Basic Physics 3A. Lecture 08.

the quiz oh actually the answers to the quiz were pretty good let me actually just draw it right here the only one i want to point out and i can draw it, i don't have to bring it back up the vector addition one A plus B equals C keep in mind so this is A plus B equals C notice the tail of A and C are lined up and C goes from the tail of A to the head of B few too many of you picked as your answer to A plus B equals C A B and then you drew C this way okay that is wrong because notice you're connecting the head of B to the tail of A that's the difference between those two like i said it wasn't a lot of you but it was enough i wanted to point that out the rest of the questions most people got right obviously it wasn't a hundred percent but most of you got right i don't think it's really worth making many comments if you're not sure why you got it right come to office hours uh... those were sort of very useful basic questions and i was glad to see how well people did but this one was just a little worrisome so i wanted to point that out any questions on that? excellent now the other thing i wanna and you know at this point this may be the last time i highlight these type of answers to what's causing the most problem but it came up so much now i picked this one just randomly trying to figure out what equations to use but this was perhaps some version of this was close to number one or number two is the answer the other number one or two problem was some version of this and i picked three that were similar but said it slightly different ways you know tackling a larger scale problem how do you break it down determine the situation how do you be patient as you try to break it down and what do you solve for first and then build from that notice this should be your biggest problem right now if you're having a problem this should not not that this isn't a problem and i understand that challenge particularly as we move on in the course notice what we're doing right now every single problem we are solving uses one of two equations pretty much and occasionally as a shortcut if you recognize it you can use this one but remember it was part of the other two so choosing the equation shouldn't be too hard at this point in the class because worst case scenario just try all three there's only three to try most of the time you need both of these that's not that will choosing the equation choosing the concept will become a much bigger problem as we go on in the class and add more of these but right now i can just start listing various types of questions where does the object land how far does it go how was it launched this could be speed and or angle how must you launch it which is a slightly different question uh... where was it launched when do you catch the other person uh... how long must you wait do that's an H do you catch the other person tired yet let's see how far away do you catch i'm already up to nine different questions all nine of these are going to be answered by those two equations that's it so the challenge isn't which equation to use you see any of those nine equations i mean any of those nine questions and there's you know at least another five six seven eight variations i can do on this you know what equations you're using it's those two what you don't know right away is where do i put in for time do i have a delay time how do i set it up what am i setting things equal to am i comparing two displacements to each other or do i just need velocity because it's the top of a throw so that's going to tell me when something's zero how do i get the information out of the given problem because for each of these questions i might give you different pieces of information that require you to use the equations differently but it's the same two equations so as you go back and review for the test the question you want to be asking yourself is not necessarily which equation i used because i used one of those two or both of them the question you want to be asking yourself is how did i piece the problem apart how did how did i tackle you know this larger scale problem how did i break it down what did i solve for first and why what were the clues i used in the problem to recognize what to solve for first if that's what you meant by what equation to use then you're asking the same as these but there are two separate issues really in any physics problem there is the equation that you're going to use and there is how you picture, break down, piece apart and look at all the pieces and this is why unfortunately it is almost impossible to do a physics class by just memorizing lots and lots of different answers to lots and lots of different problems you have to figure out the strategy and the pieces and the way it went together what you memorize are a few key equations a few key simple little pieces and then you have to think about how all those pieces go together in the particular problem you're doing any questions on that? the other thing we're going to look at today well between today and friday is another very important set of questions and I've asked the TAs to kind of go over this a little bit in discussion section you have to recognize the difference between the question does it make it and the question does it just make it and if you recall from the first lecture I said one of the most important skills in the physics class is reading you cannot overestimate how important reading is and so if you miss this one little word in a problem it can completely change how you would solve it and what your answer is and so we will do we will do two examples between today and friday that compare these two so you want to pay attention to what is the difference between solving this type of problem and solving this type of problem because sometimes you care whether you're going to just