That's super. Is that the first one? I think that was the first one. Yeah. Okay. Let's about a bit smaller over there. Well, we got time t equals zero. Standard stone vertical outwards with speed 90.6m per second. Point a, which is eight bbefore horizontal ground. Time t equals 3s. Another stone is released from rest from point b, which is also H above the same horizontal ground. Both stones hit the ground at time, 2s, two or t seconds. Most of each standard is models. That's for particle moving, frequent gravity client, the value of t and then the value of H. So by throwing up, I throw it up. The second is released from rest because that's just dropped. Basically to start at the Yeah Yeah so it's the one we've got up and one we've just like sort of dropped it from there. Yeah. Okay, watch your video first. I was thinking, just list all the stuff that we have. Okay. So should we call them Amand b? Maybe we call this the stones Amand b, because there's there's two, there's got two sets of suats ats for the two stones, isn't there? So a. So the time t is zero, then a is 19.6. And this is age day 1946. Oh, sorry, not A I mean, it's. You. And then point another. Yeah, we find up as positive or down as positive. Up is positive. Okay, that's fine, ny, it doesn't matter how you can pick either. It's just that your H, you're 9.6. Okay, so one of these will pay needs exit as well. But is 8m above the horizontal one. Shouldn't that be positive? And 19.6 is positive because it's vertically thrown upward. Oh, we know. But in the motion, if we're looking at from start to finish, the stone is going to go H meters down. So the overall displacement, like from start to finish, yes, it's going to go up a bit and then it's going to come down and full PaaS stage. Maybe I'll make up. Yeah, it doesn't matter which way you go. Yeah itwork both ways. Just you just got to be consistent with it. And then three. What else do I for? B you go here. And then which is also release. A must be 9.8 we're saying which way we're saying is positive down is positive. Yeah, okay, that's fine. Yeah okay. Other thing don't like you times. So I get I get why you said them because this is where they're released. However, whether the time for this isn't it's not the start time, it's the time of flight. So it's how long is a in the air for? How long is b in the air for? So how long does it take, a, to go from the start point to open the air and back down and land on the ground? Yeah capital two. We I thought one t is zero, so one starts at t equals zero, so a's motion starts when t is zero. When the motion starts, t is zero, but isn't it? When the motion starts. So so that doesn't work over like instantaneous time. If you use t equal zero, you'll get zero for everything else because it won't in zero seconds, the object won't move. So tiis the time it takes to go from whatever start point you've defined as the start to end point you've defined as the end. Oh, okay. Sorry. But then how can sbh? So first thing I do before I did this is I was your diagram so I look on something like this and God okay, so if this is I say this is a particle goes up and then back down turn this distance here H we've got b that goes. Okay. We said which way do we say was positive? We said done was positive. And then I'll consider these. So if we're considering the very start until the very end, so if this is the ground we've got, okay, so they're both going H down like that. A starts at 90.6 up, b starts at zero. Final velocity is we don't know they're both at the same acceleration because it says it's motion freely under gravity, and then it's the time it takes them to do this. So to go from this start point here to this enpoint a takes t seconds. Yeah but then the distance of traveled, like I get this bit where the whole time is tebit, then why the distance traveled must be more than age. So s is distance, s is displacement. Okay, so I agree the distance for a is more than H, but the displacement because whever I move up, I then move that back. Like if I go 10m up, I'll come 10m back down and then H down. So it's the overall displacement, not the distance travelled, okay? Which is why they're both age. Then we get similar other thing for be's time. So what's be's time going to be? B is going to be t minus three. Yeah, good. Yes, it starts at three. We finished the t, so it's going to take t -3s to do this motion. But now at first glance, it looks like there's not got enough information because over here we've got we know two things. We've got two unknowns. Over here we know two things. Put two unknowns. Normally you need three things to find an unknown, but because we've got two unknowns in both in the same variables, we can set simultaneous equations. Between the two sets. So while it says part a or part one, part