physics and engineering problem solving

[00:00:00]both physics and engineering are heavily reliant on problem solving so in learning physics and engineering and studying it what's also important is learning a problem-solving process and practicing that process so students at north shore are expected to use a five-step problem-solving process it's outlined in the guidelines for all work submitted in class teamwork submitted as homework as well as on the exams that problem-solving process is also outlined here the five steps what's being given what is it that you need to find what theory and analysis can you use a solution process or which we call link solve because that's the action that you're taking and then a check and i won't go over everything here but i'll give you an example you can pause if you want to make sure you read and understand what has to be represented in each of these five steps and some people ask so i need to label given fine theory and analysis you don't need to but it helps when you go back to look at your solutions in your homeworks to help you read it make it readable in the given step diagrams are also necessary sometimes it's difficult to know what the diagram is but drawing something on there will help and typically you want to clearly depict all those given that given information on the diagram you want to make sure you use the proper notations you want to make sure you use the units when you're your theory and analysis maybe at first you're coming up with a wrong theory so it might not be in this section you might be doing a little bit of problem solving a little bit of solving but you might find out that you went on the wrong path that something's missing or in fact the theories that you were trying to use would not be able to apply so when you get to that final solving process maybe you can start just plugging in numbers but make sure your notations are correct you can't just say anything is equal to anything else between equations you can use it implies notation if um you get from one's problem statement that might imply the next one if you're if you're changing the form of the of the uh of your work and then we will work in the proper si derived units we won't have to work foil things back down to the si base units the check in my mind is the most important step it's after you've completed the process you want to make sure is it complete um you know so things you can check is the magnitude and numerical value reasonable is you know plus or minus is that does that make sense that you get a negative answer or a positive answer that sort of estimation a unit check you know the unit's reasonable you're going to check using a dimensional analysis type in your check is can be very important and it's a good type of check and um you know what does it mean you know what does it mean in terms of what you have learned as well so that's the the big key that what did i learn from probably solving this problem so let's take a look at an example this is example is an experiment using a metal ball made of copper and that that in an experiment it leaves the copper with a positive charge of two micro coulombs given the mass of the ball was 50 grams we want to derive a fraction of the copper's electrons that were removed and then we're additionally given information about that which is really the theory that a copper atom has an atomic number of 29 and copper has an atomic mass of 63.5 so you know what we just what are we being given we want to take that information and we want to convert it to it's a variable form so we have a ball so the charge on that ball is given as 2.0 micro coulomb so we can write that out two point it's three that counts 2.00 so we know we have three significant digits here that are so our final answer will be rounded to three significant digits now it's a ball it's not a whole lot we can do with a ball right we don't necessarily need to say that that you know label this as q because we have this but we can like say well there's our ball and it has that charge and it's a ball it's made of copper and we know the unifor you know the symbol for copper is cu so we also have the mass of the ball so we can write that out the mass of the ball that's 50 grams um now do we need to convert that to kilograms because kilograms our base unit maybe maybe not let's leave our given in this form because if we have other things with grams we won't have to convert it um and then we want to derive the fraction of the balls copper electrons that were removed that were removed so what does that really mean so we want to know the fraction that's removed so we want to know the number we use n for number number of we can just use our subscripts to make this clear electrons removed compared to when it was neutral over the number of electron when we have a neutral if i spell neutral right i always get u and e backwards um spell check corrects me now so we're also given this information so clearly we're going to need it we're given that um the atomic number for copper is 29 so that atomic number we typically write that that's atoms per proton so um we typically our atomic number uses a z i believe so let's do z of c u that's 29 and that's um that's basically giving us the number of protons per atom so if we weren't you know really remembering this it's physics 3 that might be more of a physics 2 thing or chemistry we can look it up we've got the beauty of google now as well to look up atomic number and know what the lookup make sure we understand what is the correct um variable that we use for atomic number and then we have our um atomic mass our atomic mass here is given as equal to that's still copper as 63.5 and atomic mass in this it's given it's given in grams per mole so if we if those are going to cancel out then we can just leave things in grams we don't have