clear the bar the field goal will just clear it, your tennis serve will just clear the net and sometimes all you want to do is clear it you don't care whether you just make it over you want to just you want to make it over so we'll come to that later any questions on that? excellent oh wait, there is one, yes not at all you'll see and we're going to come to this the difference is going to turn out to be how you use those and it will all be because to answer these two different questions you'll be given different information so same equation in both cases but the information you put into it will be different and the variable you solve for will be different but same equation what I'd like to do first go back to this problem that we basically finished on monday I'm going to really quickly set it up again for you because we went a little bit fast at the end from a drawing point of view and just use it to discuss components briefly and then to ask the follow-up question to it which is what angle does the thing land at so that's what we're going to be working on first is the ball with the cliff so let's go to that if you recall I'll put the physical picture up here we had a ball on a thirty meter high cliff it got launched at an angle of thirty degrees it landed some distance away the two questions we're being asked is how far away does it land and then the new question we're being asked is what angle does it land at so keep in mind we said this on monday that answering a question about direction is a question about the velocity the velocity tells you the direction you are traveling by definition the direction of the velocity is the direction you're traveling so that's what we're going to look for to do this we know our two equations are delta x equals v naught t plus one-half a t squared and v final equals v naught plus a t to answer the first question it turned out all we needed to do was draw this picture we had v naught t we had one-half gt squared we had a right angle, we had d and we had thirty meters notice I'm drawing it a bit better today so I don't make the mistake I made on monday this little piece is one-half gt squared minus thirty meters from this picture notice we have what we're trying to solve for is d we want to find d or actually we also have another unknown we don't know t and that gives us two unknowns so we realize as you do more of these, once you realize that there's two unknowns floating around we're going to need two equations now that we're doing two-dimensional motion problems we can probably get our two equations just from the delta x but we might need to use v as well those are our sources of our equations right here, these two if I looked at my picture I saw sine of thirty was one-half gt squared minus thirty divided by v naught t and cosine of thirty was d over v naught t so I had two unknowns I now have two equations and the problem was done so that's that's the quick review I'm going to do it a little more slower now by components but I want you to see the logic I might need to use all of my delta x and vf equations, I might need to only use one remember because they're vector equations and I'm now in two dimensions, there's actually four or more equations combined in those two but I drew my picture labeled everything nicely this time I knew I wanted to find d I look over here and the only other thing I don't know is t, I know g, I know v naught, I know the angle and I know the height thirty so I am able to write down using sines and cosines my two equations I have my two unknowns and I'm done any questions on that setup? now let's ask the question suppose you wanted to do this using components what's it going to look like? well, step one if you're going to use components is always to pick your coordinate system so I'm going to have x like that and y like that now, sometimes you might pick y down as positive sometimes you might tilt it at an angle, we'll see but in this case we're doing it this way now, what I need to do is now basically go through all of my variables what I'm going to use is delta x equals v naught t plus one half a t squared but I'm going to use it in the x direction which I'll write and I know this is a little confusing notice the difference between with and without the arrow delta x is v naught in the x direction times t plus one half a in the x direction times t squared and in the y direction I have delta y is the initial velocity in the y direction times t plus one half a in the y direction times t squared so I need to start writing down all my variables what I want to find is d the distance I travel well d is related to delta x because that equals x final minus x initial which is d minus zero so delta x equals d if I want to know delta y that equals y final minus y initial which is zero meters because I'm on the ground so I draw my picture and call that zero and that's thirty this is x equals zero and then I land over here at x equals d so that zero minus my initial so notice that's negative thirty meters and that negative sign is going to become important later on so we have to keep in mind that it's negative what else am I going to need? I look at my equations I need my velocities in the x and y direction and I need my accelerations in the x and y direction and I need to know about time we already said we don't know time in this case there's only one object flying there's no talk about delaying anything there's no two things that have to be compared so I just have time, t v naught in the x direction well we know v is at an angle of thirty degrees so I use my components so that's just v naught cosine thirty and A x we know is zero because there's nothing happening in the x direction there's no acceleration that way v naught y is v naught sine thirty what's the acceleration in the y direction? what is it? ten? is it ten? or is it