A Level Maths Made Easy - Pulleys Mechanics

hey guys and welcome to another as mechanics video so this is the actual last topic before we can move on to doing some past papers for the year 12 content and then of course I'll St uploading year 13 content as well but of course we've got quite a few papers to get through at the same time now you might notice we're only going to go through four questions in this topic and the reason for that is that this comes up in every single past paper so we're actually going to get a lot more questions done than you might think and this is a fairly involved topic it while it is called pulley it's basically connected particles and also a little bit of sousat and also some resolving forces Allin one as well so we're going to be using pretty much everything and of course we're going to be drawing diagrams so if we read the question two particles A and B have a mass of 04 and8 kilos respectively they're connected by light and extensible string so remember what that means uh light means that uh their detention is the same throughout and inextensible means that the acceleration is identical so for example when if particle B moves then particle a moves with the exact same acceleration at the exact same time particle a lies on a rough horizontal table 4 and a half met from a smooth a small smooth pulley so what this means is that there's no kind of uh Force lost due to the pulley so that means the tension on both sides of the pulley is going to be identical which is fixed at the edge of the table so we can probably start drawing out some of this stuff so if we draw our little flat surface here well that's terrible let's try again there we are and then we draw that and then we have a pulley like so we have particles A and B so a is the one on the table so I'm actually going to draw it as a box um just for personal preference technically it should be a circle but it doesn't really matter that much and then obviously the other one's going to be dangling over the edge and we're going to join them by a little string like so cool and then we have this is 4 and 1/2 m so I'm am going to add this onto the thing because they wouldn't give me the information if it wasn't useful at some point in the question and we have the masses so particle a is4 kilos so that's going to be 0.4 G and particle a is8 kilos whoops which is going to be 0.8 G so because they're connected by string we're going to have tension now tension in a connect in a pulley question looks like this now again remember when you stretch something tension acts to to unstretch it so this string is being stretched over the table so mean tension is going to act like so it's going act upwards for particle B particle a is going to act to the right and hopefully this makes a bit of sense because if you look at this situation if it is accelerating then a is going to accelerate to the right and B is going to accelerate down because there's no other way that it would make sense right it doesn't make sense for B to accelerate upwards and a to accelerate to the left because there's nothing causing that so the thing that's causing that is going to be the tension because of the weight so we got that done and it says that it hangs freely string torch torque just means that it's there's no slack in it 0.5 m above the ground so what we could do is also draw this and say 0.5 M now of course with the acceleration we don't know it but if it is accelerating it's going to be accelerating down and to the right and anything else a frictional force oses the motion of particle a so that's going to be going this way which is 0.08 G that's quite nice of them to provide they release as the systems releasing rest find the acceleration okay so just like with connected particles what we can do and this is a very Nifty trick is we can actually um resolve them separately okay which is extremely extremely useful so I'm actually going to move this uh diagram over so I can actually draw on it so what I mean by that is we can actually resolve particle a and particle B separately now the key part here is I can't just combine them do you remember how in the connected particles thing because they're moving in the same direction we were able to treat them as one big particle in this case are they moving in the same direction no they're moving horizontally and also vertically but it's also like it's not in the same Dimension either so we can though resolve them completely separately which is really really useful and 99% of the time you want to resolve particle B first the one that's hanging over the edge first why well if we have a look it has fewer unknowns so in short with pulley you're going to resolve both particles anyway because generally speaking there's multiple unknowns so let's start with this the reason why I said there's multiple is because in this case there technically a resultant force a a normal reaction force over there so let's resolve particle B and again I'm going to take downwards as positive because I always take