OCR A Level Physics Paper 1 2025 Walkthrough

追加: 2026-05-13

[00:00:09]Okay. So, let's go through the 2025 paper then. Obviously, um if you need to pause it or rewind it or stuff that you can do that. Um the first thing is that for this paper, there isn't a wrong notice, which means they've made a mistake in the paper. So, um just make sure you've checked that before you try.

[00:00:23]Obviously, make sure you've done the paper before you go through it. Um and just Yeah. So, I'm just going to go like this. One sec. And um that's the notice up there for you. So just pause it, have a look at that and just make the amendments on the paper first um and then we can move on. Okay, I'll get rid of it like now I suppose. Okay. So um yeah, I'm not going to go through like content and stuff like that. It's really mainly about exam technique to have a look at what we should be doing to get you know do these questions as quickly as possible.

[00:00:56]So the first thing obviously multiple choice question um you would do this at the end of the paper by the way like so you can guess them if you need to if you run out of time but because you're just doing revision obviously will do it in just the order that the questions come in. So um first one about different number of base units. Now different is in bold and nice enough to do that for me. And really all I need to do here is just make sure I know what the base units are. So what are the base SI units? Um so some of the units that the things are going to be given in normally are going to be in base units. Um for example like jewels um when I'm using Boltzman's constant generally it's given in jewels per kilogram but I not kilogram per kelvin but I know that um jewel isn't a base unit so I'd have to break that one down. Now the tempting way that you might want to do this question is just be like oh I'll break every single one of those down um and just count then how many there are. But is there a quick way that you can do it?

[00:01:53]Obviously the exam is a bit time pressured. So is there a quicker way that you can do it? Well, let's just have a look um if that's possible. So let's start by breaking them down and just see if I haven't figured it out yet. So the first one I didn't really even need to do anything like mole. It's just one thing. So obviously that's not going to be the answer. The next one we've said um a jewel per kilogram for per kelvin.

[00:02:16]How am I going to break this down? I should know how to break down base units. If I if you don't know how to break down base units, then just I know go and do some revision on that first.

[00:02:24]I'm not going to go through it in this.

[00:02:25]Like I said, this is just about exam technique. But basically, what I should realize is that I can break down a jewel into um a kilogram m/s squared per kelvin.

[00:02:38]Um so I've done that and sorry, and then the Kelvin comes from there. Sorry. So this part is the jewel and then I've got to add the Calvin on there. So now I've got one, two, three, four different base units. Okay, gravitational constant G.

[00:02:51]So, okay, I can't remember what the units are, but I look on my formula sheet and it tells me that it's Newton me squared per kilogram squared. And again, I'm just going to break that down into the only one I need to break down is Newton because that's not a base unit. So, the Newton goes to kilogram me squar um per second squared.

[00:03:12]And then I've still got to add the same meters squared and kilogram uh per kilogram squared. So how many different base units are there? Well, I've only got kilograms, meters, and seconds. So that's only three in here. Now for the last one, the molar gas constant.

[00:03:29]I'm going to So I know that it's jewels per mole per kelvin.

[00:03:34]Um, now I don't even need to break this one down because if I know that this one is a jewel per kelvin, this one's a jewel per mole per kelvin. Um, did I actually really need to do all that thing? I could have probably saved a bit of time by just doing these two and just seeing which of those has got the most because I know that D is going to have one more than B because they're the same, but this one's just added them all. So, I could have just done B and C and not having to do the rest. And to be fair, I don't have to do D. I know the answers. Whoops. D. Where's the pen gone? Without breaking without having to break that one down. I've saved myself a bit of time. And also, I'm very happy that I've got the mark here.

[00:04:16]Okay, for question number two, then um a metal block blah blah blah that you can read the question. It says, "Which expression gives the correct value of C?" So, how am I going to start this thing off? Well, if it's saying about the specific heat capacity of the block, I know obviously the place where I should start is just do I know the form of specific heat capacity? Of course, I do. Um it is E= MC delta T.

[00:04:39]Okay. Now, what can I do from this um from this? So, it's saying I've got information about yeah the mass, the temperature, the initial temperature. It gives me a power, it gives me a time, and it gives me an efficiency. So why is it going to be power time and efficiency? Well, it's because I know that energy is going to be power time time and then the useful energy is going to be power time time efficiency. So I can change that to um power time time efficiency. And then data t is going to be because if I look at these, there's no doubt in there in any of them. I know there's a t t1 minus t and all that. So I'm trying to already get these values into my rearranging and just see which one comes out as being correct. So I'm going to say M C and then delta T is always going to be the final minus the initial. So I've done that and now I'm basically just rearranging to get C. So um that's going to be ptx over m um t1 minus t and that's going to be equal to c. So now it's time to clench my butt cheeks and hope that this is one of the correct answers. So let's take a look. Is this one of the correct answers? Thank god yeah it is at C. I can relax in the knowledge that I have got that mark in this question. So I'm hoping that question number three is going to be pretty straightforward because it says inelastic collision and it's just basically do you know the stuff about inelastic collision? Well, I know that momentum is always conserved. So straight away I'm just going to rule out C and D. I don't need to worry about them. And then if it's inelastic, I should just know that um kinetic energy is going to be um not conserved. But the total energy is always conserved. So momentum is always conserved, total energy is always conserved. Um and then in this one is the kinetic energy that's not conserved. So I know the answer's got to be B. Okay, for question number four, um again I'm seeing what the things. So what is the tension? And it says this force uh the force constant is going to be 2k. So let's have a look at how I'm going to be approaching this question. Well, I've got um two springs over here. So I've got the force constant 2k, the force constant K. I've got the F. It's it's not in the middle and it says but it's okay. So question number four um this seems like it's a little bit more tricky because there's a little bit application rather than just recall. So let's have a look at the information I've got. So I've got two springs and the um I've got the force constants K and 2K. So straight away I can see that one of them has got force constant that's twice as much as the other. So in my head already I'm thinking something about ratios. Um, and it's saying that the rod is horizontal because F is close to one of them. And I want to know the tension in this spring. That's 2K. So I'm thinking, how did Billy Hamas person know this?

[00:07:25]I've never we've not done this in the lesson or anything like that. This is far too difficult. Well, I'm just going to do what I can do then. So if it's telling me that the rod is horizontal, they're not just going to tell me that for absolutely no reason. How can I rephrase that to go, okay, this is a piece of information that's going to be useful to me. So, if they're horizontal, well, what do I know about it? I know that this spring must have stretched as much as that spring. If the thing is horizontal, they must have had the same extension. Now, can I do anything with that? Well, if they've got the same extension, then I know that they must have obviously stretched by the same amount, but this one has got a force constant of 2k, whereas that has got a force constant of K. So, if they've stretched by the same amount, what have I figured out just from this word horizontal and the fact that I've got force constants? I know that T must have had twice as much force on there as um as the other one. So this one over here has got two times as much force. And this one's just got ob one times as much force. This is going to be force. So if this one's got twice as much force as that one and I already because I can see this thing over here, I already think about ratios. Well, I'm just going to go well the ratio must have been 2:1.

[00:08:34]So if the ratio of force is 2:1, how can I then say um what the total force is going to be like how much of this force is it going to feel? Well, it's going to get two parts of um of three. So then the ratio must be 2/3 F. Is there one of these options that says 2/3 F? I hope there is. Oh yeah, it's this one over here. So therefore, the answer is going to be C. So this isn't a question that like anyone's really done. it doesn't come specifically on the specification, but just by looking at the key things in the question um and just working through it without panicking, I was able to get the right answer, like be pretty confident that yeah, eventually I've got to the right answer. So, this one just seems like another recall one just recalling the content. So, it's heated, so it changes from solid to liquid and then it changes to gas. So, what's changed? So, if something is heated up, is the mean kinetic energy going to increase? Yeah, it is. Is the order going to increase? Um, well, no, because it's become less ordered because solid's got the most order. So, that's not going to be true. And is the spacing increased? Yeah, it is. So, now again, I'm just going to hope that one of my options are one and three. And yeah, it is. It's that one. Um, I'm pretty happy that I've got the right answer there.

[00:09:47]So, we can just move straight on to question six. Um, so what's the information that I've got in there? And even before reading the question, I can see something about okay, change of momentum, average force is going to be that. So, it's the same kind of thing before that. I'm just looking at the options I've got. Um, and the change rate is either going to be 2 MV or MV. It's going to be one of those two. This one has either got um two out over V or um L overV. And then this one also, I think, only has two options. No, this one's got three options in there.

[00:10:19]Um, so, okay, it's going to be obviously one of those. So, if I'm like running out of time in the exam, this is I don't know if this is it's not a technique that always works, but it's just something that I would say maybe you can try as like a last resort. So, I'm going to do this first and then we'll do it the proper way after. So, because I know they're putting things in there that are able to like trick students into getting it wrong. If only one of these is different, I'm thinking that the answer is probably going to be either B, C, or D because otherwise um they would have had like if MV was realistic, they'd probably put it in there twice. So I think that maybe the fact they've only put in there once means that it's going to be B, C, or D. So if I'm guessing without even doing the question, I think the answer is either going to be B, C, or D. So um I'm going to rule out A straight away. Now I can use that same technique for this one because if all of them are two L over V but only one of them is only L over V. Now the I'm going to rule out C. So now the only thing that I actually care about is do I think the answer is MV^ 2 over L or do I think it's 2 M / L. Um and now I can just guess and I've got a 50% chance really if I if this was if I knew absolutely nothing about physics. So let's um actually do the question properly instead of try and cheat and let's see if it was right that the answer is either going to be B or D.

[00:11:47]So um what do I know about the change momentum of it? Well, if it's going to be hitting that the initial momentum is going to be MV. Um and then if it comes back it's going to be minus MV. So I know that MV minus minus MV is going to be 2 MV. So I was right to rule that one out in the first place. Time between collisions with the same wall. So why are they asking me about time? Why does it say the same wall? Well, um, what's the time going to be equal to? Time is going to be distance um divided by speed.

[00:12:20]So it's giving me the length, but what's the distance it's traveled? Well, it's traveled to L. So if I'm going to say time is distance divided by speed, it's traveled to L, and the speed is going to be V. So again I was right to rule C out in the first place even just by guessing.

[00:12:36]So now the last thing what is going to be the force exerted on the wall? Well I should know that force is change of momentum over time. So force is change of momentum over time. Now let's just do what I was well I know the change in momentum was 2 mv. I know the time divided by time was 2 l over v. And I'm writing it like this just because it's a little bit easier than to multiply by the inverse. So I'm going to say 2 MV * V / 2 L. And now it's a lot easier for me that that's 2 MV^ 2 / 2 L.

[00:13:10]Um, so I can just put that across there like this. And then so that I'll just write here actually. So I've got 2 MV 2 / 2 L. So how can I simplify this? Those twos can cancel out. So is there an answer that says um MV^ 2 / L? Yeah, there is. It's this one. So I know the answer is B. So I'm pretty happy again that I've got the mark with that.

[00:13:31]Okay. So let's take a look at seven then. So um the question is asking me about the distance of a star from the earth in parex. What do I know about parex? First of all I know that if I want the distance in parex um I've got to go one over the angle in arcsec. Um so that's just something I know. So if I can get this angle in arcsec um I know that I'm going to be able to get the right answer basically. Well, at the moment it's in degrees, so I'm going to have to do a conversion first. Um, so I know that one parc is going to be one par sec is 1 3,600 of a degree. So I'm just going to write that down first. So one par sec = 1 3,600 of a degree.

[00:14:14]So now how can I um convert from degrees to parex? Well, if I can't do this straight away in my head, I'll just do it a sensible way. So if I know I want to get from degrees to par sex, basically I'm trying to get from the number one over 3,600 to 1. What did I need to do? I have to multiply by 3,600.

[00:14:32]So now with the actual number, I'll just do the same thing. So I'm going to say 1.0 / sorry, * 10us 4. I'm going to have to multiply by 3,600.

[00:14:41]And then that gives me the answer of 0.36 parex. Okay. So I've got something in parex. Now, now it might be tempting to just go steaming ahead and go, "Oh, yeah, one over 0.36, but that's going to give me the wrong answer." And probably it's going to give me one of these options because they do that kind of thing to um to make students get it wrong. So, probably like one of those answers is one divided by the answer in degrees, one is one divided by this answer. Well, why is it wrong to say um to say that? Well, what's the actual question telling me? It's saying between observations made six months apart. So this might be a little bit difficult for me to imagine in my head about what's going on here. So let's just draw a quick diagram and see um and see what's happening. So basically I know that what's going to happen in six months time is going to be on the other side of the sun basically. So here's the sun um and I'll make him smile and have his hat on. There you go. Um so over here I was So this was the position for six months later it's over here.

[00:15:45]Um, and what it's saying is the apparent position of the star is going to be changing. So, first it looks like, so here's the star over here. It looks like it's at whatever angle it is. 6 months later, it looks like it's in a different position. And it says this angle here is 0.36 x.

[00:16:04]So, if that angle there is 0.36, what's the actual angle I care about?

[00:16:09]It's this over here. So, this is the angle I care about. So, um, how am I going to get that? Well, it's going to be 0.36 / 2. So, the actual angle that I care about is 0.18 x. So, I'm just going to put in my calculator 1 / 08 x. And the answer it comes up is 5.5 recurring. And therefore, the answer should be 5.6. So, the answer for that one is going to be okay. So, what's this next question about? It's very clearly a circular motion question. So when I can see that it's a circular motion question, I know that it's always going to be something to do and particularly if it's calculation F= MV ^2 / R or F= M omega^ 2 R. I've got to work out it's going to be something to do with one of these things. Um that's the first point of call that I should always use. And then I also know that the force is going to be directed towards the center. Um, and like even though this only calculation question, if it weren't a calculation question, the other thing I would need to know is that it's a um, description of the force, not a type.

