Hallo miss ss. Hello, Jackson. Great. Good. So we have these equations as our tools are soovaequations. So there's four that are given to you and we'll focus today on past paper questions, questions that can come up, which can be solved using these equations. Okay, so let's have a look at these. I'll just close my door. So the first question we look at is a multiple choice question. So it came up, I think, in 2024. So a ball is thrown vertically upwards at a velocity of 6m per second. Which of the following gives the maximum height in meters reached by the ball? Upwards acceleration and height, as if could you with velocity through vertically upwards through maximum height, final velocity is zero, initial velocity is six, and the acceleration is -9.81. And the maximum height and that is divided by a six plus zero, divided by two times nine point -9.81. S equal to. No v square equal to U square plus two as. So if v is zero and that is equal to six square plus two times, Yeah that is minus two times 9.81 times s. So that is the first one. Six square divided six square red over two times, 9.81. Well worked out. Well done. Okay, good. You could work it out as the kinetic energy will give you the gravitational potential energy. So your mouses cancel. So again, you can work out H and it rearranges to be the same thing. Here's another multiple choice question. So a ball rolling along a table, it has a, the height of the table is 0.83m. The ball has a horizontal velocity of 1.72. 1.72m per second. North Point 83m height. Yeah, so the x ration for the speed of the ball in meters per second is the floor. So Yeah, so the height is not going 83. So that is hi, Yeah, that is s distance heavy redisplacement. So the horizontal velocity one point. 72 at force of the edge of the table as it so that is initial velocity. So if that is your v square equal to U square plus two A S and an a acceleration is 9.81 so v square equal to U square plus two as and v square is 1.72 square plus two as two times 99, 9.81I choose see. Good, so c is the correct answer. Good. So those were similar. Again, multiple choice from a past paper, a ball is kicked, giving us an initial velocity at an angle of 35 degrees to the vertical. The ball remains in the air for a time t, which the following gives the distance travelled over the ground. Initial velov at an angle of 35 degrees and the ball remains air of t. So this one. Oh, Yeah. The distance travel over the ground. So. The ball remains the air ir for a time tee. Over the ground to the ground, the horizonso, this distance is what you're being asked for. So. Not one tensions. No. Yeah. Cosine 35 but. The. Travel distance over the ground. So you want the horizontal component of the velocity, velocity times, time will give you distance, won't tage. Yeah. So what's the horizontal component? Ph. This is the vertical component. This one. Will a truth be? No, no, no, no, no, not be, not be me. Yeah I see. Why is that? It's sign sigthirty 35. So vt sine 35. Yeah. Initial vertical velocity. Sign 35. I suppose. Pt sign 35. Distance travelled over the ground, I suppose, as the time spent in the air. I would have said it was a because it's this, I suppose. We're given the vertical velocity. The vertical angle. Vt. Or sign. It depends what v is, I suppose. Sign 35. Let's see if we can do something else that's similar to it. Okay, here is a longer question from the past paper from 2022. Okay, I'm still understand that question. This one. Yeah who one? I suppose we assume that v is the horizontal velocity. This is the horizontally the horizontal velocity, and it's moving at this angle and it remains for time. T. 9:35 so. So which sign is sign 35? That one. So. Sign is opposite over. Hypouse. Oh H horizontal and o Oh is so if we did it like that, this is v vertical. Here and v is the hypotenuses v. V and the opposite is vv. The vertical component is the opposite. So that's the same as that. Okay, Yeah, I'm on the now. Okay, so a longer question. In paper one might look like this. So a trepochet is like a catapult that they used to use in ancient times for attacking castles. So they put a rock here, and it was by pushing this counterway down, it flung this rock into the enemy's castle or whatever. So a tripichet, a medieval catapult to design, to project a rock over large distances, eh? So this is a. So they released, they held it there, they loaded the rock, and then they took away the peg that held it in place, and the counterweight moved down. And this was flung that way. The rock is held in a sling. When the pig is removed, the counterweight falls and the rock is projected through the air. A student assumes that all the gravitational potential energy transferred from the counterweight is transferred to the kinetic energy of the rock. So the counterways had gravitational potential energy and that they're saying is equal to the kinetic energy of the rock as is it's launched. Give two reasons why this might not be