the direction of acceleration as positive and because we don't know if it's accelerating or not it's going to be FAL Ma and this is how we're going to figure out if it is accelerating so the force at the moment we have 0.8 G acting downwards minus the tension is equal to the mass which is 0.8 * a and again maybe that's zero maybe that's not but we don't know then we're going to resolve for particle a and I take direction of acceleration so to the right is positive again I don't know if it's accelerating but if it is accelerating it's going to be going to the right and again we use f M8 and here we have T minus 0.08g = 0.4 a so we have two equations and two unknowns and if we have a look something very interesting has happened here we have a negative T and here we have a positive T So if I were to add these two equations together so if I write them on top of each other 0.8 G minus t = 0.8 a and then here we have 0.08 sorry no we don't I don't know what I'm writing that for we have T minus 0.08 G = 0.4 a if we were to add those two equations together we would be solving this simultaneous equation by elimination which is really useful hopefully that makes sense technically you can use a calculator to solve simultaneous equations by the way so there is a simultaneous equation solver but I would recommend doing it this way so add these two equations together we get 0.8 G plus minus 0.08 G so that would be 0.72 G because it's just going to be subtracting the t's cancel out which is exactly what we want equals 1.2 a so a is just equal to whatever 0.72 G ID 1.2 is so let's have a look see shall we 0.72 * 9 8 1 ided 1.2 gives me an answer of 5886 so we normally round to three significant figures so we have 5.89 m/s squared and that first part is probably worth three or four marks I do have some exam questions coming up these are exam style questions but they are made a tiny bit simpler just for the purpose of demonstration but I do have some more coming uh soon perfect it the common mistake I see is that people kind of assume that this pulley this part of the pulley is going to accelerate downwards at 9.81 but that doesn't make sense it's being slowed down by a okay perfect so now we have the acceleration that's really really useful it says find the time taken for B to reach the ground so this is where suac comes into place because our acceleration is going to be constant until it hits the ground but that's fine because it says the time taken for it to reach the ground so let's write out all the variables we know well the displacement the height above the ground is 0.5 M what's the initial velocity well it says it's released from rest so the initial velocity is going to be zero do I know the final velocity nope acceleration is 5.89 and lastly we have time is equal to well that's what we want to work out so let's write down the variables we want/ have so we have s u a t so s u t plus half a^2 would probably be the one I use half a CU again don't forget it's freeall right so it's normally going to be the S UT formula and of course for the first equation uh for the first term is going to be zero and that means we can rearrange T to be T is equal to the < TK of 2 s / a so squ < TK 2 * 0.5 over 5.89 this is going to be quite a small number which fair enough it's only moving half a meter so square root and that's going to be 1 over 5.89 0.412 seconds like so perfect now here's where things get a bit more interesting we have here a particle now it says the total distance traveled by a before it comes to rest so first of all since it's attached to a light and extensible string it is going to travel the exact same distance as B initially so it's going to have half a meter already locked in okay because again that b is pulled a half a meter before it hit the ground now when it hits the ground the string is now not providing a tension force there's no tension here because it's gone slack B is not pulling anything anymore hope that makes sense so what we have to do and this is a bit tricky is we have to work out the velocity of a and then we need to use the fact that it's going to be decelerating find out the distance and add it on this last part is normally worth five marks so it is a bit involved so so first thing I'm going to do is and again because this is an inextensible string whatever the velocity of B is is going to be whatever the velocity of a is correct because it's tra it's I mean it's joined okay so we're going to work out the final velocity now we can use any equation we want so I'll probably use v = u plus a for this one so the velocity is going to be once again that's zero so it's just literally going to be 5.89 ultip by 0.41 2 so 5.89 multiplied by whoops our answer gives us 2.43 m/s so again just to give you context B has dragged a and allowed it to get to a speed of 2.43 m/s it's then hit the ground so now is no longer being dragged but now we're going to have this situation here so after the string is slack after and again this is where pulley gets tricky we're going to