two, we're actually going to find them both at the same time. Okay, now I understand. So it's basically just a listening outguys, just like Yeah it's making sure that you get that right. So you get you definine positive. It doesn't matter which way you define as positive, as long as you consistent with if it's up it's positive and everything that's up it's positive or vice versa. And then Yeah, I think work out the tithis two minus three is the nasty bit. Okay, Yeah Yeah we should we finish off we can go up that no, no, it's okay. Okay, brilliant. Okay. I'll load in the next one whether I've got them doing that. So that was that one. So next one is this one. Again, you see that. Yeah stuff we got. So this just catching the diagram bit and the other parts of the phone. Okay, we've got two points. Amb, I'm going to do om out again so I can actually read it. Two posts. A and b afffixed the side of straight horizontal road at 860m apart shfigure. One car and a van are rest side by side in the road and level with a car and the van start to move same time in the direction. Ab car accelerates from rest of constant acceleration to reach the speed of 24. Carla moves constant speed 24, the vanic acceleration from rest of constant acceleration for 12s, it reaches the speed of v. The van then moves at a constant speed of v and the car has been moving at 24s. At 24 masecond 30s, the van draws level. With the car b, each vehicle has been traveling, each vehicle has then traveled a distance of engine 16m. Sketch on the sdiagram speed time graph for the motion of each vehicle from a to b. Okay, so speed time graph. So is got speed on here, got time on here. And the area underneath the curve will be the distance. So if we look at the car first, now see straight lines like that can lovely. So does the card do again, the car accelerates from rest constant acceleration to reach the speed of 24 meper second. So how do we show constant acceleration? That's great line. Yeah it's gonna not a Yeah Yeah I know what I mean so not that until we reached that point there which is 24 and then Carlin stays at that speed. Forever. So么了。And then what does the van do? The van accelerates from rest with constant acceleration for 12s until it reaches the speed of v. The van a moves at constant speed of v, and the current has been moving at 24 miper second 30s of anelles level with the car at b. Okay. So when they draw a level, they've done the same distance. Okay? So this one, this one's really mean actually the word. And then this is horrible. So again, we've got constant acceleration for the van. So we're going to have another straight line. They've given us a time, this time 12s until it reached aches. The speed of v Oh, because it's the speed of v doesn't actually tell us very much. You have to look at the next sentence. The vthere moves a constant speed and not next one after that. So again, constant speed has to be a horizontal on again. And then it's this bit. So if the van drows level with the car at b, what do you know about the top speed of the van compared to the car? Top speed of. So because it says the van, is it woras? I've missed it. The van draws level with the car and like v must be higher than 24. Yeah, v must be over 24. So the van is going faster. However, for it to be behind in the first place, what can you tell me about the acceleration of the vehicles? More accelerfor which one? So the vans accelerated at a faster rate. Yeah, remember, it's going to overtake too. If the van accelerates faster, the van's going to be in front of the whole way because we know it's got high top speed as well. But then but then like the band started like later than the car. They start I think they start the same time. Yeah, they start at the same time, but we know that later on. So at b, when they both get to be, the van has caught up with the car because they're both traveling at constant speed. That means the van's constant speed must be higher than the car's constant speed. Yeah and then because because the van then has a higher constant speed and only catches it up at b, that tells us the van must have accelerated slower. You see that? Yeah, one, the, Yeah. So they're both pulled off. And the car, if you think about car versus a van of car, is probably going to accelerate quicker, because it's usually going to be lighter. So the car 's sped off, and then it stayed at 24 minutes per second. The van is sped up, and it's slowly caught up with the car. Okay, so the van 's got a slower acceleration, but at higher top speed. Okay, okay. So we need to draw that on the grass, sorry. But like I'm just. Are you happy? Yeah he you drew the Yeah on the graph. Okay, so are you happy with the car? This is just arbitrary. I've just gone to on line onto 24 straight line doesn't really matter. It's