Lecture on Momentum and Impulse

let&#39;s take a little bit of time here today and talk about momentum in impulse our first unit of year we defined momentum on the very first day and kind of in everyday language is how hard it is to stop something but in the end we know that that comes from the real definition of momentum which is the product of an object&#39;s mass and velocity and really that&#39;s book that boils down to an equation P is equal to M times V the units for momentum of course we can get the unit&#39;s from here you get the units from any equation right you&#39;re ever wondering what the units for a variable is you can get the units from any equation for this one we&#39;d say because it&#39;s mass times velocity we&#39;d say its kilograms times meters per second there is another set of units that we can use for momentum as well that we learned later on when we talked about impulse does anybody happen to remember right now what that other set of units is besides kilograms meters per second not Newtons per second for your clothes yep good it&#39;s Newton&#39;s times seconds you don&#39;t need to worry about converting back and forth between those two variables between those two units because in the end one kilogram meter per second is exactly the same thing as one Newton seconds we just use usually use kilograms meters per second when we talked about momentum in Newton&#39;s seconds when we talked about impulse but really they&#39;re the same thing um listen this these are pretty straightforward right you got to be able to take the velocity you gotta be able to take the mass in the right units kilograms and meters per second and multiply them together to get momentum you also have to be able to rearrange them solve for mass we would say it&#39;s P over V and of course to solve for velocity its P over m if the entire course is this easy and straightforward we are laughing hey I don&#39;t anticipate getting any kind of crazy difficult questions on your diploma exam with just straight momentum usually the more complicated questions come from putting momentum along with impulse or conservation momentum now remember that little little things right momentum remember that when is a vector it has a direction associated with it we can remember that if we do happen to forget it by remembering by taking a look at our data sheet and noticing that there&#39;s a little half arrow over top of the momentum sign as there is with velocity as well velocities vector mass is a scalar we remember that because there&#39;s no half arrow on top of the mass here it has a direction associated with it if the velocity is to the right then the momentum is to the right if the velocity is south then the momentum is self finally I added one more little thing in here says momentum after acceleration generally we&#39;re not dealing with a problem like that but every once in a while it does rear its head we&#39;re looking for momentum after an acceleration it might be a problem like you know you drop a ball from a height of 10 meters how fast sorry what&#39;s the momentum of the ball as it strikes the ground or a car accelerates from rest for 28 meters what&#39;s its momentum after that acceleration phase it makes the question a little bit more tricky but in the end all we got to do there is still use P is equal to M times V but find the velocity some other way like maybe VF squared is equal to VI squared plus Q ad maybe a is equal to Delta V over delta T whatever equation happens to work to get the velocity the bottom line is momentum is a quantity that exists at a single moment in time what&#39;s the momentum now or what&#39;s the momentum now what&#39;s the momentum at the beginning of the problem what&#39;s the momentum at the end of the problem hey this would be this would tend to be a what&#39;s the momentum at the end of the problem and if the object is accelerating then we would have to find the velocity at the end of the problem in order to plug into that equation now one final thing before we move on to impulse which will spend more time on because it tends to to appear on an exam a lot more often we talked yesterday I think it was about the 10 physics principles the first physics principle that&#39;s on your data sheet in the bottom left hand corner is the uniform motion uniform motion is when we have of these they could have Delta D over delta T right whenever we use that constant velocity equation Hey I doubt that you would get a question that asks you for the physics principles if it was just a straight momentum question but if you do it will either be an I missed one here it will either be uniform motion if we&#39;re dealing with a constant velocity if the object isn&#39;t changing velocity which would be the majority of questions involving straight momentum but it could be accelerated motion as well it would be accelerated motion if it was one of these situations down here where you&#39;re looking for the momentum after it speeds up to a certain range again I don&#39;t think you&#39;re gonna see those two physics principles or any questions involving physics principles or it&#39;s just a straight momentum question but if you do it&#39;s gonna be either uniform motion or accelerated motion it won&#39;t be both gay it won&#39;t be both because you&#39;re not going to be going at a constant velocity and accelerating at the same time we&#39;re talking about impulse and you can see right away the physics principle is the really the only physics principle that you&#39;ll be dealing with when we talk about the concept of impulse might be accelerated motion why because what is impulse it&#39;s the change in momentum if momentum is changing then it must be accelerating doesn&#39;t mean it&#39;s speeding up could be slowing down but the velocity is changing it&#39;s accelerating therefore it&#39;s giving us a change in momentum and impulse I&#39;m just gonna say for the definition right here Delta P right you guys remember that Delta P is change in momentum and of course the units would be exactly the same thing is we have for momentum would be kilograms meters per second or it would be Newton with time seconds now we derived an equation for this we&#39;re not going to derive it again because in the end it&#39;s on your data sheet says F times Delta T is equal to M times Delta V what we usually do is tack on a little Delta P in front of it just to remind us of that equation that two-part equation or now that three-part equation if I&#39;m saying Delta P is equal to F T equals M times Delta V that whole thing is equal to the impulse F times T is a change in momentum M times Delta V is a change in momentum I&#39;m gonna circle something in red here that is meant to draw attention to a really really common mistake that people make the equation is not M times V it&#39;s M times Delta V if you guys remember on the response test that you did for momentum many of you lots of you calculated the force of impact between those two carts you guys remember that question two cars collided you had to do a whole bunch of stuff including fine in the force of impact on car one and the force of impact on car 2 you guys remember that lots of you calculated the force of impact using impulse you have the right idea but you set f times delta T equals M times V naught M times Delta V you&#39;ve got to make sure that you&#39;re using the change in velocity the other thing that I want you to be careful of here when you&#39;re solving for impulse if there happens to be two or three or four objects you&#39;re always using one of those objects don&#39;t get mixed up and say oh oh look the initial I have the initial velocity of the blue car and I have the final velocity of the red car it doesn&#39;t matter those can&#39;t go together when you&#39;re solving an impulse problem you&#39;ve got to have the initial and final velocity of the blue car or of the red car but not of you can&#39;t mix and match initial and final velocity for two different vehicles now of course we got to be able to rearrange this but this is pretty straightforward algebra right nothing complicated about this you want to solve for f usually it&#39;s going to end up being M times Delta V over delta T although sometimes it might be Delta P over delta T depending upon which two parts of the equation you&#39;re using the other thing that you&#39;re often going to have to rearrange to solve for would be Delta V and of course Delta V is going to be F times delta T or we could just say Delta P over the mouse all right I&#39;d like to take a look right now at the first of 15 multiple choice / numeric response questions that we want to do today then refer you to question number 2 it&#39;s on the second page of the book that I handed out to you I&#39;m gonna give you a couple minutes to look at that question numeric response number to see if you can solve it and then we&#39;re gonna take a look at it one thing that I&#39;ll caution you about is you look into this question pay attention to things like Delta V as opposed to V right pay attention to things like one object not two objects and another thing I should have mentioned this before pay attention to vector nature what do we mean by that the direction if we got something going to the right and then something going to the left make sure you pay attention to the fact that one&#39;s gonna be a positive one&#39;s gonna be a negative all right well set you free almost for a couple minutes here and then we&#39;ll take a look at it all right most people are finished here we&#39;re gonna use impulse to solve this problem right clearly that&#39;s what we&#39;ve been talking about for the past 10 minutes or so here guys but why if this is a question on a test and your teacher hasn&#39;t been standing up at the front of the exam room talking about impulse for the last five to 10 minutes how would you know that this is an impulse question what leads us to use impulse to solve this question it&#39;s two words in there that tell me right away that this is almost certainly gonna be an impulse question not two you know what they are is your hand going up there all right we got a change in momentum here sure sure the the speed is changing we&#39;re dealing with a mass okay if that&#39;s the case and there is gonna be a change in momentum although that&#39;s often gonna be the case we&#39;re still not necessarily gonna use an impulse to solve the problem even when there is an impulse depends on sometimes what we have given to us two words are yep it&#39;s a force at a time when we see a force at a time whether we have both of them or we have one of them then we want to solve for the other one it doesn&#39;t really matter that&#39;s we&#39;re gonna use impulse so I&#39;m gonna say F and T we&#39;re gonna use impulse to solve that problem or at least we&#39;re gonna try to use impulse to solve that problem Hey worst case scenario we try impulse it doesn&#39;t work for me it&#39;s not gonna lead me astray because we know if there&#39;s a force in a time that there it is an impulse whether it leads us to the correct to the answer or not all right Zoe what do we know about impulse we know that Delta P is equal to F times delta T and it&#39;s equal to M times Delta V looks like by our Givens it&#39;s the last two parts of the equation that I want to use and I&#39;m gonna rearrange that to solve for F it M times Delta V over delta T notice what I wrote down there M times Delta V naught V but Delta V over delta T we got a mass here it&#39;s five point seven times ten to the minus two and we got to make sure here that when we put in Delta V we take into account not only the fact that there is two velocities here an initial and a final and there always has to be in one of these questions even if one of them is zero even if it starts at rest VI is zero there&#39;s still two velocities we also have to pay attention to direction here as well and one of the things that I probably should have done when I read through this question is to circle the north and the south to reminded me when I plug my numbers in and I&#39;m not going to forget to put one of them as positive one of them is negative the final velocity is 32 meters per second so it&#39;s going to be VF minus VI 32 minus negative 35 another common mistake first of all forgetting to put the 35 in there to begin with second of all forgetting to put the negative 35 making it 32 minus negative 35 divide that by 4 times 10 minus 3 and when I do that math I end up get nine point five five times 10 to the 2 anybody else get that we&#39;d express our answer ABCD as 9 5 5 2 how many people got that good you know what that&#39;s 2% I near to poem exam right there all you need all you need is another 88% and you&#39;ve done fantastic on your diploma exam all right let&#39;s go back to our physics 30 and 16 pages for a second again we we kind of address this in the question already the vector nature of it then the direction nature of it here there is a direction associated with impulse there is a direction associated with with the momentum there is a direction associated with the velocity in the end if you have a positive velocity negative velocity right left north-south if you have a positive impulse that&#39;s gonna mean a gain in momentum if you have a negative impulse that&#39;s gonna mean a loss in momentum now you can also think of in a different way a positive momentum or positive impulse I should say a positive impulse means a gain in momentum but it can also mean an impulse to the right or an impulse to the north or an impulse in the positive direction a negative impulse means a loss in momentum but we can also think of it as I&#39;m gonna mentum to the left or I lent them to the south or a momentum down sorry an impulse I should say an impulse to the south and impulse to the left and impulse downwards and so on so really these mean the same thing when you gain momentum it means it&#39;s in the positive direction when you lose momentum it means the impulse in the negative direction and this is redundant I should cross that off because I already put that I guess the equation and being able to rearrange the equation let&#39;s go back to this predetermined impulse versus non predetermined impulse this is the whole qualitative nature of impulse remember we talked about these are my words right predetermined non predetermined impulse but they&#39;re what they mean is that in any situation in any situation we either know what the impulse is going to be or we don&#39;t know what the impulse is going to be when I&#39;m driving down the car the highway in my car I know I&#39;m gonna stop eventually I know what my impulse is gonna be I know what the mass of my car is I know what the velocity of my car is therefore I could calculate the momentum of my car and I know at some point my car&#39;s not gonna be moving my momentum of my car is gonna be zero so the impulse would be zero minus whatever the momentum of the car is right now it doesn&#39;t matter how the car comes to a stop the impulse will be exactly the same so typically we&#39;re talking about a predetermined impulse when you&#39;re stopping an object when the car is coming to a stop whether it hits the wall or whether it comes to a gradual stop at a stop sign when the baseball comes to a stop whether it stops because if it&#39;s your glove or whether it stops because it&#39;s just moving through the air and ends up hitting the ground rolling for a while and eventually coming to a stop or whether hits your bare hand the baseball is coming to a stop we know what the impulse of the car is we know what the impulse of the baseball is going to be no matter how it comes to a stop the other situation is the non predetermined impulse this is when we don&#39;t know what the impulse is going to be situation there is usually when we&#39;re starting an object the all sitting on the tea I know it&#39;s momentum is zero right now but after I hit that golf ball I don&#39;t know what its momentum is gonna be the baseball that&#39;s in my hand I know it&#39;s momentum right now but I don&#39;t know what its gonna be after I throw it its impulse isn&#39;t predetermined the car that&#39;s sitting at the stop sign I know it&#39;s momentum is right now but I don&#39;t know what it&#39;s gonna be two minutes from now I know what its gonna be I know what the impulse is when the car is stopping but I don&#39;t know what the impulse is gonna be when the car is starting that&#39;s a non predetermined impulse when an object is stopping the impulse doesn&#39;t change the momentum does but the impulse doesn&#39;t change the impulse is going to be the same no matter how it comes to a stop so if the impulse isn&#39;t going up or down then generally what we&#39;re gonna try to do is increase the time that it takes to stop the object which is going to have what effect on force smaller that&#39;s your impact cushioning thing right that&#39;s the airbag in the car your head hits the airbag your heads gonna have same impulse whether hits the airbag or the steering wheel or the pavement in front of your car your head&#39;s gonna have the same impulse but the airbag spreads the impulse out over a longer period of time it takes longer to stop which decreases the force when you catch a baseball and catch a baseball like this and kind of cradle it as opposed to sticking your hand out rigid like this and not bending your elbow right you create a lot like this to increase the time that it takes to stop to decrease the force spread the impulse over longer period of time increased time decrease force on the other hand when you&#39;re starting an object it&#39;s getting a little bit of cluttered here but try to stay within here when you&#39;re starting an object the impulse is not predetermined when you&#39;re starting an object increasing the force can increase the impulse force and time don&#39;t affect each other when it&#39;s not a predetermined impulse force and time both affect impulse when it&#39;s when you&#39;re starting something when it&#39;s not a predetermined impulse so increasing the force increases the impulse hit the golf ball harder you give it more momentum the other effect is time increase the time of contact you don&#39;t decrease the force this time rather you increase the impulse so hit the ball harder you give it more momentum hit the ball for a longer period