[00:17:08]So, it's got to be equal to something else. So, let's have a look at if there's anything that stands out to me here as being suspicious. Well, the first thing is that um, gives me the diameter. So, they're obviously trying to trick me there. So, what do I need to do with that? Obviously, I'm going to need to divide it by two to get the radius. Is there anything that stands out? Well, it gives me the angular velocity. So, the fact that it's given the angular velocity, this is leading me towards, okay, I'm going to probably use that equation. Um, and is there anything else that stands out to me again? Um, so it says, what's the magnitude of the force exerted when it's at the bottom of the spinning drum? Well, why would it say the bottom? Why didn't it just say, oh, when it's in the spinning drum? The fact that it's giving me that means that I'm going to have to talk something specific about the fact that it's in the bottom. So, maybe I don't know, like we did earlier, maybe I have to draw a diagram or something like that to help me understand it a little bit better.

[00:17:58]Well, let's just do what I can first because I know that f= m omega squ r. I can just put numbers into the equation.

[00:18:04]So, I'm going to put the numbers into the equation there and I know that f is going to be I'm not going to like do basically when I put the right numbers into the equation the answer that I get is 20.25 newtons. So, if I'm saying 20.25 25 newtons. Well, I'm actually a little bit surprised by this because I would expect that one of the options in there would be something like 20 um because they think that students just going to be like, "Oh yeah, let's stick that one down without doing anymore."

[00:18:31]The fact that it's not an option is actually quite nice because I know that this can't be the final answer now. But even if it was, the other reason why I would know it's not the final answer is because of what we just said that why is it saying bottom? Like I'm a bit suspicious about that. So why is it saying bottom? Well, I don't know. Let's draw a diagram to help and see what's going on. So if it's at the bottom, it's over here. Well, we've just figured out that this is the resultant force. So that means the force towards the center must be uh 20.25 in total. Well, what are the forces that's acting on this? I know mg is acting on it. And I know that the um reaction forces acting on it as well, but I know there's a resultant force towards the center. So the reaction force must be bigger. Well, if the reaction force must be bigger, that's the one that they're asking me about here. What do I know about the reaction force? Well, I know that the reaction force minus the um m minus mg minus the weight must be equal to the resultant force must be equal to the centriutal force, which we just worked out 20.25. And now I can just put numbers in the calculator. Um because I know that I know what mg is. So it's 0.3 time 9.81.

[00:19:42]81. So I'm just going to rearrange and say r = 20.25 plus uh 0.3 * 9.81. I'll put that in the calculator and I get the answer of um what's it say? 20 basically 23. Um so then yeah the answer is going to be B.

[00:20:03]Okay. So question nine then let's take a look. Um I've got the spheres. I can clearly see that it's a question about gravitational force. Um, and then is there anything I'm suspicious about?

[00:20:14]It's the fact that it says the shortest distance rather than just the distance.

[00:20:17]And it says the shortest distance between the surfaces. So, this is kind of standing out to me. Like, where would students normally get tricked that they forget that when we're saying F= GMM R 2, we're talking about the um R has to be from the center from the center of mass of the things. So, the fact that it says the surfaces means okay, obviously there's a trick in there somewhere. So, I'm just going to make sure that I don't fall for it. And I'm just going to basically draw the diagram again if it helps me. So, I've got that diagram down there. Now, what do I know? It says um so if I've got that distance between the um centers of them. So, this is the thing that the formula is going to give me. And then I need to know that distance over there. And it tells me that each of them have a radius 10 m. So I know that this is going to be 10 m and this is going to be 10 10 m. So I already know basically what I'm going to be doing is work out what the answer is for F, not the F for R. Once I've got the answer for R, I'm just going to have to take away 10 meters from that side and 10 meters from that side and then I'm going to be able to get the answers.

[00:21:25]So I'm not going to like put numbers in the calculator for you, but basically you work out um so F= GM over R 2 rearrange to get R and then you get R is about 100.22. two two blah blah blah whatever and then um I've got to remember to take away that 10 from each side. So the actual distance between them is going to be 80. Um so that is an option. So I'm happy that there's the right answer. Okay. So like we've been saying for all the other ones um the generally when they're giving the multiple choice answers all of the answers seem like a little bit reasonable. So if I just pick one and think oh yeah that's going to be the answer um I might actually fall for one of the tricks. So to avoid me from doing that, I'm just going to make sure I go through each of them and be like, "Oh, is this the actual thing or is it a trick there?" So, um, the first thing in there, dark matter, I don't see how that's relevant. The only reason why I can see how that can kind of be a trick here is because if students don't know the answer to something because I was safe, it's probably dark matter. So, I mean, it's not that. Okay. Electron degeneracy.

[00:22:27]Does that determine whether a super red giant evolves to a black hole? Um, no, it doesn't. It determines whether it's a white dwarf or a neutron star. So, um it's not going to be that mass of the core. I mean, super a giant to black hole. Um it's it is to do with the mass of it. So, that's possibly one of the things that it could be. Um but what's the other option of a super a giant? It could either be um a neutron star or a um or a black hole.

[00:22:57]So, maybe the mass of the core. And then the other one I'm seeing is a Chandra limit. So, does that determine whether a super red giant involves a black hole or not? Actually, that determines whether it's a red giant or a super red giant.

[00:23:08]So, it's not that. So, therefore, the one that I thought might be the answer first, it is the only option in there.

[00:23:14]And because it's the only option, I know it's right. I know I've not accidentally missed something out. Um, so I'm perfectly happy now that the answer is going to be C.

[00:23:22]Okay. So, what's this next question about? Well, I can see it says defraction gradient. So, straight away I'm going to go n lambda equals d sin theta. I don't even need to think too hard about that. As soon as I see defraction grading, I'm getting that one down. Okay, now that seems a little bit suspicious. That's a little bit too straightforward. So, is there a trick in here? Maybe. Let's have a look through it. So, firstly, okay, it says nanometers. That's not really a trick. I should just know to do the conversions.

[00:23:48]It's not that difficult at this stage.

[00:23:52]Um, it's giving me this the maximum for screen blah blah. Everything seems reasonable apart from when I see this it says the distance between the two first order maxima. Now this is what I've got to be careful about because when I'm working out the angle and when I'm working out whatever it is it's always between the central fringe and the first or second order or whatever it is um which are going to be on both sides. But what it's saying is this distance here is 1.41.

[00:24:19]So if that distance there oh sorry 1.42 if that distance there is 1.42 42. Um, what do I care about? Well, it's got to be this one, which is therefore going to be that divided by two, which is 0.71.

[00:24:31]Okay, so now can I put that um into the equation? Well, I've got n. I've got lambda.

[00:24:38]I'm looking for d. I haven't got sin theta. How am I going to get sin theta?

[00:24:42]Well, it tells me that the distance to the screen is three. Um, the one that I've just worked out is 0.71. So, I should be able to work out what that angle is now.

[00:24:52]Um, so I've got I'm looking for this angle, which is theta. I've got the opposite. I've got the adjacent. I'm going to be using tan. So I'll say tan theta is equal to opposite over adjacent. Um, and I'm going to put that in my calculator. And it tells me theta is 13.315.

[00:25:08]Now I can just put that into the um this formula over here. And yeah, you should be able to put numbers into the calculator. And the answer that I'm getting is d = 2.605 605 * 10 - 6. Is that one of the answers? Yeah, it is over here. Thank god I have got another mark.

[00:25:28]Okay. So, how am I going to get this one right? So, it says which option correctly gives a likely evolution of the sun. So, um I don't want to be thrown off by the graph and panic. Oh no, it's a Herz Russell diagram. I hate um um let's just have a look at what I do know. So, this is likely evolution of the sun. What do I know about the sun?

[00:25:45]Um, it's going to go from a main sequence at the moment where it is to a red giant to a white dwarf. Where's the red giant is going to be number one.

[00:25:53]Where's the white dwarf? It's going to be number three. So, the answer is going to have to be um B.

[00:26:01]Okay. So, for um this one over here, then I'm looking for what's the surface temperature represented um of Okay. So, how am I going to have a go at this question or identify what this question is about? Well, first of all, it's about intensity and wavelength. So, it's already kind of leading me towards something. Um, and then it tells me I've got the surface temperature of the sun and its peak wavelength. And I've got I'm being asked for another surface temperature in Kelvin of the of another star. Well, the fact that I've got these pieces of information is leading me towards proportionality. So, I do I know anything about proportionality for um for this situation? Oh, yeah, I do.

[00:26:42]because the graph is telling me about a wavelength, it's telling me about peak wavelength. And I should remember from my revision that lambda max um is proportional to 1 / t. So if lambda max is proportional to 1 / t, I can just change my fishing to an equals by putting a constant in there and say equals k over t. Now if I want because I've got this I've got this value here and I've got this value here. I can just work out k, put k in the equation and then just work it out like that. That's perfectly legitimate. Um I can't be bothered. So I can just rearrange and say k equals So I know that k equals lambda max * t. Well, if it's a constant, it's going to be the same. So that means I know that the lambda max before time t is going to be equal to lambda max after time t. Um I'm just using very quick abbreviations here obviously. Um well then again it's just a multiple choice so it doesn't really matter if you write it properly in the exam or not. Okay. So what numbers do I know? So, first the when the peak wavelength was 500 and I don't have to convert it because I know the conversions are going to cancel each other out. Um, four times 10us 9, that's 5,800.

[00:27:48]Um, what am I looking for? I'm looking for the surface temperature. So, do I know lambda 2? Well, I can get it from the graph. So, I'll do that and it's about 450. Obviously, use a ruler in the exam. Um, so 450 times um t2, whatever that one was. And then I just put that in my calculator and I'm getting the answer of 6444.

[00:28:08]Are any of these answers close to 644?

[00:28:10]Yeah, it's that one. So yeah, that's bound to be the right answer. Okay. So what's this question about? It's about simple harmonic motion. So already I'm thinking about what are the things I know about simple harmonic motion. So I know that um this is going to be um true for simple harmonic motion and I know I know what the concept of proportionality is. All of that kind of stuff. Maybe I'm going to be tested on that. Well, let's have a look about what it says. It says the correct relationship between its period and its amplitude. Now, this is kind of like a bit of a trick question because like we said, when we're seeing simple harmonic motion, we're straight away going to like this and then we're saying, oh, a= minus kx and what's the constant of proportionality minus omega^ 2 x and I'm getting all of this kind of thing. But the thing that the other thing that you need to remember because it's saying about the relationship between its period and its amplitude, what should I know about the relationship between the period and the amplitude? I should know they're independent of each other. So all of these are about time period and amplitude. Time period and amplitude. If they're independent of each other, what's the answer going to be? Well, it's going to be this because there's no relationship between them. Um, so regardless of what the amplitude is, the time period is going to be the same. So my answer is D there. Now I made a little note on my when I was doing this on the paper to say just look out for this trick basically because it is something that they like to do like testing in a sneaky way about do you know that time period and amplitude are independent of each other. So look out for it in the real exam. Okay. So for the next one it's basically testing my understanding of the cosmological principle. What do I know about the cosmological principle? First of all, I know that it's the principle that all the rules of the universe are basically going to be the same everywhere throughout the universe. There's nothing to suggest that the rules that apply here don't apply anywhere else. So now that I know I've reminded myself of the cosmological principle, which of these apply to it? Um, and I mean the only one that really applies to it is going to be B, the Kepler's laws can be applied to any solar system. Um, that's basically just a statement of one of the things in cosmological principle. Okay, so on to the um section B of the paper, the longer answer questions. So let's take a look at this. So the first thing is a graph question. Now, what should I know about graph questions? Every time that I get a graph question, I've always got to make sure I'm doing the same thing because I'm being prepared for the kind of questions that might come up. So I'm looking at axes and it says V and T. So straight away I think that somewhere they're going to ask me something about is V divided by T something. Yeah, V divided by T is the acceler acceleration. So therefore I know that the gradient is going to be the acceleration and then is v * t something. Um yeah it is v * t is distance or displacement. Therefore I know I write s for that really.

[00:30:55]Therefore I know the area under the graph is going to be the um displacement. So they're the two things that even before I've read the question I'm looking out because I've just looked at the axes and they always test you on that. Do you know that um the gradient is equal to y / x? Do you know the area end of the graph? There's a formula that says y * x. That's going to be the area under the graph. So that's going to be the gradient and that is going to be the area under the graph. Now I'm still not even going to read the first thing I have to do. I'm just going to carry on reading this and see if there's anything else I need to be suspicious about. So firstly, units are fine. So I don't need to worry about that. Those units are fine and those units are fine. And then it says during this time the force produced by the motor is constant. Now, I'm already a bit suspicious about this because I'm going, why are they telling me that the force by the motor is constant? It's probably going to come up somewhere at some point. Um, and it might even come up over the page or whatever, but I know that this is um this is going to be the case. So, now I'll just actually read what actually asking me to do. Um, it says, "Use the method you use to estimate the distance." While we've already said, we knew this was coming up. Um, I know it's going to be the area under the graph.