the case. Why might not all of the gravitational potential energy be converted to kinetic energy? Not equal to. Yeah so if that is not a core ls, so I should prove that gravitational potential energy not equal to kinetic energy. So we don't have any numbers. We just have to describe two reasons why not all the gravitational potential energy is converted to kinetic energy. Why that is not equal, I don't think. This equipment has the terminal overload terminal al energy, Oh, sorry, thermal energy. Thermal energy Yeah there would be friction in the moving parts. So that's a good reason. Some gravitation potential energy is converted to thermal energy or internal energy. Due to friction thatget you one mark. What could we say as a second reason? That is not correct. So why is all the gravitational potential energy not converted from the counterweight to the rock? That's specifically what the question is saying. Why would we expect what else has to happen in the system, in the machinery? When this goes down. Are we just lifting the rock, Jackson, do you think? The rock. Well, rock. Some seven rock. The is the rock the only thing that's been lifted. Yeah. If you think of it as. As the counterweight goes down. This bit goes up and flings the rock over. So we're not just lifting the rock, we're lifting the whole beam as well. Does that make sense? Wait, let me say that. Let me search what what means of that word. Which word? Tritriokay tribute miss. Was that? What? Tribute chat. Yes, tribute. Okay. Yeah, I know. What's that? So it's like a point point. Yeah, maybe. Okay. I think it's just through the rock. So as this enter rate goes down, the rock will go that way. But also because we take this, remove this fastening, we lift the beam as well as the rock. So the mark scheme says. The beais also lifted. As well. So some of the gravitational potential energy is used to lift this bit as well as the rock. So that would get you two Marks, okay one mark one mark. The rock was projected with a velocity of 41.5m per second, which miss, but why the thing is also lifted, as well as the rock is one mark. Why that is not correct. I mean, why that can prove this? This conclusion is not correct. If you like, we're lifting the machinery as well as the rock. When this weight goes down, this rock goes up. But also this bit because you remove that holder. So the whole energy, the whole connectic energy, the being takes part in, takes up some part. Okay? So that is not the old connecinetic energy for rock, okay? Okay. Okay, next bit of this question. So the rock was projected with a velocity of 41.5m per second at an angle of 30 degrees. To the horizontal. The diagram shows the flight of the rock after it left the trepiche. The rock rises to a maximum height and then falls back to the same height as it was released. Calculate the orihorizontal distance travelled by the rock in this. So what's the horizontal distance? How would we calculate this? Distance. So firstly, we should calculate it the horizontal, the horizontal distance. Personvelocity angle, okay. And next mahigh it so we should calculate the horizontal distance, horizontal distance, horizontal horizontal, how far it went. Yeah, Yeah, Yeah. Okay. We can calculate the vertical velocity and and we know the acceleration. And in the high, in the maximum height, the final velocity is zero. It is stationary. So the Velso, the acceleration is downwards. It's 9.81. But I should calculate the. So the vertical velocity is 41.5 times sthirty. Yes. And equto 20.75. What 1.5 times cosine 30? He equyour 3930 519. Good. So this one acceleration is downwards. So that should a equal to v divided by t so t equal to v divided by a, 20.75, divided by 9.81. That's the time. And it's 2.1s in the same time, 35.9, times 2.1 and times two. And that is the 35.9 times 2.1, times two. 150.78. Good nayeah. Meters. 115. The MaaS of the counterweight was reduced. So the last bit of this question, this MaaS is reduced. It gets less. The tripichet was then used to project the rock again, explain why the horizontal distance traveled by the rock decreased. Zona distance travelby decreases, MaaS of reduced MaaS of the counterweight was reduced, the tribute chat was then used to profy the rock again. Wait, the five Marks. One, the horizontal distance traveled by the rogue decreases. This. The MaaS of the counter Yeah the whole process is the gravitation is the connecting energy transferred to the gravitational Oh no, no, let me say again, the counterweight down moves downwards and then rock moves upwards and through it. So that is the gravitational potential energy transferred to the kinetic energy. And the equation is mgh equal to mgh transferred to the half mv square. So if the so if m decreases the gravitational potential energy with will decrease and the and there will be less kinetic energy Yeah, that is if that is equal. So that is less kinetic energy and that will be less velocity. Good. So that is the constant