of time you give it more momentum you understand the difference here how force and time are related to each other in this one increase the time decrease the force by the way you could go the other way right occasionally you might want to increase the force that&#39;s the case you&#39;d want to decrease the time bottom line is time affects force though not impulse but in this one when you&#39;re starting the object time doesn&#39;t effect force time affects impulse and force also affects impulse linearly all right let&#39;s take a look at another question this is multiple choice number three on the booklet that I gave you earlier today all right let&#39;s have a look at this one now it says airbags are a common safety feature installed in vehicles when a vehicle suddenly stops the air bag inflates um what&#39;s the purpose of that airbag we know this right it cushions an impact if any question you see involves cushing and impact we know what the answer is already whether we&#39;re talking about the airbag or whether we&#39;re talking about the baseball glove or we&#39;re talking about the hockey Pat goalie pads or are you talking about why you bend your knees when you jump off of the desk in the end what it is is the impulse which is predetermined we want to spread that over a longer period of time reduce the force let&#39;s read what it says though the force on the driver is less the impulse on the driver is greater the stopping time of the dot the driver is longer which is the state numbered above compares what happens in an automobile collision when in which an air bag inflates with which with one in which the air bag does not inflate okay so the air bag inflates what does that do does that change the force on the driver sure it does that&#39;s the purpose of the air bag right what does it do it decreases the forest on the driver the impulse stays the same time goes up force goes down time goes up right stopping time and the driver is longer yes is the impulse greater no one no the impulse is the same no matter how your head comes to a stop the impulse is the same the answer is gonna be options one or statements one and three here and that&#39;s cap take a look at one more multiple-choice question as well before we move on to our next little thing here this is number one please on the first page alright this question involves two collisions of pucks on an air table it says in the first one a stationary puck puck one is struck by a second puck puck to the massive puck two is twice that of puck one so masses are different it&#39;s a collision between two pucks where the masses are different there is no friction so we don&#39;t need to worry about that basically hey what are they telling us when there is no friction in this we have n what&#39;s the word I&#39;m looking for we have n blank system we have an isolated system there&#39;s no external forces acting yeah that really comes up later on we talk about conservation momentum okay but it kind of appears a little bit here in this question it&#39;s not really a big deal but we can at least talk about that&#39;s just a little bit here we have an isolated system there&#39;s no external forces so we don&#39;t need to worry about anything like that here a collision between two pucks one puck is heavier than the other next when we have the collision between two pucks head-on this time the pucks have the same mass they bounce back in the direction opposite to the original direction once again we have an isolated system there are no external forces all right what&#39;s the question here it says for collision one the magnitude of the impulse of the puck on the impulse on puck one a one on puck two is blank to the to the impulse of puck to unpack one what&#39;s the answer there is it equal to or is it less than because the difference in masses who says it&#39;s equal who says the magnitude of the impulses of s the magnitude of the impulse is equal okay it is equal and the reason it&#39;s equal is because ultimately if object a applies a force on object B object B will apply an equal and opposite force on object a right Newton&#39;s third law which is really just a statement of impulse really when it comes down to it here&#39;s what happens puck number one loses a certain amount of momentum or gain a certain amount of momentum let&#39;s say it loses a certain amount of momentum where does it go if it loses ten units of momentum where does that ten units of momentum go to the other guy to the other puck so if one of them loses ten the others gotta gain 10 it doesn&#39;t matter that they have different masses or anything of the sort okay in the end if momentum is gonna stay the same and then one gains momentum the others got to lose to the exact same amount of momentum opposite signs but same magnitude what about collision two for collision to the magnitude of the impulsive puck one on putt two is equal to again for the exact same reason it doesn&#39;t matter that the masses are the same here okay in the end if one guy gains ten the other guys gotta lose ten other wise momentum momentum has disappeared somewhere that can&#39;t happen this will always be the answer to a question like this if you&#39;ve got a collision between two objects and one of them gains five units of momentum the other must always lose five units of momentum even if one mass is 50 times the mass of the other one the magnitude of the impulse will always always be the same okay let&#39;s take a look at another little aspect of impulse here which is these forced time graphs I&#39;m gonna start that off by asking you those three words that I asked you way back in February when we did this for the first time when you see a force versus time graph and it looks something like this you should always think three words kind of giving it away there because area grows impulse yeah area equals impulse I don&#39;t care if you&#39;re asked to find the impulse or you&#39;re asked to find something else it doesn&#39;t matter if you see a force versus time graph you think in area equals impulse it doesn&#39;t matter if it&#39;s shaped like this if it&#39;s shaped like this if it&#39;s got some kind of crazy shape it doesn&#39;t matter you want to think area is equal to impulse now I&#39;ve made a note here about units on axes you got to be really careful with this okay you think the area should be easy right if it&#39;s a rectangle then it&#39;s base times right if it&#39;s triangle it&#39;s one half base times say a generally pretty easy shape see pretty easy equations but what we often miss is the fact that the time might be in milliseconds or whatever or it might say times ten to the minus three seconds and the force may be in kilonewtons or or millon Newtons for that matter or it might say times ten to the two Newton&#39;s or whatever the case may be make sure you pay really close attention to the axes on the units if it&#39;s in kilonewtons in milliseconds then convert it when you&#39;re finding the area to Newtons and seconds if your area is 1/2 of 200 milliseconds times 30 kilonewtons then what I want you to do is change that make it one half of 0.200 seconds times 30,000 Newtons make sure you&#39;re in standard units when you&#39;re calculating that area area close impulse make sure you&#39;re in standard units once you find the area well you&#39;ve got the impulse and then sometimes what you&#39;ve got to do after you found the impulse it&#39;s fine something else like the change in velocity or the mass and what you&#39;re always gonna do there is step this equation R equals impulse also equal to M times Delta V do not set it equal to this do not set it equal to F times delta T even though that&#39;s a valid equation do not set it equal to M times Delta V in this context here is why the force here in this graph is almost always going to be changing what value of f would we use if we wanted to use that equation in conjunction with it technically it&#39;s valid but you&#39;d have to use the average force and usually the average force is pretty hard to find on this graph so just don&#39;t even bother going there area is equal to impulse and that&#39;s equal to M times Delta V so if you&#39;ve got the impulse by the area given the mass you could find Delta V or the other way around given Delta V you could find the mass so there&#39;s three variables in here really but you have to be able to find impulse get the area mass you take the area divided by Delta V Delta V take the area divided by mass one final thing before we do an example here make sure you pay attention to what&#39;s above and below the x-axis if you got a graph that looks like this then this area all three sections of it will be positive this area below the x-axis will be negative and what we have to do to find the total impulse there is add them together plus hundred plus negative 50 or whatever the case may be yet the impulse that&#39;s what you&#39;re looking for great if you don&#39;t want to find mass or change in velocity do your thing just make sure you&#39;ve got the right value for impulse based on whether it&#39;s above the x-axis or below the x-axis sticker will get another multiple choice slash numeric response question now this one&#39;s going to be question number four on the third page of your booklet trying to take a look at it here this question says in a car safety test we&#39;ve got a 1200 kilogram car I directed towards a solid wallet it collides with the wall it comes to a rest and then we got a force versus time graph here on what&#39;s the first thing you think of when you see this graph versus time graph three words area equals impulse area equals impulse should be those three words every single time we haven&#39;t even read the question yet right we just know the context of it we know the information that we given we&#39;re thinking area equals impulse regardless of what we&#39;re asked to find okay something that I noticed here though almost missed it when I was doing my answer key gotta be always aware of this always alert for this what is it that I&#39;m talking about why did I pick up the red pen yep we got kilonewtons on the y axis right we got to be a newness and yeah kilonewton is 10 to the 3 Newtons so it&#39;s a nice easy conversion if we have 20 kilonewtons we have 20 times 10 to the 3 Newtons there anything else I should circle and right here we have seconds on the x axis but it does say times 10 to the negative 1 seconds so it&#39;s easy when you&#39;re doing the base here from 3 to 8 just to say the base is 5 so I really 5 it&#39;s 5 times 10 to the minus 1 seconds so let&#39;s be careful of both of those things we can find the area of this easily but we have to pay attention to those units as well all right what does the question actually say what&#39;s the magnitude of the impulse experienced by the car all right good and what&#39;s the initial speed of the car we&#39;re going to find two things ain&#39;t no worries let&#39;s get the impulse here first by finding the area the area is equal to Delta P which is equal to because it&#39;s a triangle 1/2 of base times height the base here goes from 3 to 8 and I&#39;m not going to forget that it&#39;s 5 times 10 to the minus 1 seconds because I circled those that 10 to the minus 1 seconds on the x-axis the height of it it goes from 0 to minus 70 so I&#39;m going to say the height of it is minus 70 times 10 to the 3 Newtons or we could say it&#39;s minus 70,000 Newton&#39;s we math the area works out to be the area works out to be negative 17,500 that right okay negative 17,500 that&#39;s the impulse here okay that&#39;s the first thing that I&#39;m looking for now I&#39;m looking for the change in the initial speed of the car it says comes to a stop so the final velocity is zero now what I&#39;m gonna do is set Delta P equal to M times Delta V and solve for Delta V Delta V works here because although I&#39;m technically finding final velocity sorry initial velocity since the final velocity is zero the change in velocity will be the same magnitude as the initial velocity so we&#39;re gonna say it&#39;s negative 17,500 over the mass here which is 1200 kilograms we do that math change in velocity works out to be negative fourteen point five eight three three meters per second so if the final is zero and it changed by negative fourteen point five then the initial must have been positive fourteen point five what do we got here impulse is one point seven five or 1.8 times ten to the four Newtons and the change in velocity is negative fourteen point five eight or negative fifteen meaning the initial velocity must have been paused at fifteen how many people get a four that good everybody missed the changing units there the kilonewtons are the ten to the minus one seconds okay one more than I&#39;m gonna get you to do before we move on to our next stop actually question number five on the next page who&#39;s gonna answer this question now who thinks they have an answer this question nobody wants to volunteer all right a couple votes for a I got a couple of votes for C here let&#39;s take a look at the question it says plutonium-239 recovered from dismantled atomic weapons undergoes alpha decay in beta decay at the same time sensors on the outside of the containers used to store it him to indicate how much radioactive material was contained by monitoring the gamma rays emitted by the material since a particular container holds this much waste it&#39;s got a half-life and it talks about the energy of the gamma ray and the speed of the alpha particle so on the question says which of the following diagrams shows the orientation of the Momentum&#39;s of the alpha particle and the gamma photon and must result in the maximum impulse of the plutonium 3239 nucleus listen when you see this question you now looking at numbers this has nothing to do with the numbers that are given to you you got to look bigger picture here you&#39;ve got blue tone &#39;i&#39;m 239 and here&#39;s the big plutonium 3239 it decays into something else okay some other slightly smaller nuclei and alpha particle in a gamma ray right now the momentum of this plutonium 239 is zero if we want to maximize the impulse and that&#39;s what the question says right to maximize the impulse then we need to give it more momentum it has zero my item we need to give it lots of momentum how are we going to give it lots of momentum maximize the amount of momentum we&#39;ll give it well let&#39;s take a look at D for a second here if the if it decays in the alpha particle goes this way with let&#39;s say let&#39;s say the value of this would is 10 if it goes this way with 10 units of momentum and the gamma ray goes this way with 10 units of momentum these two cancel each other out what&#39;s going to happen to the remaining daughter nucleus it&#39;s going to stay where it is look the momentum was zero beforehand if they&#39;re going in opposite directions afterwards and anthem there&#39;s got to be zero afterwards as well so B can&#39;t be the answer let&#39;s take a look at C now that&#39;s the one that Kari suggested there well if the alpha particle goes this way with let&#39;s say 10 units of momentum this looks like it&#39;s a little bit bigger maybe 12 units of momentum then the daughter nucleus is going to go this way with how much we could use the Pythagorean theorem to find out how much momentum of the daughter nucleus was right here what we&#39;d have to do components but we know that the daughter nucleus would go back this way with a certain amount of momentum he in the end this is the one that we that we care about here because if the momentum of the alpha particle is 10 units to the right in the gamma ray is 10 units to the right then the daughter nucleus must end up having 20 units in the opposite direction you can&#39;t get if this is 10 in 12 or 10 I think that&#39;s supposed to be the same length actually 10 and 10 you can&#39;t get 20 out of this guy by having two tens at right angles to each other you can&#39;t get 20 out of this guy we&#39;re getting two tens at funny angles here hey you get the X component of each of them combining back here which is less than ten that makes sense understand why is the answer there all right let&#39;s let&#39;s take a look at our next topic now is stopping number 15 conservation of momentum in one dimension as opposed to conservation momentum in two dimensions you can see that if you get a problem involving this it&#39;s pretty easy physics principle to identify if you happen to have one of those questions involving physics principles its conservation of momentum right hey there&#39;s nothing else really that can apply here now that&#39;s not to suggest that you can&#39;t get a question with with momentum involved like this but also something else leading up to that maybe you&#39;ve got conservation of energy where something is speeding up to a certain speed and then there&#39;s a collision taking place where it&#39;s conservation of momentum but if it&#39;s just a question like we&#39;re about to see then we&#39;re dealing with one physics principle principle number four conservation Allah mentum all right we talked a little bit about this one earlier an isolated system that&#39;s a system in which is no external forces acting no friction no no wind or anything like that in the end we can&#39;t always isolate a system from external forces by removing them for those forces so what we often see is a course we&#39;re asked for the velocity of an object immediately after or immediately before a collision that means that yeah maybe there is friction but since we&#39;re dealing with the velocity immediately before and immediately after we haven&#39;t given friction a chance to really do anything yet so we still consider them to be an isolated system because although there are external forces they they don&#39;t come into play until the object just had a chance to move after the after the collision explosion has taken place all right when do we use this this whole idea of conservation element and this whole idea that momentum stays the same when we have an isolated system well we have a collision or we have an explosion even by collision we mean literally two things hitting each other can explosion literally means one thing that it breaks apart into two or more things explosion doesn&#39;t necessarily mean a bomb breaking up it could mean something