do a lot of practice with this so it's fine it now behaves just like a regular old box which has a mass of4 it has a deceleration or sorry an um frictional force of 0.08 and there's no other forces so now what we have to do is work out the acceleration now how do we do that well it's actually quite nice and easy it's just FAL ma the only Force we have is 0.08 now which way do we want to take is positive well we could take forwards as positive still right if we wanted to because it's going to be decelerating that would make a bit of sense - 0.08 G equal 0.4 a so a is equal to 0.08 G over 0.4 let's put that in and see what we get so we get let's put in Brackets just to make sure everything goes well with the calculator whoops I pressed X not times there we are over 0.4 and we get - 1.96 I'm going to call it okay I know this is a lot so again the box has already traveled half a meter and in that time has accelerated to 2.43 m per second it's now slowing down because the string is slack there's no forward Force so if we set up suat now as it is I don't know it's not this half a meter this half a meter is before remember and this is the a key thing when we use suat we're using constant acceleration when we did this half a meter the acceleration was positive 5.89 now the acceleration has changed we have to set up new equations the initial velocity is 2.43 what the final velocity B it would be zero because we're seeing how far it travels in total a is 1.96 and T is I don't know and I don't really care so s u v and a so the equation that comes to my mind is V2 = U ^2 + 2 as s so V is of course zero so now if we push this everywhere let's say s is now going to be u^2 netive U ^2 over 2 a so that is going to be - 2.43 s over 2 * -1.96 like that cool minus 2.43 squar over 2us 1.96 gives me 1.51 m let's call it so again it travels5 Me by itself and then it slides for another 1.51 M so the total is 0.5 Plus 1.51 which is of course 2.01 M to 3sf that took 15 minutes because of what the explaining but I will point out that all of these parts together usually amounts to about N9 or 10 marks sometimes up to 12 okay so we're going to look at the other situation they can put in Pulley there's only two which is quite nice but you may notice the actual structure is the same so the first question ask you to find tension or acceleration the second one it kind of implies you need to work out some kind of speed and the third one is we're still going to do the same thing find the greatest height so otherwise we're going to find the total distance traveled which is the same thing as the first one and that's what you're going to notice with every single pulley question so although it does seem very very tricky once you kind of do it once and you get it you can do any one of these questions the final part is always going to be something along the lines of find distance traveled some kind of distance now okay there might be a Part D where it says State how you've used the modeling assumptions but that's it in terms of calculations this is the kind of structure they're going to always follow every single time so I said this is a slightly different situation because if you notice it doesn't mention any kind of table or uh horizontal surface because this is actually just a normal pulley so let me show you how it looks that is a pulley and then it's going to droop down either side and this is where I actually use normal particles and let's call them A and B it doesn't matter which way around you label them because you will get the same thing regardless and they are four and 3 Kg as long as you label them with the right measurements of course like I literally just didn't as soon as I said that I wrote the wrong number there there there we go and of course the tension is still going to be acting upwards like so now the acceleration which way does it make sense to accelerate well it makes sense for a to drop because it's the heavier one and again because we have no tension in the pulley that means that there is um there's no way for a to not accelerate right in this case CU it's smooth a is definitely accelerating downwards no question about it in the first question we had tension and tension sorry we had friction which could in theory stop it from moving but in this case nope no way 0% chance so they release from rest cool find the tension so in the previous question it said find the acceleration but if you notice we're going to be doing the same thing we're going to be resolving for A and B separately and then we're going to be solving for acceleration or tension now if I look ahead a little bit it says find speed which means we're going to need acceleration anyway so I'm just going to work out both because we will need it at some point so part A find tension well let's resolve for particle a and we're going to take downward is positive because it's my direction of acceleration FAL ma a so what forces do we have well we have 4 G Going downwards subtract the tension is equal to 4 a that is my first equation