[00:31:59]Well, that's not going to give me two marks. How am I going to estimate the area under the graph? Um, so most sensible thing to probably do would be model it as a triangle and a rectangle.

[00:32:08]Um, and then work out the area of that.

[00:32:11]So I just say model triangle as a rectangle.

[00:32:15]And then you know how to work out the areas of a triangle areas of a rectangle. Okay. Now for the next one, show the I've seen the word acceleration in there and it says show that it's about this. Um, so how am I going to work out the acceleration? Well, already I was suspicious about this. Um, and it says specifically at 12 seconds. So, what I'm going to have to do, I'm going to have to get a ruler and draw a tangent at uh 12 seconds. Um, obviously I can't bother to do it properly, but when you're in the exam, I'm sure you wouldn't mind doing it properly. So, the first thing that I'm going to do is draw a tangent. Make it very clear that I have drawn this tangent. And then on the graph, I'm just going to zoom in as well. Can I zoom in? There we go. So on the graph, obviously not with this one because the tangent is rubbish, but make sure you're very clearly showing that you are taking a large triangle. I should have done it from there because the other bits off the graph, but very clearly show that you have done a large triangle and show you're working like that because that can be a separate marking point. Then work it out. Um, and very clearly show you working. So say gradient equals blah blah blah, acceleration equals blah blah blah. Um, and because it's a show that question, it's about 1.5, you can't just put 1.5.

[00:33:24]need to show you working very clearly.

[00:33:26]Write the full number off the calculator and then the answer you should be getting is 1.5. Um on the mark scheme it says the answer you'll get is about 1.4.

[00:33:34]So um yeah.

[00:33:36]Okay. So it says use figure 16.1 to calculate the change in momentum of the car. So the fact that it says use figure 16.1 that's very nice of them because they're basically telling you just go back and use the graph. um they don't have to say this because you're expected to know that if they give you a graph in this part of the question, first part of the question, they can use that graph throughout whenever they want. So they've been nice here by saying use it calculate change momentum. Well, it should be reasonably easy. How do I calculate change momentum? Change in equals mass times change in velocity.

[00:34:07]Obviously from the graph I can get the change in velocity. I've scribbled all over it from before. So let me just get rid of that.

[00:34:14]Um, so how am I going to get the change in velocity? Well, it started at zero.

[00:34:18]And it says the question is 18 seconds.

[00:34:20]After 18 seconds, it was up here. So I'm just showing the examiner that I'm reading this thing. I'm drawing on the graph to make sure. And I'll just read that properly. So I'll zoom in there.

[00:34:29]And then that's 56.

[00:34:31]Um, always draw on the graph to show the examiner that you're taking readings from there. And we said that earlier with the triangle one as well. When you're calculating a gradient for this one, um, you have to draw the triangle when you're taking readings for this, the one that we're doing now. um just draw on the graph to show that you're doing it and mass haven't been given the mass earlier in the question it told me the mass there and it was already in kilograms or circle that so if it wasn't in a car question it's going to always be the right units in a different question um the fact that I circle the units first just means that later when I'm coming back to it and I've forgotten oh see the circles oh yeah I might have to do a conversion but here obviously don't have to so it's 1,700 times change in velocity and it went from 0 to 56. So, I'm just making sure that I'm being careful about that because that could be another trick in there that it doesn't start at zero. So, just kind of be aware of that. Put that in the calculator. And it says 95,200.

[00:35:26]Two significant figures I should do here. And um 95,000 is the answer. So, yeah, pretty straightforward to be fair.

[00:35:33]Okay, so for the next part now, I've got a new graph. So, because I've got a new graph, it's always good to get into the habit and make sure you're doing this for all your exam questions. When you see a graph, just check the units first so you don't get a trick later. Those units are fine. And then always do that same thing. Does y / x mean anything?

[00:35:48]Does y * x mean anything? So here, y / x. Force divided by time is not really a thing. Force times time is that anything? Yeah, it is. I do know that force time is the same asum change in momentum or impulse. So I know the area of the graph is going to mean something.

[00:36:03]And I might need that layer. like so I'm already prepared for going yeah that that question might come up um at some point in this um in this section of the paper. Okay. So now let's have a look what the actual question says. It says explain in terms of the for the shape of the graph. So um it's an explain question obviously. So I'm going to try to get some kind of law principle formula in there if I can. And then I do the same thing. So I'm going to get a law principle formula if I can. I'm going to try and do the state the obvious. Make sure I'm using keywords.

[00:36:33]make sure that I'm linking it to the question.

[00:36:36]Okay. So, is there a law principle formula I know about resultant force shape of the graph? I mean, it's pretty difficult for one to stand out to me from there and I can't think of one off the top of my head. But obviously, I'm not going to panic. I'm just going to go, well, okay, I can't think of one off the top of my head. So, I can't get that free mark at the moment, but what can I do? I'll just go to state the obvious.

[00:36:54]Well, for a graph question, what is the state of the obvious going to be? It's going to be describe the shape of the graph, basically. So, what can I see that's happening from this graph? I can see that the resultant force is decreasing as the time is increasing. So I'll just uh put that down on there. Um okay. Is there anything else that I know from there? Because it says in terms of the forces of the car. Um well earlier we were suspicious about something about a piece of information that they gave us and this was the thing that I put a box earlier in this question. The force provided by the motor is constant. Oh, so that's why they told me over there because they're hoping that by the time I get all the way down here, I've forgotten about it. Well, I put box around it because I was already suspicious. So, here I'm going to say the forwards force is constant.

[00:37:39]And I'm probably not going to get a mark for that because I'm just repeating the information that they've told me. But then what can I say from there? Well, if the resultant force is decreasing and the forwards force is constant, well, that must mean that the drag is going to be increasing because I know the resultant force must be equal to the forwards force take away the drag. So, um should I can kind of put a diagram here maybe to help if that's or like a formula or something if people aren't quite too sure about that. So if I know the resultant force is always is going to be the forwards force take away the backwards force basically or whatever it is. So in this case the forwards force is just F. I'm just going to call it F for this one over here. Um and we know that um it's going to be that take away the drag because the drag is going in opposite direction. But if I've said the resultant force is going to be decreasing and we've already said that this is going to be constant. I know that the drag is going to be increasing.

[00:38:36]that has to that's the only way that the front force can decrease if the force force stays constant. So I still don't really know um too much about anything about how it is linked to the question yet. All I'm doing is just saying what I know. Okay. So now that I know that the drag is going to be um increasing, is there anything else I can say about the fact that the drag is going to be increasing?

[00:38:58]Um well, I know that the drag is going to be increasing at a decreasing rate.

[00:39:04]So, I'm just going to put that in there.

[00:39:06]Um, drag is increasing uh decreasing rate.

[00:39:16]Um, and how can I uh see that that's the case? Because I can see that from the shape of the graph. Um, that it's a curved graph. And then from there, I can deduce what I've just said from that third bullet point. And then what's happened? Is there any other significant points on the graph I can talk about?

[00:39:30]Well, here the resultant force is going to be zero. And I might as well use numbers from there. Um, so it's about 15 seconds.

[00:39:39]And to be fair, on the real exam, I can zoom in. It looks about right. Um, I can't zoom in because I look closer.

[00:39:44]Should have said. So at about 15 seconds, um, the resultant force is going to be zero. I can just see that from the graph. Why would the resultant force be zero? It must be because the force force from an engine um, and drag are con are equal. Um, obviously in the real exam like write out properly. Don't abbreviate that me. But yeah, I think that's probably fair enough. And um I'm pretty confident that that's going to be enough um to get the two marks for this question.

[00:40:13]Okay. So for this next part then um explain why the value of calculator is same as the area under the graph. Well, I've already kind of figured this out earlier. I'll just select what's come over there. Um why is it going to be the case? Because area well when it says area, even if I didn't think of it before, I just know what's on that axis, what's on that axis. Is there a formula?

[00:40:32]Yeah, there is. So, I know that force times time equals impulse, which is the same as change of momentum a part three. Where's that? Well, that says change momentum there. So, that's fine. Um, and it's a one mark question.

[00:40:48]I'm pretty sure that I've got that one right. Now, even though this question did actually say area, so it made it a little bit easier for me. It kind of led me towards I've got to do axis multiply by each of that. In the real exam, it might not say that. So, just be careful.

[00:41:00]um in the real exam it might just be something like oh calculate the impulse or something like that and it's not a curve it's like straight lines and um they don't make it explicit that it's the area but because we're get always getting into this habit of saying that times that or that divided by that if it's the gradient like it was for the early question they don't have to say what's the gradient what's the area they'll just have to say like what's the impulse or what's the acceleration of the previous one and it's your job to know so always just be on the lookout for that this one was a bit easy but it might not be the real Okay, so this question is asking me it says use Newton's third law. So again, this is quite a nice one because if it gives me the law principle formula, I just know, okay, I'm just got to apply it to the question really. So what do I know about Newton's third law? I know it's got to be I'll put it up here to um so I save space answer line. So it's going to be two different objects.

[00:41:50]The forces are going to be of the same magnitude.

[00:41:54]That's magnitude.

[00:41:56]um same type and opposite direction.

[00:42:01]Okay. So, how can I apply this to this question? So, it says, how does the frictional force cause a car to accelerate? So, I've got two objects.

[00:42:07]The two objects are the tire and the road. Obviously, um the same magnitude is fine. Same type is going to be friction opposite directions. How can I find this here? So, I'm going to say the tire exerts a force on the road. A frictional force on the road.

[00:42:28]And which direction is this going to be in? Um, so the direction of the frictional force from the tire on the road is going to be that way. So I can say to the left or backwards.

[00:42:39]Therefore the road exerts so same uh same type. So frictional force um on the tire.

[00:42:55]So now I've said two objects um of the same magnitude um forwards and I've incorporated all of those things from Newton's third law. So I've applied to the question I'm definitely getting um two marks here.

[00:43:16]Okay, so next one then calculate the resultant force on the car acting at 12 seconds. Use your answer to a part two.

[00:43:23]So if it says at 12 seconds, um I'm kind of a bit suspicious why it's saying 12 seconds and also saying a part two. So let's just have a look at what we said for a part two. And what was the answer I got for that one? It was about the show. The acceleration was whatever. So it says about 1.5. Like we said before that um we I couldn't be bothered to do it fully but how to do it the tangent gradient all of that kind of stuff. Um and we got the answer of 1.4 the mark of 1.4. So that's the one that we can use.

[00:43:51]Um if you didn't get that one then obviously in a show that question they give it you and you're thinking okay they give it you because you might need to use it later. So if I couldn't get an answer that was close to 1.5 I would just use the 1.5 they told me. But because the answer that we pretended to get was close to 1.5. That's the one that we can use. Okay. So then how do I know about it? So resultant force. I know that resultant force is going to be the mass times acceleration. So I'm just putting numbers into an equation. Um we got the mass earlier. That was 1.4. And then the answer that you should be getting is just like whatever that one is. So I just put that in the calculator. And um it's that.

[00:44:32]Okay. In order to get this question right, um you have to use the answer from the previous question. So if you put this in the calculator, it's um 2 380 newtons. But obviously your number might be different if your gradient was slightly different or if you use 1.5 or whatever it is, but for the sake of this, we'll just use this one. But bear in mind that yours might be slightly different. Okay. So um calculate the total frictional force um which was a thrust on the car. Well, what do I know about this situation? We've already said that the resultant force is 2380.

[00:45:04]So I no I can do it here I suppose because there's more space. I know the resultant force equals 2380.

[00:45:12]And I know that all the resistive forces are uh 29 kons. So how am I going to get the forwards force? Well I know um if this is the resultant force I know what the forwards force is. I know that the Ford's force has to be um 2,380 plus the resistive forces which is 29* 10 3.

[00:45:35]Okay. So if I know that this is the case, I'm going to put that in the calculator and I get around 31ish to this scale. It's fine. So I'm going to get 31 kons.

[00:45:45]Now I've just got to worry about my scale. So let's just have a look at how what the scale they're using. They haven't labeled it. So what scale have they used? This is obviously zero over here. And then what are they using? If there's if this is 29, then this is going to be here. That's going to be 10.

[00:46:01]That's going to be 20 and that's going to be 30. And then that's just touching there. So I know that each little square means one. Well, now I figured out that scale. Now I can figure out what 31's going to be. So that's going to be 10 here, 20, that's going to be 30. So 31 is going to be this little square there.

[00:46:17]And I'll just use a ruler and go up to that. And then yeah, job done.

[00:46:23]Okay. Okay, so after all the hard work and stress of all the previous difficult questions, um we get to be rewarded with a nice two one. How exciting for everyone. Everyone's favorite question.

[00:46:31]So um it's projectile motion and it's saying what is the um conditions for projectile motion.

[00:46:39]So um one of them is constant vertical acceleration. I just need to know the other one and that is no horizontal acceleration.

[00:46:47]Horizontal acceleration or you can say constant horizontal velocity. That's fine. Okay, so now we get to the fun part of it. So, show that the time taken is about 23 seconds. So, let's just make sure I'm reading the question properly.

[00:47:00]Even though I'm very, very excited, I can't let my excitement get the better of me.

[00:47:05]Um, I've got to just do all the sensible things I'm doing and just make sure I'm not making any mistakes here. So, what information do I have? It starts at 10,000 and it says a constant speed of 220. It controls are adjusted. in projectile motion. It tells me that's in projectile motion there until the altitude is 7,500.

