directions-questions-1-2-and-3-are-short-free-response-questions-that-require-about-13-minutes-to-an

Question

Directions: Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.
A car of known mass $m_{1}$ will collide with a second car of known mass $m_{2}$. The collision will be head on, and both cars will only move linearly both before and after the collision. In a clear, coherent, paragraph-length response, explain a method for determining whether the collision is perfectly elastic, perfectly inelastic, or neither. If the collision is perfectly inelastic, include at least one possible cause of energy loss.

Linda Winkler · Accepted Answer

in this example, we are going to look at collisions and review some of the things that are true about collisions. Um and we&#x27;ll start off with two cars that are going to run into each other with a one dimensional head on collision and we&#x27;ll presume that we know the mass of each car and the initial velocity vector along the line for each of those. And they&#x27;re headed towards some head on type situation. Sorry, that looks a little bit more like a flying saucer, but you get the idea it&#x27;s a car. Yeah. And there interacting on a straight line. So one dimension just to make things easy. Um and the first thing to realize is that as long as the external forces are negligible, they don&#x27;t have to be identically zero. But uh very small forces such as friction, much smaller than the collision forces that these interact with. Um what is true is the sum of the momentum, which is mvs initial have to equal the sum of momenta afterwards after the collision. And that&#x27;s known as momentum conservation. We could actually write a formula for it. In this particular example, we would say that M1 if you want initial minus M. Two uh B two initial equals and then we&#x27;d have to have some picture of the collision afterwards. So let me kind of show a picture after the collision. So we can have a little bit more clear picture of what&#x27;s going on. Let&#x27;s pose that the first car continues after the collision. To the right with some new velocity fee to final Be one final, sorry. And our second car gets bumped backwards. Um Okay, that looks a little bit more like ufo again, but be too final. Our momentum conservation then would have both of those momentum going in the same direction to the right, which I&#x27;m calling positive in this case. Um And so we would have momentum conserved. Um So that&#x27;s true for any collision, that you&#x27;re going to let momentum conservation hold true. Uh It&#x27;s a little bit different and how collisions get classified from there on it has to do with energy conservation. And in particular kinetic energy, a perfectly elastic collision is one in which the some of the kinetic energies beforehand totally add up to the kinetic energies afterwards. So that&#x27;s known as a perfectly elastic collision. And the sum of kinetic energies before identically add up to the sum of kinetic energies afterwards. And we could write another formula for our little example here, We&#x27;d have 1/2 and one V one initial squared plus one half. I am too. The two initial squared is equal to the same sort of some but with the final velocities uh huh. After the collision coming into play, I was and I&#x27;m just about out of space there. Uh Let me see if I can get a little bit more space. There we go. Okay, so we could write down a relationship for that. And if that relationship holds then you have a perfectly elastic collision. A perfectly inelastic collision is one in which you&#x27;re absolutely guaranteed that a lot of energy in the form of kinetic energy will get lost two other forms. So the kinetic energy afterwards is going to be much, much less than this. Total kinetic energy before the collision. Yeah, so energy is never created nor destroyed. It just converts from one form to another. Um But what&#x27;s also true about perfectly elastic is not only is kinetic energy lost, but you also have the condition that the cars stick together. And I can say that simply their final velocities have to agree with each other. They&#x27;re traveling together afterwards, uh stuck together in some way, shape or form. Um Whether there could get all tangled up in each other um Could happen with cars. Uh huh. But with people they could just be holding on to each other. And the question is, where does the energy go? Um How does the energy convert? Uh can again and again and g. That the system had beforehand can get into converted into a number of forms. And we&#x27;ll just write down some possibility possibilities. One is in the form of heat and there are several ideas for how the heat could happen on his friction. Mhm. Another idea is that the atoms and molecules, let&#x27;s just say Adams to make life easy in the cars can start to var vibrate and eventually those vibrations die down. Um There&#x27;s all sorts of ways for that vibrational energy can get transferred into heat. Um Or sound even and that&#x27;s another way that sound, the sound could come out of the collision and that&#x27;s another way for the energy to get lost. Um There&#x27;s also the understanding of the atomic bonds getting broken or actually absorbing some of the energy. Just like if you had to compress the spring um you have to put energy into that spring to make it compress. So atomic bonds are kind of like springs and they can absorb energy or they can break. And that of course takes energy uh to transform atomic bonds. Mhm. Okay so that&#x27;s different between perfectly and elastic uh and perfectly elastic and sometimes it&#x27;s good just to think um Kind of phenomena logically perfectly an elastic collisions are what you might call sticky things stick together whereas perfectly elastic are bouncy. Yeah the objects have a tendency to bounce off of each other but that&#x27;s not a very good quantitative discussion