then we have particle B and we're going to take upwards as positive and this is actually why I like being consistent so technically speaking there's nothing stopping me from taking downwards as positive again realistically but it makes it means that I won't be able to eliminate it as easily I'll have to use substitution or I'll have to subtract or do something else this way all I need to do is add adding with simultaneous equations is less prone to error because when you subtract you might need to do like minus t minus minus t or something which I know it doesn't sound super hard but in an exam lots of students make mistakes so again that's just another reason why I always always take um the direction of acceleration to be positive so here again we can see all we need to do is add and this is actually quite easy 4 g - 3 G is just G minus t and t cancel so we just have G is equal to 7 a so in other words a is just equal to a 7th G so we've got our acceleration which is we're going to use for Part B nice perfect but that's not what we were here for we wanted tension so if I sub this into equation number two and move 3G to the other side I get T is equal 3 * a so 3 a 7 of G plus 3 G so you could work this out in terms of G so in some mark schemes you might see them give answers as fractions of G so in this case it be 3 over 7 + 3 so that would be what 24 over 7 G so you could have it as 24 24 over 7 G and you'll find that in quite a few mark schemes or you can and that's just because it keeps it nice and um a whole number like it makes it a precise value or if you just did 24 / by S Times by G that is also perfectly fine either way you want to do it so I'm going to just leave it as a three significant figure number because it really doesn't make much of a difference if you take three sigfigs but that is our value for a so we have two different values well they're the same value there written differently that would probably again be for about three marks when a is Trel a distance of 2 m it strikes the ground and immediately comes to rest so again it does tell you which one's actually going to be accelerating it's a ex um accelerating downwards dropping find the speed of a when it hits the ground okay so if we do suat we have S is equal to 2 m u is equal to zero it's released from rest V is equal to don't know and that's what we want a is equal to a 7th of G and it actually tells me doesn't tell me the time cool so v = u + 2 s is probably what I would use and once again U goes to zero nice and easy so V is just equal to the square < TK of 2 * a 7th G multiplied by S which is again two excuse the really poor handwriting at times soqu < TK 2 and we have G so 9.81 / 7ti 2 gives me a pretty hideous number 2.37 m/s so let's keep it as that that would be a two marker pretty solid and then it says assuming that b does not hit the pulley find the greatest height so they've actually done a lot of work for us a lot of work but we still need to do some of it ourselves so again for C I like to write that there is an initial 2 meters we have the Velocity but once again the acceleration is not the same otherwise it would actually keep accelerating forever right because it's just going to Ping off into the space it's going the acceleration should be going downwards now and why is that well it's actually just a freeall question right it's actually not too bad so if we look at particle B now we have particle b as a downwards force of 3G and that's actually it right so that means the acceleration is also going to be down now the really useful thing about writing in terms of 3G let me show you is because FAL ma we have 3G equals 3 a so in other words a is just going to be G so that's actually one of the reasons we use um we use G all the time because it actually tells us what the acceleration is straight away now remember all objects on Earth accelerate with G if they're under freeall so in this case it's under free in the previous case it wasn't under freeall it was being dragged along a surface so in this case it's of course going to be G because it's falling so let's work this out so s is don't know and again we're actually going to take upwards as positive for all of this I know that that means this will be negative and and yada yada yada U is going to be upwards 2.37 we're going to look at when V is equal to Z a is going to be negative negative G why because again it should be slowing down this object and T I don't really care so again we can use V2 = U2 + 2 as s oop get rid of that and if we rearrange this for S we get U ^2 over 2 a so - 2.37 S I just had Deja Vu I've done so many of these questions in the past that it's um incredible perfect so if you want we can actually use an exact value and whenever you can you may as well use an exact value because it's it's fun uh 2 * 9.81 yeah that's all good 2 over 7 wow that's a really precise number 0 286 M so in total and again you want to make really clear that you're doing a total 2 + 0.286 so I'm going to do that to 3 SF so 2.29 M so you can