a-0800-kg-ornament-is-hanging-by-a-150-m-wire-when-the-ornament-is-suddenly-hit-by-a-0200-kg-missile

Name: a 0800 kg ornament is hanging by a 150 m wire when the ornament is suddenly hit by a 0200 kg missile
Uploaded: 2019-06-24
Duration: 6 min 14 s
Channel: Numerade
Description: A 0.800-kg ornament is hanging by a 1.50-m wire when the ornament is suddenly hit by a 0.200-kg missile traveling horizontally at 12.0 m/s. The missile embeds itself in the ornament during the collision. What is the tension in the wire immediately after the collision?

Question

A 0.800-kg ornament is hanging by a 1.50-m wire when the ornament is suddenly hit by a 0.200-kg missile traveling horizontally at 12.0 m/s. The missile embeds itself in the ornament during the collision. What is the tension in the wire immediately after the collision?

Ben Nicholson · Accepted Answer

problem. 8.45. We&#x27;ve got a ornament hanging off of a wire and a missile of some sort approaches it with the given speed and ah, embeds in it. And what we want find is what is the tension in this wire immediately after this impact occurs? So this is going to combine what we know about mo mentum with what we know about circular motion, because the ornament yeah, it&#x27;s going to start moving along a circular art. And so there has to be some centripetal acceleration on the ornament that comes through the tension in the wire. So before getting to that, we even know how fast it&#x27;s going because we&#x27;re going to use that this a trip. It&#x27;ll acceleration this v squared over R. So we would like to know what he is, and we&#x27;ll call the speed of the bullet a missile. It&#x27;s called bullets. We don&#x27;t have too many m&#x27;s. The Be even will call the speed of both combined after the impact. Just just be m o for ornament feet. So we find the speed of the ornament missile together immediately after the impact is going to be the ratio of the mass of the bullet to the total mass times the initial speed of the bullet. And this is 4.5 meters per second after. Consider um, a free body diagram for for the ornament going to have attention in the wire acting upwards and the weight of the ornament acting downwards. And so the sum of the vertical forces you call that say the tension is directed in the positive wide direction is going to be the tension. And since we have both the ornament and the bullet moving together, both of their masses contribute to the wave. And so this forms the centripetal acceleration. There&#x27;s some. Is this in trip? It&#x27;ll forces minus w here. Yeah, excuse because we need the mass of the bulletin ornament to make a force instead of instead of Justin acceleration and then are, of course, is just length of wire that the ornament is suspended from. And so the tension in the wire is going to be the mass of the ornament and bullet. I&#x27;ve done that again. I start reading be its toe and B Okay, they have a free spirit over our the centripetal acceleration, plus the acceleration due to gravity and then putting all of these numbers in, we find that our attention is 186 Newtons.

Sri Datta Vikas Buchemmavari · Answer

So in this problem they have a five kg ornament. This argument as a matter of five kilograms and it&#x27;s hanging from a thread off 1.50 meters length. This length is 1.5 centimeters. No, a bullet off mass. Three kilograms incorrectly. Grams is moving with a velocity 12.0 meters per second on embedded into this ornament. So after the collision, this new mask this bullet Lis Arnold has a mass off eight kilograms and little It has a velocity. See now because it has a velocity of on because this lamp is fixed, this most in a circular motion like a kangaroo. And because of the circular motion, there will be a centrifugal force acting on the object. Big magnet, your M V Square by our in addition, really afraid which already is already acting downwards. So this is the total downward force that is acting on this mask on has to be contacted by the stream so that tension in the string will be equal. So this this some off the gravitational force to rest the centrifugal force. So if we find the velocity, then we have the answer now, initially before the collision. The initial moment on P I is just the momentum off the bullet because the argument is not moving and it is mass times velocity, which is three into 12 which is 36 18 meters per second. Now after the moon after the collision, the ornament and the bullet more together with the velocity V hands then Windham finally would be eight kilograms. Multiply way the velocity V. This velocity is the general velocity, whether a let&#x27;s call you or something So eight times we is to finally went. Um but because there are no other external forces in the horizontal direction, remember that gravity is only downwards, which is what the moment of his guns out and hand. The final moment of also must be 36 kilogram meters per second. From this we get the velocity is 36 by aid is a little 4.5 meters per second. Now that we know this, the tension we know in the spring is mg less and be square by our additional weight less the centrifugal force. We know that the mask is eight kilograms on. The gravitational acceleration is 9.8 meters per second squared unless we know that the master is again. Eight kilograms multiplied by the square would be for 15 squared. Divided by the radius would be the lender off the string, which is 1.5 meters, because when the object moves, it moves about this point as its origin. Because the string is fixed here, this point will become the origin of the circle which this object traces. Hence the radius will be 115 meters. But like this into a calculator, you can see that this is 186 Newtons. I know that because all our numbers are given after three significant digits, we have rounded it down to three significant digits.