like a gun being fired bullet goes one way the gun recoils the opposite way or maybe maybe a balloon filled with air okay you have the balloon filled with air and you got the end of it held with your fingers and then you let it go the air goes one way and the balloon goes the other way right if the it here&#39;s a good question not so much concentration momentum but takes it&#39;s back to something that we did here earlier today if I&#39;ve got that balloon in my hand filled with air and then I let it go the air goes one way the balloon goes the other way if the balloon happens to gain 10 units of momentum to the north then what&#39;s the momentum of the air that&#39;s going out the back of the balloon if the balloon was in my hand and now when I let it go the balloons going forward with 10 units of momentum then the air must be going backwards with 10 units of momentum because the impulse experienced by the balloon must be the same as the impulse experienced by the air even if the air is lighter balloon is heavier or vice versa doesn&#39;t really matter the impulse is always going to be the same all right back to this whenever we&#39;ve had a collision or an explosion we&#39;re gonna do this whole constellation momentum thing which says P AI equals PF count up your objects put an envy tournament for every object maybe it&#39;s M 1 V 1 I plus M 2 V 2 I plus and so on and so on and so on depending upon how many objects we have equal and 1 V 1 F plus M 2 V 2 F and so on if they stick together then the mass that&#39;s combined and we typically see it written like M 1 V 1 I plus em 2 V 2 I equals MV F and being a combined mass if they bounce apart then we typically see it written as em 1 V 1 F plus m2v2 F literally just count the objects that you have before and after the collision and/or explosion and put an M V term in for each of them pay attention to positive versus negative in your calculations hey that&#39;s important that&#39;s so important again take you back to that written response test that we had several weeks ago on momentum weighed the collision between the two vehicles the car in the truck one of them was going to the north one of them was going to the south and lots of you nailed that question lots of you got it correct but what if you didn&#39;t get it correct because lots of you failed to take into account that one of the objects was going to the south before the collision so you forgot the negative okay when you see that question north-south-east-west circle in red to remind yourself of signs they circle north circle south to remind you that it&#39;s got to be opposite signs okay units what do we know about units with conservation momentum typically we&#39;re gonna be using kilograms and meters per second for mass and velocity but in a conservation momentum problem it doesn&#39;t actually matter as long as you&#39;re consistent how come how come if I&#39;m looking at this equation right here how come I could use grams for mass if I wanted to even though kilograms of the standard units al you look like you&#39;re ready to give me the answer but yeah they cancel it right if they appear in every term then they cancel oh no it&#39;s not the same instead of the same mass in every term but it&#39;s the same units in every term so the units could cancel it if in doubt what should you do if in doubt you&#39;re not sure what units you should be using what should you do go to the standard right go to the standard units kilograms in meters per second feel that you almost always have to do that it&#39;s just the odd situation like conservation of momentum we can get away without doing it all right I&#39;m gonna give you a question to work on before I talk about what I mean by a progressive collision we&#39;ll talk about that in just a couple minutes the one that I want you to work on now is gonna be question number six please of course number six on the same page is the question you just did it&#39;s a skateboarder question all right a skateboarder has a mass of 50 kilograms is running at a speed of 4.5 meters per second skateboarder jumps onto an initially stationary skateboard that has a mass of 4.2 what&#39;s the speed of the skateboarder and scape a skateboarder and skateboard after the after this interaction takes place it&#39;s like a collision right it&#39;s not literally a collision where two cars hit each other but it&#39;s one object hitting another object or or becoming one with the other object so we&#39;re gonna say it&#39;s a conservation of momentum problem P I is equal to P F we&#39;ve got two objects before this collision takes place so we&#39;re gonna say m1 v1 I plus m2 v2 I and then we have one object afterwards because the skateboarder and the skateboard become one object so we&#39;re gonna call it MV f we&#39;ve plugged in our numbers here we don&#39;t have to worry about signs in this everything is in the same direction units are all standard units so it&#39;s a rather straightforward example of 150 times 4.5 plus 0 equals the combined mass of fifty four point five kilograms times VF and we do the math on question number six and I think it works out to be a four point one five how many people get that all right good what did I mean by that progressive collision we don&#39;t see that very often but we did see one of these on a unit test that we had I think was a test or a quiz or something that we had at one point I&#39;m gonna give you that same question again in a few moments here but essentially what I mean by this is a progressive collision is when when there&#39;s a collision that takes place which goes into another collision so let&#39;s say it&#39;s kind of like a domino effect right you&#39;re you&#39;re running you&#39;re you&#39;re playing football and you&#39;re running down the field and you hit somebody okay and that and you knock that person out of the way and then that person hits somebody else that&#39;s a progressive collision now how do we deal with that well we can deal with it as two separate collisions we can say P I equals P F for the first collision and then for the second collision P ICO&#39;s PF but p f for the first collision would be pi/4 the second collision that make sense so basically what I&#39;m saying is let&#39;s not call it INF because there&#39;s three different momentum there&#39;s two different initials and finals here let&#39;s say P 1 equals P 2 so the initial momentum equals the final momentum and then the the font that momentum from this one becomes P 3 so we&#39;ll say P 2 equals P 3 because the final momentum for the first collision becomes the initial momentum for the second collision and then P 3 equals P 4 these are my two collisions these are my two collisions here okay here&#39;s the final momentum for the first collision here&#39;s the initial momentum for the second collision but the final momentum for the first collision equals the initial momentum for the second equation right hey I collide with you you bounce off with a momentum of 10 okay you hit somebody else you hit them with the momentum of 10 your final momentum in my collision is your initial momentum in the next question take a look at what I just wrote down though if P 1 equals P 2 and P 2 equals P 3 and P 3 equals P 4 then how can we simplify this P 1 equals P 4 so if you&#39;ve got a progressive coalition which we&#39;re about to see in a second um you can you can go through the process of finding the initial and final momentum a couple of times a couple different problems but you can simplify it by simply looking at the initial momentum before anything happens in the final momentum after everything happens you don&#39;t have to look at each individual collision you can but it&#39;s twice as much work if you want to do that let&#39;s take a look at this question please it has that you have seen before but you&#39;re not like we&#39;re gonna remember the answer for okay let&#39;s have a look now we get three cars so your car - and car three are at rest before anything takes place car one is moving at four meters per second car one collides with car two then car one and car two both combine and collide with and stick to car three so after before all this is said and done okay car one is the only one moving after all is said and done car one car two and car three I&#39;ve all stuck together and they&#39;re moving off as one object yeah you can treat this as two collisions if you want P I equals P F for the first collision and then P I equals PF for the second collision or you can just look at it as P I equals P F to the entire thing the initial momentum would be m1 v1 I because there is no momentum for the second object to the third object now if the second or even the third object was moving before this collision took place then we&#39;d have to add a term for that even if there was all three objects moving we would say m2 v2 i plus M 3 V 3 I whatever the case may be they&#39;re not so we can ignore those terms okay but it&#39;s possible that down the road they could be moving before the collision you have to include those if they are afterwards they stick together we&#39;re gonna call it MV f and being the total mass here my the mass of object number 1 is 1,500 kilograms v1 I is 4 meters per second the total mass I got 4500 is that right 4,500 kilograms times V F we&#39;re gonna solve for V F there and what we do it works out to be one point three three meters ii would express that it&#39;s kind of an odd one to express 1.33 on our numeric response answer sheet we&#39;d put it as one decimal three three notice three significant digits right three digits one point three three not scientific notation nothing like that just the number good all right we&#39;re gonna do another one here now this question number nine but I&#39;m gonna do this one with you this one talks about a lab which basically we did we pretty much did this exact lab and cross except we didn&#39;t use ticker tape or a spark timer we use our we used our sparkvue app along with our motion sensors in order to detect the motion but in the end we did the same thing essentially where we took two carts car one car two cart number one was initially at rest v1 I was zero meters per second cart one was was moving we sent moving at a certain speed this should have been the two iv1 i with a certain speed that we manipulated that we changed through the experiment he they stuck together we had this little plasticine okay next year based on some of your sources of air we needed that lab make sure i&#39;m gonna stick some velcro to both of these carts they&#39;re going to stick together a little better regardless they stick together they go off as one object we prata graph a VF the final velocity of both objects combined versus the initial velocity of car number one car number two doesn&#39;t have an initiative law so there&#39;s nothing to plot there and this is the data that we get just a bunch of data points now normally we draw a line of best fit for this right most times we don&#39;t actually have to on an exam draw the line of best fit for this one we actually do have toon you&#39;ll see why in just a second let&#39;s skip to the question not question number eight but question number nine and see what they&#39;re asked to find we see what I asked fine based on a point on the line of best fit cart one has a speed of six centimeters per second before the collision then the combined speed of the two cards after the collision would be eight as will be a centimeters per second okay based on a point on the line of best fit we don&#39;t have a line of best fit so you&#39;re going to have to draw one okay what you don&#39;t want to do is just go over to six seconds and then go over here and say all right at six seconds we&#39;re at two point two point four centimeters per second that&#39;s not a line on the line of best fit take a ruler draw your line of best fit we don&#39;t usually have to do this but occasionally we do hey lion investment might look something like this now the good news is if you&#39;re just reading a point off of the line of best fit you&#39;re probably going to get this right if you&#39;re on invest videos even remotely close to be reasonable if you draw out crazy then you&#39;re gonna get the wrong answer but they&#39;ll accept a range of answers here as long as it&#39;s consistent with a reasonable line of best fit okay we want to find we wanted to find here the speed of the combined mass if the speed of cart number one was six meters per second or six centimeters per second I should say so we go up we go over and it looks like we get about 2.5 2.6 this would be two point five is fifty two point six two point five five or two point six centimeters per second we didn&#39;t have to do anything crazy difficult there right no calculated work no analysis of the graph really other than up and over we&#39;re gonna fill in by the way the a V with two and six the combined speed of the two currents after the collision would be two point six centimeters per second now based on the slope of the line of that&#39;s that the combined mass would be what this is where it gets trickier right but that&#39;s okay we know how to deal with this based on the slope of the line of best fit what&#39;s the combined mass let&#39;s figure out what this slope means y is equal to MX plus B the y axis is VF m is the slope the x axis is v1 I now the y-intercept is pretty close to zero looks like it&#39;s close to zero I don&#39;t know if it&#39;s exactly zero but I think it&#39;s gonna be close enough we&#39;ll wait and see what the other equation looks like first before we decide for sure whether the y-intercept is going to be zero what other equation from my data sheet or that I can that I can derive might go with this don&#39;t necessarily go straight to your dadís you just think about what&#39;s going on this is a collision between two carts what do you usually do when you have a collision between two cards Carrie good P I equals P F so let&#39;s do that okay let&#39;s try that that seems to be the relevant equation here P I equals P F we&#39;re gonna say m1 v1 I plus zero because the second card is at rest equals M V F M being the total mass what&#39;s rearranges to solve for V F becomes M 1 over m T times V 1 I this works right carry we said that we have to have an equation that has the same variables in it VI equals P F has the same variables V F in V 1 I we cross off the F we cross off II 1 I and we&#39;re left with the slope is equal to m1 over Mt if we want to find the total mass and all we&#39;re gonna do is take the initial map the first mass and divided by the total I&#39;m sorry divided by the slope the first mass is one point five four kilograms we don&#39;t know what the slope is so let&#39;s do this together on our calculator so game everybody pull your calculator and let&#39;s go through this process we&#39;re gonna say stat edit when you go there it should probably be blank if it&#39;s not just go up to just go up to l1 and press clear now let&#39;s enter data points now you got to be careful here okay if you look at make sure you get these data points correct okay I got the first data point to be one point four in two point four one point four in zero point four what&#39;s going on here there we go now I got my second data point here to be at two point four because each of the x-axis intervals is point two and looks like zero point nine two point four and zero point nine my next one looks to be three point two and 1.15 that one&#39;s tricky 1.15 look at this yeah this is one point 1 this is 1 point 2 I&#39;m halfway in between so I&#39;m 1.15 my next at a point is 4 point 3 and I&#39;ve got 1.75 my next data point is 5 point 4 and I&#39;ve got 2 point 3 on my y-axis 6 on my x-axis in 2.4 on my y-axis 6.4 on my x-axis and the last one here is 3 on my y-axis so there I&#39;ve got 1 is it 1 2 3 4 5 6 7 data points here now remember you hit stat edit now it&#39;s going to be stat calc option for linear regression keep hitting enter until you get a number here my slope is point 4 8 5 so I&#39;m gonna take the first mass that I got which is 1 point 5 4 divide it by zero point 4 8 5 when I do that I get 3 point 2 so my total mass here my total mass my combined mass which is what they&#39;re looking for here it&#39;s 3 point 2 2 6 3 2 would be my answer yep we&#39;re getting there now we&#39;re talking two dimensional collisions and or courses in two dimensions though we got really three kind of interactions that we were going to see here there&#39;s the t-bone collision which looks like this and then typically we have them going off as one object afterwards we have what we call glancing collision which is something like this eight two objects hit each other but not straight on but kind of at the edge and sends one one way and sends one another way and then of course we have the explosion which is like you know got one object it splits up into three or four pieces like this sometimes in a collision they stick together sometimes they bounce apart that&#39;s no different than a one-dimensional collision if they stick together it&#39;s m1 v1 i plus m2 v2 i equals MV f if they balance apart its m1 v1 F plus m2 v2 F how do we do these questions though right how do we do the first one the red one the blue one or the green one keep on collision the glancing collision or the 2d explosion well the good news is if you can do one you can for the most part do any of the three we break them up into x and y-components and for both the x and y components we say P I is equal to P F in this case it would be m1 v1 plus m2 v2 I equals MV F because they stick together for both the X and the y in the second case it would be m1 v1 plus m2 v2 I equals they bounce apart m1 v1 F plus m2 v2 F if we&#39;re talking about the third case over here then it would be zero initial momentum because it&#39;s an explosion right that objects at rest equals m1 v1 F plus m2 v2 F plus m3 the because we&#39;d have three objects after the explosion takes place do the same thing for the Y component solve for usually it&#39;s gonna be VF for both the X and y components get both of them the X component and the y component of VF and then do the Pythagorean theorem simple as that sometimes especially knees glancing collisions here we got a funny angle what do we do what do we do if we have a funny angle not what do we do for you have a