a sketch. I'll do the van we'll do the van in blue. So the van so we know the van has a higher top speed because at b the van catches up with the car. Yeah Yeah. However, for it to have caught up in the first place, it must have been behind at the start. Because they both start at the same time, the van's acceleration must be lower, which means we'll have something like this. Okay. So be itbe a less steep line because its acceleration is lower, but it's going to go higher up on this because v is above 24 and then something like that, and then theyboth get to be at the same time. So we need them to stop at the same time. So my that so this is v. What's that time there going to be? Oh, good. And then I could also mark on. The whole thing is dirthe, whole thing is dirty. Well, not quite. So that is good, that Green, but it's going to be the 30. Because when the cars been moving for 24, 24 for 30s, that's when they catch up. So that bit it's going to be. 30 okay I'm going units less from a units there in seconds 9m per second. So just have we missed anything without that? Highlights? Because highlights of it. There he is. All right. So let's go through what we used. Car accelerates from rest, constant acceleration. We've used that. We've used the 24, then move ves, constant speed 24. If you use that, van accelerates from rest, comes to acceleration. We've used that for 12s, yet we've got twelve on there to reach the speed of v. We've done that. But within our constant speed, we use that camera as been moving at 24 for 30s vdraws level. Each vehicle traveled then traveled a distance of 860. So we haven't used the distance. We could Mark T the distance. No, you can't really Mark T the distance on, because obviously distance would be the area. If you look at our ketch, I would say roughly the area underneath that blue line is about the same. I'd say this bit here about the same as this top bit. Not not perfect. We could have gone a bit higher with this, I reckon a bit a bit of like a lower slump, but it's a sketch. Okay. And then am I right saying then you got the part b? Yeah, okay. Part c is just okay. Do you want to do part c happy with part c? I'm okay. Does that make sense for our diagram? Yeah, it's a mean one because there's again, there's a lot of information. But it doesn't actually tell you that much. Select like the bit about inferring that v has a lower acceleration, but a hard top speed. All you get for that is that they start the same time and the van driws level with the car b. It's a lot. It's a big jump to go. Oh Yeah. Well, that means the van has a higher top speed and the van has a lower acceleration. It's a Yeah, it's a bit mean. Should we load in the third one? I want to so thatbe that one. This projectiles again. Yeah ball is projected vertically upwards on the ground with the speed of U. Point p is temis with the ground. Ball is modelout is particle most freely under the influence of gravity. The ball passes through the point p. Find the range of values of U, given that U is 20 and the length of time for which the ball is 12m. Tabove for guokay Yeah pretty pretty standard nasty question for one dimensional suver. So again, I would start the. Diagram, okay, starts on the ground, goes up. So Oh no, don't want that nice highlighter. I am I took the highlight off. Okay, so we're going straight up from the ground as our ground we go up and then it comes back down. Okay, that's it. We know it goes through the point p so let's say p is there. Thatbe 10m. We know it's got an initial velocity that is an awful arrow initial velocity of you. Okay. So the diagram, as you can see, I'm definitely not an artist. I don't teach art. It doesn't have to be much. It's just to help with you. See, that's. Right. And then it's, so do I opto be positive or down to be positive? Down okay so again, it literally doesn't matter at all. You can have whichever you like and what we want. So given the ball passes through the point p on the range of values of U you case I use what you want to find. Okay, what do we know then? No. Finally. So we we know here is you. What should I find? A is 9.8 good. Okay. So for it to PaaS through p, how high does it need to get. Negative ten, it needs to go ten because it's a range of values of U. Can you see that if it were to get to exactly ten, it would that would be the minimum value of you. Can you see that if I threw something in the air and it got exactly 10m up in the air, whatever initial velocity it had would be the minimum amount of initial velocity I need. Oh, Yeah, Yeah. So if I throw any Harif, I get you. Let's say that you was 15, not what it is. Let's say it's 15. Anything over 15 it's always going to get it's always going to PaaS through point p. Okay, so let's look weapons if we go exactly 10m up. Just realiyou