A Level Mastery - A* SUVAT Question

hey guys welcome to my a level maths Mastery course so what we're going to be doing in this course and through this series of videos is going through some of the hardest possible questions that can turn up in each chapter of a level maths So today we're going through suat now again these questions are found from similar exam boards from University exams from whatever but essentially these are the hardest questions you can possibly get in a level and I'm going to focus on Ed XL maths okay so as you can see these are suac questions that we're going to go through now sometimes I'm may go through individual questions if they're really big so if those of you on the Discord I put a 14 marker logarithm question in there now again the idea behind this is that if you can do these problems and you can work through these problems you get so good at maths that you'll be able to do the stuff that will appear on your exam so this is how you can get an AAR in a level maths that's the whole idea of this so again they may be harder than what will actually appear the idea is if you focus on completing questions and working out answers to really really hard hard problems the stuff that's still hard is going to be easier for you think of it like lifting weights if you have to LIF lift 60 kilos then why don't you practice lifting 100 kilos and eventually 60s going to be really light for you anyways without further Ado I'm going to go through exactly how you can get an AAR in a level sou so here we have Andrew hits a tennis ball vertically upwards towards his sister barar is leaning out of a window 7 and 1 half MERS off the ground to try and catch it when the ball leaves Andrew's racket it is 1.9 m above the ground and traveling at 21 m/s bar fails catch it on the way up but succeeds as the ball comes back down modeling the ball as a particle and assuming the air resistance can be neglected find the maximum height above the ground which the ball reaches now this is actually uh deceptively a bit simple because you don't actually need all that much information for part A Part A is actually not that bad a lot of this information is redundant because if you think about it it's just a normal I throw a ball upwards how high does it reach question and we know how to solve that and so the key part with a level maths but especially mechanics is drawing out diagrams okay so if you think about the diagram for the first part I don't actually care about his sister Barbara because remember we're just looking for the maximum height above the ground which the ball reaches so if you think about it we have a ball here it's being projected upwards and the speed the initial speed is 21 m/ second now of course gravity is going to be acting to try and stop that at 9. 8 1 m/s squared okay that's all the information we really need now you might think what about the 1.9 m above the ground well that is going to be very useful in this question because think about it what we're going to do is work out how high the ball gets and then we're going to add 1.9 M because it starts off 1.9 m above the ground the question does say Max height above the ground so now this is just a suat question now I'm going to take upwards as positive for this question and again the way I do sua is I write out the variables in order okay so s do we know what s is no we're trying to work it out again we're going to ignore the 1.9 and just add it to our answer to find the maximum height U well U it tells me is 21 V well if it's the maximum height we're going to work out how high it has traveled how far upwards it's traveled when it reaches zero velocity cuz the idea is if you imagine the acceleration it's going to slow down this 21 it's going to go down down down down down it's going to hit zero then it's going to start coming back down so the maximum height is when the velocity is zero because after that it's going to start falling again then we have a which is of course 9.81 because remember I'm taking upwards as positive all of these values are vectors with the exception of time and time we don't know and I don't really care about it because what's the equation I always use when I want s u a and um sorry and V so Su u v and a well with freefall and this is a freeall question we tend to use the equation s = u + a t^2 um but in this case what we're going to be using is slightly different we're going to be using B ^2 = u^2 + 2 a s okay so again just keeping on your toes which equation we're going to use so B ^2 = U ^2 + 2 a s and this is quite useful because V is zero in this question now if I rearrange this for S we get U ^2 over 2 a subbing the values we have- 21 all 2ar over 2 * - 9.81 now in terms of my value for gravity I know some of you use different values so sometimes 9.8 sometimes they say to use 10 the way you decide is that in the paper itself it will tell you this is actually an old spec question where you had to use uh a is 9.81 but again it's not going to make much of a difference but in the question it will tell you which value to use so really don't worry about that or it'll tell you at the beginning of the paper depending on what exam board you're doing and what exam and what they decide to to do for this year but either way you will know what value to use so don't worry too much about it and it tells me 22.5 let's do it to 3sf just to make it a bit easier for ourselves 22.5 M so the max height and again you want to write this very explicitly the max height equal 1.9 + 22.5 remember the goal here is to make the examiner's life easy I know you may not like them but the easier their life is the easier your life is they can just give you marks without really thinking about it and that is our final answer for part A now if we look at the let's read over the question one more time so Andrew hitting a tennis B upwards and his sister is leaning out of a window so what we're going to do is we're going to redraw our diagram again we have this 1.9 M that's initial and then we have that Barbara is 7 and 1/2 m above the ground so if we go up here let's say that our 7 1/2 M again a straight is line and actually wait let me reread that again