funny angle yeah we don&#39;t like funny angles right so we get rid of the funny angles okay if I have something like this that&#39;s at 25 let&#39;s make it 30 degrees because it&#39;s easier math if I have something like this that&#39;s 10 at 30 degrees and just take it off to the side over here and solve for an X and solve for a y the X component would be cosine 30 degrees is equal to x over 10 and the y component would be sine cuz it&#39;s opposite so 930 degrees is equal to Y over 10 solve for X solve for y and then whatever value you get for X goes into VF VF for object number one okay don&#39;t worry too much about units here just be consistent eight kilograms usually and that&#39;s the standard unit if you&#39;re confused just go to kilograms meters per second but in the end you know that you can use grams or kilometers per hour whatever you want make sure that you resolve the components if we got a funny angle don&#39;t ever don&#39;t ever go with a funny angle they get rid of it get rid of the funny angle into an x and y-components know how to deal with those remember to always and I&#39;ll tell you like you wouldn&#39;t think of this to be a problem but it is remember to always break it into X&#39;s and it&#39;s too often people see a collision like this and then they just have this brain freeze this brain fart where they forget to break it into x and y&#39;s and they just and they just say m1 v1 i plus m2v2 i equals whatever you can&#39;t do that if there&#39;s if it&#39;s a collision like this or like this or any kind of two dimensions anything like that you see above you have to break it up in exelon components pay attention to the pauses and negatives this is kind of repeat right cuz we talked about most of this stuff with one dimensional the only thing that really separates this from the one-dimensional collisions that we&#39;ve already done roni that&#39;s right it&#39;s really the only thing that makes it different then and these questions that we&#39;ve already done earlier today well talk about J and K in just a couple of minutes what I&#39;d like you to do right now is have a look at a couple of questions here first of all question number ten in the booklet that I gave you just now all right the question says after the explosion we want to find a piece of speed or the speed of P see here look don&#39;t just say P I equals P F for this right we went we just went through this we just said that when you&#39;ve got something like this or a glancing collision okay like this or you&#39;ve got an explosion like this a or a t-bone collision like this you can&#39;t just do P I equals B F it&#39;s got to be bi crispy F twice on the X and the y axis here so let&#39;s let&#39;s do that now let&#39;s break it up into X component and Y component let&#39;s say for both of them P I equals P F for both of them the initial momentum is zero because it&#39;s at rest right zero is equal to m1 v1 F plus m2 v2 F plus M 3 V 3 F and for the Y component same deal zero is equal to m1 v1 F plus m2 v2 F plus M 3 V 3 F good so far plugging in numbers here but be careful be careful you you look at either X or Y at one time here X components we&#39;d say object number one has zero momentum even though it&#39;s moving it&#39;s moving on the y axis object number two is two kilograms times forty four point six and object number three is four kilograms times V 3f we solve for V 3 F there oh except to be negative 22 point three meters per second and you do the same thing for the Y component but this time there is a y component it&#39;s 3 times 39 point to know what&#39;s initial know what&#39;s final and know what is X and know what is y and don&#39;t mix those up the y component of the second one is zero because it&#39;s all in the x axis solve for V 3 F here and V 3 F for the Y component work so it could be negative twenty nine point four and then we got to combine them we got negative twenty two point three negative twenty nine point four you guys solve for V there we don&#39;t need to solve for theta okay but we could use in the tan function V is just going to be the Pythagorean theorem and it&#39;s going to work out to be thirty six point nine meters per second if all these masses were the same what could we do to simplify this if they were all four kilograms we could cancel the mass would you have to cancel the mass of course not but you could if you wanted to save yourself a little bit of time and effort we go back to this down in J within topic number sixteen says problems given an unknown angle a lot of times a lot of times you&#39;ve got a question where we have to solve for that angle right we&#39;re given the unknown angle this time right if we&#39;re given the unknown angle and then we don&#39;t have to treat any differently we can do it the way we just did but notice what we just did it we didn&#39;t pay any attention to that angle because oftentimes we don&#39;t have that angle so I did it as if we don&#39;t have the angle so that you can do any question like this whether you have the angle or not but not you&#39;re right there is a shortcut here if you recognize that you have that angle let&#39;s get rid of all of this for the Y component and then let&#39;s say the X component is twenty two point three the angle is thirty seven point two okay we can find this speed just by using the sine function now sine thirty seven point two degrees is equal to twenty two point three over V so you can save yourself some time if you&#39;re given that quote unknown angle it&#39;s not really an unknown angle right but it&#39;s the angle that we&#39;re often trying to find if we&#39;ve got it there&#39;s a shortcut but you never have to take that shortcut it&#39;s just a little bit a little bit of the time savings if you spot it um but you gotta be careful sometimes people right before I&#39;m looking for shortcuts right you take five minutes looking for a shortcut that&#39;s gonna save you - well you haven&#39;t really taken much for a shortcut then right you spot her right away great if you don&#39;t just do it there just do it the regular way oh you&#39;re gonna love this one okay this one let&#39;s do this one together okay ozone in the upper atmosphere protects life by absorbing dangerous photons of ultraviolet electromagnetic radiation produced by the Sun when such a photon collides inelastically with the stationary initiation ozone molecule the molecule is broken apart into an oxygen molecule in an oxygen atom as shown below before the collision after the collision listen we&#39;re talking about photons here we&#39;re talking about oxygen molecules but when you look at this I think it&#39;s pretty clear that this is a collision right and this is a conservation of momentum problem we got to do X and y components because of that but what are you going to do first what&#39;s the first thing you got to do I don&#39;t like funny angles get rid of the 74 degrees look I don&#39;t care if it&#39;s photons or oxygen or atoms or molecules or ozone I don&#39;t care anything about that listen I know what to do here my velocity here was 220 meters per second I could find the X and I can find the Y by using some trigonometry cosine and sine and when I do that I&#39;m not gonna show all that work here because I think you can do that I end up getting 60 point six four zero two for X and I end up getting two hundred and eleven point four seven seven eight for the Y component now I break it into my X&#39;s and Y&#39;s now that I&#39;ve never dealt with that funny angle that I don&#39;t like I do my X&#39;s and Y&#39;s PII equals PF I&#39;m gonna state look look I&#39;ve got the momentum of the photon if I if I had the wavelength I could find the momentum by saying H over lambda by the energy I could find the momentum by saying a over C Y over C using those equations in the top right hand corner of your data sheet but I&#39;ve got the momentum here the length of the photon is nine point three times ten to the negative 24 plus zero because the ozone is at rest plus the mass of the oxygen which is sorry equals I should say the final momentum the mass of the oxygen times you guessed it the the X component of the final velocity of the oxygen sixty point six four zero two and then we add to that two point six six times ten to the minus twenty six times bf so momentum of the photon plus the momentum of the ozone momentum of the molecule plus the momentum of the atom we solve for VF there we accept to be two hundred and twenty eight point five seven one six now that if I had we know the oxygen atom is gonna go down here somewhere right if I had this angle given to me already what what could I do yeah yeah I probably end up using cosine because I found the X component I&#39;ve got the I&#39;m looking for the hypotenuse cosine theta is equal to x over two twenty eight over the hypotenuse right but I don&#39;t have that unknown any go but I&#39;m looking for it so I&#39;ve got to go through that process now of finding in Y components we&#39;re gonna say zero initial minute them on the y-axis because everything&#39;s on ax right is equal to here&#39;s the molecule five point three one times 10 minus twenty six times two hundred and eleven point four seven seven eight this is the Y component the final momentum of the molecule plus two point six six this is the atom solve for V F and we get negative four twenty two point one six oh six listen this is an easy question now right do the same thing as we always do there now do the same thing as you did for the last question same thing as we do for every other 2d question 228 for twenty two point one six zero six I could find theta by the inverse tan function and we can find V by using the Pythagorean theorem I&#39;m through that because I think you&#39;re okay if you need me to go through it I will but otherwise I&#39;ll just give you the answers here me is 480 and fader works out to be 61 point 6 okay we&#39;d say that would be if this is a north-south east-west thing then we&#39;d say that would be 61 point six degrees south of East because we went east then we went from the East Gary we went south from the East self after the East or self from these okay look at the angle here it&#39;s measured not from the self because this is self it&#39;s not even touching South right it&#39;s measured from the east axis but it&#39;s self from the east axis it&#39;s below the east axis so it&#39;s self ities alright let&#39;s take a look at our last thing here which is you&#39;ve asked it inelastic collisions now couple quick little reminders an elastic collision remember is a collision in which kinetic energy is conserved remember if they stick together it&#39;s always inelastic if they stick together if they bounce apart could be elastic could be inelastic don&#39;t know when we see a question we hope for them to stick together because then we have a definitive answer without doing any calculations if they stick together it must be inelastic if they bounce apart it&#39;s probably still inelastic but it could be elastic scalar nature always pause if no angles right this is when even in a question like this we would not break it up into x and y components why I cuz momentum is a vector but kinetic energy is a scalar so all you&#39;re gonna do is say whether it&#39;s a one-dimensional or two-dimensional whether it&#39;s north and south or north and north you&#39;re gonna say EK i equals one-half m1 v1 i squared plus however many objects you have get a number do the same thing for the final you know those two numbers are the same it&#39;s the elastic if EK i is bigger than EK f its inelastic EK f won&#39;t be bigger than EK i if it is you&#39;ve made a mistake units joules always want to use joules here we&#39;ve used the electron volts a lot now since we&#39;ve gone through atomic physics don&#39;t use the electron volts here can use joules and remember again just with a scale and H are always positive no components Hey no funny angles okay this is the one time when you can say listen I don&#39;t like funny angles but they&#39;re not really bothering me right now if you&#39;re dealing with the scalar funny angles don&#39;t matter you can hate them all you want but just to leave them alone when you&#39;re dealing with a scalar and finally before we do a quick example here finally in an inelastic and inelastic collision kinetic energy is conserved so it&#39;s momentum in an inelastic collision kinetic energy is lost momentum is still conserved doesn&#39;t matter what kind of collision you have momentum is conserved example I want to do if this goes back to question number eight goes back to this information on the graph that we already did question number eight says the collision of the two carts is classified as blank because blank man I&#39;m telling you I&#39;m really hoping these two things stick together because I don&#39;t want to calculate e ki and e KF I&#39;m not sure what the I and VF I&#39;m gonna use because there&#39;s seven different data points for VI and VF when you look back at the information though what would be hold they do stick together and because they stick together we know that this is what kind of collision inelastic we know it&#39;s inelastic and why do we know and in what happens in an inelastic collision kinetic energy is not conserved is momentum conserved yes it is momentum is conserved kinetic energy is not conserved because it&#39;s inelastic the other one that I want to do is multiple choice number 12 which is going to be towards the vacuum well could if I can find it here this isn&#39;t across and station two identical well friction Democrats are pushed toward each other such that they&#39;re traveling at the same speed when they collide they said two cars collide two sick carts same mass same speed collide they come to a complete stop teacher challenges students to describe in schoolers and some of the students descriptions are provided below before the collision the carts have the same momentum before the collision they have the same kinetic energy this is an example of an elastic collision this is an example of an inelastic collision shawne&#39;s are correct do they have the same momentum before the collision all this is a great question do they have the same momentum before the collision they have the same mass they have the same speed do they have the same momentum before the collision who says yes who says no why do you say no Allen good they have the same magnitude of momentum they do not have the same momentum momentum is a vector the momentum of one of them is whatever value to the right the momentum of the other one is the same value to the left those are two different Momentum&#39;s they do not have the same momentum they have the same magnitude of momentum but momentum is a vector we have to pay attention to direction do they have the same kinetic energy yes or no who says they have the same kinetic energy who says they don&#39;t have the same kinetic energy they do have the same kinetic energy they have the same mass and they have the same speed momentum depends upon velocity kinetic energy depends upon speed they do have the same kinetic energy elastic versus nos clearly can&#39;t be both is it elastic or inelastic we don&#39;t know how fast they go on here so I&#39;m not sure how we&#39;re going to calculation elastic or inelastic who says he last a few objects same speed same ass hit each other come to a stop who says elastic who says inelastic Hugh I do you say interesting okay okay if in doubt go with inelastic because nine times in a ten or even more than that it&#39;s gonna be an elastic truly only subatomic particles have our our elastic rang kiss they don&#39;t squish but you get it it&#39;s conceivable to get the collision it&#39;s really really close to us it could be technically classified as you elastic this is anyone a stick though it is inelastic how come what&#39;s the initial kinetic energy I don&#39;t know but something what&#39;s the final kinetic energy zero something does not equal 0 if obvious number one is 10 object number two is 10 combined it&#39;s 20 not zero because they&#39;re scalars right 10 and 10 is 20 afterwards the total moment total kinetic energy is zero we lost kinetic energy since we were asked in attic energy it&#39;s any other aspect so what we got two and four there are three more questions that we&#39;re not gonna do here right now these are three questions that aren&#39;t in order they don&#39;t really go at all in order with the with the topics that I&#39;ve listed in that 16 page book up just kind of look at the end of everything after we&#39;ve done gone over the unit maybe you want to take a look at three kind of random questions within one scenario we&#39;re not gonna do those right now you want to take a look at those yourself and then come back and ask me how to do something if you have trouble that&#39;s great if you want to ask me Monday morning where you guys come in for your next extra help session then you can do that as well I will give you the answers though right now for questions 13 14 and 15 number 13 the answer is C number 14 the answer is C and number 15 the answer is B