isn't going to be you, is it? Be mindless you? B is just zero, b is than zero Yeah. We just don't know. We don't know. We don't care for time though because we've got we've got three values now. You need to be careful with which ones we're defining as positive though. So if we're going down positive ten, these to be negative, negative and then you I've just rubbed you out so you would also be negative, although that doesn't actually matter. I wouldn't use negative because it will come out as a negative answer. Oh, no change now. I'll use negative. Number one, go up as positive. Ach. So Yeah, in general, as you've just done there, you want to have it so that you've got as few values that negative as possible. But ultimately, it doesn't. I know what doesn't matter. It might be eight. That's to see. Weird if it was now I wonder. Yeah, now it's suvax. We've got suv and a so that gives us v squared equals U squared plus two as and v squared is zero. That's -9.8. So we get. U squared equals two times ten times 9.8. Oh, look how nice this is. So use squone hundred and 96. So you is what's you going to be? Oh, sorry, 14. Yeah, 14. Okay. I would always encourage you to give a direction for this. I know, I know, we've defined up as positive. So positive 14 should imply it's going up. However, just to make it really because the examiner, I'd like 40 mesecond upwards, because it might not always be clear, particularly if weto said it downwas positive and wehave got a negative answer. At first glance, that might look like we think 14 down. So I'd just just be really clear. You mean 14 upwards? Okay. Oh, actually I haven't finished the question. I haven't finished question. I haven't read it. So that isn't the answer. U equals 14 isn't the answer. She asks for a range of values. One week. Right? So what did we say about if it gets to exactly ten. Then it's a thank you it's it's minimum Yeah yes that's our minimum value so we need you be over 14I've gone with over Grace than not Grace than or equal to because it says it passes through. So I would argue at exactly 14, it doesn't PaaS through it. It just touches it. You see that? So again, it's really weird. Like it's really nasty language in these questions because like that is really so it might not matter. It might send the microscreen me. You can have either, but I would interpret passes through as it has to go over p and at 14 it only gets to p. Okay, part b. So this is the nice thing about this part b is in theory, you can do this part b without the part a. You don't need the answer to a because they then change the value of U. So given that U equals 20, find the length of time for which the ball is 12m above the okay, they change p as well as p was ten, there's twelve. So I would draw the diagram again. It's going like this. We've got 20 upwards. We've got some point here that's 12m above the ground where I put the twelve there. And we want to know how long it's in the air for. So how do we do this? Really as well with the suof as we didn't do it on this one, we should be defining where we're looking to. So for this one, we did start to 10m up. If we go back to this one, what do we do for this? We did. This was just start to finish. Okay. So just to make it really you don't have to. But I think sometimes you do that, it helps you to understand what you're doing and particularly if you've got a question where you end up doing. Like multiple difference, like for example, for this, for a particle like this that goes starts at zero, ends at zero, you could look at the whole motion, you could look it to the maximum point, you could look it to a certain height. So it all all depends on the question. I would always pick like where you start if I'm wanting to finish it. And so we won't start until when. When do we want it to stop? We need to reach p when the distance Yeah start to 12m off. Okay. And then it's just the case of what do we know? So s is twelve s twelve positive Yeah setup is positive Yeah and then v is one, which is one. B V was V1. No, sorry, sorry, I mean zero. So v would be zero at the peak. But we're going to 12m off. So we don't really know. V, do we know you? No. You was 20Oh Yeah you were told back so you was 20 we don't know age of an la. Again, negative 9.8. You actually have well remembered on the negative. And then t is what we're after. Yeah. Okay. So we've now got Suu a one t. So that's going to be s equals U T plus half 80 squared. And then we're just puking up. We know. So we've got twelve equals 20t -4.9t squared. Youprobably rearrange to something like that, 4.9t squared -22 plus twelve equals zero, no quadratic. How you solve quadratics? A nasty. Oh. The first zz time. And the phone, the nine and then just remember it's 4.9, not four times nine. Sorry, sorry, sorry, sorry. Have point. So when you've got horrible Quadra like this, how would you solve it? Just use the quadratic. Yeah just use the formula. So Yeah minus b, plus minus square root of. B No, that's cubed. B squared minus the four times. Nine times twelve, maybe two lots of. Okay. So I've got two answers. So t comes to I've got 4.012 and then it keeps them going. And then I've got. 0.06955. And again, it keeps on your own. So what do I do with my two answers? And that is not it. We're done. Why do I get three point something? Yeah, so I have to see my diaper. I had an extra negative in love just it's at a 4.9, add 0.4 minus, but add four -0.9. So no, I didn't get that. I got 0.73. 