[00:47:23]Um, and throughout it, the passengers experience apparent w. So, essentially it's like dropping. Um, so vertically it's going to be dropping, but horizontally it's going to be carrying on that constant speed of 220. So, that's why it's projectile motion. Um, okay. So, the time taking reaching new altitude is about this. So, the first thing I've got to do is put a question mark. Well, because it's to show that question. Um, I'm looking for T even though I know that's going to be the answer. So, my question mark goes next to T. Now, for all of these, I should be able to fill in um three of them, and I should have one X.

[00:47:55]So, let's just go down in order, I suppose. Um, or actually, to be fair, I should be reading the question properly and going in that order. So, it says the new altitude of 7,500. Well, it started at 10,000 and it's going to 7,500. So, it has dropped by um 2,500.

[00:48:10]So, that's my answer.

[00:48:12]U is going to be zero. So don't get tricked and think it's 220 because the horizontal speed was 220. The vertical speed was zero to begin with because it's essentially dropping. Um V I don't think I know it. So hopefully I've got a Have I got a? Yeah, I have because it's dropping. A is going to be 9.81. So I put my X next to there and I'm pretty happy now that I've not made any mistakes here. Um and I've just double checked. It is going to get faster when it drops. So A is going to be positive rather than negative. Okay, what's the formula without V in there? Um the formula without v is s= u t + half a t^2. Um now whenever I'm getting this I'm just hoping that u is zero. If I can get rid of that u is zero. Fantastic news. So I know that I can get rid of that because u is zero. So I'm getting half a t ^2. And now I'm just going to put the numbers into the calculator. Um so I'll put it here first just in case people struggle um with rearranging. So half * 9.81 81 * t ^2 and then rearrange it to say t ^2= blah blah blah. So that and then put that in the calculator and you should get t = 22.576 something something something. Um and that's yeah so then I can say that's roughly equal to 23 because it was a show that question. I need to just make sure that I'm making that statement.

[00:49:27]Yeah, it is 23.

[00:49:29]Okay. So the next question I'm just going to kind of get rid of that and just remember that I did it. That's fine. Okay. So, determine the magnitude and the velocity. That's bad enough because it looks like I haven't done it.

[00:49:39]I'm just going to put it back, but I'll put it up here. T= 22.576.

[00:49:45]Therefore, it's going to be 23. Okay.

[00:49:49]So, determine the magnitude and the direction of the velocity of the aircraft at time two. Okay. Now, this is one that um it it comes up like quite often because if it's talking about the magnet and the direction, the first thing I'm suspicious about is there's probably going to be a horizontal one and there's probably going to be a vertical one and then I'm just going to need to resolve and get whatever it is. So, I'm going to need this one. I'm going to need like a horizontal one. I'm going to need the vertical one and then just use trigonometry to get it. So, because it's talking about the velocity, what do I know about the horizontal velocity? So, the horizontal velocity was constant.

[00:50:23]Well, it told me it was 220 at the start. Therefore, after however many seconds, who cares? It's still going to be 220.

[00:50:30]Okay. Now, I've got to find out the vertical velocity. So, for the vertical velocity, I'm still going to use sata.

[00:50:35]So, I'm going to put s down because it's vertical and it's accelerating vertically. This is the one that I can use. So, um I now have remember the rule was that I need one x um and then I have the others being question. Oh, sorry. one question mark one x and then the other's fill in.

[00:50:53]Well, in this one, I've actually got all of the values. So, what am I looking for? I'm looking for v. And actually, I know t because I've just worked it out.

[00:51:00]However, just in case I've worked out wrong, t is going to be my x and then the rest of them I know. So, 2500, u is still zero, a is still 9.81.

[00:51:11]Um, so I should be able to So, what's the formula without t in there? Um, and that's the easier way to look at it.

[00:51:17]Rather than looking what has got this, look at what hasn't got the one with x.

[00:51:20]And straight away I can see is v ^2= u ^2 + 2 a s. I'm not going to do the calculation like you should be able to do the calculation for yourself and you get v = 221 something or so. So 221.472 blah blah blah. Okay. So I've done that one. Um, and now I know um what the how I'm going to get the um overall magnitude and well the overall magnitude I'm going to use uh Pythagoras.

[00:51:55]So the overall V um so the V overall is going to be the square root of these two together. So that was 221.4 4^ 2 plus I'll have to roll that thing out the way now. I should have thought of that first plus 220^2.

[00:52:16]I'll put that in the calculator and v comes to being um 312.169 blah blah blah. So two significant figures. The velocity is going to be 310.

[00:52:28]Angle to the vertical. How am I going to get the angle to the vertical? I'll move this out the way so I can redraw the diagram. Um, so that one I knew was the horizontal one was 220. The vertical one I know was uh what was it? 221.

[00:52:46]Um, so now what is that one going to be?

[00:52:47]So it says angle to the vertical. This is the one that I care about. So I've got opposite and adjacent. So when I've got opposite and adjacent, I'm going to use tan.

[00:52:56]So I'm going to say tan theta equals opposite over adjacent 221 over 220. And the answer that I'm getting is 44.87 blah blah blah. And I can see that that is pretty close to 45. Um which is what I would expect. And actually it does round to 45 when I'm doing two significant figures because those two values are basically the same. Um so therefore I'm expecting that I'm getting an answer that's pretty close to 45 degrees and I have got that. So I'm happy. Oh, look. Four marks. That's nice. So, that's essentially a free four marks, isn't it? Because it's the same question that's going to come up pretty much every year. Okay. So, this next part is an explain question. So, I'm going to do the law principle of formula first if I can. And it says um why do the engines have to produce thrust during projectile motion? Well, when it's projectile motion, what's the law principle for? I know about that. We've already said the horizontal velocity has to be constant.

[00:53:51]And if the horizontal velocity must be constant, there must be um the resultant force horizontally must be equal zero.

[00:53:59]And obviously you're going to be writing this out for um when you're doing it. So that's the law formula I've got. So now how can I kind of apply it to this question? Well, is there a state of the obvious thing I can say? Well, the aircraft is moving. If the aircraft is moving, there must be drag. Um because aircraft moving and again I can't bother to write out but write out properly in the exam. Um, so because there's going to be drag, therefore there must be a thrust from the engine to equal drag.

[00:54:27]Okay, so what's this next question?

[00:54:29]Well, I'm very excited that I can see the word ratio. And just remember that ratio always means divide. So don't ever like when you see ratio, students will think, oh, I need to give like two to one or something like that. That's not the case. Whenever you see the word ratio, it just means it's got to be something divided by something. And you would just always give it as like a number, whatever the answer is, you want to calculate. Sometimes you might leave it as a fraction but generally um you would just give it as like a number. So it says here um it actually gives me the formula I've got to use gravitational field strength at this distance or that altitude the gravitational field strength of the surface. Well, what information have I got here? Um well the first thing it's telling me that radius of the earth is this. So obviously there's a trick and I've just got to make sure that I am converting it. And the other thing that students will get mixed up on is um like at 10,000 me altitude obviously you need to add the radius of the earth to the altitude and then that's going to be sorted. So when it says the altitude I've got to convert this first. So six um 6,400 and then times 10^ 3 add 10,000 and the surface of the earth is just going to be 6,400* 10 3. Well, because it's a ratio, let's have a look at how we should do it because I've only got the like altitudes here. So, basically, I know because I'm looking for gravitational field strength, I need a formula that's got G in there, and I needed to have R in there as well. Well, what's the form of G and R in there? Is G = G M R 2. Now, there's a couple of ways that you can do this. I'm going to go through it the slightly longer way first and then just see like how that relates to the other way that we could have done it. Um so because it's saying I know that G equ= GM over R squ basically I can just do it from the um top and the bottom. So for the top one I'll just put that equals there. So it's going to be GM um divided by and then I've got to add those things together. So, it's going to be 10,000 plus 6,400 * 10^ 3. And I've got a room to square that.

[00:56:29]And then divided by GM / um six uh 6,400 and then times in three.

[00:56:38]And I've obviously got room to square that as well. Now, I've got fractions of fractions um here. So, I just need to know how I can make sure that I'm um like simplifying this. So I I will write it the other way actually. Obviously if you're more confident with maths you can just like kind of forward through this bit. But if you write it as the other kind of fraction. So I'm just going to put GM over I'll just put a for altitude there.

[00:57:04]And that's divided by GM over surface. I know that I can do the thing where I multiply by the inverse. So I'm going to say GM / altitude multiplied by surface divided by GM. Well those GMs cancel out. So, it's going to be whichever one the surface one was divided by whichever one um 10,000 m altitude one was. So, basically I'm going to end up with um that squared divided by the adding together one squared.

[00:57:37]And then if I put this in the calculator, the answer that I get from the calculator is 0.99688 dot dot dot. Um, and then the question says is greater than 0.99. Well, then I'm just going to make that statement.

[00:57:52]Um, this is greater than 0.99 just to show the examiner that I've linked to the question. Now, what's the other way that you could could have done it? Um, well, I didn't even need to write about the GM things like this because you might be asked in terms of different relationships and people students might get confused. Well, the easy way that I can do it is I know that if two things are proportional to each other, so if x is proportional to y, I'm just going to rub these things out the way so I've got a space to write.

[00:58:24]So if x is proportional to y, I know that the ratio of x1 / x2 is going to be equal to y1 over y2 because they're proportional. That's fine. And that if they're inversely proportional, I know the ratio is going to be x1 ratio of x1 over x2. Well, the inversely proportional now is going to be y2 over y1. And then in this situation, I could see that the actual relationship that I care about is that um well, we know that g= gm / r 2. So I know that g is inversely to r 2. So basically because inversely I can just turn it upside down. Um, and then with the squared, I've got to put the squared in there, which is how I've got to this. Well, it's not how I got to it, but it's another way that I could have got to that stage a little bit quicker and not worrying about having to cancel the GMs or stuff like that. It's just knowing this stuff about proportionality that they're going to test you on. Either way, obviously, I got to that stage and then I'm going to write down the answer.

[00:59:19]So, I'm going to get the same answer and then just make that comparative statement to make sure the examiner knows that I know and just there's no point throwing away simple marks like that. Now, for this next one, two mark question. as long as I take a pretty sensible approach and I'm going to get the marks as one of those ones that it's like just testing your knowledge and understanding of the content. So what's the word that stands out? What is the word that stands out is weightless and it's an explain question which means is there some kind of law principle formula. So the first thing what do I know about if something is described as weightless? Well I know if something is described as weightless I know that there's going to be no like reaction force or no contact force. So I'm just going to get that mark first. I'm just going to say because I know that has to be the answer that I'm working towards.

[01:00:00]So, passengers with feel no contact force.

[01:00:04]Um, now I know there has to be a reason for that to be the case. So, no contact force, no reaction force. Um, and now I just need to think of a reason why that would be the case. But why would it be that they're feeling no reaction force? Well, the only reason why they would feel no reaction force is if they're accelerating at the same rate as the um whatever the thing is the aircraft here. So that means so I'll say because um the passengers are accelerating at the same rate as the aircraft.

[01:00:42]Now, this is one that you could possibly get wrong because it looks like, oh, how am I supposed to know they're weightless or whatever, but I'm just because I'm seeing that word firstly, I'm showing the examiner that I know um I'm showing the examiner that I know what is meant by that word and I know that every situation is going to be this. And then that second bullet point is now I've just used my brain a little bit and just think how can I apply to that situation? They must be accelerating the same rate as the aircraft. I keep on stopping to write to talk. Um, but yeah, that's going to be that. I mean, that's the answer. There's nothing else you can say really for that. Okay, so I've got a whole lot of information for the next question. I'm sure that you've read it and checked your conversions and all that kind of standard stuff. So, let's just go straight on to the question saying, so explain why the weight of the water displace must be equal to the weight of the drum. Well, firstly, when I'm seeing stuff like that, water displaced weight, whatever, this is quite clearly telling me it's our community's like it's obviously a question of Archimed's principle. So why does it have to be equal to the weight of the drum? Um well what is Archimedes principle? Archimedes principle is that the up thrust is equal to the weight of water displaced. So why does that have to be equal to the weight of the drum? Well, is there something in there that's showing me that the force of balance is stationary? There's no resultant force. Yeah, it's floating. So I know that it's floating. So I'm just going to put it's floating.

[01:02:01]Therefore, no result force.

[01:02:04]And I'm using the word no resultant force rather than forces are balanced because this is the one that um is going to get you the mark properly like it always gets you the mark. Um, so I'm going to say therefore up thrust must be equal to the weight. And just to make sure that I know the weight of the drum, I can put up thrust equals weight of water displaced and just show that that I know our community's principle. So even though this is only a one mark question, I've written those three things. um because I don't know which one's going to get the mark or how much of I need to say. So I'm just showing the examiner that I've worked through this properly.

[01:02:49]That's the um state of the obvious thing that I said and then again that kind of kind of link or the state of the obvious. That's the law principal formula. Um I'm happy that I've got the mark now for this question and obviously even as a single mark question I want to make sure that I get it. So I don't want to I don't know miss out saying Archimed's principle or miss out using the keyword up thrust or something like that. Why would I risk it? Even, you know, just make sure you're going for every single mark. Okay, so I've got a calculation question. Now, normally obviously calculation questions hopefully have marks in the bag, but if it's something like, okay, calculate D, it's not exactly a formula that I can just plug the numbers into. But what are the things I'm suspicious about here about how am I going to do this? So, it seems pretty difficult to begin with, but as long as I'm taking a sensible approach, I should be able to get the answer right. even if I don't know from the outset what I'm going to do like what the final thing is going to be. It doesn't matter. You're just going to do what you can. So, let's take a look at the um information I got. Now, hopefully when I read this information at the start of the question, there was something that should have stood out to me. That's this over here, cross-sectional area, because when do you ever really need the cross-sectional area? And particularly because it's giving like a distance and stuff over here. Um really, what's that telling me?