let&amp;#39;s take a little bit of time here today and talk about momentum in impulse our first unit of year we defined momentum on the very first day and kind of in everyday language is how hard it is to stop something but in the end we know that that comes from the real definition of momentum which is the product of an object&amp;#39;s mass and velocity and really that&amp;#39;s book that boils down to an equation P is equal to M times V the units for momentum of course we can get the unit&amp;#39;s from here you get the units from any equation right you&amp;#39;re ever wondering what the units for a variable is you can get the units from any equation for this one we&amp;#39;d say because it&amp;#39;s mass times velocity we&amp;#39;d say its kilograms times meters per second there is another set of units that we can use for momentum as well that we learned later on when we talked about impulse does anybody happen to remember right now what that other set of units is besides kilograms meters per second not Newtons per second for your clothes yep good it&amp;#39;s Newton&amp;#39;s times seconds you don&amp;#39;t need to worry about converting back and forth between those two variables between those two units because in the end one kilogram meter per second is exactly the same thing as one Newton seconds we just use usually use kilograms meters per second when we talked about momentum in Newton&amp;#39;s seconds when we talked about impulse but really they&amp;#39;re the same thing um listen this these are pretty straightforward right you got to be able to take the velocity you gotta be able to take the mass in the right units kilograms and meters per second and multiply them together to get momentum you also have to be able to rearrange them solve for mass we would say it&amp;#39;s P over V and of course to solve for velocity its P over m if the entire course is this easy and straightforward we are laughing hey I don&amp;#39;t anticipate getting any kind of crazy difficult questions on your diploma exam with just straight momentum usually the more complicated questions come from putting momentum along with impulse or conservation momentum now remember that little little things right momentum remember that when is a vector it has a direction associated with it we can remember that if we do happen to forget it by remembering by taking a look at our data sheet and noticing that there&amp;#39;s a little half arrow over top of the momentum sign as there is with velocity as well velocities vector mass is a scalar we remember that because there&amp;#39;s no half arrow on top of the mass here it has a direction associated with it if the velocity is to the right then the momentum is to the right if the velocity is south then the momentum is self finally I added one more little thing in here says momentum after acceleration generally we&amp;#39;re not dealing with a problem like that but every once in a while it does rear its head we&amp;#39;re looking for momentum after an acceleration it might be a problem like you know you drop a ball from a height of 10 meters how fast sorry what&amp;#39;s the momentum of the ball as it strikes the ground or a car accelerates from rest for 28 meters what&amp;#39;s its momentum after that acceleration phase it makes the question a little bit more tricky but in the end all we got to do there is still use P is equal to M times V but find the velocity some other way like maybe VF squared is equal to VI squared plus Q ad maybe a is equal to Delta V over delta T whatever equation happens to work to get the velocity the bottom line is momentum is a quantity that exists at a single moment in time what&amp;#39;s the momentum now or what&amp;#39;s the momentum now what&amp;#39;s the momentum at the beginning of the problem what&amp;#39;s the momentum at the end of the problem hey this would be this would tend to be a what&amp;#39;s the momentum at the end of the problem and if the object is accelerating then we would have to find the velocity at the end of the problem in order to plug into that equation now one final thing before we move on to impulse which will spend more time on because it tends to to appear on an exam a lot more often we talked yesterday I think it was about the 10 physics principles the first physics principle that&amp;#39;s on your data sheet in the bottom left hand corner is the uniform motion uniform motion is when we have of these they could have Delta D over delta T right whenever we use that constant velocity equation Hey I doubt that you would get a question that asks you for the physics principles if it was just a straight momentum question but if you do it will either be an I missed one here it will either be uniform motion if we&amp;#39;re dealing with a constant velocity if the object isn&amp;#39;t changing velocity which would be the majority of questions involving straight momentum but it could be accelerated motion as well it would be accelerated motion if it was one of these situations down here where you&amp;#39;re looking for the momentum after it speeds up to a certain range again I don&amp;#39;t think you&amp;#39;re gonna see those two physics principles or any questions involving physics principles or it&amp;#39;s just a straight momentum question but if you do it&amp;#39;s gonna be either uniform motion or accelerated motion it won&amp;#39;t be both gay it won&amp;#39;t be both because you&amp;#39;re not going to be going at a constant velocity and accelerating at the same time we&amp;#39;re talking about impulse and you can see right away the physics principle is the really the only physics principle that you&amp;#39;ll be dealing with when we talk about the concept of impulse might be accelerated motion why because what is impulse it&amp;#39;s the change in momentum if momentum is changing then it must be accelerating doesn&amp;#39;t mean it&amp;#39;s speeding up could be slowing down but the velocity is changing it&amp;#39;s accelerating therefore it&amp;#39;s giving us a change in momentum and impulse I&amp;#39;m just gonna say for the definition right here Delta P right you guys remember that Delta P is change in momentum and of course the units would be exactly the same thing is we have for momentum would be kilograms meters per second or it would be Newton with time seconds now we derived an equation for this we&amp;#39;re not going to derive it again because in the end it&amp;#39;s on your data sheet says F times Delta T is equal to M times Delta V what we usually do is tack on a little Delta P in front of it just to remind us of that equation that two-part equation or now that three-part equation if I&amp;#39;m saying Delta P is equal to F T equals M times Delta V that whole thing is equal to the impulse F times T is a change in momentum M times Delta V is a change in momentum I&amp;#39;m gonna circle something in red here that is meant to draw attention to a really really common mistake that people make the equation is not M times V it&amp;#39;s M times Delta V if you guys remember on the response test that you did for momentum many of you lots of you calculated the force of impact between those two carts you guys remember that question two cars collided you had to do a whole bunch of stuff including fine in the force of impact on car one and the force of impact on car 2 you guys remember that lots of you calculated the force of impact using impulse you have the right idea but you set f times delta T equals M times V naught M times Delta V you&amp;#39;ve got to make sure that you&amp;#39;re using the change in velocity the other thing that I want you to be careful of here when you&amp;#39;re solving for impulse if there happens to be two or three or four objects you&amp;#39;re always using one of those objects don&amp;#39;t get mixed up and say oh oh look the initial I have the initial velocity of the blue car and I have the final velocity of the red car it doesn&amp;#39;t matter those can&amp;#39;t go together when you&amp;#39;re solving an impulse problem you&amp;#39;ve got to have the initial and final velocity of the blue car or of the red car but not of you can&amp;#39;t mix and match initial and final velocity for two different vehicles now of course we got to be able to rearrange this but this is pretty straightforward algebra right nothing complicated about this you want to solve for f usually it&amp;#39;s going to end up being M times Delta V over delta T although sometimes it might be Delta P over delta T depending upon which two parts of the equation you&amp;#39;re using the other thing that you&amp;#39;re often going to have to rearrange to solve for would be Delta V and of course Delta V is going to be F times delta T or we could just say Delta P over the mouse all right I&amp;#39;d like to take a look right now at the first of 15 multiple choice / numeric response questions that we want to do today then refer you to question number 2 it&amp;#39;s on the second page of the book that I handed out to you I&amp;#39;m gonna give you a couple minutes to look at that question numeric response number to see if you can solve it and then we&amp;#39;re gonna take a look at it one thing that I&amp;#39;ll caution you about is you look into this question pay attention to things like Delta V as opposed to V right pay attention to things like one object not two objects and another thing I should have mentioned this before pay attention to vector nature what do we mean by that the direction if we got something going to the right and then something going to the left make sure you pay attention to the fact that one&amp;#39;s gonna be a positive one&amp;#39;s gonna be a negative all right well set you free almost for a couple minutes here and then we&amp;#39;ll take a look at it all right most people are finished here we&amp;#39;re gonna use impulse to solve this problem right clearly that&amp;#39;s what we&amp;#39;ve been talking about for the past 10 minutes or so here guys but why if this is a question on a test and your teacher hasn&amp;#39;t been standing up at the front of the exam room talking about impulse for the last five to 10 minutes how would you know that this is an impulse question what leads us to use impulse to solve this question it&amp;#39;s two words in there that tell me right away that this is almost certainly gonna be an impulse question not two you know what they are is your hand going up there all right we got a change in momentum here sure sure the the speed is changing we&amp;#39;re dealing with a mass okay if that&amp;#39;s the case and there is gonna be a change in momentum although that&amp;#39;s often gonna be the case we&amp;#39;re still not necessarily gonna use an impulse to solve the problem even when there is an impulse depends on sometimes what we have given to us two words are yep it&amp;#39;s a force at a time when we see a force at a time whether we have both of them or we have one of them then we want to solve for the other one it doesn&amp;#39;t really matter that&amp;#39;s we&amp;#39;re gonna use impulse so I&amp;#39;m gonna say F and T we&amp;#39;re gonna use impulse to solve that problem or at least we&amp;#39;re gonna try to use impulse to solve that problem Hey worst case scenario we try impulse it doesn&amp;#39;t work for me it&amp;#39;s not gonna lead me astray because we know if there&amp;#39;s a force in a time that there it is an impulse whether it leads us to the correct to the answer or not all right Zoe what do we know about impulse we know that Delta P is equal to F times delta T and it&amp;#39;s equal to M times Delta V looks like by our Givens it&amp;#39;s the last two parts of the equation that I want to use and I&amp;#39;m gonna rearrange that to solve for F it M times Delta V over delta T notice what I wrote down there M times Delta V naught V but Delta V over delta T we got a mass here it&amp;#39;s five point seven times ten to the minus two and we got to make sure here that when we put in Delta V we take into account not only the fact that there is two velocities here an initial and a final and there always has to be in one of these questions even if one of them is zero even if it starts at rest VI is zero there&amp;#39;s still two velocities we also have to pay attention to direction here as well and one of the things that I probably should have done when I read through this question is to circle the north and the south to reminded me when I plug my numbers in and I&amp;#39;m not going to forget to put one of them as positive one of them is negative the final velocity is 32 meters per second so it&amp;#39;s going to be VF minus VI 32 minus negative 35 another common mistake first of all forgetting to put the 35 in there to begin with second of all forgetting to put the negative 35 making it 32 minus negative 35 divide that by 4 times 10 minus 3 and when I do that math I end up get nine point five five times 10 to the 2 anybody else get that we&amp;#39;d express our answer ABCD as 9 5 5 2 how many people got that good you know what that&amp;#39;s 2% I near to poem exam right there all you need all you need is another 88% and you&amp;#39;ve done fantastic on your diploma exam all right let&amp;#39;s go back to our physics 30 and 16 pages for a second again we we kind of address this in the question already the vector nature of it then the direction nature of it here there is a direction associated with impulse there is a direction associated with with the momentum there is a direction associated with the velocity in the end if you have a positive velocity negative velocity right left north-south if you have a positive impulse that&amp;#39;s gonna mean a gain in momentum if you have a negative impulse that&amp;#39;s gonna mean a loss in momentum now you can also think of in a different way a positive momentum or positive impulse I should say a positive impulse means a gain in momentum but it can also mean an impulse to the right or an impulse to the north or an impulse in the positive direction a negative impulse means a loss in momentum but we can also think of it as I&amp;#39;m gonna mentum to the left or I lent them to the south or a momentum down sorry an impulse I should say an impulse to the south and impulse to the left and impulse downwards and so on so really these mean the same thing when you gain momentum it means it&amp;#39;s in the positive direction when you lose momentum it means the impulse in the negative direction and this is redundant I should cross that off because I already put that I guess the equation and being able to rearrange the equation let&amp;#39;s go back to this predetermined impulse versus non predetermined impulse this is the whole qualitative nature of impulse remember we talked about these are my words right predetermined non predetermined impulse but they&amp;#39;re what they mean is that in any situation in any situation we either know what the impulse is going to be or we don&amp;#39;t know what the impulse is going to be when I&amp;#39;m driving down the car the highway in my car I know I&amp;#39;m gonna stop eventually I know what my impulse is gonna be I know what the mass of my car is I know what the velocity of my car is therefore I could calculate the momentum of my car and I know at some point my car&amp;#39;s not gonna be moving my momentum of my car is gonna be zero so the impulse would be zero minus whatever the momentum of the car is right now it doesn&amp;#39;t matter how the car comes to a stop the impulse will be exactly the same so typically we&amp;#39;re talking about a predetermined impulse when you&amp;#39;re stopping an object when the car is coming to a stop whether it hits the wall or whether it comes to a gradual stop at a stop sign when the baseball comes to a stop whether it stops because if it&amp;#39;s your glove or whether it stops because it&amp;#39;s just moving through the air and ends up hitting the ground rolling for a while and eventually coming to a stop or whether hits your bare hand the baseball is coming to a stop we know what the impulse of the car is we know what the impulse of the baseball is going to be no matter how it comes to a stop the other situation is the non predetermined impulse this is when we don&amp;#39;t know what the impulse is going to be situation there is usually when we&amp;#39;re starting an object the all sitting on the tea I know it&amp;#39;s momentum is zero right now but after I hit that golf ball I don&amp;#39;t know what its momentum is gonna be the baseball that&amp;#39;s in my hand I know it&amp;#39;s momentum right now but I don&amp;#39;t know what its gonna be after I throw it its impulse isn&amp;#39;t predetermined the car that&amp;#39;s sitting at the stop sign I know it&amp;#39;s momentum is right now but I don&amp;#39;t know what it&amp;#39;s gonna be two minutes from now I know what its gonna be I know what the impulse is when the car is stopping but I don&amp;#39;t know what the impulse is gonna be when the car is starting that&amp;#39;s a non predetermined impulse when an object is stopping the impulse doesn&amp;#39;t change the momentum does but the impulse doesn&amp;#39;t change the impulse is going to be the same no matter how it comes to a stop so if the impulse isn&amp;#39;t going up or down then generally what we&amp;#39;re gonna try to do is increase the time that it takes to stop the object which is going to have what effect on force smaller that&amp;#39;s your impact cushioning thing right that&amp;#39;s the airbag in the car your head hits the airbag your heads gonna have same impulse whether hits the airbag or the steering wheel or the pavement in front of your car your head&amp;#39;s gonna have the same impulse but the airbag spreads the impulse out over a longer period of time it takes longer to stop which decreases the force when you catch a baseball and catch a baseball like this and kind of cradle it as opposed to sticking your hand out rigid like this and not bending your elbow right you create a lot like this to increase the time that it takes to stop to decrease the force spread the impulse over longer period of time increased time decrease force on the other hand when you&amp;#39;re starting an object it&amp;#39;s getting a little bit of cluttered here but try to stay within here when you&amp;#39;re starting an object the impulse is not predetermined when you&amp;#39;re starting an object increasing the force can increase the impulse force and time don&amp;#39;t affect each other when it&amp;#39;s not a predetermined impulse force and time both affect impulse when it&amp;#39;s when you&amp;#39;re starting something when it&amp;#39;s not a predetermined impulse so increasing the force increases the impulse hit the golf ball harder you give it more momentum the other effect is time increase the time of contact you don&amp;#39;t decrease the force this time rather you increase the impulse so hit the ball harder you give it more momentum hit the ball for a longer period of time you give it more momentum you understand the difference here how force and time are related to each other in this one increase the time decrease the force by the way you could go the other way right occasionally you might want to increase the force that&amp;#39;s the case you&amp;#39;d want to decrease the time bottom line is time affects force though not impulse but in this one when you&amp;#39;re starting the object time doesn&amp;#39;t effect force time affects impulse and force also affects impulse linearly all right let&amp;#39;s take a look at another question this is multiple choice number three on the booklet that I gave you earlier today all right