to convert them to kilograms so now we have in fact what we're given and i'll just jump to the next page here where i've written that out a little bit more condensed so we can fit everything on the same page so this is our given information but we've also worked out what it was that we wanted to find and we worked through the first two steps we really did that just by analyzing what was being said but now we've got to do some theory and how do we find this how do we find the number of electrons removed from the ball how do we find the electrons in a neutral ball so when we're doing our theory analysis we have to understand that that number of electrons in the neutral ball is equal to the number of protons in the neutral ball so that's our theory right if it's if it's neutral that means the charge is equal to zero and the things that have charge in matter are electrons and protons so we have the number of protons per atom so we just need to determine what are the number of you know total number of um protons right then we'll get the total number of electrons so in doing that here we have a protons maybe we don't have a formula because we're not studying this necessarily with a formula but we could use a dimensional analysis approach we want to end up with protons we have protons per atom so we need to get rid of atom so we might need um something else ah we have a mole and we know avogadro's number is number of atoms per mole right so we need this this avogadro's number here 6.02 times 10 to the 23 atoms per mole we didn't realize it it wasn't given to us but in attempting a dimensional analysis you know that's what came out so we got we want grams grams per mole that gives us a mole then we have atoms per mole so the moles cancel and then we have protons per atom and we end up with our number of protons so instead of using a theory here what we used was our dimensional analysis now we can come up with the theory after the fact and um instead of trying to find a problem that does this exactly and what we have is our our mass divided by our um our molar mass right mass divided by our molar mass times our avogadro's constant times are times our atomic number right and that's what our formula the formula that we ended up using to find the number of protons so once we have that number of protons we basically have the number of electrons here in the neutral ball so now we have to work on finding the number of electrons removed from the ball so we have the charge of the ball if we divide that out it's positive right it's a positive charge that means electrons were removed um if if it was a negative charge electrons would be added to the ball it could have a negative number of electrons removed i suppose um but then we could just use the charge divided by the charge per electron which we know that's the elementary charge so we have the number of electrons um in the neutral ball that's the number of protons the number of electrons removed from the ball charge of the ball divided by the elementary charge so here we did use this formula right whoops um we use this formula in our in our um analysis process so that's what we call a theory and analysis so there's the formula and then it might be better to work vertically i'm working horizontally so it fits on this page but we substitute it in exactly so we can we can look for here we have cue ball 2 times 10 to the negative 6. i didn't have to use 2.00 in the end i'm going to round it to this should say 2.00 um right but we'll check we'll check on that at the end um and then we can directly see that our elementary charge we can check that here additionally we could write if we want up here that we're using e is equal to 1.602 times 10 to the negative 19 coulomb and i'll use the assuming biology e minus for electron um so now we have a number for that so basically we're done with our analysis we know what we want to find we found these two numbers so we can jump right to our solution process solving we'll use the the variables that we came up with in our find and plug into that formula and we get um our final result we put a box around that and it doesn't have any units we have electrons removed divided by electrons so electrons removed divided by electrons is unitless well we're just looking as we're asked to find a fraction so that is not going to have the units it's electron per electron or electrons for electrons so now we can check you know so check we talked about this at the beginning the charge is positive so the electrons have been removed so you know the positive number on our solution is correct we can go back through again with our dimensional analysis and verify that our units are correct so you know grams per branch now that's per mole so it brings mole up so these moles end up canceling and then the atoms cancel and we end up with our units of proton so we do can do a unit check as well but then further understanding what we have here the number of electrons removed if we look at this number so it's good to calculate it that's very large 10 to the 13th where um but the total number of electrons being 10 to the 25th on the order of 10 to the 25th that's much larger it's a factor of 10 to the to the um to the 12th larger than that so our fraction is actually on the order of 10 to the 13th but so our fraction here 10 to the negative 13 that's very small so if we're going to try to subtract out and determine how the how in fact many electrons were removed we'd have a problem we'd have to carry out 13 significant figures in order to get this number right so that's why we're looking for the fraction removed so that and that's how we find the charge in the ball the number of electrons associated with that charge of the ball to begin with all right so that's it for this video and hopefully that helps you understand the five step process