Salamat Ali · Answer

we can do to mind the final velocity in the party using the law of conservation of momentum. So love conservation of momentum Compare today&#x27;s for the given collision is must one times the last one before collision is equal to mass one. Running out with the last one, plus a master going out with lost e +32 then using since it&#x27;s elastic collision so they can take energy before and after collision will be saying which is canceling out one over half on both sides. Kim Burton is m one. The square is equal to in one way we want Sorry, we one square, uh, plus and two of the two square than expanding this storm and rearranging we get here him two. The two square is equal to m one into remind us we won into re plus me month. We won. Um, Then we can be arranged. Let&#x27;s call it this A We can arrange a and then we can get there. The equation in one into B minus. The one is equal to And two, uh, V two. Let me call this excursion one. So dividing in question one by a Then we get so here, we cancel out. And to me too. Here we have inclusion on in one, the minus me one. So basically this part and this part will cancel out this part. This part this was because Well, so we&#x27;re left with a duty to that is equal to we. Plus, we want then, uh, again substituting the value awfully, to encourage one and rearranging we get everyone. Is it called to we into moss one minus moss too divided by mass one plus muss two in part B Ah, substituting We want in this equation. So we&#x27;ll substitute this Be one that we got will just put it here Then we confined to me to hear so we too will be cool too. We plus me into M one minus and two and two divided by m one plus in two and two. Then further simplifying gives us two times off him one. We divided by in one and one plus in two. So this was but be in boxy, uh, artsy from a B. We, too will be always positive values. Um, the two masses will be in the same direction. Well, the same direction if he won is also opposed to 04 We won in part A. This is true if the mass one is greater than the master you if muss one is less than mask too must too The two masses will be travelling in opposite directions so one will go here and one

Ze-Han Lee · Answer

So in this problem, we have two cars and one and them too. So want to know if this Ah, Clyde on da. This equation is last, Eko. Inelastic. So one method obey. You can simply detected the velocity before the collision. The velocity after the collision. Ah, so if this ah lost the equation than energy would be considered and he was going to see that initial kind of energy to the final kind of energy. If this is a pure a last tick, then you can see that the two cars, they would just stay together, right? They will. They will, they will. They will not be apart anymore. So that will be the totally inelastic. If neither like, ah, after the cliche, the two cars, they&#x27;re separated. But to the energy is not conserved. Then this is just a state between the elastic and elastic. All right. And the so the energy off the if this is a long stick than the energy off the that. I mean, the kind of energy will be transferred to the ah to the two cars to make them dented. Right? So So for example, this is this is the first car and the second cars often cliche. It&#x27;s something like this. Maybe it&#x27;s like this. So as you can see that this is somehow vented and if this is vented, that unit to coast energy to make invented right, so this is the way the energy goes.