above the ground so that's actually the the total length out here so from this point all the way to the ground is 7 and2 m um when they try to catch it so we still have the U being the exact same value that doesn't change in fact I'm going to get rid of this that's a bit confusing now we have U is is 21 m/s and we still of course have the acceleration downwards of 9.81 m per second squared now if we have a read of what it says it says when the ball leaves and trets on this it's 1.9 me yeah yeah yeah Barra fails to catch the ball on its way up but succeeds as the ball comes back down now this does seem barely tricky but if we think about it let's write out our variables what we know so first of all uh I'm going to take upwards as positive once more now s we know what s is if we're starting from the ground or from from this height over here we know that s is going to be 7.5 subtract 1.9 because remember in this case we took s to be zero and then we added on the 1.9 so we need to take away the 1.9 now let's think about why let me go through it because down here there is actually no acceleration right the acceleration actually changes right it starts off here and then it begins accelerating downwards so we can't actually include that bit so the actual uh displacement we're looking for is when it's 7 1/2 m above the ground now the way we can work that out is by doing 7.5us 1.9 which is going to be 5.6 M because what we're effectively looking for is when the ball's gone up so if I add that back in it's gone up by 5.6 M or so the ball if I draw the trajectory of the ball it's going to go up she misses it on the way up but then it comes back down and she catches it over here so can you see the displacement will be 5.6 M either way she misses it the first time catches it the second now some of you may have already realized what equation we're going to use we're going to use S is UT + half a s now because it's a quadratic you're going to have two times one of them is when she misses it on the way up the next one is when she catches it on the way back down but that's when it's 5.6 m above its starting position it's displacement remember hope that makes sense U yet again is going to be 21 V I'm going to put don't know for now a is going to be 9.81 and T is exactly what we're working out so if I write out the equation s = u + a t^2 we we actually have all of the information we need for this one and then all we're going to do is solve the quadratic and that gives us our answer for six marks 10 mark question in total this didn't take me very long and this is not going to take me very long so let's Sub in um there's two ways you can do it since we know it's a quadratic what we could do is move everything to one side so we know it's going to be half a t^ 2 plus UT minus s equals z and then we can sub in our stuff so we have a half 9.81 T ^2 + 21 T minus um- S I thought I said- 5 for a second I was very confused equals zero so that should be an S let me write that out again because that does look a bit wonky I did move the s to the other side and what we can do now is basically just use our calculator to solve so I've shown you how to do this before go down to equation Funk go to polinomial because remember we have a power greater than one polinomial degree is two and the number in front is going to be a half we can actually type in an equation like this and then when we press equals it does it for us so we can be extra lazy and you know me I love being lazy and then - 5.6 it gives us two times now I like to do it to 3sf so what we could do is do T is equal to 3.99 dot dot dot which is equal to 4.00 seconds to 3sf or technically it's and not or um but in so I have had comments say Well when do you write or when you write and well technically it's and because there's two solutions but in this case what I'm trying to illustrate to The Examiner is the time is either 4 seconds or a different value and I'm deciding which one there isn't two answers to this specific question that make sense um then we're going to press equals one more time and it gives us 0.28 blah blah blah blah blah so 0 285 dot dot dot so to 3 SF it's 0.286 seconds to 3sf so let me just put that on the screen for you as you can see now it's a good idea to go a look back at what the questions actually asking because this isn't our answer yet and you're not just going to circle 4 seconds now if we look at this it says find how long she has to wait from the moment the ball first passes her until she catches it so we're basically working out the time between these two red marks now when the ball gets to the first red mark it's going to be 0.286 seconds logically and when it gets to the next one it's going to be 4 seconds so the time difference is going to be 4. subtract 0.286 okay which is what 3.7 14 actually I don't know if that's right let me just double check that quickly um but hopefully you can see the kind of logic there they word things in very difficult ways to make it a bit harder for you but it is normal okay that is what they do but this is a SE in total a 10 mark question but in this particular part six marks and it's very little working out but can you see how the diagram is essential for this a lot of students when I teach them mechanics they read the question they go I have no idea what to do and the first thing I ask is have you drawn a diagram and they say no or they've draw they've done a tiny one but they're like but I don't know what to draw and I so we'll read it and draw it out draw it as it goes along again I didn't know what I was going to do until I drew the diagram and saw what was happening and I wouldn't have got the right answer if I didn't draw the diagram because I wouldn't know that s was 5.6 so common mistakes would be doing this whole process but writing s is equal to 7.5 instead and that would give you an answer it would give you times it's not a math error but you wouldn't get the right answer and that's kind of the key thing here mechanics is testing your application it's testing can you draw out situations and if any of you want to do kind of engineering physics or anything like that you're going to have to get good at doing this