negative ten? it's negative ten because remember it's g the acceleration in the y direction is g which is down so it's negative ten meters per second squared notice do I have to worry about whether v naught is plus or minus at this point do I have to worry about a plus or minus sign on the velocity not really but you do have to be careful if I had said I was throwing it down at thirty degrees below the horizontal right now you have to be careful if you pick that to be x and y you need to recognize that the velocity in the y direction is negative and you need to put that in you would need to write negative v naught sine of thirty because the sine of thirty is going to give you a positive number and so you would need to know based on the problem what was going on so that's just a little aside you've got to watch for any questions on our givens? notice this is the more traditional way to do a problem right you write down all your variables in a big long list you have your equations in a big long list and you move forward and that's why I say you want to look at the two methods and see which makes you happier uh... now we write down we plug in so in the x direction we have d equals v naught cosine thirty times t plus zero and in the y direction we have minus thirty equals v naught sine thirty times t minus one half times ten meters per second squared times t squared if you looked up above these are the same two equations we had on monday this one becomes one half gt squared minus v naught sine thirty times time minus thirty equals zero that gives us our t of minus one point five seconds or four seconds we choose the positive one we take the positive one and plug it in here and d equals v naught cosine thirty times our four seconds and we get our eighty six point six meters any questions on that? I really encourage you for this problem because it is about the most straightforward type of projectile motion problem you can have it's one object just moving nothing really fancy look at the two solutions and decide which one you like better so you start practicing that now the initial velocity, that's why I didn't write it in, I couldn't remember I'll go to the problem twenty five meters per second yeah so v naught is twenty five meters per second so that's what gets plugged into both here and here any other questions? Yes. so notice right here our ball was launched up at thirty degrees so there's no negative this, what you want to do right here this is not for this problem it was an example of if the velocity was down I was just warning you about possibly needing a minus sign because I can either throw a ball up at an angle or down at an angle if you're drawing the picture drawing it down takes care of that plus or minus and you wouldn't need to worry about it any other questions? the second part to the question is what angle does it land at? now for this one now you know the answer to clicker one but I don't know how we got there uh... da-dun-da-dum restore? No, that's not going to work how about we do that you all memorized that answer right there, right? you had your chance let's see what happens if I start the lecture over, oh good now it's normal okay where does it land? I just want to look at this real quickly what angle does it land at is all about v final so it's fairly straightforward we draw v naught so that's equal to v naught plus at so there's gt there's v final v naught's at thirty degrees v final's at some angle theta which is also that angle there so the reason I mention this question is you do have to remember a little bit of your geometry when you do physics you have to remember that if I have two parallel lines like this and I have a line going between them the alternate interior angles are equal you're going to have to remember things like that uh... so if those aren't coming back to you that's a good thing to come to office hours with either the TA or myself to ask about but once I've drawn this let's remake it a little bit neater so I have v naught I have gt this was thirty degrees so right here is v naught sine of thirty this is the theta I want for v final because it tells me the direction I'm going in and so this little piece here is my gt minus v naught sine thirty this piece here is my v naught cosine thirty so if I look at that angle theta the tangent of theta is the opposite over the adjacent so it's gt minus v naught sine thirty over v naught cosine thirty and then if I solve for that because I know t from part one I know that already I'm going to get that it's fifty one point eight degrees down from the horizontal in doing this there's one physics thing I knew I knew my final velocity had to point down why is that the only direction the thing I threw could be going at this point is there any way I can be moving up and hit the ground? no, that's just physics you hit the ground when you're going down so that's why I knew to draw it down now what I'm going to do is I'm also going to do this by components but I'm going to do that as a separate little video that I'm going to put up on the web for you to watch so right now I'm just leaving this here and then we'll follow up with that as a video any questions on this because all I wanted to point out two things I want you to get from this if you're asked direction you're going to have to solve for something about the velocity and often you're going to have to know it helps to know its direction ahead of time in this case to know that it's going down now we will have a clicker question which you may have memorized the answer to but here's our problem and I better get my safety goggles we're out trying to capture and tag monkeys well I'm not going to wear my safety goggles because I don't know how to get these things open in the future always wear your safety goggles when hunting monkeys so here's our monkey hanging