Paul A. · Answer

once again welcome to a new problem at this time. We have two cars will given two cars, driving in opposite directions. You have a bigger car with the mass going to call it em. A happens to be seventeen fifty kilograms, and then it has of last e v a, which you call the initial velocity moving in the, uh in the right direction. It&#x27;s a positive velocity, so that&#x27;s gonna be one point five meters per second. Positive. That&#x27;s what we&#x27;re given. And then we&#x27;re also given a second car. I&#x27;m going to call it B aunt. It&#x27;s driving towards the left, has a mass off one thousand four hundred and fifty kilograms, and then it has an initial velocity in the left off one point one meters per second. But I&#x27;m gonna have it is negative because it&#x27;s moving in the opposite direction. So this is positive and that&#x27;s negative. But the other thing that were given is that thes two cars, they&#x27;re gonna bump onto each other so there&#x27;s gonna be some bumping going on, though they&#x27;LL collide. Okay, so this is kind of like the before on then, after you know, after what&#x27;s they collide So this is the second car. When they collide, the final velocity off the fast car is still in the positive direction, but this time it&#x27;s reduced to zero point two five meters per second for the other car after collision. So we&#x27;re saying this is what happens after collision. It&#x27;s going to be going. So both of them are now traveling in the same direction. So the final velocity of the second car happens to be. It happens to be if you check that, you know, we don&#x27;t know what the fine of lost is going to be. So that&#x27;s what we need to find out. Look at want. I want to find out what happened afterwards. Ah, and then also, we want to find the change in kinetic energy. Okay, remember, there&#x27;s no friction in this scenario, so that means we don&#x27;t have any losses, losses of energy in the system. So the the fact that there&#x27;s no friction, the momentum is conserved. Okay, we have, um we have conservation of momentum, you know, that&#x27;s that&#x27;s what&#x27;s going on in this case. Change in Connecticut and you simply is the final kinetic energy of the system, minus the initial kinetic energy of the system on DH. That means we have to account for both the kinetic energy off the two, the two cars. So gonna kick off one part, eh? And the sea that we&#x27;re applying the so momentum mentum after a collision happens to be exactly the same as the momentum before condition. And so the momentum after collision is the Marcel&#x27;s eight times the final velocity off, eh? Plus the muscles. The times, the final velocity of B, which is a second car and then before condition. We have mass off eight times the initial velocity of a plus the Marcel&#x27;s B times. The initial blast. You&#x27;LL be or goes to find the final velocity of B, which is a second car, and we know it&#x27;s it&#x27;s moving, going to see how it&#x27;s moving. It could be moving towards left. It was right. Just depends on the sign off off the solution. So this is m A ve a plus M B of the FBI minus. So moving this one to the right m a the a final. We have all that information and we want to divide everything by the Marcel&#x27;s beat that gives us our final velocity of B for purposes of writing the numbers in the problem. We&#x27;re going to jump into the next page to get the V fi final velocity of B of the second car. And if you can see from that, we have the mass off a seventeen fifty kilograms. Okay, we have seventeen fifty kilograms times the velocity off a which is one point five meters per second. The initial velocity of a one point five zero meters per second on DH then also want to add the muscle B fourteen, fifty kilograms times the velocity initial velocity of B one point one zero meters per second on DH. Then there&#x27;s also another portion ofthe problem which leads to be subtracted and that happens to be the mass off A. On the final velocity of hay, the mass of a and the final velocity of sa was subtracting that seventeen fifty kilograms times. The final velocity of a happens to be if you go back on this side zero point two five meters per second, zero point two five meters per second on all of that is going to be divided as you can see by the mass will be always a good habit to leave the variables up until the end. So this is fourteen fifty kilograms. When you plug in the numbers, you find that the velocity of light a car ends up being zero point for zero