i-a-110-kg-tackler-moving-at-25-ms-meets-head-on-and-holds-on-to-an-82-kg-halfback-moving-at-50-ms-w

Question

(I) A 110-kg tackler moving at 2.5 ms meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

Kai Chen · Accepted Answer

So this quadrant we have two learning to each other. We have one hundred kilogram tackler satis moving at two point five meters per second, and it means heads on. Teo, a twenty eighty two program halfback, says movie at five meter per second, it&#x27;s a head on collision. So if they should be abstraction and then the cloud together end, they hold onto each other. So now they&#x27;ve become just wear. Huge mess moved in together and we&#x27;re trying to figure out how this moves. So firstly, let&#x27;s decide a positive direction because things are moving ops directions. One will have a positive velocity. And one where has an actual plastic and this just cause It&#x27;s passed too. So Andi, let&#x27;s cause checker a hands have back B. So now we have m A equals two massive sea Steckler. This slander tanking around cans of velocity, is two point five meters per second. For half back we have M B because eighty two kilogram and its philosophy is active since it&#x27;s moving opposite direction. So it&#x27;s this is what happens and this is all before collision, so ignition. What about finals? That is object collision after collision. There is really Just this white you dress because so I hold onto each other so we can think of some as a whole. And that is just one bite to kill. Now we&#x27;LL give us good Bostic and in section two or we need to know is the conservation for mentioned, Which means for any isolated system without external forces, some momentum will be the same before and after his collision. Hey, and this this equation will be beauty here before the creation. The momentum is just the true momentum is addition of their separate momentum class M DVB And of course, he&#x27;s velocity, which is also a factor. And after the collision, it&#x27;s just big. And he&#x27;s a total mass because, say, really become one particle and the total velocity And we already know M v a with no envy in Vienna, and we&#x27;re looking for me. So we got all we need. Then we just we arranged this equation. So we have a b a close Andy baby over them, and we plug in all of the numbers and we have one hundred ten kilogram times two point five feet of the second. Um, Plus, did you kill him? I was five meter per second open. This is connected. It&#x27;s and divided by one hundred two children. Sad actually gets us. I just tried him existed. A cure snacked it by because this is an abstraction that gives us an active Cyril point seven liter percentage. What does this knocked me? The slutty minx that after? Because this is a positive direction and this is an active direction. So heavy and inactive direction is after the collision. It will actually moves this way. So for the tackler, he would actually turn and go back inwards concert. So after the collision they will be moving in the direction so half back before the question. So if you just happen to define some direction, is the other way. If you originally defined how packed it is Aha, a positive direction then you&#x27;LL get a positive zero point seven. But it also the sign is just a way to tell you which direction it&#x27;s moving. It&#x27;s moving in how in directions that the halfback wass moving and the matatu is our point seven. So that is see, we just be after his collision

Farhanul Hasan · Answer

Well, so in this one, we use momentum. Conservation. Initial momentum will be called final momentum. Initial momentum. Ah, will be the combined momentum. Um, off the halfback in the tackler and final momentum is the, uh our initial momentum of the some of their momentum. Momenta Final momentum will be their combined momentum. So er was called 1/2 back to know the Halfback by H. Liston Order Tackler Buy teas Considering the direction off the half back Positive, we have MH VH mass times velocity of half back minus cause tacklers moving in the opposite direction M t v t for the tackle on That&#x27;s equal to MH plus m t times Ah, final velocity VF Ah! And so final velocity V f is just m h v h minus empty view t over the sum of the messes. So now we plug stuff in. Uh, we have 82 kilograms times five meters per second for the halfback minus 1 30 kilograms times uh, negative are, um 1 30 kilograms times negative 2.5 uh, meters per second for the tackler over 82 plus 1 30 killer pants. And this If you work it out, it will give you points 40 meters per second, 222 decimal places and that&#x27;s it