Maria Gabriela Cota Moreira · Answer

So in this exercise, we have a ball off mass and B that is hanging by a mess. Lis wire off length are and this must be is just about to get hit by a mass A that has initial velocity V a one. Okay, in the velocity is on the horizontal right after the collision between the particles A and B. The collision is perfectly elastic. And the particle be games velocity V B to deposit to the negative X direction and the particle, um, A will have a final velocity that points, um to the opposite direction off its initial velocity. So what we want to know in this exercise it what it is what is going to be the tension on the wire right after the collision. Okay, so So let me race this. So it is easier to see. We have that there is the gravitational force, so I&#x27;ll call it F g f g. Here. We also have the tension that I&#x27;ll call t one. Okay. In this case, 21 is equal to F. G. So the information we have is the massive particles, A and B. We have the initial velocity off particle a and we have the length off the wire. Okay, so, um, since it is a lasted collision, we can use, uh, the books equations 8.25. And this equation relates, uh, the velocity off particle be with the initial velocity off park. Oh, wait by, uh, two times a m a plus m b. So by this factor, this is equal to V A one. Okay. And why do we want the B? Since Because we want to find t. So why calculate vb? Well, we have to calculate, uh, vb because, um because in Newton&#x27;s second Law, so second law, we have that the net force eyes going to be the mass times, the acceleration and in our case, since the forces on Lee point into the radio direction. So we&#x27;ll look to the net force in this case. So because the forces only point point to the radio direction, we can express this by so and be times the velocity off B squared over two. Sorry. Over our on which this is the radio acceleration off particle be and this is going to be equal to the tension on the wire that we want to find t minus the gravitational force which is equal to M B times G. Okay, so we can isolate t in this case, and we have that tea is going to be equal to be, uh, times v b squared over r plus g. So this is why we want to find VB. Okay, So using this expression and the values we have, we can calculate VB So we have that this is equal to two times to over two plus eight times the velocity, the initial velocity off particle A which you know it is equal to five. So we find that the initial velocity off particle we that is the velocity of particle. We right after the collision is just equal to 2 m per second, so we can use this. And so the students in the expression from detention and we have that attention is going to be equal to the mass of particle piece. So eight times the velocity squared, so two squared over the radios 0.135 plus 9.8, and this is going to be called to 102 Newton&#x27;s. And with this, you conclude the exercise

Julie Farhm · Answer

in this problem of all is a suspended from wire first horizontally, and then it&#x27;s allowed to release, and it falls down a semi circular path and hits a block. So the two parts of the question aren&#x27;t to find the velocity of the ball just before it hits the block and then to find the velocity after it hits the block. And they specify that the collusion between the ball in the bloc is elastic, so we need to know the lengths of wire, which is 1.2 meters. That&#x27;s the same whether it&#x27;s vertical or horizontal, the same wire. So that&#x27;s ultimately the height from which it falls. And even before we analyze any collision, we can say that the potential energy before the balls released is going to equal the kinetic energy when the ball is just about to hit the block. So I used to mg times the height is 1/2 and b squared. The mass is just the massive this fear, so that cancels. It&#x27;s the same on both sides, and I can find the velocity I multiplying both sides by two and then taking this square reach, I can find the velocity of the ball just before it hits the block, so we can do that to get an answer for part A. So two times 9.8 meters per second squared times the height, which is 1.2 meters. Take the square root of that. You will get 4.85 meters or second. So that&#x27;s a basic conservation of energy problem. The second parts a little trick here because they wouldn&#x27;t know the velocity of the ball after it hits the blocks that were sort of analyzing this system here, and it&#x27;s an elastic collision. So will say that the mass one is given. That&#x27;s the massive this fear and that is 1.6 kilograms. The velocity of the ball just before it hits the sphere hits the block. I&#x27;m sorry is 4.85 meters per second. We just I&#x27;m not accusing energy. The second mass is the block, and that was a heavier massive 2.4 kilograms. But its initial velocity iss zero. Okay, so we do know that energy is conserved P before is it looks api after of the whole system so we can write out what we know and symbols. First on one, the one be not one plus in to be not too. So these are the initial masses, and velocities are equal to the masses which haven&#x27;t changed. But the velocities have. Okay, But we now we can plug numbers in because we can get to a point where you&#x27;ll see we have two unknowns. So, um, first masses won&#x27;t wait. Six, the first velocity was 4.85 This whole term is zero because the initial velocity of blood to zero. And on the right hand side, we have 1.6 times the final velocity of block one, which I don&#x27;t know what it above 100 what it is. And second mass times the final two. So we have two unknowns and without another relationship can&#x27;t solve this problem. Um, the book does it one way. It&#x27;s on page 1 99 in 200 of your textbook. You can certainly do it that way. If you look at problem 33 of this solution guide um problems 33 In chapter seven, I derive our relationship, which I find more helpful than the one that the book uses, which is Onley True for elastic collisions and that relationship. Feel free to go back and see that if you haven&#x27;t already. But the elastic collision relationship is that the velocity of Block one, the sum of its initial and final velocities of object one. Whether it&#x27;s a block, Baltar&#x27;s matter is equal to, uh in your show and final velocities of walk to remembering that these are vectors. So if anything is moving in an opposite direction, then you have to the Sunday negatives into it. So in this case again, I have the final one, and the final to both is my unknowns. But if I plug in my numbers here, I have another relationship that we can use to solve the problem. So V not one is for 25 meters per second. The final one. I don&#x27;t know. We not to zero and the final two, I don&#x27;t know, but because the zero is here and it doesn&#x27;t do anything I haven&#x27;t expression for. The final to the final two is 4.85 meters per second, plus the final one. This whole thing can go into this equation where the final two is, and then we&#x27;re left with only one on No. So we&#x27;ll move a little bit more to the left so I can have more space. So 1.6 times 4.85 And I suppose I could have multiplied that out. Do it in the next line. So the left hand side hasn&#x27;t changed the right hand side. I&#x27;m gonna keep the first term, as is the final one. And then I&#x27;m gonna plug in instead of the final two. Been going to plug in 4.85 meters per second plus the final one. So now I can simplify 1.6 times 4.85 this 7.76 So that side is done from now and then, I&#x27;m gonna be able to combine this, but you can&#x27;t yet. So 1.6 times the final one plus 2.4 times 4.85 is 11.64 kilogram meters per second, plus 2.4 kilograms. The final one. So now I have two terms with the final one so I can combine knows that would be exactly four. The final one. And then if I subtract this from both sides or move it to the left hand side and change the sign of 7.76 minus 11.64 which gives me negative 3.8 h kilograms meters per second. Now it&#x27;s equal to four times piece. I know one. So just dividing both sides by four. The final one, with just the final velocity of the ball after it strikes the block is negative. 0.97 leaders for a second and that&#x27;s what I asked for. They don&#x27;t ask for what the velocity of the block is, if they did, would be finding the final to the easiest way to do that would be to come back to this equation. Employees get this thing back in. We don&#x27;t need it in this case. So our final answer is that the sphere bounces back with a velocity of 0.9 seconds per second.