see how we're doing like the same thing again and again and again it's quite nice and we're definitely building up to that A2 knowledge which is fun and again this kind of question will come up every single year 12 exam period um I don't think I've seen a paper that hasn't had it in there um but again we're going to do every single Paper anyway so I'll be able to prove that to you hello there check this out seven 10 13 14 marks you see what I mean so they're between 10 and 14 sometimes even 15 marks these questions pretty beefy indeed but they give us a diagram which is very nice of them so uh which one's going to be accelerating downwards well if we look how here it says Q reaches the plane so Q is heavier so again that's fine all I'm doing this for is to make my diagram accurate so it's not actually like it's cheating at all you can't really cheat in uh in math sad L uh and that would be K mg and this would be 3 mg we have tension going up that's all our forces a k is less than three oh okay so we could have just looked there never mind it does tell us in the question particles connected by light in extensible blah blah blah pass smooth light fixed pulley yeah cool the systems held at rest with the string Tau and blah blah blah blah blah the same height above a horizontal so that's a really key Point here is that they're at the same height which is probably why they gave us the diagram by the way they've been quite nice systems released from rest and oh hello it tells us the acceleration 1.3 1 over 3G that is very nice of them and part A it says calculate the tension so once again all we're going to do is we're going to resolve for both of these particles so let's resolve uh q and we're taking downwards as positive um whoops there we are FAL ma the we have three mg going downwards we have a tension going upwards m is 3 m that's also a thing that make if some people make mistakes they write 3 a instead of 3 ma a in this case m is 3 m which is very confusing I do 100% wholeheartedly agree with you let's do the same thing for Mr P and we have K actually no know we have tension minus kmg equals km by the way there's actually one thing that I want you to I want you to know and I think it's really important for teachers to tell you this I don't know how we're going to do this straight away so what do you do when you don't know and bear in mind there's a lot of mechanics questions where I don't know kind of the full method that will get me to the answer so what do you do you always resolve every particle in every direction that you can as at the moment we're only really dealing with one dimension right we're doal with up and down okay but when we come on to A2 you'll find that you can resolve up down and left right sometimes so in general with mechanics if you're stuck if you just look at this and you go I have no idea what I'm going to do just resolve everything that you can and that sometimes leads to a um realization of what you can do so let's have a look at this shall we um it says we need to calculate the tens in the string so that probably means we we may need to work out acceleration actually no we know we know what acceleration is so we have two unknowns we have two equations don't we so what we can do is we can actually add these two things together which I know it's going to get rid of the tension but let's have a look 3 mg minus kmg so let's write that out I know what you're screaming but sir isn't there three unknowns you're a very clever person there is three unknowns but I'm going to show you why I don't care and this again is why you should just do it because there's m in every single term I can divide everything by m now there's only one unknown cool right so that's the kind of um I want to say it's the kind of skill that you need to develop but you just need to have that confidence okay resolve things in every direction and then you know start trying to eliminate stuff and speaking you'll come to a solution hopefully that makes sense we can divide everything by n so we have 3 G minus kg equal 3 * a 3 is just going to be G Plus K over 3 G so what I want to do is I want to move this kg over to this side and this G over to this side so 3 G minus G gives me 2G then we have K over 3 + G Plus kg so that would then be um should we do that it would be 4 over three 4K over 3G get rid of the G's so then uh we can work out what K is just by doing the opposite so 2 * 3 is going to be 6 6 / 4 so half it gives me three half again 1.5 now again that's not the question however look at Part B show that K is equal to 1.5 good so we've actually done Part B which is fine what we can do now is we can actually label that as Part B and that counts so this is another thing that's I really really adore about the mechanics exams is you have a lot of freedom so you can do Part B before you do part A and use Part B to do part A and I have no quals about that so this is now my part B working out so how would you do uh part A without doing Part B first well what you'd have to do is