eight, eight six, eight, seven. So let&#x27;s let&#x27;s dislike sure that this is this is the right the right process. So we have seventeen, fifty time sworn point five zero plus fourteen fifty no times negative one point one zoo, then minus seventeen fifty times point two five. So we and then we have to divide by fourteen fifteen. That gives us a zero point force, your eight six. That&#x27;s the final lost yield off the smaller car. And it&#x27;s positive, meaning that it&#x27;s moving towards the right meters per second Member was still under the law ofthe, ah, conservation of momentum. Okay, and so that&#x27;s That&#x27;s what we&#x27;re using in this case in the second part of the problem. That&#x27;s but be I&#x27;m going to see how the change in Connecticut and happens. That&#x27;s just the Connecticut and the final off the system minus Connecticut initial off the system and so the final kinetic energy it is um, the Connecticut. The final kinetic energy. Both the bigger car, which we&#x27;re calling a final plus the final kinetic energy off the smaller car minus the final. Our initial kinetic energy off the big car, plus the initial Connecticut Angie off the a smaller car. So that&#x27;s that&#x27;s the process right here. So this is one half Emmy V a final squared plus one half M b v be final squared minus Always used rockets when you have to one half em the initial squared plus one half em be Phoebe initial squared. So that&#x27;s our change in Connecticut. Do we can simplify the whole thing by moving things around. So we have one half emi The final squared this one right here. I want to put them together minus one half Emmy be a initial squared and then bring this on Plus one half em be the the final squared minus one half em be be nation squared. So what we&#x27;re doing right now we&#x27;re doing algebra, were doing a distribution of the miners to both of them and then putting all the airs and all the bees together. So these are there&#x27;s a big one half outside And then we have Emmy of the final squared minus the initial squared. Plus Oh, you have to change this part right here because it&#x27;s a single we pulled. We pulled out the one half outside because it shows up in all of them. So we want to have that. So this is plus em be vehbi. Um, I should have called this V b final. So, you know, it looks like something a little bit different. It&#x27;s a subscript Oh, so we have to bring it back down their final squared like that. Get rid of this. This one right here. This square will need a single one. So in this case, Vehbi final squared minus V b initial square. Don&#x27;t forget that square right there. And that&#x27;s our whole problem. Next part is we just need to plug in numbers. We could use the difference between two squares if you wanted to. But again, that&#x27;s optional up up to you. So the change in kinetic energy becomes equivalent to one half. I will plug in the numbers muscle, a hiss seventeen fifty kilograms and then the plug in the final velocity of A, which is zero point two five meters a second. You know you want to square. I want to square that results. So we have to be very focused on how we simplify this. So you know this this whole thing is spread and then minus the initial velocity was one point five. That&#x27;s for a could see. The initial velocity was one point five meters per second on That&#x27;s also going to be squared and then go back. We see we made a plus must be which is forty two fifty kilograms on then the final velocity off B. We need to get the final blast of future zero point four zero eight six, zero point four zero eight six meters per second, squared minus. If you go back initial velocity off of B, that&#x27;s at the beginning. Where we see it&#x27;s negative one point one zero meters per second, only one for it, one zero me this second and we also need to square that one. We plug in the numbers, plug in the numbers, you get to see that the final final. Um so we have seventeen. Fifty. My start was simplify. This is one point five and so the final final becomes negative. Twenty six. Seventy round, About point three jewels. Okay, so this is the change in kinetic energy. That&#x27;s the change in kinetic energy That happens when you have to, Because moving in the opposite directions and then they bump into each other. Okay. Uh, so negative. Two six, seven point three. Jewels are kilograms. Hope you enjoyed the video. Feel food to share any questions. Comments? Looking for tea, uh, talking to you in the next video and have a wonderful day, okay.

