Hello, miss. Hello, Jackson. So today I thought we would look back over everything we've learned about materials. We'll check that you can recognize the equations a paper, and then we look at some questions which are not just practicing one equation. So we'll look at a question, we'll see which topic it's about and how we'll solve it. Okay? Because this is what you will have to do in may. So paper one is mechanics and materials. So we learned about density Stokes law. So density Stokes law, Young motulus stress strain. So the Young modulus is the stress divided by the strain. That's what we were practicing yesterday. And hook law f is equal to K X. So these questions, I would imagine youget one or two multiple choice questions and probably one longer question on a paper. Okay, so these are questions with answers. So we look at this question. Okay, are we ready? I need to Yeah, ready. Okay. So a student is investigating a Cartesian diver. The diver is made from a plastic pipette. When placed in plastic bottle full of water, the diver rises to the top and touches the lid. So a Cartesian diver is like, it looks like this. So this is water. Show that the downward force of the lid on the diver is about 2.0 to nutions. So we know the volume of the diver. So we know the volume of this space, the MaaS of the diver, the density of water. So density is MaaS over volume and weight is MaaS times g. So you calculate the upthrust. So the upthrust is equal to the weight. So density of water times volume. Volume of water. MaaS over volume. So that's the first thing you do. So the volume times the density will give you the MaaS. So MaaS equals density times volume. Then weight equals up thrust. Equals m times g. So the. Weight equals up, thrust equals m times g so the lid force. Is 0.0785-0.57. The upthruthe weight, so the resultant force on the lid is 0.2 newtons. When the pressure is squeezed by, when the pressure is increased by squeezing the bottle, water is forced into the diver, increasing its weight. The driver sinks, then remains at rest in the position show. So if you squeeze the bottle, more water rushes in here, so it will fill it. So itbe heavier the volume of air in the empty bottle. In part a was this show that the volume now occupied by the air is this so. F is equal to phi rho g, so. F is equal to. The row g. The volume of air is that the volume of air so up thruis V G, so the new volume of air is that okay? This next question is a bit more like what we've been practicing. So this is question two. A school dynamics trolley has a plunger inside to a spring. When the plunger is pushed in, the spring is compressed. So we're compressing a spring. When the plunger is released, it is back. It is pushed back out by the spring. A student investigates the spring to determine whether it obeys hooks law. In compression, the trolley was placed vertically in front of a scale, and weights were added in turn to the top of the plunger. As shown, the position of the end of the plunger was recorded each time, so the more the weight, the more the spring should compress. So to work out compression, how do you think we would do this, Jackson? So. Half of weight against the compression. So I think the weight is equal to compression. Maybe not so remember with hook law. When you add a weight. You don't measure the length of the spring, you measure the extension of the spring, don't you? So we want to know the extension of the spring when we add a weight to a spring normally. So this is the extension. But this time we are compressing a spring, so this time the spring is in here. And it is getting shorter as we add weights to it. So this is the position. Without any dy waits on the spring. No compression. If we add two nutions boshes the compression. If we subtract this from this, we get the compression 0.3. And our units will be centimeters. Again, we subtract this from this so six by 64. So this will be 0.7. This is the uncompressed position. So to work out how much the spring has decreased in height, we subtract this from this, then this from this. What is the compression of this one, Jackson? 1.1. Yeah. Good. And this one. So 1.4Yeah and 35.5. 5:36. 1.8, 1.8. So it's changing up at about 0.3 or 0.4 each time. To no point four. Each time use your results to plot a graph of weight against compression. You may use the additional column for your data. Okay, so compression, I didn't use the right bit of information when I plotted this, so we might need to plot graph. Anyway, you should get a straight line. 12.3. If we made this 0.2, 0.4, etcetera. So I just plotted this graph using weight, using the compression, and you can see the student concluded that the spring obeys hooks law with the spring constant of about 600 nutrions per meter, determine whether the student's conclusion is justified. So f equals K X. So instead of using the graph, how could we prove if f equals kx is obeyed for our spring? How could we prove that k is about 600 nutions per meter? So this is percent measure. So we should. You Yeah. So if we take. If we substitute some values equals k times extension 1.8 by ten to the minus two. So if we get a value for k, so ten divided by 1.8x minus two is so gives me 555. Nutrients herniche. So I use this data, choose another set of data. And see what value of k you get, Jackson. My answer is. 556. Yeah 556 okay. You're asked to prove that the spring constant is about 600 nutrions per meer. So if we repeat this with a different set of data, and we get about 600. So if we take this set of data, so six divided by 1.1x minus two is five, four, five. So using. This set of data I get 545. Mutions per. So. So. Yeah. So we can say it's justified. Okay? Another trolley was adapted by placing a tube around the plunger so it would be used to launch marbles. So