Ben Nicholson · Answer

problem. 8.83. We have a ball with a massive m moving at a speed were given and I was colliding with the block suspended from a wire that has a massive three M given the length of the wire. And now the question we want to know the answer to is what is the maximum angle that the block swings through? First things to a suppose whatever, um, before it, you know what? Well, what&#x27;s the maximum ankle before it starts coming back down? So block is initially at rest. Let&#x27;s call it Be this, eh? So equation 8.25 which is for elastic collision, where one thing is addressed speed immediately after the collision. For the resting thing is to m a over and a bus and b times e A. And so this is two to fight it by four em time. Four meters per second and this is two meters per cent Now. The height the goes to when it swings through an angle theater. Why equals l times one minus. Who&#x27;s that? And you can work out using trigonometry where this comes from. Basically, l minus y is this length that this is L. Why is the height that it goes through? You use the so kowtow to figure out the relationship between these and the So now, knowing what the height is that it goes through. We have the conservation of energy. So 1/2 times three m b b squared. I was going to eat wall three em. I&#x27;m g plans the height which the block ascends. Three ends obliging Cancel out for us. So the coastline of the angle is equal to one minus B squared. Fight it by two g, no. And so if you take you evaluate this and take the article no sign of it defined that our angle the 3.7 it means

Vishal Gupta · Answer

hi. In the given problem, uh, 0.400 kilograms Object swung, You know, vertical circular parks as shown in the figure at the top, Most wine talked a circular track attention in the screen and weight off This object boat will be acting What really don&#x27;t work and centrifugal force Maybe acting on the object in worked really upward direction. If the length of the string is l the expression for this centrifugal forces and the square divided by l So balancing these forces, we get m re square by l mean centrifugal force is equal to wait off the object Last tension in the string. So the tension in the string will come out to be centrifugal force minus bitch off the object. So putting his values we know the mass of the object is 0.4 k g. The speech at the top most point is Eunice full need a per second. So it becomes for swear lend off. The stream has been given as 0.5 and the value off acceleration to gravity is 9.8. So it comes out to be 0.4 16 by 0.5 minus nine point h Or we can say is your bank four into 30 to minus 9.8 or finally it comes out to be 8.8 it Newton. This is the well you off tension in the stream at the top. Most point. Thank you.

Problem 45 Hard Difficulty

A 0.800-kg ornament is hanging by a 1.50-m wire when the ornament is suddenly hit by a 0.200-kg missile traveling horizontally at 12.0 m/s. The missile embeds itself in the ornament during the collision. What is the tension in the wire immediately after the collision?

Answer

$13.6 \mathrm{N}$

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