go to us? The dress enough. And one on this on adversity and the manager of 2.79 inches per second to this instance Service into this question.

We saw this question by using moment in preservation. The total momentum before the collision is because they're total momentum after the collision. Know that after the collision they stick together and not dressed for after the collision. The momentum is the cost zero because the momentum remembers proportional to p Times V. So if the is equal to zero, then the moment will cost 202 before the momentum before the collision. Waas zero already but the momentum before the collision has two contributions one from player one and another from player to true. For the first player, we have m one 31 divided by the correction factor one minus the one divided by C Square, plus the second players momentum and to the two divided by the square. Root off one miner's V two divided by C square on the course to zero. Because off the conservation of momentum, now we have to blink in the vials that were given in the problem. So em one is close to 88. The one is because two plus two meters by second letter say that, and this is divided by the square root off one minus two divided by three Because the problem tells us pose. That seat is close to three meters by second squared plus m too, which is 120 kilograms times. Me too, which is not no divided by the square root off one miners be too divided by a squared in. These is close to zero. Now we have to solve this wall problem or be true to simplify your solution. Let us call this war term you won then 120 not plus it will be minus because we too will negative note that one player is going to the right and another one is going to the left. So 120 times be true. Divided by the square root or phone miners v choose divided by the square. He's a ghost to be want. We begin my squaring both sides of the equation to get the following 120 squared virtual square divided by one minus ritual divided by see where it goes to be one squared. No, we continue by doing the following saying this term to the other side. 120 square kinds of each square is equal to be one square minus you One squared times the true divided by C squared. No, we're saying this term to the other side. So 120 square V two square into coastal Be one squared. Uh, forgot your order term. So these plus few one squared V two divided by C squared is equal to P one square. So we have we have since just this term to the side of the equation. Now we can write it as v Chu Square times 120 square plus be one squared, divided by seas where you close to be one square Then we can stand this term to the other side To get reach you squared is equals to be one squared divided by 120 squared plus be one divided by sea squared Now take the square it off both sides to get me to as being close to be one divided by the square root off 120 squared close be one divided by sea Where now we have to plug in the values that we had before Now let us remember what waas be one The one was equals two 18 kinds. True, Divided by the square root off one miners. Two words square. These is approximately 236.129 kilograms meters per second. Then the two he's equals. Two 236 0.129 divided by the square of off 120 plus 236 0.129 divided by tree square, which is approximately one 0.6 meters by second. But remember that this player is going to the left.

Welcome back to New Murad. My name is Kevin Chirac, so let's take a look here where we have a football player whose 110 kilograms so write that down is the Mass. The Mass is 110 kilograms and has a head on collision with a 150 kilogram defensive end. So we're gonna write. That is M one and then m two. We said M two was equal to 150 kilograms. Okay, no problem. After they collide, they come to a complete stop. So before the collision, which player at the greater momentum and which player had the greater kinetic energy offense? That's interesting. Let's consider the momentum. First. We know that the the sum of all of momentum's initially has to be equal to the sum of all of the momentum's at the end. Okay, so let's first figure out what Thean initial momentum was. Well, it said that we had the first kilogram football player had a head on collision with the 2nd 1 and doesn't look like we have anything here about uh then moving or not moving. So let's figure out if we can deduce that So we'll dio m one v one plus m two v two to be that some of all of the momentum's of both players in the beginning. And then all right, that is m one V one prime plus m two. The two prime where the primes represents the velocity is after the collision. Oh, and those are going to be a zero, right? Each of those lost eases zero because it all comes to rest. So that gives us some useful information. M one v one plus m Teoh v two. It would make sense that these would be opposite signs and we could begin to plug in some information. Here's weeks a 110 V one is equal to negative 150 v two, all kinds that works out great, because we know that the second object second player here is gonna be heading in opposite directions. So those negatives should cancel out pretty well. We could get a ratio so we could say that V one to V two is equal to negative 150 to one time. Okay, so that works out just great. Now notice. We did answer one of the questions in the process. This is before the collision. Which player had the Greater Mo mentum? If they came to rest, then there moment was actually would have had to have been equal. So we do know they have equal mo mentum. We can use this ratio here to determine that V one was bigger in magnitude than V two, right, because this fraction is greater than one in magnitude. Well, if the velocity for the one is bigger, let's consider what the kinetic energies would be. Kinetic energy for the first player, 1/2 M. V. One squared within as a result, Be greater than 1/2. I m, uh, to be to squared, right? Well, let's let's confirm that that's that. That that might not be entirely obvious. Let's let's scroll down and give that one a second look. This is my hypothesis. Will circle this and come back to it. All right, let's consider the kinetic energy of the first would be at 1/2 m one V one squared right? But we could solve out for V one squared using this to say that V one squared would be the same as negative 150 divided by 1 10 V two squared. So I just saw that out for V one and plugged it in. So now I've got 1/2 m one V one or the equivalent of the one squared. And I'm trying to compare that. I'll do a box here with a question mark. We don't know how it compares to 1/2 m two. The two squared. Well, hopefully now we can start to see that we've got 1/2 uh, m one, which we do know. And one over here is going to be 100 intense. We can write that in. So we have 110. Then this would be negative. 150 quantity squared. You just write it. Is that? And then in the bottom, we could write 110 quantity squared. The two squared again. We're gonna be comparing that Teoh 1/2 m two or where m two is that 150 right? So we got that 150 V c. This word. Now look what we can do. We can simplify out this 1/2 on the left and on the right. We could simplify out of 1 50 on the right and ah, 1 50 on the left. Right. This would be a positive. As a result, um, we can simplify out a 1 10 and one of these on the bottom. So it looks like we're gonna result in saying that we have on the left a remaining 1 50 divided by 110 V two squared. Oh, and even the V two is canceled. I don't That would be two squares. Cancel away. OK, so now we've got 150 divided by 1 10 and we're simply trying to compare that to what's left on the right hand side. What's left on the right hand side is just one. And so we know that the inequality must face this way. Okay, So that means that the kinetic energy