let&amp;#39;s have a look at this one now it says airbags are a common safety feature installed in vehicles when a vehicle suddenly stops the air bag inflates um what&amp;#39;s the purpose of that airbag we know this right it cushions an impact if any question you see involves cushing and impact we know what the answer is already whether we&amp;#39;re talking about the airbag or whether we&amp;#39;re talking about the baseball glove or we&amp;#39;re talking about the hockey Pat goalie pads or are you talking about why you bend your knees when you jump off of the desk in the end what it is is the impulse which is predetermined we want to spread that over a longer period of time reduce the force let&amp;#39;s read what it says though the force on the driver is less the impulse on the driver is greater the stopping time of the dot the driver is longer which is the state numbered above compares what happens in an automobile collision when in which an air bag inflates with which with one in which the air bag does not inflate okay so the air bag inflates what does that do does that change the force on the driver sure it does that&amp;#39;s the purpose of the air bag right what does it do it decreases the forest on the driver the impulse stays the same time goes up force goes down time goes up right stopping time and the driver is longer yes is the impulse greater no one no the impulse is the same no matter how your head comes to a stop the impulse is the same the answer is gonna be options one or statements one and three here and that&amp;#39;s cap take a look at one more multiple-choice question as well before we move on to our next little thing here this is number one please on the first page alright this question involves two collisions of pucks on an air table it says in the first one a stationary puck puck one is struck by a second puck puck to the massive puck two is twice that of puck one so masses are different it&amp;#39;s a collision between two pucks where the masses are different there is no friction so we don&amp;#39;t need to worry about that basically hey what are they telling us when there is no friction in this we have n what&amp;#39;s the word I&amp;#39;m looking for we have n blank system we have an isolated system there&amp;#39;s no external forces acting yeah that really comes up later on we talk about conservation momentum okay but it kind of appears a little bit here in this question it&amp;#39;s not really a big deal but we can at least talk about that&amp;#39;s just a little bit here we have an isolated system there&amp;#39;s no external forces so we don&amp;#39;t need to worry about anything like that here a collision between two pucks one puck is heavier than the other next when we have the collision between two pucks head-on this time the pucks have the same mass they bounce back in the direction opposite to the original direction once again we have an isolated system there are no external forces all right what&amp;#39;s the question here it says for collision one the magnitude of the impulse of the puck on the impulse on puck one a one on puck two is blank to the to the impulse of puck to unpack one what&amp;#39;s the answer there is it equal to or is it less than because the difference in masses who says it&amp;#39;s equal who says the magnitude of the impulses of s the magnitude of the impulse is equal okay it is equal and the reason it&amp;#39;s equal is because ultimately if object a applies a force on object B object B will apply an equal and opposite force on object a right Newton&amp;#39;s third law which is really just a statement of impulse really when it comes down to it here&amp;#39;s what happens puck number one loses a certain amount of momentum or gain a certain amount of momentum let&amp;#39;s say it loses a certain amount of momentum where does it go if it loses ten units of momentum where does that ten units of momentum go to the other guy to the other puck so if one of them loses ten the others gotta gain 10 it doesn&amp;#39;t matter that they have different masses or anything of the sort okay in the end if momentum is gonna stay the same and then one gains momentum the others got to lose to the exact same amount of momentum opposite signs but same magnitude what about collision two for collision to the magnitude of the impulsive puck one on putt two is equal to again for the exact same reason it doesn&amp;#39;t matter that the masses are the same here okay in the end if one guy gains ten the other guys gotta lose ten other wise momentum momentum has disappeared somewhere that can&amp;#39;t happen this will always be the answer to a question like this if you&amp;#39;ve got a collision between two objects and one of them gains five units of momentum the other must always lose five units of momentum even if one mass is 50 times the mass of the other one the magnitude of the impulse will always always be the same okay let&amp;#39;s take a look at another little aspect of impulse here which is these forced time graphs I&amp;#39;m gonna start that off by asking you those three words that I asked you way back in February when we did this for the first time when you see a force versus time graph and it looks something like this you should always think three words kind of giving it away there because area grows impulse yeah area equals impulse I don&amp;#39;t care if you&amp;#39;re asked to find the impulse or you&amp;#39;re asked to find something else it doesn&amp;#39;t matter if you see a force versus time graph you think in area equals impulse it doesn&amp;#39;t matter if it&amp;#39;s shaped like this if it&amp;#39;s shaped like this if it&amp;#39;s got some kind of crazy shape it doesn&amp;#39;t matter you want to think area is equal to impulse now I&amp;#39;ve made a note here about units on axes you got to be really careful with this okay you think the area should be easy right if it&amp;#39;s a rectangle then it&amp;#39;s base times right if it&amp;#39;s triangle it&amp;#39;s one half base times say a generally pretty easy shape see pretty easy equations but what we often miss is the fact that the time might be in milliseconds or whatever or it might say times ten to the minus three seconds and the force may be in kilonewtons or or millon Newtons for that matter or it might say times ten to the two Newton&amp;#39;s or whatever the case may be make sure you pay really close attention to the axes on the units if it&amp;#39;s in kilonewtons in milliseconds then convert it when you&amp;#39;re finding the area to Newtons and seconds if your area is 1/2 of 200 milliseconds times 30 kilonewtons then what I want you to do is change that make it one half of 0.200 seconds times 30,000 Newtons make sure you&amp;#39;re in standard units when you&amp;#39;re calculating that area area close impulse make sure you&amp;#39;re in standard units once you find the area well you&amp;#39;ve got the impulse and then sometimes what you&amp;#39;ve got to do after you found the impulse it&amp;#39;s fine something else like the change in velocity or the mass and what you&amp;#39;re always gonna do there is step this equation R equals impulse also equal to M times Delta V do not set it equal to this do not set it equal to F times delta T even though that&amp;#39;s a valid equation do not set it equal to M times Delta V in this context here is why the force here in this graph is almost always going to be changing what value of f would we use if we wanted to use that equation in conjunction with it technically it&amp;#39;s valid but you&amp;#39;d have to use the average force and usually the average force is pretty hard to find on this graph so just don&amp;#39;t even bother going there area is equal to impulse and that&amp;#39;s equal to M times Delta V so if you&amp;#39;ve got the impulse by the area given the mass you could find Delta V or the other way around given Delta V you could find the mass so there&amp;#39;s three variables in here really but you have to be able to find impulse get the area mass you take the area divided by Delta V Delta V take the area divided by mass one final thing before we do an example here make sure you pay attention to what&amp;#39;s above and below the x-axis if you got a graph that looks like this then this area all three sections of it will be positive this area below the x-axis will be negative and what we have to do to find the total impulse there is add them together plus hundred plus negative 50 or whatever the case may be yet the impulse that&amp;#39;s what you&amp;#39;re looking for great if you don&amp;#39;t want to find mass or change in velocity do your thing just make sure you&amp;#39;ve got the right value for impulse based on whether it&amp;#39;s above the x-axis or below the x-axis sticker will get another multiple choice slash numeric response question now this one&amp;#39;s going to be question number four on the third page of your booklet trying to take a look at it here this question says in a car safety test we&amp;#39;ve got a 1200 kilogram car I directed towards a solid wallet it collides with the wall it comes to a rest and then we got a force versus time graph here on what&amp;#39;s the first thing you think of when you see this graph versus time graph three words area equals impulse area equals impulse should be those three words every single time we haven&amp;#39;t even read the question yet right we just know the context of it we know the information that we given we&amp;#39;re thinking area equals impulse regardless of what we&amp;#39;re asked to find okay something that I noticed here though almost missed it when I was doing my answer key gotta be always aware of this always alert for this what is it that I&amp;#39;m talking about why did I pick up the red pen yep we got kilonewtons on the y axis right we got to be a newness and yeah kilonewton is 10 to the 3 Newtons so it&amp;#39;s a nice easy conversion if we have 20 kilonewtons we have 20 times 10 to the 3 Newtons there anything else I should circle and right here we have seconds on the x axis but it does say times 10 to the negative 1 seconds so it&amp;#39;s easy when you&amp;#39;re doing the base here from 3 to 8 just to say the base is 5 so I really 5 it&amp;#39;s 5 times 10 to the minus 1 seconds so let&amp;#39;s be careful of both of those things we can find the area of this easily but we have to pay attention to those units as well all right what does the question actually say what&amp;#39;s the magnitude of the impulse experienced by the car all right good and what&amp;#39;s the initial speed of the car we&amp;#39;re going to find two things ain&amp;#39;t no worries let&amp;#39;s get the impulse here first by finding the area the area is equal to Delta P which is equal to because it&amp;#39;s a triangle 1/2 of base times height the base here goes from 3 to 8 and I&amp;#39;m not going to forget that it&amp;#39;s 5 times 10 to the minus 1 seconds because I circled those that 10 to the minus 1 seconds on the x-axis the height of it it goes from 0 to minus 70 so I&amp;#39;m going to say the height of it is minus 70 times 10 to the 3 Newtons or we could say it&amp;#39;s minus 70,000 Newton&amp;#39;s we math the area works out to be the area works out to be negative 17,500 that right okay negative 17,500 that&amp;#39;s the impulse here okay that&amp;#39;s the first thing that I&amp;#39;m looking for now I&amp;#39;m looking for the change in the initial speed of the car it says comes to a stop so the final velocity is zero now what I&amp;#39;m gonna do is set Delta P equal to M times Delta V and solve for Delta V Delta V works here because although I&amp;#39;m technically finding final velocity sorry initial velocity since the final velocity is zero the change in velocity will be the same magnitude as the initial velocity so we&amp;#39;re gonna say it&amp;#39;s negative 17,500 over the mass here which is 1200 kilograms we do that math change in velocity works out to be negative fourteen point five eight three three meters per second so if the final is zero and it changed by negative fourteen point five then the initial must have been positive fourteen point five what do we got here impulse is one point seven five or 1.8 times ten to the four Newtons and the change in velocity is negative fourteen point five eight or negative fifteen meaning the initial velocity must have been paused at fifteen how many people get a four that good everybody missed the changing units there the kilonewtons are the ten to the minus one seconds okay one more than I&amp;#39;m gonna get you to do before we move on to our next stop actually question number five on the next page who&amp;#39;s gonna answer this question now who thinks they have an answer this question nobody wants to volunteer all right a couple votes for a I got a couple of votes for C here let&amp;#39;s take a look at the question it says plutonium-239 recovered from dismantled atomic weapons undergoes alpha decay in beta decay at the same time sensors on the outside of the containers used to store it him to indicate how much radioactive material was contained by monitoring the gamma rays emitted by the material since a particular container holds this much waste it&amp;#39;s got a half-life and it talks about the energy of the gamma ray and the speed of the alpha particle so on the question says which of the following diagrams shows the orientation of the Momentum&amp;#39;s of the alpha particle and the gamma photon and must result in the maximum impulse of the plutonium 3239 nucleus listen when you see this question you now looking at numbers this has nothing to do with the numbers that are given to you you got to look bigger picture here you&amp;#39;ve got blue tone &amp;#39;i&amp;#39;m 239 and here&amp;#39;s the big plutonium 3239 it decays into something else okay some other slightly smaller nuclei and alpha particle in a gamma ray right now the momentum of this plutonium 239 is zero if we want to maximize the impulse and that&amp;#39;s what the question says right to maximize the impulse then we need to give it more momentum it has zero my item we need to give it lots of momentum how are we going to give it lots of momentum maximize the amount of momentum we&amp;#39;ll give it well let&amp;#39;s take a look at D for a second here if the if it decays in the alpha particle goes this way with let&amp;#39;s say let&amp;#39;s say the value of this would is 10 if it goes this way with 10 units of momentum and the gamma ray goes this way with 10 units of momentum these two cancel each other out what&amp;#39;s going to happen to the remaining daughter nucleus it&amp;#39;s going to stay where it is look the momentum was zero beforehand if they&amp;#39;re going in opposite directions afterwards and anthem there&amp;#39;s got to be zero afterwards as well so B can&amp;#39;t be the answer let&amp;#39;s take a look at C now that&amp;#39;s the one that Kari suggested there well if the alpha particle goes this way with let&amp;#39;s say 10 units of momentum this looks like it&amp;#39;s a little bit bigger maybe 12 units of momentum then the daughter nucleus is going to go this way with how much we could use the Pythagorean theorem to find out how much momentum of the daughter nucleus was right here what we&amp;#39;d have to do components but we know that the daughter nucleus would go back this way with a certain amount of momentum he in the end this is the one that we that we care about here because if the momentum of the alpha particle is 10 units to the right in the gamma ray is 10 units to the right then the daughter nucleus must end up having 20 units in the opposite direction you can&amp;#39;t get if this is 10 in 12 or 10 I think that&amp;#39;s supposed to be the same length actually 10 and 10 you can&amp;#39;t get 20 out of this guy by having two tens at right angles to each other you can&amp;#39;t get 20 out of this guy we&amp;#39;re getting two tens at funny angles here hey you get the X component of each of them combining back here which is less than ten that makes sense understand why is the answer there all right let&amp;#39;s let&amp;#39;s take a look at our next topic now is stopping number 15 conservation of momentum in one dimension as opposed to conservation momentum in two dimensions you can see that if you get a problem involving this it&amp;#39;s pretty easy physics principle to identify if you happen to have one of those questions involving physics principles its conservation of momentum right hey there&amp;#39;s nothing else really that can apply here now that&amp;#39;s not to suggest that you can&amp;#39;t get a question with with momentum involved like this but also something else leading up to that maybe you&amp;#39;ve got conservation of energy where something is speeding up to a certain speed and then there&amp;#39;s a collision taking place where it&amp;#39;s conservation of momentum but if it&amp;#39;s just a question like we&amp;#39;re about to see then we&amp;#39;re dealing with one physics principle principle number four conservation Allah mentum all right we talked a little bit about this one earlier an isolated system that&amp;#39;s a system in which is no external forces acting no friction no no wind or anything like that in the end we can&amp;#39;t always isolate a system from external forces by removing them for those forces so what we often see is a course we&amp;#39;re asked for the velocity of an object immediately after or immediately before a collision that means that yeah maybe there is friction but since we&amp;#39;re dealing with the velocity immediately before and immediately after we haven&amp;#39;t given friction a chance to really do anything yet so we still consider them to be an isolated system because although there are external forces they they don&amp;#39;t come into play until the object just had a chance to move after the after the collision explosion has taken place all right when do we use this this whole idea of conservation element and this whole idea that momentum stays the same when we have an isolated system well we have a collision or we have an explosion even by collision we mean literally two things hitting each other can explosion literally means one thing that it breaks apart into two or more things explosion doesn&amp;#39;t necessarily mean a bomb breaking up it could mean something like a gun being fired bullet goes one way the gun recoils the opposite way