[01:04:00]It's basically saying I'm going to need volume basically. So I know that this question is probably going to be something to do with volume because I know that okay cross-sectional area times the length is going to be the volume. Um so maybe that's something that I can do. And the cross-sectional area time d is going to be um the volume below the water. So that's one place I can start with. Maybe that might help me. What's the other clue that I've got?

[01:04:24]Well, the fact that it's part two to a question and I just had a very easy question that they were basically like shoving Archimedes principle down my throat by using the terms like weight of water displace and all that kind of stuff equilibrium that they're talking about Archimedes principle. So, it's going to be any of those things. So, I've got in my head like, okay, that's the approach I'm going to be taking. I still don't really know what I'm doing. I'm still stuck or whatever it is. I'm panicking in the exam. I can't see how those things link together. Well, I'm just going to do what I can do. So, because I'm talking about all of the stuff, the clues that they gave me first, I'm just going, well, what do I know straight away, but I can work out the weight of the drum because that's going to be that's some kind of force and it's probably going to be something relevant. So, I'm going to go weight of the drum is equal to and I've got the numbers off the start. So, if the mass was 1,200, G is 9.81. And I'm just going to put that in the calculator and I know that that is um I've got the answer here. Okay. So now if I know the weight of the drum, I know the weight of the water displaced must be because we've just said that or why they're equal. So therefore the weight of the water displaced is going to be that and opus is going to be that. So I can write the oxus is equal to 11772 newtons. The weight of the water is also going to be equal to 1172 newtons. Um now I'm getting somewhere because if it's saying the weight of the water, what do I know about the weight of the water? Well, that's going to be the mass times the gravity of the water.

[01:05:49]Um, so is there a way that I'm just going to write that for now. Mass times gravity of the water. So now let's take a look. Have I got the mass of the water? Well, no. Is there anything that stands out? Yeah, because it's giving me the density of the water. It's secretly telling me that there's going to be something to do with volume. Well, now I've remembered, oh, actually, yeah, mass is going to be density time volume.

[01:06:08]So, I know the weight of the water is the density times the volume time g. Um, so the d So, I can just put the numbers in now that I've got them and just see what's missing. So I know that the weight is 11772. The density is going to be 1,000. I don't know the volume, but I know G is 9.81. Well, now I know the volume is a cross-sectional area times the height. So again, I can kind of do that same thing. So 11772 1,00 instead of volume, we say cross-sectional area times the height or the distance below the water. So 0.79 * D * 9.81. And now all I need to just do is rearrange this, put it in my calculator, and the answer that I should be getting is this. Write down the full numbers and then round to two significant figures unless there's reason to give otherwise. There's no reason to give otherwise here because that's two significant figures even though this is three. That's the two.

[01:06:58]Um, so I'm going to give it to two significant figures. 1.5.

[01:07:04]And I'm pretty happy that I've got full um full marks for this question.

[01:07:08]So, they're so kind that they've decided to put the word equilibrium in bold on there as if I shouldn't um like be able to spot it for myself. So, they've seen the word equilibri I've seen the word equilibrium. Sorry. And again, I don't really know what the answer is to this question. I'm stuck on whatever concept.

[01:07:23]I can't conceptualize anything like that. Who cares? Because I've seen the word equilibrium. So, I just know what I'm going to say straight away. I know that equilibrium means that there's no result. I'm going to start with the law principle formula and I'm just going to show the examiner that I know that. So equilibrium means that there's no equilibrium means no resultant moment, no resultant force.

[01:07:48]Um, and the reason why I'm kind of leaning toward leaning towards the moment here is because it's saying why can't it be on its side? So I'm going well it probably because it's going to be turning. So I know I've shown the examiner that I know the definition of that. Whether or not that's a mark, I'm not too sure. Um, it probably is going to be because I've said the laws performative. But again, even if it's not, I'm just saying it to give myself.

[01:08:13]Actually, I'm looking at the mark scheme and yeah, it is. So, that's good. I've just checked now.

[01:08:18]Um, so, okay. So, now basically because saying why it can't do that, we've done this thing before that we're saying, well, if I say, oh, why it would do that, I can put it in the negative and say, okay, this isn't the case that's going to be happening. So if I'm saying that equilibrium means there's got to be no result force, no result moment. I'm saying it can't float on its side. It can't float in equilibrium. Well, that means that on its side there must be a resultant force. There must be a result of moment one of those. And now I know in this situation I'm kind of I've already said we're leaning towards moment here because like we can see it's going to like tip or something like that. So I'm imagining what's going to happen. So now I need to know why there's going to be moment. So B, I'll just say that first. So I when horizontal.

[01:08:58]Now remember, even if I don't know this and I don't know why, I just know it has to be the answer because it said it can't float on that side. So I'm just going to act like I knew this all along and just say, yeah, there's going to be a result of moment and when horizontal, that word mention horizontal. So why is there going to be a horizontal moment?

[01:09:17]Well, I'll just show you the examination point by moment. So I'm going to say moment equals force times perpendicular distance.

[01:09:24]So is there going to be a force? Well, there's always a force anyway. That means there must be a perpendicular distance. And again, I'm just going to act like I knew this all along. So, there will be a perpendicular distance.

[01:09:34]And now, I've just got to look at the information I've got to figure out why there's going to be a perpendicular distance between the force. Well, let's take a look at a picture. And it shows me here that the center of mass isn't in the center of the drum.

[01:09:48]So, because the center of mass isn't in the center of the drum, the drum is 2 m and the center of mass is 0.5. So when I put it on its side, it's obviously not going to be in the middle. So it's not going to be um right there. So I'm going to do everything distance between the weight and um and the force. Oh no, sorry. The weight is the force between the weight and the pivot.

[01:10:16]And when horizontal.

[01:10:21]So, um, because the pivot is kind of like you can take that as being like the upr and they're not going to be at the same point. So, there's going to be distance there's going to be a distance between them that's going to create a pivot. Um, so therefore the um that's going to cause a net moment. So therefore, they'll cause a net moment. And just to be extra safe there, I can say W and U are not acting in the same line which causes it to pivot.

[01:10:55]Now I know again this is only a two mark question but I don't know when I'm sitting the exam what's going to be on the mark scheme. Um so I've started here with a law principle formula and then I've just made sure I've worked my way through it and um like even if I didn't know the answer like I said I said I knew the principle first because it says cannot I just knew that had to be the case. I'm showing the examiner what I know by moment. I'm linking it to the question and therefore you know I've definitely got those marks.

[01:11:26]Okay. So before we actually answer this question, this is one of the one where the words was that should have corrected at the start. So that D over here that's meant to be X. So I'll just change that on there now. Okay. Now let's have a look at this question. So what's the thing that stands out to me first? It's saying by considering the magnitude of the additional up thrust. So what how am I going to get the additional up thrust?

[01:11:47]I know that well because I know the formula for up thrust I also know that the additional up thrust is going to be basically the additional weight of the water displace the weight of the additional water displaced additional weight and I'm just going to put W there for water displaced and that's fine um obviously in the exam right I'm just rushing so the video is a bit shorter oh now that I've said all about I could have just written oh too late um okay so how am I going to get the additional weight reward this place. Well, we already kind of um worked out some of the stuff before. So, I know that the um well, actually, to be fair, you could just rewind the video if you wanted to um if you're not too sure on how to get it, but just from what we were doing on the previous question, I know that it's going to be 0.79 * 1,00* 9.81. And then that's going to be times x because that's the additional distance that it's gone down. So, I've got the volume times the distance again. Um what can I do now? So I can just put those numbers into my calculator and I get 7749.9x two significant figures 7,700x and what does it say? Show that k is about 8,000. And I'm just going to still make that statement. So I'm going to say um so if f= kx so that's the force.

[01:13:06]Um, I'm basically going to say therefore K is roughly equal to 8,000 um because K equals 7,700 something like that. So just make sure that I've um showing the examiner basically that I'm linking back to the question always just in case. So what stands out to me here?

[01:13:26]Well, I can see the word oscillation and they're saying about simple harmonic motion. So straight away I should be able to identify that they basically ask me to define some harmonic motion. They do this all the time. It's like a question that comes up every year. Just different ways of phrasing basically what is simple motion. So in this case explain why they're demand. Okay. So what do I know about simple manic motion straight away? I should know the defining equation is force is proportional um to the displacement in the opposite direction.

[01:13:54]So I've got a fishy minus x there. Um obviously write in words to make sure that you're getting the marks there.

[01:13:59]Don't just abbreviate like I am. You've got to write a full sentence. Um so the full sentence in here would be that the acceleration is proportional to the displacement and it's acting in opposite direction. So I've got the proportional bit and I've got the minus bit. Um so it's going to be one mark for the proportional bit one mark for the minus bit and that would be fine. You can say the force or the acceleration well if it's an acceleration that's going to be a force. So I can say the force is directed towards the equilibrium position or the acceleration is always towards the equilibrium position. Both of those will be fine. I just can't bother to write that. Okay. So calculation questions when it comes to simple human emotion they're always going to be like along the same theme.

[01:14:36]So as soon as I see it as a calculation they've already told me it's simple and emotion basically I know that um because we've already said a fishy minus x I know that the um constant of proportionality is omega squ. So it is probably going to be something to do with that. They've already given me and I should already note as well that omega= 2 pi f. So I know it's going to be something to do with that. Again, they've already told me fals or f= minus kx. In this situation, they've told me that. So why would they tell me it for no reason? They're saying I was given by that. I'm probably going to have to use that here. That might be relevant. What else do I know? I know that fals ma. So these are all things I know. How can I end up getting f from here? Well, I'm going to say if f= minus kx, f= ma. Therefore obviously um I can say that minus kx is equal to ma. So that might be something relevant to me here. Well if this is the case can I do anything with this?

[01:15:36]Um well what's a going to be equal to? I can say that a is going to be equal to minus k / multip x. Well why is this relevant? Because if we take a look at this equation that we said at the start, we know that I'm always kind of aiming for something like that. So I needed to know what omega squared was. And the way to do it was like that. So I'm always kind of looking out for that kind of thing. It's always the same technique.

[01:16:02]Well, now that I've got that, I can just make that statement that because these two are the same, I know that minus k is going to be equal to omega^ 2 minus omega^ 2, sorry. And now I can just get rid of the minuses and say K M is going to be equal to M^ 2. Well, do I know what K is? Do I know what M is? Yeah, I do because I worked those things out earlier. So the K is the full number that I worked out here. M is 1,200. And I'll just put that into the calculator.

[01:16:30]So omega is going to be the square root of K over M, which is um 7749.9 / 1,200. So omega is going to be equal to 2.5 5413.

[01:16:49]Keep the full number in your calculator.

[01:16:51]Well, I know that therefore that number because we've already said that's equal to 2 pi f. So therefore 2.5413 blah blah blah is going to be equal to 2 pi f. And then you're just going to rearrange that to get f is equal to 0.444 4044 dot dot dot and two significant figures unless they're reading together otherwise. So that's going to be 0.4.

[01:17:13]And yeah, I'm happy that I've got the right answer for this one. Um, but what you need to just make sure that you're doing like I just want to recap the simple hormon this is defining equation and then when it came to the calculation I just made sure that I used the defining equation over here. I tried to get it in a format based on the things I know so far. So, they already gave me this kind of thing and I had to get it get an A in there.

[01:17:41]Um, sorry. Yeah, get an A in there to make it into uh to make it look similar to this one. Um, and it's very likely that a very similar question is going to come up again. So, just make sure that you've got this technique down.

[01:17:55]Okay. Now, for this question, you can probably get a lot of it from the M scheme. This is like a lot of content.

[01:18:00]Um, but I'm just going to go through like the main things and then I'll maybe put the mask in the video or just direct you to or whatever. Um, but I mean let's just have a look at the main things that should stand out to me that I should always be suspicious about thinking about it. Well, first of all, it's saying that the natural frequency of the two machines are identical. Graph show the results but incomplete. Um, so the describe part is just can I remember the procedure? And that's the kind of thing that mainly I mean just get it off and asking just what's the procedure? I'm not going to write it all down. Um the main thing that um students might lose marks on about this part over here and this is the thing that comes up over and over again. So because it says complete the graph with suitable labels and annotations I can see that neither of the axes have been labeled. So I know that the axes here are going to be amplitude and over here it's going to be frequency because that's what the um information in the question is giving me and I know it's this type of graph just by looking at it. Anyway, um what have I missed here? I've still missed the unit.

[01:18:58]So I've got to say the amplitude is going to be in meters and the frequency is going to be in hertz. And again, is there anything that I should notice or significant points on the graph? Well, it tells me that the natural frequency is identical over here. Um and then I know the natural frequency is going to be the high the um basically yeah where the where there's maximum energy transfer. So it's going to be that point over there, the highest amplitude. Um now is there anything that I can go when I draw a line down from here? I know that's a natural frequency. I'm going to write f not. But have I actually got a value for f not? Well, yeah, I have because I just worked out in this previous question over here. And the reason why I'm going through this part first is because it's that principle that we're mentioning about if I have numbers, I should be able to put the numbers on a graph. I don't really care about the scale. I don't need to mark all of these off. If this is 0.4, I need to figure out what these are going to be. I don't care. Um I just know that that is 0.4 and I'm labeling that one on there. Then the scale is not my problem and it's never my problem. So even if it's an awkward number and it they've put the line on an awkward number. So just say I worked out the amplitude and it was whatever for B or something like that and I'm going, "Oh, that's an awkward number here." It doesn't matter.