This question we are given, uh ah, American football context. So, uh, 90 k g fullback is running at ah to the right at five minutes the second. And this is 90 k g. And then there's another, uh, opponent eyes running north. Uh, we speak heart three meters per second and this person is 95 kg. So this is the before situation, okay? And after this stick together and then they move. Ah, they're going to move some way in this manner. OK, so this is the after case silly imprints A, um we want to explain why successful taco is a perfectly inelastic collision. OK, so this collision is over a very short time in Togo. Keys are outside. Forces have no time. Two in part, significant in post. So the indirection, it's a collision. Okay, the opponents grabs the food bank, and that's not let go. So two clears most together at the end, both there in direction. Okay, which means that Ah, the collision, huh? Eyes completely in elastic. Yeah, que Dannion prats be We want to complete Ah, the final velocity of the players que doing practice b s so just to redraw the diagram. Um, so food bank going to the right beside me just a second. And then the opponent is moving up with treatments for second. And then after the condition very go together. If some speak, be with some velocity V s strong. So, um, you have a perfectly inelastic collision. So conservation of momentum gives, um, and one really want i class. And to me, too. I, uh, equals two on one class m two, um, b k calling this and one and the opponents and to Okay, so, um, putting in the numbers so be would be, uh, 1/1. It's five kinds. 19 times five, I class 95 times tree G. And so this is equal to ah 90. Divided by 37 I plus 57 divided by 37 g. Okay, so that many to after velocity, it's go to, uh, the X Square, plus many white square. I swear it. And this is equal to 90. Divide by 37 square plus 57 divided by 37 square square it and this is equal to 2.88 meters per second. And then the direction. So this is so given that the final velocities in this manner. So I'm going to define data here. So this tech data, uh, is Ken Jinan? Wes attention, we Why were the X So this is equal to attention 57 derived by 90. And this is equal to 32.3 degrees. Case of the answer for parts B are 2.8 meters per second, uh, at 32.3 degrees above the horizontal. Okay, then in practice E, we went to calculate the loss off mechanical energy. So the loss. Mm, uh, mechanical energy. Okay, so we know that are in power fairly inelastic collision, That is, um, decrease in the mechanical energy. So the initial kinetic energy is going to be larger than the final kinetic energy. So I'm calculating the magnitude off the loss in Connecticut in mechanical energy. Okay, so, uh, this is the formula and then putting in the numbers. Okay, So for the full back, this is the union shirking at the energy opponents. Uh, the injured kinetic energy is given a such and then the combine, uh, kinetic energy after the collision. It's given us such and then when you calculate, this is 786 juice. Okay, so accounts for the loss. So the loss won't you? Uh, yes. Convert it in two. The internal energy off the system after, uh, to, uh, play your system. Okay, so this is the answer for practice E.


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