or maybe maybe a balloon filled with air okay you have the balloon filled with air and you got the end of it held with your fingers and then you let it go the air goes one way and the balloon goes the other way right if the it here&amp;#39;s a good question not so much concentration momentum but takes it&amp;#39;s back to something that we did here earlier today if I&amp;#39;ve got that balloon in my hand filled with air and then I let it go the air goes one way the balloon goes the other way if the balloon happens to gain 10 units of momentum to the north then what&amp;#39;s the momentum of the air that&amp;#39;s going out the back of the balloon if the balloon was in my hand and now when I let it go the balloons going forward with 10 units of momentum then the air must be going backwards with 10 units of momentum because the impulse experienced by the balloon must be the same as the impulse experienced by the air even if the air is lighter balloon is heavier or vice versa doesn&amp;#39;t really matter the impulse is always going to be the same all right back to this whenever we&amp;#39;ve had a collision or an explosion we&amp;#39;re gonna do this whole constellation momentum thing which says P AI equals PF count up your objects put an envy tournament for every object maybe it&amp;#39;s M 1 V 1 I plus M 2 V 2 I plus and so on and so on and so on depending upon how many objects we have equal and 1 V 1 F plus M 2 V 2 F and so on if they stick together then the mass that&amp;#39;s combined and we typically see it written like M 1 V 1 I plus em 2 V 2 I equals MV F and being a combined mass if they bounce apart then we typically see it written as em 1 V 1 F plus m2v2 F literally just count the objects that you have before and after the collision and/or explosion and put an M V term in for each of them pay attention to positive versus negative in your calculations hey that&amp;#39;s important that&amp;#39;s so important again take you back to that written response test that we had several weeks ago on momentum weighed the collision between the two vehicles the car in the truck one of them was going to the north one of them was going to the south and lots of you nailed that question lots of you got it correct but what if you didn&amp;#39;t get it correct because lots of you failed to take into account that one of the objects was going to the south before the collision so you forgot the negative okay when you see that question north-south-east-west circle in red to remind yourself of signs they circle north circle south to remind you that it&amp;#39;s got to be opposite signs okay units what do we know about units with conservation momentum typically we&amp;#39;re gonna be using kilograms and meters per second for mass and velocity but in a conservation momentum problem it doesn&amp;#39;t actually matter as long as you&amp;#39;re consistent how come how come if I&amp;#39;m looking at this equation right here how come I could use grams for mass if I wanted to even though kilograms of the standard units al you look like you&amp;#39;re ready to give me the answer but yeah they cancel it right if they appear in every term then they cancel oh no it&amp;#39;s not the same instead of the same mass in every term but it&amp;#39;s the same units in every term so the units could cancel it if in doubt what should you do if in doubt you&amp;#39;re not sure what units you should be using what should you do go to the standard right go to the standard units kilograms in meters per second feel that you almost always have to do that it&amp;#39;s just the odd situation like conservation of momentum we can get away without doing it all right I&amp;#39;m gonna give you a question to work on before I talk about what I mean by a progressive collision we&amp;#39;ll talk about that in just a couple minutes the one that I want you to work on now is gonna be question number six please of course number six on the same page is the question you just did it&amp;#39;s a skateboarder question all right a skateboarder has a mass of 50 kilograms is running at a speed of 4.5 meters per second skateboarder jumps onto an initially stationary skateboard that has a mass of 4.2 what&amp;#39;s the speed of the skateboarder and scape a skateboarder and skateboard after the after this interaction takes place it&amp;#39;s like a collision right it&amp;#39;s not literally a collision where two cars hit each other but it&amp;#39;s one object hitting another object or or becoming one with the other object so we&amp;#39;re gonna say it&amp;#39;s a conservation of momentum problem P I is equal to P F we&amp;#39;ve got two objects before this collision takes place so we&amp;#39;re gonna say m1 v1 I plus m2 v2 I and then we have one object afterwards because the skateboarder and the skateboard become one object so we&amp;#39;re gonna call it MV f we&amp;#39;ve plugged in our numbers here we don&amp;#39;t have to worry about signs in this everything is in the same direction units are all standard units so it&amp;#39;s a rather straightforward example of 150 times 4.5 plus 0 equals the combined mass of fifty four point five kilograms times VF and we do the math on question number six and I think it works out to be a four point one five how many people get that all right good what did I mean by that progressive collision we don&amp;#39;t see that very often but we did see one of these on a unit test that we had I think was a test or a quiz or something that we had at one point I&amp;#39;m gonna give you that same question again in a few moments here but essentially what I mean by this is a progressive collision is when when there&amp;#39;s a collision that takes place which goes into another collision so let&amp;#39;s say it&amp;#39;s kind of like a domino effect right you&amp;#39;re you&amp;#39;re running you&amp;#39;re you&amp;#39;re playing football and you&amp;#39;re running down the field and you hit somebody okay and that and you knock that person out of the way and then that person hits somebody else that&amp;#39;s a progressive collision now how do we deal with that well we can deal with it as two separate collisions we can say P I equals P F for the first collision and then for the second collision P ICO&amp;#39;s PF but p f for the first collision would be pi/4 the second collision that make sense so basically what I&amp;#39;m saying is let&amp;#39;s not call it INF because there&amp;#39;s three different momentum there&amp;#39;s two different initials and finals here let&amp;#39;s say P 1 equals P 2 so the initial momentum equals the final momentum and then the the font that momentum from this one becomes P 3 so we&amp;#39;ll say P 2 equals P 3 because the final momentum for the first collision becomes the initial momentum for the second collision and then P 3 equals P 4 these are my two collisions these are my two collisions here okay here&amp;#39;s the final momentum for the first collision here&amp;#39;s the initial momentum for the second collision but the final momentum for the first collision equals the initial momentum for the second equation right hey I collide with you you bounce off with a momentum of 10 okay you hit somebody else you hit them with the momentum of 10 your final momentum in my collision is your initial momentum in the next question take a look at what I just wrote down though if P 1 equals P 2 and P 2 equals P 3 and P 3 equals P 4 then how can we simplify this P 1 equals P 4 so if you&amp;#39;ve got a progressive coalition which we&amp;#39;re about to see in a second um you can you can go through the process of finding the initial and final momentum a couple of times a couple different problems but you can simplify it by simply looking at the initial momentum before anything happens in the final momentum after everything happens you don&amp;#39;t have to look at each individual collision you can but it&amp;#39;s twice as much work if you want to do that let&amp;#39;s take a look at this question please it has that you have seen before but you&amp;#39;re not like we&amp;#39;re gonna remember the answer for okay let&amp;#39;s have a look now we get three cars so your car - and car three are at rest before anything takes place car one is moving at four meters per second car one collides with car two then car one and car two both combine and collide with and stick to car three so after before all this is said and done okay car one is the only one moving after all is said and done car one car two and car three I&amp;#39;ve all stuck together and they&amp;#39;re moving off as one object yeah you can treat this as two collisions if you want P I equals P F for the first collision and then P I equals PF for the second collision or you can just look at it as P I equals P F to the entire thing the initial momentum would be m1 v1 I because there is no momentum for the second object to the third object now if the second or even the third object was moving before this collision took place then we&amp;#39;d have to add a term for that even if there was all three objects moving we would say m2 v2 i plus M 3 V 3 I whatever the case may be they&amp;#39;re not so we can ignore those terms okay but it&amp;#39;s possible that down the road they could be moving before the collision you have to include those if they are afterwards they stick together we&amp;#39;re gonna call it MV f and being the total mass here my the mass of object number 1 is 1,500 kilograms v1 I is 4 meters per second the total mass I got 4500 is that right 4,500 kilograms times V F we&amp;#39;re gonna solve for V F there and what we do it works out to be one point three three meters ii would express that it&amp;#39;s kind of an odd one to express 1.33 on our numeric response answer sheet we&amp;#39;d put it as one decimal three three notice three significant digits right three digits one point three three not scientific notation nothing like that just the number good all right we&amp;#39;re gonna do another one here now this question number nine but I&amp;#39;m gonna do this one with you this one talks about a lab which basically we did we pretty much did this exact lab and cross except we didn&amp;#39;t use ticker tape or a spark timer we use our we used our sparkvue app along with our motion sensors in order to detect the motion but in the end we did the same thing essentially where we took two carts car one car two cart number one was initially at rest v1 I was zero meters per second cart one was was moving we sent moving at a certain speed this should have been the two iv1 i with a certain speed that we manipulated that we changed through the experiment he they stuck together we had this little plasticine okay next year based on some of your sources of air we needed that lab make sure i&amp;#39;m gonna stick some velcro to both of these carts they&amp;#39;re going to stick together a little better regardless they stick together they go off as one object we prata graph a VF the final velocity of both objects combined versus the initial velocity of car number one car number two doesn&amp;#39;t have an initiative law so there&amp;#39;s nothing to plot there and this is the data that we get just a bunch of data points now normally we draw a line of best fit for this right most times we don&amp;#39;t actually have to on an exam draw the line of best fit for this one we actually do have toon you&amp;#39;ll see why in just a second let&amp;#39;s skip to the question not question number eight but question number nine and see what they&amp;#39;re asked to find we see what I asked fine based on a point on the line of best fit cart one has a speed of six centimeters per second before the collision then the combined speed of the two cards after the collision would be eight as will be a centimeters per second okay based on a point on the line of best fit we don&amp;#39;t have a line of best fit so you&amp;#39;re going to have to draw one okay what you don&amp;#39;t want to do is just go over to six seconds and then go over here and say all right at six seconds we&amp;#39;re at two point two point four centimeters per second that&amp;#39;s not a line on the line of best fit take a ruler draw your line of best fit we don&amp;#39;t usually have to do this but occasionally we do hey lion investment might look something like this now the good news is if you&amp;#39;re just reading a point off of the line of best fit you&amp;#39;re probably going to get this right if you&amp;#39;re on invest videos even remotely close to be reasonable if you draw out crazy then you&amp;#39;re gonna get the wrong answer but they&amp;#39;ll accept a range of answers here as long as it&amp;#39;s consistent with a reasonable line of best fit okay we want to find we wanted to find here the speed of the combined mass if the speed of cart number one was six meters per second or six centimeters per second I should say so we go up we go over and it looks like we get about 2.5 2.6 this would be two point five is fifty two point six two point five five or two point six centimeters per second we didn&amp;#39;t have to do anything crazy difficult there right no calculated work no analysis of the graph really other than up and over we&amp;#39;re gonna fill in by the way the a V with two and six the combined speed of the two currents after the collision would be two point six centimeters per second now based on the slope of the line of that&amp;#39;s that the combined mass would be what this is where it gets trickier right but that&amp;#39;s okay we know how to deal with this based on the slope of the line of best fit what&amp;#39;s the combined mass let&amp;#39;s figure out what this slope means y is equal to MX plus B the y axis is VF m is the slope the x axis is v1 I now the y-intercept is pretty close to zero looks like it&amp;#39;s close to zero I don&amp;#39;t know if it&amp;#39;s exactly zero but I think it&amp;#39;s gonna be close enough we&amp;#39;ll wait and see what the other equation looks like first before we decide for sure whether the y-intercept is going to be zero what other equation from my data sheet or that I can that I can derive might go with this don&amp;#39;t necessarily go straight to your dadís you just think about what&amp;#39;s going on this is a collision between two carts what do you usually do when you have a collision between two cards Carrie good P I equals P F so let&amp;#39;s do that okay let&amp;#39;s try that that seems to be the relevant equation here P I equals P F we&amp;#39;re gonna say m1 v1 I plus zero because the second card is at rest equals M V F M being the total mass what&amp;#39;s rearranges to solve for V F becomes M 1 over m T times V 1 I this works right carry we said that we have to have an equation that has the same variables in it VI equals P F has the same variables V F in V 1 I we cross off the F we cross off II 1 I and we&amp;#39;re left with the slope is equal to m1 over Mt if we want to find the total mass and all we&amp;#39;re gonna do is take the initial map the first mass and divided by the total I&amp;#39;m sorry divided by the slope the first mass is one point five four kilograms we don&amp;#39;t know what the slope is so let&amp;#39;s do this together on our calculator so game everybody pull your calculator and let&amp;#39;s go through this process we&amp;#39;re gonna say stat edit when you go there it should probably be blank if it&amp;#39;s not just go up to just go up to l1 and press clear now let&amp;#39;s enter data points now you got to be careful here okay if you look at make sure you get these data points correct okay I got the first data point to be one point four in two point four one point four in zero point four what&amp;#39;s going on here there we go now I got my second data point here to be at two point four because each of the x-axis intervals is point two and looks like zero point nine two point four and zero point nine my next one looks to be three point two and 1.15 that one&amp;#39;s tricky 1.15 look at this yeah this is one point 1 this is 1 point 2 I&amp;#39;m halfway in between so I&amp;#39;m 1.15 my next at a point is 4 point 3 and I&amp;#39;ve got 1.75 my next data point is 5 point 4 and I&amp;#39;ve got 2 point 3 on my y-axis 6 on my x-axis in 2.4 on my y-axis 6.4 on my x-axis and the last one here is 3 on my y-axis so there I&amp;#39;ve got 1 is it 1 2 3 4 5 6 7 data points here now remember you hit stat edit now it&amp;#39;s going to be stat calc option for linear regression keep hitting enter until you get a number here my slope is point 4 8 5 so I&amp;#39;m gonna take the first mass that I got which is 1 point 5 4 divide it by zero point 4 8 5 when I do that I get 3 point 2 so my total mass here my total mass my combined mass which is what they&amp;#39;re looking for here it&amp;#39;s 3 point 2 2 6 3 2 would be my answer yep we&amp;#39;re getting there now we&amp;#39;re talking two dimensional collisions and or courses in two dimensions though we got really three kind of interactions that we were going to see here there&amp;#39;s the t-bone collision which looks like this and then typically we have them going off as one object afterwards we have what we call glancing collision which is something like this eight two objects hit each other but not straight on but kind of at the edge and sends one one way and sends one another way and then of course we have the explosion which is like you know got one object it splits up into three or four pieces like this sometimes in a collision they stick together sometimes they bounce apart that&amp;#39;s no different than a one-dimensional collision if they stick together it&amp;#39;s m1 v1 i plus m2 v2 i equals MV f if they balance apart its m1 v1 F plus m2 v2 F how do we do these questions though right how do we do the first one the red one the blue one or the green one keep on collision the glancing collision or the 2d explosion well the good news is if you can do one you can for the most part do any of the three we break them up into x and y-components and for both the x and y components we say P I is equal to P F in this case it would be m1 v1 plus m2 v2 I equals MV F because they stick together for both the X and the y in the second case it would be m1 v1 plus m2 v2 I equals they bounce apart m1 v1 F plus m2 v2 F if we&amp;#39;re talking about the third case over here then it would be zero initial momentum because it&amp;#39;s an explosion right that objects at