3.35 so what you got. But Yeah minus b plus minus b squared minus four, eight Yeah there we go. Sorry. Like Yeah I that was weird. Okay. So now what? Because it asks for the length of time, which is twelve basabove ground. Only 3.3, though if we go back to the diagram, we see that it's it exactly twelve. Twice. Oh Oh Yeah Yeah. So this one's the point of seven. So again, sorry. I said then do we want like the 0.73. So it depends what the question are. So let's go ask the question. So we want the length of time. Oh, sorry, never mind. Which is twelve minutes ago. So we need 3.35-0.7Yeah guess the difference. Yeah the this worden's a bit rubbish actually because the length of time when it's exactly twelve minutes above the ground is instantaneous but they want the length of time for which is twelve or more above say it take them away. I'm right and highlights ter David it point 623.35-0.73 would you get sorry, 2.62 there you go in the air 12m above for two, a bit seconds. Happy Yeah. I've noticed the common theme. All these questions you pick in the language is really, really subtle. It's really, really mean for the soup. A lot of the soup that's are quite nice, but the ones you're picking here are really mean in the language they use. Sometimes sometimes I just find it really confusing to Stein, especially when they just use it. I don't know. I feel like I'm not really good at growing diagrams. Yeah, I think in mechanics, it's one of the underrated things. Pretty much every mechanics question you can draw a diagram. It's not always useful. But I think I think like for that last one, it's quite nice to visualize because we got two answers. And initially a lot of kids will go, I got two answers, I only want one time. And then they just sort of pick one. But if we look at our diagram and go, Oh Yeah, it should be about twelve for it should go twelve twice. You go back to reading the question, it's like, Oh, how long is it above 12m? Oh, it's just a difference. Like the extra step. It's worth a mark. But it's not not really much to it, but it's just it's just to take eaway like social math involved is easy. One, let me move this. There we go. But what we got the next one small ball is projected vertical puts from the ground speed you takes 4s to return to ground level. Draw it in the space below a velocity time graph to represent the motion of the ball during the first 4s. And I have like the right answer here, but it's just like. I was like not very sure why did them be Internet? That's me. All right, all right. So. Thank you. So a couple of things. So so it is this year. So the first bit that always annoys kids is that Oh no, at the start the ball is in a hand and it's not moving. So loads of children want to start at zero. However, that that move from it being at rest, sting your hand to it moving freely in the air happens instantaneously. As soon as I got that with the object, it goes from zero my hand to the second leaves of my hand. It's it. What do we call it? You. That's why we start at you. We're then accelerating due to gravity downwards. So it's going to be constant acceleration, but negatively because we've said U is positive. And then because it's starting and ended at the same height, it's then got some symmetry. So that's why its final velocity is minus U and that's why it's at its highest point. It's got speed of zero at halfway. So that's why the time is two when it's got of lost to zero. Because any object that I throw into the air and then catch at the same height, it will hit my hand at the same speed it was going when I released it, and it will take half of its time to go up and half of its time to come down. Okay. So there's there's a lot to it. So essentially, it's a straight line because gravity's acceleration is constant. It's negative because we set up as positive. It starts at you because that's the speed it is projected upwards with. It then gets to minus U because there's always symmetry in projectile motion. So if it starts and ends, it's the same with we go back to this question. If we wanted the velocity at that point, let the different color. If we wanted the velocity