[01:20:06]Just whatever number you've got, put it on there because it's only really ever going to be one. Um so that's going to be fine. Like don't worry about going, "Oh, it's got to be 10 or something like that. A nice number on my scale. Who cares?" Just um do it on there. Just make it clear to the examiner that you know what that is. Okay. So then um is there anything else that I can really say which is the most effective at generating electrical energy? Um I'm just going to do the I'll put the mark scheme on there now and we can kind of go through because I don't want to write anything down. I can't be bothered. Um so I I'll do the click and let's hopefully it'll work otherwise I'll look very silly. Um hopefully that's worked otherwise I'm and if if it doesn't I'm just masking but I don't really want to look silly. So hopefully it has. Okay.

[01:20:47]So what are the things that saying here?

[01:20:48]It's just going to be the standard stuff. So, because it's a frequency against amplitude one, obviously I'm going to need some kind of way to measure the frequency. So, um they're saying they're using oscilloscope, stopwatch, whatever it is. So, it's like, oh, count the number of waves per 10 seconds and then divide by 10 or something like that. Um, and then if you take a look at the last bullet point in that first section, this is something that comes up quite often as well about video analysis, take a video, um, slow it down, that kind of thing. That's always in there. None of the other things I think are particularly controversial in that first section.

[01:21:22]It's just something that just s standard things that you should be saying on the annotation of the graphs. I think we've covered all of that. So that's okay.

[01:21:32]And then um the discussion part again there's nothing too controversial here.

[01:21:36]I don't think um I should obviously see that B has been more damp because the peak has been shifted towards the left and also it's lower down. I'm just going to make sure I'm training someone that I know that um and then which is going to be most effective. Well, I can see that A has got um a greater amplitude and therefore that that last thing I'm just linking to that question. So, I don't think there's anything particularly difficult in here that should um trip you up. But it's just about making sure that you go through that main thing that I said first about going through the graph and annotating what you need to annotate and then the rest of it as long as you're going through the standard procedure. Um saying the things that come up quite often about use this equipment, use the video, whatever to slow it down and then link to the question as the very last thing. I think it should be pretty uh pretty okay hopefully.

[01:22:25]Okay. Even before reading the question, um, just go through the stuff about the information that you've got and just see if there's anything that you're suspicious about. So, I've got a few things in there. What am I suspicious about? Well, firstly, kilopascals. I've just got to make sure that I'm being careful with that. And there's kilowatts, kilogjles, all of that. So, there might be conversions in there. And the main thing that I I'm possibly suspicious about is um the fact that they've given me this in degree C because I know if I'm going to do a calculation that isn't to do with a change in temperature absolute temperature. It's always going to be in Kelvin and that tends to be the one that students forget quite a lot. So I don't know how or if or why or whatever any of those relevant but they might be coming up in the in the future questions. I've circled them. I've made notes on it on my paper so that if it comes up I'm not going to get tricked now. um later on if I let my guard down by getting too complacent.

[01:23:16]Oh, the other thing this is involved.

[01:23:18]So, why does it have any liquid oxygen?

[01:23:20]Um, okay. So, that's something else I'm suspicious about. Okay. Why does thermal energy pass into the tank? Well, it's always like this is just basically can I remember what the law principle of formula is. And the principle I just need to remember is that energy is always going to transfer from high temperature to low temperature.

[01:23:37]Um, from high temperature to low temperature. So I'm just going to abbreviate like that.

[01:23:45]Um and then in this case if I know that it's passing into the tank I know that the tank must have a lower temperature then the surroundings whatever is there um tank must have lower temperature than the surroundings and I'm just making that link even though it's only a one mark question and I possibly just got it from that one anyway just in case I've got to always play it safe I link to the question and then yeah get that mark in the back. So just look how I'm being very careful in my exam because I just want to pick up every single scrap of mark that I can.

[01:24:20]Okay. So for this next one, it's saying explain in terms of partic I'm going to go look into the formula somewhere in there if I can. And it says the temperature of the oxygen is remaining constant despite the fact that thermal energy is going into the tank. So this could be like kind of a tricky question like going oh well because I'm expecting thermal energy is going in there. I'd expect the temperature to increase.

[01:24:41]Well, what is the um thing that I need to realize here? And the thing that I need to realize is that yeah, when energy goes into tank, so thermal energy is going into the tank, there's one of two things that can happen. And the one that students will always kind of think of it, oh, the temperature is going to increase. But what's the other thing that can happen when thermal energy goes into a substance, it can change state instead. So if it's thermal to kinetic, it's going to be changing temperature.

[01:25:05]If it's thermal to chemical, it's going to be changing state. Well, this has told me that the temperature stayed constant. So therefore, if the temperature stayed constant, I'm going to say that that thermal energy must be transferred to chemical energy.

[01:25:16]Obviously, right, I'm full in the exam and that's a C or that's mass changed it for me. Um, so thermal energy is going to chemical energy. Um, and I'm just going to show the exam that I know I can make those links. So therefore, it's changing state, not changing temperature.

[01:25:32]Um, and is there anything that else I can say? I can maybe say, oh, the kinetic energy stays constant.

[01:25:37]Therefore, the temperature stays constant. Um, just again just to be extra safe. Okay.

[01:25:45]Constant.

[01:25:47]And I think I've covered pretty much anything everything that I can say in here. So, the fact that they have said energy in their question. I've just made sure that I've talked about energy here.

[01:25:56]I've talked about energy here. I've linked it to the question. Um, and now, yeah, I've probably got the two marks for this question.

[01:26:03]Okay. So, I've got to show that question, which means obviously my working as always, it's very important anyway. But I've got to make sure that it's very very clear um to make the examiner's life as easy as possible to give me the marks rather than having to make a work through and miss something something like that. But what stands out to me here first the fact that it says 1 hour. I know there's going to be some kind of conversion there. And it shows the volume of liquid oxygen that evaporates. Well, I was already suspicious about that in the first place. That's giving me the density. It might be something to do with that.

[01:26:30]because it's saying that evaporates. I know that's going to be something to do with specific latent heat. Um so, and they tell they've g me some information here. Specific lat heat of vaporization.

[01:26:39]Yeah, that's the one that I care about because vapor means um a change between liquid and gas and it's evaporator.

[01:26:46]That's fine. So, I know that this is going to be something to do with the equation equals ml. The fact that it says 1 hour, there's going to be some time in there. Well, actually, it tells me here that thermal energy is going through and it's giving me a power over there. So, I also know that energy equals power times time. And now I'm pretty happy that I can do something here. Um, I can do enough here to at least get myself started. So, if I'm saying energy equals power time, well, I've got the power and I've got the time. So, I can work that out. So, the power is 1.3 * 10^ 3 time the time. It says 1 hour time 60 to go into minutes time 60 to go into seconds. I put this in the calculator and the answer is 4.68 68 * 10 6 Jew. Well, if I know that's energy, now I can put that into the equals ml part. I'm just going to put that out from there and put it down here. So, I'm clearly working in order.

[01:27:38]So, 4.68 * 10 6 is going to be equal to the mass times the specific latent heat, which it tells me over there. And look, I underlined the kilogjles earlier because I might be now so excited. Oh, look, I'm I'm getting somewhere. That's easy for me to kind of miss that out by accident. So, um, times 214 time 103 gives us kilogjles. And now, if I work that in the calculator, I'm going to get mass as equal to being 21.869.

[01:28:14]Well, I'm pretty happy with myself because I've worked all of this out. But what's the question asking me? It's it's asking about the volume. So, because it's asking about the volume, I know there's going to be volume in there.

[01:28:25]Well, I've got mass. So, I'm looking for volume. I've got mass. What haven't I used yet? Density.

[01:28:31]So, I've got um is there an equation that's going to link them all together?

[01:28:35]Yeah, density equals mass over volume.

[01:28:36]I'll put that over here. So, density equals mass over volume. And I'm just going to rearrange that now. So, is it the right density that I care about?

[01:28:43]Yeah, it's liquid oxygen. So, that's what I care about. And is any of the densities anyway? No, there aren't. So, um like it kind of has to be that one anyway. And I'm just going to put the numbers in and rearrange that equation.

[01:28:55]And now I should be getting the volume as being 0.019183 dot dot. And it's a show that question.

[01:29:05]Well, I'm still going to round two significant figures in 0.019.

[01:29:09]Is that close to what it said? 0.02. And I'm going to put roughly equal to 0.02 m cubed. I've shown it and I've made it very clear the steps in my work in I've got four marks here. I'm very happy.

[01:29:22]Okay. said, "Let's have a look at how I can figure out the equation of the approach for the next part of the question." Well, first it's saying calculate the pressure of the gas. Um, so, okay, I'm going to need an equation with pressure in there. And the thing that stood out to me first is the fact that it says moles. So, when it's pressure and stuff like that, I know that I'm going to be using one of these two equations.

[01:29:44]PV= NKT or PV equals NKT or PV= NRT.

[01:29:51]Um, which one is to do with moles?

[01:29:53]Couple of different ways that you can try to remember. So, the way I always remember is when it's moles, I'm going to go R. Look how cute those moles are.

[01:30:00]And then, oh, it's going to be this one over here. So, I know that I'm going to be using PV equals NRT. Um, so that's going to be something I'm going to be using. Well, I haven't really got enough information yet, and I'm still stuck on what I can do.

[01:30:12]Um, so, you know, because I've got, okay, I've got the number of moles, but I've not really got anything else. So, is there any information in the question I haven't used yet? Let's have a look.

[01:30:23]So, it's saying that I used to have um 9.8 m cubed of liquid oxygen. The volume of the tank is 10 m cubed. I've had all of these things about the pressure of the all of those kind of things. I haven't used any of those things yet.

[01:30:35]So, is there anything that I can do with it? Well, if I know that I started off with 9.8 me cubed of liquid oxygen and and this much has evaporated 0.02 not two um I know that I can basically do something from that. So the volume of oxygen remaining and still remember I might not even know what I'm doing with this but I can do some inensible so 9.8 um take away 0.02 not to um or you could I mean to be fair you should be using the um the number that you worked out but um I I haven't I just did it quickly and I can't bother to do it again. Um so 9.78 meters cubed that's going to be left. So if I know that there's 9.78 meters cubed of oxygen remaining, does that really help me? I don't know if I can do anything with it yet. But what other information do I know? I know that the volume of the tank was 10. And if I know that there's 9.78 liquid oxygen, I should have said, that means everything else in the tank must be gas. So I'm going to say the gas in the tank must be equal to 10, which is the total volume take away 9.78. So therefore, I know that there's 0.22 me of volume of gas in the tank. Okay. Is there anything I can do with that now? Well, I've got a volume. Um, I'm trying to look for pressure. So, this is sorted. This is sorted because I'm looking for it. I know N. I know R. I know T. And I did the conversion earlier just so that I don't um or I put a box around it earlier just to make sure that I don't get messed up. So, now I'm just going to put these numbers into the equations.

[01:32:17]I'm going to use PV= NRT.

[01:32:20]Therefore, um P is going to be equal to NRT over V. And I can just put those numbers in. So, that's equal to 710. R is going to be 8.31.

[01:32:31]the um temperature I've got to convert to Kelvin because it's not about temperature difference and absolute temperature. So that is minus 183 oh sorry min put a point minus 183 plus 273 and then divided by the volume of gas that I just worked out 0.22 22. I put that in the calculator and the answer that I'm getting is this. Um, so, oh sorry, the answer that I'm getting pressing kilopascal, sorry. Um, and I put that down there in kilopascal in the calculator be this.

[01:33:11]So, completed missing value in the table. So, um, what am I looking for here? So, number of particles per unit mass. So, what information have I got?

[01:33:18]Okay, molar mass and all that. So it kind of leads me towards yeah I'm going to have to do um something about using a chemistry formula. So if so I know a lot of you would do chemistry so you might be more confident with it but if you're not we're just going to go okay what do I know about this situation?

[01:33:33]So I know it's going to be something to do with number of moles equals number of moles is going to be equal to mass over the molar mass.

[01:33:44]Okay. Um and then it's saying here okay the number of particles per unit mass.

[01:33:48]So what's the unit mass going to be? So the unit mass is going to be 1 kilogram.

[01:33:53]So when I'm saying mass over mass, so it's going to be 1 kilogram divided by the molar mass, which it tells me over here is 0.084.

[01:34:03]Um, and then if I want to get the number of particles from there, I've got to times it by Avagadro's constant. So I'm just going to say number of particles.

[01:34:11]I'm just calling it P because I've not got permit space. So it's going to be Avagadro's constant multiplied by the number of moles which was whatever that's going to be 0.084. I put that in the calculator and I get 7.16 recovering time 10 24. This is giving me units of time 10 24. So that's okay. So I can just put 7.16 or well to be fair I should think 7.2 because everything else is two significant figures. So if I don't give my answer two significant figures um it's going to be wrong. Um, the only thing I really need to be careful about here is the fact that it says in time 10 to 24. So, this was nice because it came out as time 10 24. But if it comes out as like a time 10 to 25 or 10 to 23 or something like that.