rest equals m1 v1 F plus m2 v2 F plus m3 the because we&amp;#39;d have three objects after the explosion takes place do the same thing for the Y component solve for usually it&amp;#39;s gonna be VF for both the X and y components get both of them the X component and the y component of VF and then do the Pythagorean theorem simple as that sometimes especially knees glancing collisions here we got a funny angle what do we do what do we do if we have a funny angle not what do we do for you have a funny angle yeah we don&amp;#39;t like funny angles right so we get rid of the funny angles okay if I have something like this that&amp;#39;s at 25 let&amp;#39;s make it 30 degrees because it&amp;#39;s easier math if I have something like this that&amp;#39;s 10 at 30 degrees and just take it off to the side over here and solve for an X and solve for a y the X component would be cosine 30 degrees is equal to x over 10 and the y component would be sine cuz it&amp;#39;s opposite so 930 degrees is equal to Y over 10 solve for X solve for y and then whatever value you get for X goes into VF VF for object number one okay don&amp;#39;t worry too much about units here just be consistent eight kilograms usually and that&amp;#39;s the standard unit if you&amp;#39;re confused just go to kilograms meters per second but in the end you know that you can use grams or kilometers per hour whatever you want make sure that you resolve the components if we got a funny angle don&amp;#39;t ever don&amp;#39;t ever go with a funny angle they get rid of it get rid of the funny angle into an x and y-components know how to deal with those remember to always and I&amp;#39;ll tell you like you wouldn&amp;#39;t think of this to be a problem but it is remember to always break it into X&amp;#39;s and it&amp;#39;s too often people see a collision like this and then they just have this brain freeze this brain fart where they forget to break it into x and y&amp;#39;s and they just and they just say m1 v1 i plus m2v2 i equals whatever you can&amp;#39;t do that if there&amp;#39;s if it&amp;#39;s a collision like this or like this or any kind of two dimensions anything like that you see above you have to break it up in exelon components pay attention to the pauses and negatives this is kind of repeat right cuz we talked about most of this stuff with one dimensional the only thing that really separates this from the one-dimensional collisions that we&amp;#39;ve already done roni that&amp;#39;s right it&amp;#39;s really the only thing that makes it different then and these questions that we&amp;#39;ve already done earlier today well talk about J and K in just a couple of minutes what I&amp;#39;d like you to do right now is have a look at a couple of questions here first of all question number ten in the booklet that I gave you just now all right the question says after the explosion we want to find a piece of speed or the speed of P see here look don&amp;#39;t just say P I equals P F for this right we went we just went through this we just said that when you&amp;#39;ve got something like this or a glancing collision okay like this or you&amp;#39;ve got an explosion like this a or a t-bone collision like this you can&amp;#39;t just do P I equals B F it&amp;#39;s got to be bi crispy F twice on the X and the y axis here so let&amp;#39;s let&amp;#39;s do that now let&amp;#39;s break it up into X component and Y component let&amp;#39;s say for both of them P I equals P F for both of them the initial momentum is zero because it&amp;#39;s at rest right zero is equal to m1 v1 F plus m2 v2 F plus M 3 V 3 F and for the Y component same deal zero is equal to m1 v1 F plus m2 v2 F plus M 3 V 3 F good so far plugging in numbers here but be careful be careful you you look at either X or Y at one time here X components we&amp;#39;d say object number one has zero momentum even though it&amp;#39;s moving it&amp;#39;s moving on the y axis object number two is two kilograms times forty four point six and object number three is four kilograms times V 3f we solve for V 3 F there oh except to be negative 22 point three meters per second and you do the same thing for the Y component but this time there is a y component it&amp;#39;s 3 times 39 point to know what&amp;#39;s initial know what&amp;#39;s final and know what is X and know what is y and don&amp;#39;t mix those up the y component of the second one is zero because it&amp;#39;s all in the x axis solve for V 3 F here and V 3 F for the Y component work so it could be negative twenty nine point four and then we got to combine them we got negative twenty two point three negative twenty nine point four you guys solve for V there we don&amp;#39;t need to solve for theta okay but we could use in the tan function V is just going to be the Pythagorean theorem and it&amp;#39;s going to work out to be thirty six point nine meters per second if all these masses were the same what could we do to simplify this if they were all four kilograms we could cancel the mass would you have to cancel the mass of course not but you could if you wanted to save yourself a little bit of time and effort we go back to this down in J within topic number sixteen says problems given an unknown angle a lot of times a lot of times you&amp;#39;ve got a question where we have to solve for that angle right we&amp;#39;re given the unknown angle this time right if we&amp;#39;re given the unknown angle and then we don&amp;#39;t have to treat any differently we can do it the way we just did but notice what we just did it we didn&amp;#39;t pay any attention to that angle because oftentimes we don&amp;#39;t have that angle so I did it as if we don&amp;#39;t have the angle so that you can do any question like this whether you have the angle or not but not you&amp;#39;re right there is a shortcut here if you recognize that you have that angle let&amp;#39;s get rid of all of this for the Y component and then let&amp;#39;s say the X component is twenty two point three the angle is thirty seven point two okay we can find this speed just by using the sine function now sine thirty seven point two degrees is equal to twenty two point three over V so you can save yourself some time if you&amp;#39;re given that quote unknown angle it&amp;#39;s not really an unknown angle right but it&amp;#39;s the angle that we&amp;#39;re often trying to find if we&amp;#39;ve got it there&amp;#39;s a shortcut but you never have to take that shortcut it&amp;#39;s just a little bit a little bit of the time savings if you spot it um but you gotta be careful sometimes people right before I&amp;#39;m looking for shortcuts right you take five minutes looking for a shortcut that&amp;#39;s gonna save you - well you haven&amp;#39;t really taken much for a shortcut then right you spot her right away great if you don&amp;#39;t just do it there just do it the regular way oh you&amp;#39;re gonna love this one okay this one let&amp;#39;s do this one together okay ozone in the upper atmosphere protects life by absorbing dangerous photons of ultraviolet electromagnetic radiation produced by the Sun when such a photon collides inelastically with the stationary initiation ozone molecule the molecule is broken apart into an oxygen molecule in an oxygen atom as shown below before the collision after the collision listen we&amp;#39;re talking about photons here we&amp;#39;re talking about oxygen molecules but when you look at this I think it&amp;#39;s pretty clear that this is a collision right and this is a conservation of momentum problem we got to do X and y components because of that but what are you going to do first what&amp;#39;s the first thing you got to do I don&amp;#39;t like funny angles get rid of the 74 degrees look I don&amp;#39;t care if it&amp;#39;s photons or oxygen or atoms or molecules or ozone I don&amp;#39;t care anything about that listen I know what to do here my velocity here was 220 meters per second I could find the X and I can find the Y by using some trigonometry cosine and sine and when I do that I&amp;#39;m not gonna show all that work here because I think you can do that I end up getting 60 point six four zero two for X and I end up getting two hundred and eleven point four seven seven eight for the Y component now I break it into my X&amp;#39;s and Y&amp;#39;s now that I&amp;#39;ve never dealt with that funny angle that I don&amp;#39;t like I do my X&amp;#39;s and Y&amp;#39;s PII equals PF I&amp;#39;m gonna state look look I&amp;#39;ve got the momentum of the photon if I if I had the wavelength I could find the momentum by saying H over lambda by the energy I could find the momentum by saying a over C Y over C using those equations in the top right hand corner of your data sheet but I&amp;#39;ve got the momentum here the length of the photon is nine point three times ten to the negative 24 plus zero because the ozone is at rest plus the mass of the oxygen which is sorry equals I should say the final momentum the mass of the oxygen times you guessed it the the X component of the final velocity of the oxygen sixty point six four zero two and then we add to that two point six six times ten to the minus twenty six times bf so momentum of the photon plus the momentum of the ozone momentum of the molecule plus the momentum of the atom we solve for VF there we accept to be two hundred and twenty eight point five seven one six now that if I had we know the oxygen atom is gonna go down here somewhere right if I had this angle given to me already what what could I do yeah yeah I probably end up using cosine because I found the X component I&amp;#39;ve got the I&amp;#39;m looking for the hypotenuse cosine theta is equal to x over two twenty eight over the hypotenuse right but I don&amp;#39;t have that unknown any go but I&amp;#39;m looking for it so I&amp;#39;ve got to go through that process now of finding in Y components we&amp;#39;re gonna say zero initial minute them on the y-axis because everything&amp;#39;s on ax right is equal to here&amp;#39;s the molecule five point three one times 10 minus twenty six times two hundred and eleven point four seven seven eight this is the Y component the final momentum of the molecule plus two point six six this is the atom solve for V F and we get negative four twenty two point one six oh six listen this is an easy question now right do the same thing as we always do there now do the same thing as you did for the last question same thing as we do for every other 2d question 228 for twenty two point one six zero six I could find theta by the inverse tan function and we can find V by using the Pythagorean theorem I&amp;#39;m through that because I think you&amp;#39;re okay if you need me to go through it I will but otherwise I&amp;#39;ll just give you the answers here me is 480 and fader works out to be 61 point 6 okay we&amp;#39;d say that would be if this is a north-south east-west thing then we&amp;#39;d say that would be 61 point six degrees south of East because we went east then we went from the East Gary we went south from the East self after the East or self from these okay look at the angle here it&amp;#39;s measured not from the self because this is self it&amp;#39;s not even touching South right it&amp;#39;s measured from the east axis but it&amp;#39;s self from the east axis it&amp;#39;s below the east axis so it&amp;#39;s self ities alright let&amp;#39;s take a look at our last thing here which is you&amp;#39;ve asked it inelastic collisions now couple quick little reminders an elastic collision remember is a collision in which kinetic energy is conserved remember if they stick together it&amp;#39;s always inelastic if they stick together if they bounce apart could be elastic could be inelastic don&amp;#39;t know when we see a question we hope for them to stick together because then we have a definitive answer without doing any calculations if they stick together it must be inelastic if they bounce apart it&amp;#39;s probably still inelastic but it could be elastic scalar nature always pause if no angles right this is when even in a question like this we would not break it up into x and y components why I cuz momentum is a vector but kinetic energy is a scalar so all you&amp;#39;re gonna do is say whether it&amp;#39;s a one-dimensional or two-dimensional whether it&amp;#39;s north and south or north and north you&amp;#39;re gonna say EK i equals one-half m1 v1 i squared plus however many objects you have get a number do the same thing for the final you know those two numbers are the same it&amp;#39;s the elastic if EK i is bigger than EK f its inelastic EK f won&amp;#39;t be bigger than EK i if it is you&amp;#39;ve made a mistake units joules always want to use joules here we&amp;#39;ve used the electron volts a lot now since we&amp;#39;ve gone through atomic physics don&amp;#39;t use the electron volts here can use joules and remember again just with a scale and H are always positive no components Hey no funny angles okay this is the one time when you can say listen I don&amp;#39;t like funny angles but they&amp;#39;re not really bothering me right now if you&amp;#39;re dealing with the scalar funny angles don&amp;#39;t matter you can hate them all you want but just to leave them alone when you&amp;#39;re dealing with a scalar and finally before we do a quick example here finally in an inelastic and inelastic collision kinetic energy is conserved so it&amp;#39;s momentum in an inelastic collision kinetic energy is lost momentum is still conserved doesn&amp;#39;t matter what kind of collision you have momentum is conserved example I want to do if this goes back to question number eight goes back to this information on the graph that we already did question number eight says the collision of the two carts is classified as blank because blank man I&amp;#39;m telling you I&amp;#39;m really hoping these two things stick together because I don&amp;#39;t want to calculate e ki and e KF I&amp;#39;m not sure what the I and VF I&amp;#39;m gonna use because there&amp;#39;s seven different data points for VI and VF when you look back at the information though what would be hold they do stick together and because they stick together we know that this is what kind of collision inelastic we know it&amp;#39;s inelastic and why do we know and in what happens in an inelastic collision kinetic energy is not conserved is momentum conserved yes it is momentum is conserved kinetic energy is not conserved because it&amp;#39;s inelastic the other one that I want to do is multiple choice number 12 which is going to be towards the vacuum well could if I can find it here this isn&amp;#39;t across and station two identical well friction Democrats are pushed toward each other such that they&amp;#39;re traveling at the same speed when they collide they said two cars collide two sick carts same mass same speed collide they come to a complete stop teacher challenges students to describe in schoolers and some of the students descriptions are provided below before the collision the carts have the same momentum before the collision they have the same kinetic energy this is an example of an elastic collision this is an example of an inelastic collision shawne&amp;#39;s are correct do they have the same momentum before the collision all this is a great question do they have the same momentum before the collision they have the same mass they have the same speed do they have the same momentum before the collision who says yes who says no why do you say no Allen good they have the same magnitude of momentum they do not have the same momentum momentum is a vector the momentum of one of them is whatever value to the right the momentum of the other one is the same value to the left those are two different Momentum&amp;#39;s they do not have the same momentum they have the same magnitude of momentum but momentum is a vector we have to pay attention to direction do they have the same kinetic energy yes or no who says they have the same kinetic energy who says they don&amp;#39;t have the same kinetic energy they do have the same kinetic energy they have the same mass and they have the same speed momentum depends upon velocity kinetic energy depends upon speed they do have the same kinetic energy elastic versus nos clearly can&amp;#39;t be both is it elastic or inelastic we don&amp;#39;t know how fast they go on here so I&amp;#39;m not sure how we&amp;#39;re going to calculation elastic or inelastic who says he last a few objects same speed same ass hit each other come to a stop who says elastic who says inelastic Hugh I do you say interesting okay okay if in doubt go with inelastic because nine times in a ten or even more than that it&amp;#39;s gonna be an elastic truly only subatomic particles have our our elastic rang kiss they don&amp;#39;t squish but you get it it&amp;#39;s conceivable to get the collision it&amp;#39;s really really close to us it could be technically classified as you elastic this is anyone a stick though it is inelastic how come what&amp;#39;s the initial kinetic energy I don&amp;#39;t know but something what&amp;#39;s the final kinetic energy zero something does not equal 0 if obvious number one is 10 object number two is 10 combined it&amp;#39;s 20 not zero because they&amp;#39;re scalars right 10 and 10 is 20 afterwards the total moment total kinetic energy is zero we lost kinetic energy since we were asked in attic energy it&amp;#39;s any other aspect so what we got two and four there are three more questions that we&amp;#39;re not gonna do here right now these are three questions that aren&amp;#39;t in order they don&amp;#39;t really go at all in order with the with the topics that I&amp;#39;ve listed in that 16 page book up just kind of look at the end of everything after we&amp;#39;ve done gone over the unit maybe you want to take a look at three kind of random questions within one scenario we&amp;#39;re not gonna do those right now you want to take a look at those yourself and then come back and ask me how to do something if you have trouble that&amp;#39;s great if you want to ask me Monday morning where you guys come in for your next extra help session then you can do that as well I will give you the answers though right now for questions 13 14 and 15 number 13 the answer is C number 14 the answer is C and number 15 the answer is B

Transcript for:Lecture on Momentum and Impulse

Transcript for:
Lecture on Momentum and Impulse