Ryan Williams · Answer

So here we have a schematic of our problem, where we have one player with any faction against a mass moving at a certain velocity, chasing another player, a smaller velocity there allegedly when they collide, they will act as one object with a mass of each of their messes individually, some moving at some final velocity which we want to calculate. And do you? This was gonna simply use the conservation of momentum which says that the initial momentum asked equals final momentum of the system. Well, this is our initial and this is our final situation. So our initial momentum will be the mass of the first player, plus the mass of the second player for the momentum of both players. So m one a one plus m two g two you should be the mass of the first player is 85 kilograms and his velocity, where their velocity is 5.5 meters per second and when these two are multiplied together, you&#x27;ll get the units of kilograms meters per second plus the mass of the second. She&#x27;s 95 times their velocity, which is 4.1 meters per second, adding these together plug amusing calculator will give 857 kilograms meters per second and all this fast equals the final momentum which will be the mass of both players times the velocity of both players asked equal the final for the initial momentum. 857 kilogram meters per second. If we&#x27;re solving for people&#x27;s well, give us 857 divided by the mass of both which would be 85 kilograms plus in 95 kilograms. And since the units of the But and the Dominator be kilograms, it would just be left with meters per second and plugging this again into a calculator gives approximately 4.76 Neither is per second.

João Gabriel Alencar Caribé · Answer

we saw this question by using moment in preservation. The total momentum before the collision is because they&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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.

Averell Hause · Answer

we have two balls will just say m someone equal in 0.50 kilograms. I&#x27;m so to Equalling the 1.0 kilograms so easier. Sub scripts on dso weaken Apply the conservation of momentum M sub one visa born initial plus M sub two visa to initial equaling EMS of one these of one final plus m sub two multiplied by These are two final, so we can then solve for the visa to final. This would be equal to M sub one initial or other times of one visa one initial plus m sub two visa to initial minus I&#x27;m someone be someone final divided by himself to and so we can solve and say that then visa to final is Equalling 0.50 kilograms multiplied by 6.0 muse per second plus 1.0 kilograms multiplied by negative 12.0 meters per second minus 0.50 kilograms. Most applied by negative 14.0 meters per second, all divided by 1.0 kilogram. And we find that then he&#x27;s up to final is Equalling negative 2.0 meters per second and here here the negative sign indicates that the direction of the second ball after the collision is opposite to the initial direction and again the the magnitude of the velocity of the second ball. 2.0 meters per second. That is the end of the solution. Thank you for watching.

Averell Hause · Answer

here Weaken job the before picture. And we just have essentially three football players and we&#x27;ll just label them A, B and C. And we know that A has initial velocity to the right. B has initial velocity to the left as well a c and we can say that if this was the positive X direction after, we simply have a three team, a three person system, football players, A, B and C and we have then some final velocity to the right or to the left tour. Essentially, we would need to figure it out before, um, so I&#x27;m just gonna put it to the right for now and we can doesn&#x27;t be party for Part B. We want to find the velocity of the football players after the collision. So this velocity final, we can say that the initial momentum equals the final momentum. And so we can say that I m sub a visa a initial plus M c b visa be initial plus, um, subsea, the subsea initial Equalling I m. C A plus and be plus M sub sea quantity multiplied by the final velocity. So the final velocity is then going to be equaling two again the some of these impulses divided by the sum of the masses. And so we have 92 kilograms multiplied by 5.0 meters per second. Then we have 75 kilograms multiplied by negative 2.0 meters per second. Given that he is acting in the opposite direction, this will be plus 75 kilograms multiplied by negative four 0.0 meters per second. And we find then rather, this would be, uh, divided first by 92 plus 75 plus 75. Of course, the units, our kilograms. And we find then that the final velocity of the of all three players would be 0.0 for one meters per second. So this is positive. Which means that here this is a vector is in the correct direction. However, it za Satan, this must make it much smaller. And so for part C, does the fullback score a touchdown? We can say yes. The final velocity ah is greater than zero. And, uh, the positive X direction is to the right and based off of the two diagrams again, we can say that Yes, the fullback continues on with the ball and the football eventually crosses the goal line and the fullbacks scores a touchdown. That is the end of the solution. Thank you for watching.

Problem

(II) Calculate the force exerted on a rocket when…

Problem 4 Easy Difficulty

(I) A 110-kg tackler moving at 2.5 ms meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

Answer

$0.70 \mathrm{m} / \mathrm{s}$

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$0.70 \mathrm{m} / \mathrm{s}$

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Christina K.

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Christina K.

Andy C.

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