[01:34:54]Well, then I'm going to have to do a little conversion and go, oh, if it was time 10 to 25, what's that going to be the same as? Or 10 to 23, what's that going to be the same as? Like here, for example, they've um it's not in standard form because it's um obviously it didn't go out to 24. like if I'm not too confident with it and I've got a little bit of time obviously because I've got all these values I can just double check my answer or my method with any of these um and then if I do that thing I know that I'm going to get those answers out of there and then I'm happy yeah my method is going to be correct. Okay. So plot the missing data point on the graph. So even if I didn't know this answer this first one here that we've just worked out and I didn't know what to do um I'm still going to kind of give it a reasonable go um so then I can get an error carried forward for this one.

[01:35:40]Now, obviously, I've worked out the right answer. I'm going to use the right answer. But if I was trying to cheat or something in the exam, I could kind of do it in the way where I could kind of even draw a line of best fit without those points. Obviously, you would use a ruler. And then because it says 250, um, I can just go from 250. So, I just got to be careful over here and go, okay, well, that's 300 here. So, I know that 250 is going to be halfway there. So, 250. Obviously, just always be careful on scale axis. So, I know that my point is going to be like somewhere here. And then I could kind of go and act like I knew that all along. Um that oh yeah, the answer is whatever that says and I might get away with it. I probably won't, but I mean it's worth a try. Um because it's likely that it wouldn't be the exact right number there. It's going to be pretty difficult to read. But I mean it's a one mark question. So if you get it right, you know, you might as well give it a go like that. But how do I do it properly? Again, just be careful with the units. So this is 10, which means that this is going to be five over here. So that's going to be six. That's going to be seven. So 7.2 two is going to be around here. And then we did the other thing. I think it was about there.

[01:36:42]The 250 was there. And then yeah, I've done that. And then line of best fit. So obviously you would use a ruler. I'm not using a ruler because um like I don't want to. Um so it looks like it's going to be a straight line through the origin. Obviously that's not a straight line, but you're going to be using a ruler. And this is actually very very bad because um I've got not the same amount of points above and below the line. If it doesn't go through every point, that's fine. As long as the same number of points are above and below the line, then that's okay.

[01:37:11]Um, yeah. So, that's hopefully just a a pretty simple um pretty simple mark there. Okay. So, the first part of this question should be pretty easy because I've just I've literally just drawn straight line through the origin. So, when it's a straight line through the origin, what do I know straight away?

[01:37:25]It's got to be that is proportional to this. So, I'm not going to write that in full because again, I can't be bothered.

[01:37:31]that those two things are going to be um proportional to each other. So the specific heat capacity has got to be proportional to the number of particles per unit mass. Right? I'm following the exam. How do I know that? I've just got to show the examiner that I know that the rule is always that when there's a straight line through the origin, um those two things are proportional to each other. Um make sure that you mention the fact that it's straight.

[01:37:52]Make sure you mention through the origin. If it's not straight and it's not through the origin, then it's not going to be proportional. It's got to be both of those things to be proportional.

[01:38:00]Okay. Okay, now it's a format question.

[01:38:01]Is that going to be format? Obviously not. I've only said two things. So, what do I need to suggest an explanation? So, really, I've just described it here. So, what's the reason that it could that could be the case? Um, well, I'm not too sure about this. So, is there anything that I do know that I can say? Well, first of all, something about specific heat capacity. So, I'm just going to show the examiner that I know what specific heat capacity is. So, specific heat capacity is the energy required um to increase temperature of 1 kilogram of substance by one degree Kelvin. And obviously this could also be degree C. Um because a change it doesn't matter. They're going to be the same. So I didn't really know.

[01:38:42]Okay, I can't think of an explanation, but I'm just going to do that and show the examiner that I knew that. Well, now that I've done that, I might have given myself a little clue here. So if I'm saying I'm increasing temperature, what am I doing by increasing temperature?

[01:38:54]Well, increasing temperature means an increase in kinetic energy.

[01:39:01]Okay, so I'm giving myself clues as I'm going along. So the kinetic energy must be increased. Therefore, kinetic energy of particles must be increasing.

[01:39:11]And again, I'm abbreviating you have full. Okay. So then how can this link to um any of these things? Well, if I'm saying number of particles and I want to increase the connect particle, have I done enough now to make a link? Well, I think I can figure this out now. If I've if I know it's got to have more kinetic energy and there's more particles connected into particles. Well, now I can say that more particles per unit mass must mean there's more energy required.

[01:39:43]Therefore, it must therefore it's directly proportional. Something should be directly proportional.

[01:39:48]Um and yeah, I'm pretty happy now that I've said enough to get these marks here. Now remember, I didn't really know first about what's the answer, but I didn't panic. I just did the things that I know has to be true about this thing over here. I know that's got to be true because it's straight line. I show the examiner I know why it's true. Now, even if I can't figure anything else out, at least I've got two marks rather than panicking to get zero. But still, what did I do? I just showed the examiner, okay, it's something with specific heat capacity. I'm going to write down the law principle of formula. This is the definition. I've just got that down. And then from there, I was able to carry on going and end up getting four marks for this question, which I'm very, very happy about, as you can probably tell from my tone of voice. Okay, this is one again where I don't really want to write it out in full um write too much down because I can just get the marks given up after, but let's just break the stuff down about how I should be doing it. Um the things that I'm going to be suspicious about to get the marks there.

[01:40:41]Well, firstly, obviously, I've got to mention about um the dark lines, the cause of dark lines. So I should know this is about the dark plans are on an absorption spectrum. I should say something about what that's going to be.

[01:40:51]So the those particular wavelengths are absorbed. Um why are there lines there?

[01:40:55]Because the star is going to produce a continuous spectrum. Um and I've got those particular waves are going to be absorbed and they get reitted in uh different directions.

[01:41:06]Um and they get absorbed by the atmosphere of the star or that kind of stuff. So I'm mentioning and again the mark's going to be there so you can just have a look at the full detail for that but just make sure that I know that first. Describe and explain at least one difference between the spectra. Now it's saying at least one difference. So this is kind of an indication that um I mean there's going to be maybe a couple of differences I can say. What's the most obvious difference I can see from there is that um I mean the Tali one, whatever that one, however that's pronounced has got more lines. Um Vega has got fewer lines. I'm just again I'm just going to say that first. Okay. So, if it's saying explain it, why has like why would one have more lines and why would one have fewer lines? It's because what if it's got more lines? It must be absorbing more different wavelengths. It's if it's absorbing more different wavelengths, it must have more um elements in its atmosphere in order to absorb them.

[01:41:59]Okay, so I've got that right now. Okay, that I'm pretty happy with that. I can also say there are certain ones that it's got in common as well. Maybe even though that's not saying about one difference, I'm just mentioning that just because I can see it. So, um I might as well say that I can say they both got whatever um element they've got. So, I mean they're probably going to have hydrogen helium. Um and then after that, yeah, um Tali has got extra elements as well. Then when it's asking me um afterwards um which star is older, well, how do I know which one's older? Well, if I've just figured out that Vega's only got the basic elements and Tal's got more elements, well, it must have the heavy elements. That means it must have had more time to fuse them. And then that's telling me that um it's probably got the it's probably going to be older. Now, there's something on the marking when you see it as well. I'll just go like this. Wait. Um hopefully that's made it come up now. Um that is saying except the idea the very bottom bullet point is saying except the idea that Vega may be older than TA if has more mass. And yeah, that's I mean that's true, but it would just seem like a really out of order thing that they're going to like put on you and saying, "Oh, the reason why it's got more elements is because it's fused more because it's heavier and fair enough. That's a reasonable explanation." But the fact that they're asking you analyzes to determine which one is older, that's the only reasonable reason why that one's going to be older.

[01:43:28]Okay. So, what's the indication in this question about how I'm going to have to do it? Well, the fact that it says use Kepler's third law. So I mean that's a pretty nice indication straight away that was kept third law I know that t ^2 is going to be proportional to r cubed and now they're doing something very nice here which I wouldn't expect to be fair is that they're saying object orbiting the sun at a distance of 1 a year has a period of one year that's something that you should know um but I mean they're nice enough to tell us that because that's giving you a reminder that oh actually so if I'm going to do this I can say t ^2= k r cubed because they're saying this is one and one if I keep my distance in AU and my time in years. Well, I can just say 1^ 2 = K * 1 cubed. Therefore, K is going to be equal to 1. And therefore, if K is equal to 1, I can just say T ^2 is equal to Rub.

[01:44:17]Now, I know that T ^2 is equal to R cubed, I can just kind of work out the next part of it as well. So, it says the distance of 3.5. Um, how can I do that now? So um I'm just going to say t ^2 is going to be equal to 3.5 cubed. I can say* 1 if I want. Um therefore t is going to be equal to square root of 3.5 cubed and therefore t is going to be equal 6.547 blah blah blah and I'll just call that 6.5. Um so hopefully this was a pretty easy question for you.

[01:44:52]But I just want to point out like they might not make it this nice in the real exam. So whenever they say use this law principle of formula, the first thing you do is just write down whatever that is and just get that mark. Um and when it's something about Kepler's third law, something like this, I mean you should just know this thing about that 1 AU is one year. Um if it's not around the sun, then obviously they're going to have to give a bit more information. But if it is something around the sun, you should just know that yourself.

[01:45:20]Okay. So when it says calculate kinetic energy, I'm all I would just go to the same thing. Firstly, I'm just going to say K= half MB² and just hope I've got those values. Have I got those values?

[01:45:29]Well, let's have a look. K is what I'm looking for. So that's fine. Half is a half m. Have I got the mass? Yeah, I have. There it is. Have I got V squ? No, I don't. Well, obviously it wasn't going to be that easy. So now I'm going to have to do something else to get V ^2.

[01:45:42]Well, I know it's in circular motion.

[01:45:45]Um, so how am I going to get V? Well, I know speed equals distance over time.

[01:45:49]And if it's in circular motion, the distance is going to be 2 pi r.

[01:45:54]And um what's the time? Well, I've just worked out here. That's why they made me work out the period. So in one orbit, it's taken 6.5 years. Um well, that's obviously in years. I've got to convert it into uh seconds. So 6.5 * 365 time 24 * 60 * 60. And I always just do it like step by step. I just go years.

[01:46:18]So that's going to be um days, hours, minute, seconds, job done. I just do that in my head to make sure I don't accidentally miss out at 60 or miss out at 24 or something like that. So I put that in the calculator and I get V is going to be equal to 1,692 m/s.

[01:46:36]Okay, so now I know this. I can just put that number back into this equation over here. So K equals a half * 10 times the number that I've just worked out. and remember to square it. I put that in the calculator and the answer I'm getting is 1.2948 blah blah blah * 10^ the 9. So round to two significant figures unless there's reason to think otherwise and that should hopefully be a reasonably simple five marks on that page. Okay, so I've got a graph for the next question and I'm using that same general principle that I'm not going to be happy until I've like taking reading from the graph and take the graph or something like that. So am I going to be able to do that here? when it says use the graph to calculate change GP. So firstly I'm looking at that the change in gravitational potential energy is it something that's on axis of my graph. Um well over here I've got gravitational potential not gravitational potential energy. So maybe that's a little bit of a trick in there that I've kind of got to be aware of. Um and we can have a look at that later. So even if I miss that we can see how I should be you know putting new checks and balances and stuff in to make sure that I'm getting it right all the time. Okay. So let's have a Look, I know I'm going to take reading from the graph. Um, and it says from like from the moment of collision until it crossed the Earth's orbit. So, I can see the change in whatever. So, change is always going to be telling me something take away something else. And then these are my two situations. Moment of collision until it crosses Earth's orbit. Well, what's the moment of collision? I've got earlier that it was um orbiting around at 3.5 AU.

[01:48:06]Um, so if it was orbiting around at 3.5 AU, I know that's where it was to begin with. So, I'm going to go down from 3.5.

[01:48:13]Make it very, very clear that I'm drawing those lines on there. It crosses here. Again, make it really obvious that you're going all the way across. We would obviously do it with a ruler.

[01:48:22]And just check my scale properly. So, if that's minus one, I've just got to be careful. That's going to be minus0.5, which means this is minus0.25.

[01:48:30]So, that's going to be fine. And the next one crosses the Earth's orbit.

[01:48:35]Well, like we said earlier, what do I know about the Earth's orbit? I know it's got to be at a distance of 1 AU.

[01:48:40]So, it's kind of a hidden piece of information, but really you should just you should know that. So, I've gone down from one over here. It's hit this line over here. I'm going across there. And again, just read the scale properly. And that's -0.9.

[01:48:54]So, now I can go from here and say, okay, well, I've got the change in this over here. So, the change in the VG is going to be equal to um -0.25 - -0.9 and that is going to be equal to 0.6.

[01:49:10]65. Now I might think, oh yeah, great.

[01:49:12]I've got the answer now. So let's just have a look. If I was going to put answer on here, so I can say, oh minus, sorry, the change is just positive anyway. So 0.65 is fine. Um, now let's check the units. That says gigles or these units over here. They're not the same. So obviously that can't be the right answer. So what have I forgotten to do in this situation? Well, this question is asking me about gravitational potential energy. And I've just worked out the change in gravitational potential, not gravitational potential energy. So, how am I going to get the energy from there?

[01:49:42]It's going to be the change in GP is going to be equal to the change in gravitational potential times the mass.