rearrange for K and sub it in pretty ugly right so instead I've made my life a lot easier uh we could sub this back in um actually can we no we can't sub it back in all right I was trying to be a bit lazy so what we can do is sub this into this equation here and try and eliminate M instead so doing that will give us the correct answer but just so you're aware you could have done um rearranged for k or rearranged for M and subed it in and you would have gotten uh the tension in that way but for this case we need to wipe out the tension so what I'm going to do is I'm going to subing the K is 1.5 into this equation just so we have it so minus 1.5 mg equals uh so we have 3 over2 time a thir which would be 1 / 2 G so it' be 1 over 2 kg not K mg damn it all right here we are G move the mg to the other side we get T is equal to um a half plus 1.5 so that's 3 over2 plus a half so that be 4 two which be 2 mg 2 mg like so cool so from here we could either Sub in that tension is equal to this work out what m is and then work out what G is and would that actually help us I think it would um or we could rearrange this for M and sub that in so what I could do from here is I could say that m is now equal to t2g and sub that into one of our two equations probably the top one because we rearranged the second one so three T 2 G um minus t = 3 d 2 G time the thir G so gather all of the t's to one side so here we have if I factorize this out factorize both these out we can get T Gus one factor this T out we get T um minus it to move it to the other side so I'm trying to do this all in one step cuz I'm being a tad lazy I'm not going to lie but we could move the T over and then uh solve from there so let's go through that so quick note uh we were doing it correct this is the actual final answer so all we need to do is just mention that we've done part A down here and we should be done so just erase it back to here um back to k equals this do this first step and then that will be our answer we need to record a little bit which says T = 2 mg is the answer by the way so if you notice uh we can't if I were to sub this back in I can't actually work out what the tension is as a number and that's because in mechanics we can leave our values as Expressions just like this so This is actually completely and utterly fine I can keep it as is so that would be the answer to Part B now for part C it says show how uh sorry State how you've Ed the information that the pulley is smooth well we use that because the acceleration on both sides of the pulley is the same so acceleration both sides of pulley is the same lastly we need to go through part D which is a beefy seven marker so let's get started with Part D so after sending for 1.8 seconds the particle Q reaches the plane is immediately brought to rest by the impact with the plane and the initial distance blah so basically um these last sentences that they write here it just means that it doesn't hit the pulley you know and in 99% of cases it doesn't hit the pulley um they will say like will it hit the pulley kind of thing so if it's 2 MERS away and it travels 2.1 M then yes it would hit the pulley that kind of stuff so it also has the answer in terms of G which is good because our acceleration is in terms of G so we're just going to leave everything in terms of G if we can so it's the same question but look at that seven bloody marks that's a lot okay so what we want to do is we want to work out the final speed of Q and also the distance that it travels so if we look at just Q for a second so uh let's look at particle Q then it has the tension going up but realistically I care about the acceleration which is a third G and we also know that so s is we don't know and we do care we do want that U is zero released from rest let's zoom in a little bit V is we also don't know a is a third G and T was it told me I just can't remember it was 1.8 seconds so from here there's two things we want I want its final velocity and the displacement because that's going to be kind of the um beginning distance so let's work out let's work out V first because that'll give me V is equal to U plus a and this is quite nice because U is just zero so it's just a thir G time 1.8 and again because we want everything in terms of G I'm actually just going to do a third of 1.8 which is 0.6 G okay so then for S what should we use what would be a good one um you could use S U UT plus um UT plus a half a^ 2 um yeah why not U is zero and that means we get a half third G t^2 1.82 so what I'm going to do is on my calculator is I'm going to put in a half so 0.5 time a thir times uh bracket 1.8 or squar and it gives me 27 over 50 uh is that all correct it is okay it's a bit of an ugly number 27 over 50 g let's leave it as that so 27 over 50 g for S so this is like the first part of the the distance okay so if we look back at the question the greatest height reached by P above the planes 1.26 G well we have it's gone up by 27 over 50g now there is a bit of a stealthy nature to this so if you remember