in the tree and now I'm just going to tranquilize it, it's just a tranquilizer tennis ball okay it's not a deadly tennis ball uh... notice right now we are currently aimed at the monkey, those of you on this side can confirm that our laser sight is aimed right at the monkey's chest that's where we're aimed but I don't know if that's the right place because here's what's going to happen it's a very smart monkey as soon as the gun fires the monkey's going to let go and drop so once the monkey hears the gun firing okay where did where did our clicker uh... uh... there it is okay we'll start the clicker question so as soon as I fire he's going to drop so do I want to aim right at the monkey? do I need to aim above the monkey? do I need to aim below the monkey? or does it depend on something like I don't know how fast my bullet's going now keep in mind the monkey doesn't push off or anything the monkey just lets go and falls as the bullet is fired at the monkey or sorry not the bullet the tranquilizing tennis ball this looks like a demonstration this is exactly the type of homework problem you might get you need to throw something at something and hit it okay so believe it or not this is actually a question that could show up on an exam okay very good this is the best the class has ever done I wonder why okay let's see if it works this always makes me nervous because there are so many things that can go wrong the monkey can forget to let go uh... because they're not a very smart monkey it's not much in the head but ready let's see ah no let's try again oh shoot I'm not hitting the fire button this has been a long week ready, set, go nailed the monkey it worked I'm not even going to try that again notice we were aimed right at the monkey and we hit exactly where the monkey was now be honest how many of you actually thought that was going to work how many of you only answered because you had seen the answer excellent but now you know it works right and I was talking long enough that you could have done the following you had enough time to do the following let's draw our displacement vectors for this so here's the monkey we know for the monkey that the only acceleration is gravity and v naught equals zero so the displacement of the monkey looks like this and it's equal to one half gt squared now if I'm going to hit the monkey and I am starting over here somewhere I need to have my delta x look like that that's the delta x of the ball because what is our rule for a collision you should know this by now same place same time and I also know I only use a single t because it's same place same time so how do I get a delta x here well notice for me since I have to use the same t my one half gt squared vector for the ball has to be exactly the same as the monkey because g is the same and t is the same so the only way I can draw the one half gt squared because delta x for the ball is one half gt squared plus v naught of the ball times t actually let me that's going to be my way let me draw it down here delta x of the ball is v naught t plus one half of the ball plus one half gt squared so that vector goes there and now the only way since this is a plus b equals c and I know how to draw that the only way to make it work is for that to be v naught t so the only thing I can do is aim at the monkey if I aim anywhere else I miss because I don't end up at the same place at the same time so without putting any numbers in if the question was on the test where do you aim to hit the monkey if the monkey lets go at the moment you shoot all you'd have to do is draw this picture and say I aim at the monkey because this gets me there at the same place at the same time and again this is one of the reasons drawing the pictures are really helpful if the monkey pushes off with an initial velocity okay before when you fire then you draw a different picture then I'm drawing for the monkey v naught t of the monkey plus one half gt squared and that's where I end up now for me I still draw the same delta x t is still the same so my one half gt squared has to be exactly the same as it was before and as it is for the monkey so now in this question where do I aim do I aim at the monkey below the monkey or above the monkey and I know I hid the picture real quickly but I want you know hopefully you can think about it so now if the monkey pushes off down with a velocity do I aim at the monkey below the monkey or above the monkey five more seconds four three two one excellent below the monkey almost everybody got that right because if I come and look at my picture the only place I can aim I've got to connect my delta x with my one half gt squared with my v naught t and that's where I'm aiming and the monkey started up there and now if I asked you if the monkey jumps up off the ledge you would also know where to aim because you can work it all out then it gets a little trickier because then you've got to kind of figure out you know where all of these are going but you'll have to aim above any questions on that there is great great power in physics in drawing the picture and drawing it correctly that's the lesson to learn from here that's the vector for the ball in this picture red is the ball and the black ink is the monkey okay now let's move on to our next so there's at the object our next problem so we have two more example problems to do this week two more main ones the first one is this football players trying to make a field goal the field goal post is thirty five meters away and ten meters high they're going to kick the ball at twenty five meters per second at an angle thirty five degrees from the horizontal and our question is do they make the kick? that's our question we need to know what that means and so let's answer this clicker question what does it mean to make the kick does it mean that the final vertical displacement