velocity. So s equal to s equal to V T minus a half vt minus a half. You a half A T square. Yeah. So if the final veloif, the if if the velocity in the maximum height is zero, so that is minus a half 80 square. Maybe not, maybe, maybe not, this this equation, maybe not, not in this equation. Well, you can mention time, because if the velocity is less itspend less time in the air, but I should use another equation, for example, v equal to v equal to s divided by t maybe, but that is in constant velocity, U plus A T. Remember, we consider the horizontal velocity as constant. Horizontal velocity. Well, Yeah. The horizontal that is constant. So Yeah, v equal to s divided by t so s equal to U T. So if v so if v decreases, the t will be. Longer I mean t will increase and then the horizontal distance will decrease. I'm a flight will be less, horizontal velocity will be smaller, so the counterwage travel transfers less gravitational potential energy, so the transfer of kinetic energy to the rock is less. Rock has smaller vertical velocity, less so smaller vertical motion, less time in the air, so horizontal distance will be less. So that one, two, three, four, five. So what you were doing very well, we would at least have got three or four Marks because you were working through what you knew. Five, six, seven, eight, nine, ten, eleven, twelve. So that was the twelve mark question. Jackson, okay. We were looking at this question. Last day as well, yesterday, we had just started this question. This is also at the back of paper one. In the game of badminton, a shutleccock is hit by the racket. The graph shows how the vertical displacement and horizontal displacement of the shutlecock vary from the moment it leaves the racket. Positions are recorded every 10s. Every point one of a second. So we're talking about a shuttle cock. So this is. Vertical displacement, horizontal displacement, and this graph shows how the velocity in the vertical direction varies with time. So the velocity is highest initially, and then it reduces and then it acts in the opposite direction. Explain how the velocities have been calculated from the success essive vertical positions of the shuttle lecock. Remember, we had this, that instead of every point 1s, what they did at 0.05s. So they took the hm. They calculate the difference in the position on the y axes used. Velocity is distance divided by time. Two to 3.1 so that's 1.1 divided by 0.1, 0.1 divided by 0.1, Yeah. But is and where's 1.1 and not point 12 to 3.1? So that's 1.1 there. From there to there is 1.1. And this was at time t is zero time t is 0.1s later. 1.1 divided by 0.1, but how we can get one 0.1. I'm from the question. Every note point 1s Oh Yeah okay. And when velocity so they get eleven. That's eleven. Let me see that. But once that is 1.1. Okay. I didn't think this was a very nice question. Explain how the velocities have been calculated from the success of vertical positions. So they took the change in vertical displacement over time. Delta y over t statewide. These velocities have been plotted at a mid range of the time interval. So we did this yesterday. They were taking the average velocity. So there of the time average. State with the reason two pieces of evidence from the graphs that show that the shutlecock does not follow the motion of a projectile moving freely under gravity. Why, Jackson, is this not evidence of projectile motion? Like this is a perfect projectile motion, isn't it? What's the evidence that this shutlecclock is not moving in perfect projectile? Don't know. Now look at this, Jackson. Yep, I'm looking look at the graph ss. And look at this graph. What's the difference? The difference is this not Yeah not this graph that graph Yeah this graph this object does not back to the ground. Maybe there's another floor or grounds can get it can. Okay, please. Maybe it's higher place. I don't understand, Jackson, what you mean when you kick a football, it goes in a parabola shape. It follows. This path. So this so now this questions object gives you a higher place this doesn't okay. Yeah it goes to a high place and then. See, I mean, the nature of this object, it's not like the rock from the last question, the rock, which the trepechet threis a uniform solid MaaS, whereas this has greater MaaS here than here at the feathers. So it's a non uniform object. So its flight pattern is a bit different to a rock. The rock would show. This path, whereas our shuttle cock shows this path. Is this symmetrical, Jackson? Well, Yeah, I know some the object should ot back to the original, or maybe not the original place, maybe they were back to the ground. And Yeah, but I'm not very understandfor this graph. Two pieces of evidence from the graph that show that the shutlecock does not follow the motion of a projectile moving freely under gravity. So. This is a non uniform MaaS, isn't it? It's a non uniform object because this bit is heavier than this bit. These feathers are affected