at that orange point, and then we worked at velocity at that other orange point, theybe the same, but a negative sign because they're at the same height. Okay, so this is just a fact of all of that projector motion. Same with if we want it. So we found the time it is at 12m, tup. If we wanted the time it's at that midpoint, at its maximum height, it would be the middle of two these two times. Okay, so so the fact, so the symmetry, if you like, is just the fact of trajectile motion when we ignore our resistance. Okay? So that's why it's a zero velocity at halfway and why its final velocity is the same speed as its initial velocity. Yeah ends up looking like a really weird graph. But when you talk through it, you go, Oh Yeah, that makes sense. But the graph always catches people off card d for this because it looks looks really, really, really weird. Okay, does that make sense though? Yeah. And then you got part b. So what what does this but say? Acceleration is constant. Grain is constant. Straight line. Yeah. So Yeah straight line. It's negative because of gravity. And then this is minus U and this is two because of the symmetry that we get. Yeah. Okay. That was was that the fourth one? One, two, three feet? Yeah was the fourth one. So I've got some pulleys now. Still be a good one to finish her. I all need I think I think this one I'm okay with now. Happy with this one. Get that one. Yeah. It's just like the other belily question. Okay, fine. Yeah. We could look at the next one. Quickest one. Yeah that one, one, two, three, four, five makes this one so puover the corner. Yeah. Then then I'm big of these for like puover the corner. I actually prefer to puthese over the corner. I think these ones are easier. So we got two particles, a and b, masses 35, respectively. Collby light in a sensible string. A lies on rough hobs onto table string. Pasover a small, smooth, smooth, small, smooth pulley, which is fixed. The edge of the table b hangs freely. Frection between a and the table is 24.5. System is released from rest, fine acceleration of the system, tension in the string, magnitude of the force, exst on the poly bar string. Okay, first thing, what should we do first in a mechanics question? Draw a diagram. Draw a diagram. If they give you one like this, just annotate it. Okay? With projectiles and soua diagram isn't 100% necessary. Anything like this, you've got to draw a diagram because of all the forces. Okay, so I would annotate this one. So this book here is a so that's got a massive five. So what force will it have going straight down? Wait, Yeah. What's the weight of it? Five, 5G. Yes, we've got 5G down. We'll then have some reaction force upwards. Well, I have tension in the pulley pulling it that way. Like so because we know it moves. Are you happy that a is going to slide towards the table like the edge of the table? But. If we set this up in real life, one of two things would happen, either wouldn't move or b would fall down and a would slide across the table. Yeah because of the nature of the question and they're asking for the acceleration of the system, we can guess that it's probably going to move. Yeah Yeah it might not but if it doesn't the question doesn't really work and they don't really do trick questions in maths. Okay, we'll we'll see if it doesn't anyway, so it's probably gonna to move and it's going to move. So a is going to go towards the pulley, b is going to go away from the pulley. So this is our tension and then we had friction which always opposes the motion or opposes the way motion would happen going that way. Oh no, the tolest friction, didn't they? They did four, 24.5. So that's particle a. And then we can do a similar thing for particle b. So what's what's happening into particle b? It's just weight and tension. Yeah, good. It's got weight straight down and it was up three was three. So 3G put 3G down and then is it the same tension? Can I use the same letter? So we can, so that was me asking you a trick question. I'm sorry. And we can because of a few bits, because of the string is light and inextenensible, and the pulley is smooth. So with these questions in elepha math, they will always, if the tension always be the same, though sometimes therebe a part of the question. So the reason I ask you that is because sometimes therebe a part of d that says, explain one limitation of the model that allows you to have constant tension throughout the throughout the string. And it's always because we've said it's lighand and extensible and it's smooth puly. If it's not, other things happen. However, to make the mouwork we have like technically as well got what is that? That was rubbish. So why is it what. How did I do that? Do that? But there we got to fix that, fix it. So technically, there's tension here as well. And