is greater than the bar height does it mean that the final vertical displacement equals the bar height does it mean that the final vertical displacement is less than the bar height or do we need to know both the bar height and the direction of the velocity that the ball is going I'd like you to talk to the people next to you about this one now as you discuss about half of you are going with you need the velocity and half of you are not thinking the velocity is relevant try and converge on that is the velocity relevant? do you need to know if it's going up or down? or do you not need to know it? that's the question you need to decide right now I mean going up or down yeah obviously I have to kick towards the bar to have any chance of making it I said I kicked it at the field, I mean what football player is going to kick it that way to make a field goal over there a little common sense is needed in solving physics problems okay this is not a trick this is reality so if you wish to change your answer now based on that if that was your reason direction of velocity I mean up or down this is interesting okay everybody vote it looks like there's one vote missing or did a clicker leave five four three two one we're still pretty split somebody make a case for needing to know the velocity okay half of you think you need to know the velocity somebody make a case for it, why would we need to know whether the velocity was up or down to know if we made the kick see now that you have to say it out loud it's getting you nervous huh okay they don't want to commit those of you who don't feel you need to know the velocity anyone want to commit to that? yes if you're above the pole you're above the pole I mean I thought that was said very well if you're above the pole you're above the pole it doesn't matter it's not like if I'm going down and I'm above the pole suddenly magically I hit it and the field goal doesn't count I mean how many people have ever watched the football game only a few of you okay let's pick a different sport next time that's clearly the problem it's the sport if you're above the pole you're above the pole that's why you have to be now let me ask you this again this is the problem with physics we do sometimes ask you to know about stuff that you may not have watched and it is it is perfectly reasonable if you have never seen a football game or have never seen a basketball game to ask a question if you're a little confused let's suppose I was shooting a basket and I want to know if you made the basket then would it matter whether the ball's going up or down yeah because to make a basket the ball goes down through the hoop unless you're standing under it and you shoot it up but they don't count that so there are times where the direction of the velocity matters this is not one of them this is just being over and this would work if I was a high jumper going over a bar if I was a home run trying to go over the fence if I was a car trying to make a jump over a bunch of other cars right if I'm doing an evil stunt it doesn't matter which direction I'm moving I just have to get over the stuff so that's clearing something so there we just have to find a greater than the bar now let me ask you this suppose I asked a slightly different question and there's going to be a homework problem you have to do that's like this and we'll do this maybe as an example I think the last example may have a little bit of this in it suppose I wanted to know what it would take to just make the kick what does it mean to just make the kick in this case the final vertical displacement is greater than the bar it's equal to the bar height it's less than the bar height again the velocity is relevant now at this point now the velocity is relevant okay ten more seconds five, four, three, two, one a lot of people have another class to go to there we go a little more variability here a couple people still think the velocity might matter a couple people just wanting it higher this is really b this type of question you want that's what I warned you at the beginning of the lecture in physics when something is just, if you see the word just just stop just clear just make it just fall anytime in physics we use the word just we want two things set equal to each other that's what we're going to do the reality is to just clear the heart the bar you have to be infinitesimally teeny teeny teeny teeny tiny amount higher than the bar or you hit the bar but you can't put a number to that and so when we say just we solve for the case of being equal and then you know as long as you do anything more than that you're clear and one of the key ways you'll know that things are different in this problem right here i was given the speed and the angle so i know i know v naught and what i'm trying to find is how high i am i know the angle and i know how far away i am so i know everything about the beginning and i'm being asked what's the height and i'm going to have to ask the question is it bigger or less than the ten meters if i was asked the problem about just clearing the difference would be you wouldn't know something about one of these you might not know v naught you might not know the angle you might not know how far and you're going to put in the ten meters because you're going to be solving for it to exactly hit the ten meters and you're going to use that to find the missing piece of information to find how you make it just clear so this case you're given all the stuff about the beginning that's this problem and you're asked how high it is basically does it make it is asking you how high is it here if you're going to just make it you know what height you want and you're being asked something about the beginning so work this at home see if you can get the answer by friday if it makes it or not and we'll solve it friday see you then