by air resistance more. Then the front bit. So this is none the meat maybe this so miss is that so this graph is a shateau coke. Yes, Yeah, maybe this shayeah okay, I understand this graph well, this graph maybe. Means that in the projis, that projective projecle projecle Yeah projectile, it should be on curve and and the distance and the distance before and after getting high, the maximum height is same. But maybe this graph is different because the horizontal distance after maximum height is lower than before. Okay okay, this one okay, so maybe there's air maybe there's as air resistance or maybe the kinetic energy not equal to the to the gravitational potential energy maybe because of the thermal energy caused by the feather and air resistance. Yes. So the curve should be symmetrical if it's projectile horizontal distance after to maximum height is which is not the same due to air resistance caused by feathers. Perhaps. The horizontal distance between the points sphere is a graph is not symmetrical around the vertical velocity, time graph is not a straight line or gradient of velocity, time graph is not constant, so the gradient is non uniform. It's steeper here than here, so it's not got the same velocity going up as going down. The horizontal component at velocity should be constant, acceleration in the vertical direction should be equal to g. Yeah, so. So acceleration due to g should be constant, though it's not. So there's a few possibilities we could have for that. But the key thing is it's not a symmetrical graph. G is not uniform. So it must be due to the air resistance, due to the non uniform nature of a shuttlecock. Okay, Jackson. Yeah show using the velocity time graph that the maximum height gained by the shuttlecock is about 3m. So the velocity time graph. So this graph show that the maximum height. It's about 3m. Meters. Maximum height. How do you get distance from a velocity against time graph? Well, that is a able variable, sorry, variable velocity. Yeah maybe there is maybe we should use sweats. So. The velocity time. They in the question, they want you to use the graph, okay, so we have to use the graph and the distance. From a velocity time graph. Will be the. Area equals to the area under the graph line. Okay. Sorry, distance times equal to. Yeah distance. So I'm just going to get rid of this. So when you don't have a perfect rectangle or triangle, what you do is you count the full squares and then the part squares. Okay, so 1234567. So we have seven full squares. And each square is 2.5 but by 0.1. And then we count the part squares and make them half squares. A, B, C, D. He has. Six. You know maybe we kind of shift to a. Shift some parts to together and make the whole triangle. Yeah, well, that's what you do. A, B, C, D, E, F. So I have six half squares and seven full squares. So three and so so that's ten times. Point 25 times 0.1. So I guess. Oh, it's 2.5. Isn't it? 2.5 that makes it better? Yeah, two point ten times 2.5. Times 0.1. Ten times two point, five times point one gives me. 2.5m. I could have taken this square as well. So maybe 2.5. Ten times three times 0.1Yeah. So you do a rough, you count the full squares, then you count the part squares as half ones, and then you estimate ated that way. But the distance travelled is the area under the graph line. So we can get 2.5m, which is nearly 3m. Okay, Jackson. Yeah so Yeah, maybe I should. Yeah 2.5m. But why we use six divided by two? These purple numbers. I mean, six divided by two. Why we use that? That's a full square in purple. Give me a minute, please. In orange, these are my part squares or square. Okay, so A, B, so a should be like that. A is just cut off there. A, B, C, D, E, F. And I could have had maybe g there too. So I could have had the part squares are multiplied by a half. So one, two, three, four, five, six, seven. I have times seven is 3.5. I'm getting a different number now is however, you drew your your curve. Let's get rid ash one when you're estimating something. You have clearly full squares here. But then you have the part squares, which you have to make up. So I count the part squares and multiply the number of these by a half. Multipses by half so we should so that six let me see. So they are six. So is there a. So Yeah. There. So when I meet, so how do you draw this triangle? So and I know and I know I see if there and. I can connect it to the y axis. Yes. How to? Yeah, I'd do it like you were doing. Just okay. So Yeah they are different, so I can amount amount. So. We want the area under this graph, but because it's not a rector triangle, it's a rectangle. The method you use with graphs is to count the whole number of squares and then the part squares. So. Hosquares 123456. Seven okay seven hole squares and then the part squares. One, two, three, four, five, six and I could take this one step Yeah. So six divided by two. Okay. Yep, got it. Okay right Jackson, we'll continue these tomorrow. Okay okay bye bye.