p and Q were at the same height okay so let me just draw out the situation that's happening that's P that's q and there's a floor down here now Q has dropped 27 over 50g yeah that means P must have started off at 27 over 50g correct but now it's moved up so Q has ended up down here and P has ended up somewhere up here so it's gone up by another 27 over 50g so it's an additional part I can imagine that potentially being a bit confusing so realistically this distance is actually doubled because remember what it's saying above the plane it starts off already 27 over 50g and then because Q has gone down 27 over 50g this side must go up 27 so in total it's two lots of 27 over 50 I hope that makes sense I hope that makes sense so if we double that um we get 1.08 G you can probably see why it's like seven marks now so now what happened is p is no longer uh being dragged up so we need to now work out the new acceleration so we don't have a value for tension uh but that's fine because tension has disappeared but we have kmg and if you look our acceleration was in terms of G beforehand anyway so if we write so let's draw out the situation for p now it's going down by kmg but if you remember we worked out K as 1.5 so it's 1.5 mg and that's it so if we do fals ma again we want to take upwards as positive can so we have - 1.5 mg equals um I mean 1.5 M times a those two things cancel out and we left with a isal to G now you do see that once more so s is equal to don't know U is equal to 0.6 G our final velocity is equal to uh z a is going to equal G yeah this should be fine and T is equal to don't know and we want s so uh v = u + 2 a s just like we've used before so V is zero so our value for S is going to be negative U 2 over 2 a that's - 0.6 g^ 2 over uh 2 * negative G so -2 G uh what I can do here is I can cancel a g and one of those G's out so let's do that it's going to be canceling that and technically the squared so it's going to be 0.6 s g over -2 cancel out the negatives so that would be 0.36 0.18 G and okay that works perfectly because if you notice now the total height or the max height is 1.08 G Plus 0.18 G which gives us the 1.26 G that they were talking about earlier 1.26 G now here's where the mistakes would obviously be made is the timesing this by two um I've taught stats and Mechanics for six years and I have seen this question before but uh I did actually kind of forget about it but then when I went through it again before the lesson of course I did realize very quickly so with mechanics the wording is very careful it's very particular now this for seven marks isn't actually that bad if you think about it but again the timesing this by two is definitely where people mess up and another thing that people mess up on is when the answer is in terms of G they don't do all their working out in terms of G so what they do is they'll let's say for example instead of 27 over 50g they'll put it as a decimal so I actually don't even know what that would be as a decimal uh times 9.81 and the problem is now they're dealing with numbers that are quite long and so they might round it so then when they then try and put it in terms of G by dividing by G they've got the wrong answer like then instead of getting 1.26 they might get 1.24 1.25 or 1.27 and then they go where where have I gone wrong it's like you haven't gone wrong it's just that you've rounded so keep that in mind if they give you the answer in terms of G do all your working out in terms of G it's not that bad if anything makes it slight easier maybe but I hope that makes sense that was a very long question 4 marker not bad we've got one more and then we'll be done once again we have a pretty annoying uh in terms of length question if you look at the whole paragraph yeah you need to get used used to this unfortunately uh they do give you quite a bit of information but what I'll do is I'll start adding some of the forces that you know are going to be here so 1.5g it's obviously going to be accelerating downwards because again it it can't really accelerate left so two two and a half kilos so let's put that for Force there it's a rough table so let's put the friction there uh the tensions of course going this way and this way is K Newton so that frictional force here is K system released from rest okay let's make sure we haven't done anything else and then after release Q descends a distance of8 m in 0.75 seconds by modeling p andq as particles calculate the acceleration of Q now there are a few things you can consider uh one of which is of course using suat to work out the answer to this so we have S we have t and we have U because it's from rest so we could work out what a is from there bit cheeky but you can absolutely do it um by doing v u plus oh by doing a is the the change of velocity or something or v = u + 2 a s any one of these things um but I think it again it makes sense that even if you're not sure you may as well just resolve for p and Q and see what happens so for this question because it's specifically asking