[01:49:49]Have I got the mass? Yeah, I got it earlier. It says on the question earlier that it's 10. So, it's going to be 0.65 * 10. And then the answer is going to be 6.5. And now, even if I weren't too sure about this formula anyway, well, I knew I was trying to get rid of a kilogram from that unit because I was going from gigles per kilogram and my unit was just in gigles. So, I would have had to multiply by some kind of kilogram, some kind of mass. What's the only mass that I had? It was 10. So, it must have been that anyway. Um, but I know from the formula anyway, it's got to be that. So, either way, um, I've got the right answer here. So obviously I said at the start like there's different checks and balances that you put in and the units is always something that you should be checking. So don't go steaming ahead.

[01:50:32]Always just double check the units because they are places where tricks come in. Um like they're popular tricks that they put in there. Okay. Um I've just got to remember this content basically because it's saying Kepler's first law. So it's saying what do I know about Kepler's first law? I know that um Kepler's first law is that they're all going to be elliptical. All the orbits are going to be elliptical and the sun is at one of the two folky. So if it's already told me it's elliptical, the other one is just going to be the sun is at one of the two folky two folky. And really um this is just Whoops. I'm just I've got to make sure that I remember it and I've remembered it, written it down and got a mark. I'm very very happy.

[01:51:14]Okay, so I've got a question here that's a potentially difficult one. I'm getting towards the end of the paper. I'm seeing a recession velocity. Okay, I'm seeing a graph that doesn't look particularly nice and I'm still using a general principle. Okay, I've got to take reading from a graph. Obviously, if they give me a graph, but um let's just see what I can do here. So, the first thing when it's saying recession velocity, that's kind of making me think of red shift. It's giving me the fact that the spectrum of hydrogen showed the emission line to be here. Um okay, so what can I do with this now? Is there a reading I can take from the graph? Well, in the um lab it showed that it should have been this. What's it showing from the graph?

[01:51:52]It's showing that I'm just this is meant to be the ruler, but it's showing that it's 654.

[01:51:57]So, I've got that number over there.

[01:51:59]I've got something about recession velocity. Is this leading me towards anything? So, I've got something about a wavelength. I've got original wavelength and and a new wavelength. I've got something to do with the velocity. Oh, yeah. Actually, this has reminded me that the formula that I've got to be using is change in velocity over um sorry, change in wavelength over wavelength is going to be velocity over the um the wave speed. So, I can just write that formula down.

[01:52:24]So, I've just identified the formula here, but just by looking at what I've got using the general principle of I need to take readings from the graph, taking that reading over here, even if I didn't know what I was doing to be fair, I would still make clear that this answer over here. So I'll say the wavelength is equal to 654 nanometers or* 10us 9 meters, however you want to do it because even if I can't go further with it, I'm thinking that maybe I'm going to get a mark for like reading the graph or something like that. I'm not sure. Well, from here, can I put the numbers in? Um I mean, yeah, I can. So the change in wavelength is going to be um 656.3.

[01:53:03]And I'm not going to put in the times 10 minus 9 because I know they're going to cancel out. that if you're not sure why, then just always do it. Like it's not going to hurt you to put them in. Um take away the um original wavelength and it's always divided by the original sorry, take away the um one I've got and it's always divided by the original wavelength which is um the one in the lab. So 656.3.

[01:53:27]Um I'm looking for V. So that's okay.

[01:53:29]And then have I got C? Well, this is a EM wave. So it's going to be 3 * 10 8.

[01:53:35]But what you need to be careful of is that if it's not an EM wave, you can't use 3 * 10. So don't just always assume that C is 3 * 10 8. Students do assume that, but C just means wave speed. So it's just whatever the speed of the wave is. Now, if I rearrange this, um, I should be able to get V is equal to 1.0513 blah blah blah 10 6.

[01:53:57]Um, what's the units here saying? It's in kilometers/s.

[01:54:01]So that is going to be to two significant figures. One point one I was going to give it to three nearly. Can I give it to three? Oh, I can give it to three maybe. Um yeah to be fair I can.

[01:54:15]So 1.05 time 103 km/s.

[01:54:22]Okay. So this next question, this was something that was in the arat as well at the start. So um it just said change this time 10 the -19 over here to time 10^ theus8. Um so that I mean you should have done that at the start anyway but we've just got it here now for the sake of it. Okay so I'm estimating the distance in light years um using Hubble's law and again they've told me use Hubble's law. So that's quite nice.

[01:54:46]I don't even have to figure out the formula but even if they didn't tell me well I'm looking for a distance. I'm going to need something with D. They've given me Hubble's constant H. What's the only form of D and H? Particularly because I've just worked out the velocity. It's going to be V= H N D.

[01:55:01]Okay, what am I looking for? I'm looking for distance. So, distance is going to be V over H. I've put those numbers in.

[01:55:09]So, V is um 1.0513 dot dot dot * 106 divided by H. I'm just going to check the units and that's standard units S minus one. So, I don't need to worry about that. So I can just put that in as normal * 10us8 and put that in the calculator and I get the answer 4.205 blah blah blah * 10 23 and because I checked all of my unit seconds and I made sure I converted that back into this one back into the proper units over here. I know that this is going to be meters but it's asking me for it in light years. I'm always like we said before always checking those units. So, I look at my um formula sheet for the conversion of light years, and then I know I've just got to divide the number that I've just worked out by um 9.5 * 10 15. And I'll put that in my calculator.

[01:56:02]And the answer that you should be getting is 4.4 * 10 7.

[01:56:07]Um so, yeah, I'm happy that I've got that. And again, I didn't get tricked by the units because I'm in that habit of always checking the units on the thing.

[01:56:15]I think I've identified the formula correctly by doing exactly what they've told me. I'm just doing all these little things that's going to add up over the course of three papers, it's going to add up to a lot of marks.

[01:56:25]Okay, so I've got a percentage uncertainty question here. So, I know this is always going to come up every year and there's different ways that you can do it, but if we're just consistent with the same way that we always been doing it, um I'm going to get the marks for it and that's what matters. So the way that we're always a to do it just make sure that when we're doing percentage uncertainty always do the biggest take away the real value um the biggest take away the accepted um divided by the accepted. Now there are different ways that we said like I said you can do one but just always go that way and you'll be fine. So biggest takeway accepted. Now what am I looking for the percentage uncertainty in? It says percentage in uncertainty in the distance and it says using the width of the emission line in the diagram. Well, I've got to be careful here because I'm looking for percent into the distance.

[01:57:09]But what does the information in the diagram give me the wavelength? So, how am I going to get the wavelength into a distance? Well, how did I manage to calculate the um the distance earlier?

[01:57:21]Well, I from the wavelength the change I did the change in wavelength to get um speed from this formula over here. And then once I had speed, I used this formula over here in order to get the um the distance. So let's have a look at how I'm going to be doing it. So I how am I going to get the biggest one is what I should be saying. So in order to get the biggest distance, I'm going to need the biggest speed. And then in order to get the biggest speed, if I rearrange that this equation up here to say V equals this change in wavelength over wavelength time C is going to be equal to V. So how am I going to get the biggest speed? I need the biggest change in wavelength. So now that's helped me with okay, that's how I'm going to get the biggest. So how can I do that? Well, let's just take a look at the information I've got.

[01:58:05]So um I need the biggest change basically. So this is the real value that I've got over here. And then it says using the width of the line. So basically I took the middle one when we took the real value. This is the one that I think it is. But that line is going here. So that's like 653.6.

[01:58:23]And then over here I'll zoom in as well.

[01:58:24]So if we go down there, that's like 654.

[01:58:30]Yeah, that's right. Um, so I've got 653 Whoops. 653.6 here and 654.4 here. Well, this is my real value over there. So, if I want to get the biggest difference, these are both smaller than that value. So, I'm going to need to take the smallest one basically. So, to get my biggest change in wavelength, I'm going to have to go 656.3 take away 653.6.

[01:59:00]six. Um, so let's do that down here.

[01:59:04]Okay, so my biggest V basically is going to Oh, I might have to rub some of this stuff out to have space. Um, so let me rub all that out actually.

[01:59:14]So I know that change in wavelength over wavelength time C is going to be equal to V. So my biggest V, like we said earlier, is going to be 656.3.

[01:59:25]my real one divided by the smallest one.

[01:59:28]It's not divided by take away the smallest one divided by the real one and then multiplied by v. And then I put this in my calculator and I get my v as being 1.23419 blah blah blah * 10 6 m/s.

[01:59:45]Okay, so this is my biggest V. Now I've got my biggest V. How am I going to get my um biggest distance? That's why I know that V equals sorry distance equals V over H kn. So I'm going to put my biggest V into that equation over here.

[02:00:01]If they gave me like uncertainty with H knot because sometimes you have to work out H knot from like a gradient and stuff like that. And then you might have to work out the biggest H knot and the smallest H. Well, if I want my biggest D, I would have to take my biggest V and my smallest H. But in this case, they just give me one value of H. So I'm just going to use the one they've given me.

[02:00:18]And now I know that the maximum d is going to be I'll put it in the calculator uh 4.9 36 * 10 23. Okay. Okay, so now I've got my maximum D.

[02:00:34]And remember when we're doing the percentage uncertainty, I'll do that. Percentage uncertainty is always going to be the biggest take away the actual or the acceptance take away the um actual value divide by the actual times 100. I might have forgotten to say times 100 earlier. I hope I did. Well, I think I did, but times 100. Okay, so I'm just going to put these numbers in here.

[02:00:56]So the biggest is this one over here. I haven't really got space to write it, but the biggest is that one over there.

[02:01:01]Um, the actual value is this one over here that I worked out earlier. No, it's not. It's this one over here that I worked out earlier. It's still not, is it? Um, because that's in light years.

[02:01:13]No, it's the one that I did first.

[02:01:14]Sorry. The one in meters. So, it's this one over here.

[02:01:18]Um, so I've got to put those numbers in there. I might as well do it now just in case I confuse everyone by doing that. I don't want to record this part again. So 4.936 * 10 23 minus 4.205 * 10 23 divided by the accepted value which is this one. So divided by the 4.205 * 10 23. Now like before we've said because they're all times 10 23. I know it's just percentage. I don't really have to put those in. But I put this in the calculator. I'm going to have to rub that out to make space. And the answer that I get is 17.4023 blah blah blah. So I'll have to write down I suppose 4023 blah blah and then round to two significant figures. So the answer that I should be getting is 17% and yeah I'm happy that I've got that one right now. So like I said there are different methods for percentage uncertainty but always go with biggest takeaway actual um and identify how you're going to get the biggest in the way that we've done. So this one's maybe slightly harder than some of the percentage uncertainty questions because there was more than one formula I had to use and each step of the way I just kept in my head how am I going to get the biggest? How am I going to get the biggest? how I'm going to get biggest.

[02:02:26]Okay, so how am I going to get this next one over here? Um, suggest why M87 might contra Hubble's law. Well, the first thing it's saying is Hubble's law. So, what is Hubble's law? I I know that it's something to do with um V equals H, not D. So, Hubble's law is saying V is proportional to distance, velocity is some processional velocity proports that. So, there obviously must be something that's saying it's not that.

[02:02:51]Now I'm not too sure what it is yet. I'm finding this pretty difficult. So I'm just going to go. So this was the law principle of formula. I'm still finding it hard. So I'm going to go to the state of the obvious. So what can I say from the situation? Well, this is the um all the stuff that they've been leading me towards. That's what I've been working out. And this was the actual wavelength over here. So the actual wavelength was 653 uh sorry 656.3.

[02:03:16]The real wavelength that I got was 654.

[02:03:20]the real wavelength is shorter. Well, what do I expect the real wavelength to be? I expect the real wave to be longer.

[02:03:24]It should be moving away. It should be redshifted. So, the obvious thing that I can say here is the um the wavelength has been blue shifted, not redshifted.

[02:03:33]Now, straight away that's thinking okay maybe I am you know that is contradicting Hubble's law now because Hubble's law says that everything is moving away. The recessional velocity proportional to the distance recessional means moving away. Well, if it's blue shifted, it's not moving away. Well, I've just got to make sure that I'm linking that here. So, that's been blue shifted, therefore moving towards because Hubble's law says recessional velocity. I'm just going to make it clear there. Kind of link it back. So, Hubble's law says everything moving away.

[02:04:08]Okay. Now, it says suggest why. So, I just need to give any sensible reason why that might be the case. And normally the reasons that you're going to have is like local gravitational influences um because it's so close that it's actually being attracted from the force of gravity and that um it's not moving through space as much with space expanding. It's more moving um you know with gravity as in through space rather than with space. And the further something gets away the less likely you are to see that because the further something gets away the more space is expanding. So even if there are like gravitational influences, it's not going to really be anything compared to how fast these things are moving away. Okay.

[02:04:49]So I've written a little bit over this last question, but I mean we can hopefully read it anyway. So suggest one reason why the Big Bang Theory is still the accepted model of the origin of the universe. Oh, this is nice. This is just a one mark question to finish us off.

[02:05:00]Why is it accepted model? And this is even a GCC question to be fair and it's content that I just need to remember.

[02:05:06]Um, I know that because we CNBR is the thing that made us accept the Big Bang Theory. Um, so that's going to be fine.

[02:05:14]That will get me the mark. But if I just want to make it um linked to the thing that I said above as well, it's like very few um galaxies are going to be redshifted. It's only the closest ones that are redshifted. All of that kind of stuff um are blue shifted.

[02:05:31]It's only the closest ones are blue shifted. Most of them will be redshifted. And yeah, I've definitely got the mark with this one. And now look, we are at the end of the question paper. How exciting.