here, because of every reaction, has an equal and opposite reaction. Okay. However, this question doesn't ask us anything about the pulley, so I don't care. There's one thing I've forgotten. So what have I forgotten? What part I ask us for. Eleracceleration, Yeah. So which way is it going to accelerate? Which way is are you going? A is going that way. So like to the to the right. Yes. Oh, that's cool. I just realiit mirrored you. So so for me, that that way, that way towards the pull. Is that not right for you? No, no. Oh, that's really weird. Yeah, that's cool. Yes. Aa, towards the pulley. And then for b, it's going to be this again, it's going be the same acceleration downwards. Okay. Again, this is because the string is inextensible that they have to accelerate the same rate because the strings the same length doesn't change, okay? So with every connected particles question, there's always two different outlooks you can take. So the first thing you can do is you can look at each particle individually. So essentially what you do is you go, let's go that color, let's use the diet. So you go like this, you go, I'm going to look at just that in isolation, or I'm going to look at just that in isolation, or sometimes just that in isolation, then if you've got all your particles moved in the same direction, so we can't for this one. So that makes this a bit easier. If you've got say a car pulling a trailer and they're both moving in the same direction, you can also then consider a whole system as one particle together, okay? You can only do that if they're moving in the same direction, okay? So for us, it's quite straightforward. What we need to do initially, we need to look at each particle individually. Okay? I don't really care about the pulley, doesn't mention the pulley. I only care about what's happening to a and what's happening into b. So if you want, you can draw another diagram, but separately for particle a and separately for particle b, but it will literally be what's in this box. Okay. So we're after the acceleration of the system and the tension in the string. So we're going to look at each box individually. So what should should we start with a or should we start with b? Start with a? Okay. Well, let's get me a stupid domode. Okay. So should we look at a's motion horizontally or vertically? We look them like through, look them like together. So for this question, normally, Yeah like if if if we're like working up a slope, definitely look at both. However, we want the acceleration. We only care about the acceleration. The acceleration is going horizontally. So the vertical doesn't really matter. We can't look at vertical, but particle a isn't moving off the table or into the table. So all we get for that is that R equals 5G, that's all we'll get. Yeah. Okay. Okay. So it doesn't doesn't really give us anything but the question doesn't ask for the reaction force. So it doesn't doesn't really add anything, but we can doesn't matter. Then we're looking at horizontal. So horizontally, what we got. To what's the resultant force? The resultant force is f because ma by Yeah so what equals five a. So what's our force? So this is Newton's second law. Oh, team, that is 24 point good. Yeah. The overall force must be to the right because that's the way we're accelerating. It must be 22 minus to 24.5. That was five a. Annoyingly, we've now got an equation we can't solve because we don't know a and we don't know t. We've got two simultaneexactly. Yeah Yeah we going do the same thing for particle b. So I should have I should have done this. I should have heard this. Missed that minute. So a lot of students will jump straight into this. I would again encourage you to get, okay, well, this is a, we should have labeled these a and b really as well A, B and then I'd have gone, I am resolving horizontally just to make it really clear what you've done, particularly for when you get to year two because in year two. You'll be resolving in two different directions for the same particles. Okay. And that might be horizontal vertically like we're doing now. It might even be parallel and perfpendicular to some slipe, but that's a future problem. Okay? We'll do the same thing for b. Which way should we result solve for b? No, it's not political, obviously what you said. Oh, late tical like that. Yeah, that's sorry. Sorry, I missed that. Sorry. I paused and it sounded like I was dying. Yeah, I just missed what he said. So it's on vertically. Okay, so what have we got vertically? What's the resulting force that way? Three a. So yes, so the MaaS acceleration is three a. But what's our results? Force in terms of our force arrows? 