Zachary Warner · Answer

our question, says the look at Figure 9-73 which shows an overhead view of two particles that are sliding at a constant velocity on a frictionless surface. The particles have the same mass and the same initial speed, which is four meters per second. They collide with their paths intersecting. An X axis is arranged to bisect the angle between their incoming paths, such that the angle fate, as you call the 40 degrees the region to the right of the collision is divided into four letter sections by the X axis and four numbered by dash lines. In what regions, um in what region? Or along what line to the particles travel. If the collision is a completely in elastic, be completely elastic, see an elastic and what are the final speeds of the collision? Is di completely an elastic and E elastic says it&#x27;s kind of a conceptual question, so I&#x27;ll walk you through it. So for part A, this is a highly symmetric collision, and when we analyze the white components of momentum, we find that their net value is zero. That&#x27;s the stuff together. Particles are going to travel along the X axis so we can go ahead and right here along the x axis. Alright. Eh? Okay. And for part B is says to treat it as completely elastic this time around instead of completely in elastic. And since the elastic collisions with identical particles, the final speeds are the same as the initial values and then conservation of momentum along each access then assures that the angles of approach are the same as the angles of scattering. Therefore, one particle travels along line two and the other particle is going to travel a long line three. So if you go ahead and write that as well, Part B All right now for part C were asked to consider the same scenario. But this time it&#x27;s in elastic. Okay, And since it&#x27;s an elastic here, the final speeds are less than they were initially. The total X component cannot be less, however, by momentum conservation. So the laws of speed shows up as a decrease in there y velocity components. This leads to similar angles of scattering and consequently, one particle travels through region B, the other through region. See, the paths are symmetric about the X axis. We know that this is on intermediate between the final states that are described in a part B as well as part A. Okay, So for part C, we can go ahead and write the region between X axis and lines two and three. Okay. Now, for part D were asked to find the final speed if the completion complete. Excuse me. Collision is completely an elastic eso. Conservation will minimum along. The X axis leads to two simple observations that the ex components of each particle remained constant because there are identical particles. Okay, so, um, we can write the of FX here. Excuse me. This stylist back out the of f of X is equal to the co sign data where V is four meters per second and data is equal to, uh 40 degrees. So therefore, via beth is equal to two meters per second. So right that here, two commuters for a second boxing and his solution for party get a new page going here for Part E were asked to find the final sweet speed if it is a completely elastic and as noted above in the in the case, the speeds are unchanged of both particles are moving at four meters per second and the final state since they&#x27;re unchanged so we could go ahead and write. The final here is equal to four leaders for second. Boxing is your solution for party.

Kristela Garcia · Answer

in this problem. You have one car traveling south with a 1200 kilogram mess and a speed of eight meters per second and another car traveling west at 17 meters per second with a massive 850 kilograms. And it tells us that these two cars collide and then travel off at an angle. So this is going to be a perfectly in the last two collision. But remember that moment, um, are vectors, so we can actually treat this as vectors before anything else so we can solve for our mo mentum in the UAE direction, we can assault for our momentum in the ex direction and then we can use that to determine our initial mo mentum. It&#x27;s at an angle, and that momentum at the beginning at an angle would have to be equal to this moment at the end when they have collided so we can say our moment in the ex direction squared, plus our moment of the Y direction squared Square rooted will give us this moment. We&#x27;re gonna kind of do this someone big Chuck. So our momentum of the ex direction is just 17 times 8 50 mass times velocity I swear. Plus that 1200 times eight squared gives us a momentum of 17,348 kilograms meters per second. Again, this is our moment before and after. And since it tells us they stick together than this, momentum is equal going to be equal to mass times velocity. So are mass is the total mess. So that&#x27;s that 17,348 divided by my total mouse of 1200 plus 8 50 gives me velocity in this direction of about 8.46 meters per second. Now it also wants us this offer the direction of this velocity. So our direction of this velocity will be in the same direction as its moment. Um so we can actually go back and use moment for this. So the inverse tangent of my why component of momentum divided by my ex component of momentum. So again, inverse tensions opposite over adjacent gives me an angle of 33.6 degrees. Now this angle&#x27;s measured from 180 so that&#x27;s 33.6 degrees downward. If this is 180 degrees. So if we were to measure from zero that would be 1 80 plus 3 36 3 33.6 And the next part asks us to determine how much kinetic energy is lost. So for that, we simply need to calculate my kinetic energy at the beginning, my kinetic energy at the end, and then compare those two values. So my kinetic energy at the beginning, because kinetic energy is a scaler we don&#x27;t need to include direction doesn&#x27;t matter. That one is traveling South and one&#x27;s traveling list. We can just calculate its total kinetic energy. And I get a kinetic energy of 161,000 225 jewels and then at the end there, both traveling together so I can add their masses together and then use the velocity that I found that 800 that 8.46 and this gives me a kinetic energy of 73,360. And then the difference between these two kinetic energies my final minus initial will give me in negative 87,000 800 and 64.1. Jules

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