me for acceleration and they gave me information in terms of suat let's do suat so a part I we have S is equal to 0.8 U is equal to Z it's released from rest V is equal to don't know and I don't care a is equal to don't know but I do care and T is equal to 3 3/4 of a second so s u a and t the equation that comes to my mind is s = u + a t^2 which again because it's in free four this is a equation we use all the time that goes to zero we want a so let's quickly rearrange times both sides by two gives me 2 s uh so be 2 s over t^2 so 2 * 0.8 over 0.75 2 let's have a look see and see what happens two 0.8 close bracket uh 0.75 all squared that gives me an answer of God um 2.84 m/s squared three marks three bloody marks it's not bad is it not bad at all part two show that the tension is okay so again this is where we're going to actually start resolving now again I would always recommend resolving p and Q if you're unsure but if we have a look at this P has two unknowns T and K we know what a is now whereas Q only has one unknown tension so let's resolve for Q and we I like to take the direction of acceleration which is downwards as positive so again we're using classic fals ma as usual so the downwards force is its mass which is or sorry it weight which is 1.5 G minus t equal Mass which is 1.5 * 2.84 so if you were to rearrange this you move the T to the other side move this thing over to this side you get that t is equal to 1.5 Gus 1.5 * 2.84 and we may as well put that in Brackets because we obviously mean to work that out first not too bad I hope so 1.5 m by 9.81 subtract 1.5 MTI by oops 2.84 of 10455 let's say it's 10. five I should actually put that these are 3sf because they are answers so they should realistically be given um estimates for accuracy so that's four marks oh hang on it says 10.4 to three significant yeah sorry 10.4 then that's totally fine if you're wondering why I got um 10.5 instead of 10.4 it's because they used um 9.8 as their value for G so we did get the right answer just in case you're wondering uh the reason why I always use 9.81 is because in some exams and some exam boards they use 9.8 some they use 9.81 so it looks like um they haven't mentioned it but they mentioned it at the beginning of the paper the annoying thing is is in some papers some years they say 9.81 or in some questions they say use 9.8 or use 9.81 so in this question obviously they meant for us to use 9.8 so that's why I got a slightly different answer but I promise you in the question or the paper they will tell you to use G is equal to something 9.8 9.81 whatever it is plus four marks four marks come on it's like three lines of mats and it's easy mats too not bad not too shabby at all and that's again why I love mechanics it's my favorite Topic in math find the value of K well again we're going to be resolving P again we always resolve p and Q eventually anyway and we only have one unknown which is K nice and easy and you might also notice that in these mechanics questions they give you a lot of information for free so for example our questions show that here so that means if you didn't get this right part correct you can do part C still so we have T minus K so T is 10 well it says 10.4 to 3sf let's just use the value they give us minus k equals mass I can't remember the mass it was 2.5 times the acceleration which was 2.84 yeah got the memory of a Civ all right so then K if we just rearrange this is going to be 10.4 actually K = 10.4 minus 2.5 * 2.84 not bad and this was an actual exam question by the way so I know I normally say I I won't put like all of the exam questions in there I haven't put all of them by the way but I just wanted to let you know and in the textbook by the way there are previous exam questions if you're interested but after this this is 3.3 we are doing every every single P paper in existence Part B how have used the information that the string is inextensible simple the acceleration oh sorry both particles accelerate at the same time that's it and that's it for pulley my goodness and that's it for as mechanics and stats we're done so there it's not that long which is really good uh the bright side is A2 stats is three chapters it's really small which is really really nice mechanics is a bit longer what we're going to do for now is we're going to do a bunch of past papers and what I will also do and you might notice we're going to finish really really early so what I'll do is I'll do year 12 stuff as in the the past papers every single one that should take us until about January because there's quite a few papers then we're going to start A2 but before your actual exams in May June July time I'm going to go back and do some predicted paper some um specimen papers Solomon papers basically papers are extremely difficult I'll even do some of the really difficult questions in the textbook The Challenge questions just to prepare you for your exam okay so we will go back to year 12 stuff other than that hope you have a wonderful day I'll see you in the next one