3G minus t good J, it goes the other way. This time 3G minus t equals three a and now it's just source ine equations. Okay, easiest way to solve is to add the two equations. Yeah so one out, two minus five, 12 gives us that game 3G -24.5Yeah and the one put equals eight a of late. And then we just go from there to 3G. But we know J Don't make something makes like of Asia. So we've got 29.4-24.5 that is apparently 4.9. So we've got a 4.9. It was eight a and then we actually divide it by eight. It is 0.6125Yeah. So a equals 0.6125. That's the dodgiest zero I've ever drawn. But let's go with it. And then how do we find tefrom that. And then you don't which one would you sub into a or b of one or two? Like one? Yeah, it doesn't really matter. Theyboth work. So if we subit into one, we've got five a at 24.5. So we've got 58 at 24.5. I would then just pick up into a calcular half, 27.5625 of them plus 27.562. That's what I got. Yeah, 27.5625. One thing, obviously we've got to four decimal places just because of have other masses worked with suver stuff because we've took gtp 9.8. You can't technically go more than two significant figures or one decimal place depending on how you wanted to find. I would go, I would round them, I'd write the full value and then and I'd would use the full value to work out other things. I would then go, Oh Yeah, that's 0.61 and that is 28. Okay. Because we can't we can't actually say that it's 0.6125 because the acceleration due to gravity isn't we haven't gone that accurate. We've only gone to 9.8. And then so that's part a and part b done. And then c, Oh no, we doing got part c lovely. So part c, this is what's open to c, magnitude of the force exerted on the pulley by the string. So that is where this third box comes into play. So part c is usually the part that confuses everyone. They're used to doing this. Part ab, this is this is a really, really common type type question for this one. Part c is a bit more unusual, and you've got to remember this bit. So so you have to know tension comes out both ways from that. Like imagine if you were that pulley, if you were holding that string coming around, you yoube pulled across the table and down the table like this. And this comes back to Newton's dudlaw of every reaction as an equal and opposite reaction. If there's tension pulling the box towards the pulley, there's tension pulling the pulley towards the box. So we just need the resultant force of this. So imagine so I joined the triangle. So we've got, if you're the pulley, you've got tension that way. You've got tension that way. What is the resultant force going to be? And you can then judoing this way you go, or I can move that triangle there, move that arrow there to get my resultant force like that. Do you remember doing this? Yeah. And then we just wanted, was it just the magnitude? I think it said, Yeah the magnitude of the force exists in the puly body string so that it's just going to be you just use Pythagoras ground, you actually. Is that one, too? So 38 point like 3939. So I agree it's quite did it long or pull one over 16 squares? What did you get? Sorry. 8.97. Yeah accounwhat Yeah, Yeah. So 38.97 again, I would run at two significant figures. Go 39. Guess who experiences a full touch? Ching, I need sometimes theyask you for a direction for this one. Probably not because it's 90 degrees. It's just going to be like that. We can see it's nice othis trigle. That's just 45 in some of the more creative ones. So particularly year two, you might get a pulley that's being pulled in different directions that aren't like horizontal vertical and then it's it's just your resulting forces stuff that you'll have seen before. Does that make sense? They're happy with that one. Yeah, they're hard. But the good news is if you get that, it's the same question now, like like that is such a common question where you've got particle on a rough table, particle hanging freely down the table. And it's it's always that which is which is the nice part. If you if you can set if you can set the diagram up, the bath doesn't change. Okay you you're very welcome. I've got you for 20 eighth one Sunday Yeah, Yeah so there's quite a big gap now but then the next five are all really close. I've got you down for 20 eighth, thirtieth second, third, fourth. I think that's it. Yeah. Okay, so not not nice nice finish to look actually you've got three weeks of Christmas haven't you? No This time we've got four weeks weeks I've only got two which is why so so I go back on the fifth which is why I've I've squeezed them normal into second, third, fourth so it's a short one for me. See sorry about I know you wanted them a bit more spread out, but annormonly just because of Christmas. But no theybe nice, productive. I mean, we won't forget massive amount between them as well. Yeah. So enjoy the next week and I'll see you on the 20 eighth queue. Yeah, thank well to Christmas and gia. Bye bye, bye bye. 拜.