you compress the spring and then you release it and the marble shoots out. A marble was placed in the tube while the plunger was depleted. And the pluncher was released. It launched the marble, determined the maximum possible launch velocity. When the screen is compressed. This one, maybe we can use simultaneous equation to solve this. We can firstly, we can use a spring, we know spring constant, that is k MaaS of the marble. That's the m MaaS of. Wait, wait, let me see the structure. Yeah, and I know that it's hooks low f equal to kx and. Tim marvel. Terrible and plger so maybe the m is the sum of marble and plger. So if you think of it, we're told how much the spring is compressed. So it's compressed 5.4. We know the spring constant, so we know the force. So 610 times 5.4. I tend to the minus two. Will give us the force. So 61 zero times point zero 54 is. There's she 2.94 nutions. So Jackson, when we compress the spring, we've used this force. And when we've compressed the spring, we give it elastic potential energy. So we store elastic potential energy in the spring. Look slow. Elastic strain energy, see this equation a half f delta x. So. So the straight energy stored a half f times delta x. So that's how much energies will be stored and how much of that will be converted to kinetic energy. And we can work out the speed then. So a half times 32.94. Times. North Point north five, four, 5.4 cm. So when we work this out, we get 0.9 joules. So this will be equivalent to the kinetic energy. Equals 0.9. Equals. A half. We know the MaaS of the marble. How much velocity will the marble get? Times naught. Point naught. Naught for one. The MaaS of the plunger. Let say v square plus v square red. Plus the MaaS of the thing that pushes the spring down. So that will be North Point north. 35.43 35.4. Times V I squared, as you said, times V I squared so we can work out the launch velocity. Do you see why we've had to add we're launching the marble and the plunger. The plunger pushes down the spring, compresses it. So. B will be equal to. Point nine times two. Divided by going zero zero 41 plus point zero 354. 4.9m per second. I get 6.75. So 0.9 times two divided by all of this is v squared, then square root, my answer, so I get 6.75m per second. Seven, five. Okay. The launch velocity was measured using a light gate and a data logger. This produced a smaller value for the launch velocity than calculated in b. If a reason why this method produced a smaller value for the launch velocity. So if you had light gates here. It's a sensor. Why do you think the launch velcity? Might be less. So the answer says that there will be some. Friction between 800 ger and sides des. Is air existance? The marble. Will decelerate. As it is, watch. Launched perhaps. Okay. The last one is the marble will accelerate as it's laoh Yeah as it's going up, it's already slowing down. So those are three possible answers you could put there. So that's quite a nice question. Instead of stretching a spring, we're compressing it. Okay. This next question is about viscosity of a steel sphere as it's falling through glycerol. So if you have a container with glycerol, which is like oil, it's quite thick. It's like engine oil. The student drops a steel sphere with this radius into a cylinder of glycerol. The sphere reaches terminal velocity and takes 3.9s to fall half a meter. So you could. Half a meter so we can measure the speed. Calculate the viscosity of oil, density of steel, density of glycerol. So which equation do you think you're going to use here? Reason maybe density zero equal to m divided by v. And we need to use stoke law because this quantity eater is viscosity. So eter, okay, Yeah, maybe it's ducslow. The different intensity of the materials. Density minus density of. Steel. Over. Six. My. Or it. So. It's just hard to make out. So steel minus glcerol or the liquid. So and we need fee equals distance over time. And we need the volume of a sphere is. Four over three pi R cubes. So. So we know the radius, so we can calculate the volume of the sphere. Is four over three pi R cubed. The sphere reaches a terminal velocity and takes, so we can work out the speed, so distance divided by time. And then we can use this relationship, which we've come across before. Let me find the fluids. So two R squared g of density of steel minus density of the fluid terminal velocity. So the viscosity is. So. V is equal to. Vg. Density of steel minus density of liquid over. Six I. Hr. So this will give you the velocity. We work out the velocity there and we can get that. So. And this is the volume, this is the density of steel which we know there. Here's the density of glycerol and put it over six pi l R and we get e ter will be equal to 1.8. Ask Cal seconds. Ask. Cal seconds. So that's quite sure. There are two cylinders available for the student to use. One cylinder has a diameter of 1.5 cm, the other has a diameter of 5 cm. State and justify whether the cylinder, which cylinder the student should use in order to gain a more accurate value for the the viscosity of glycerol. Is it better to use a wider cylinder or a smaller cylinder? What do you think if we're dropping our sphere? Is it better to have a bigger diameter cylinder or a smaller diameter cylinder? What do you think. I think. We want to drop this. The bigger one, yes, because. You get if you get the sphere hitting the sides of the container, it could lead to non laminar flow. It could mean you get some turbulence. The the sphere will be affected by the sides of the container if it's falling through glycerol. So it might not flow with laminar with a. A laminar flow. So the 5 cm diameter is better because you get a laminar flow in lares. So this one is better more. Laminar flow. Okay. More leminar flow, that means more liquid flow. It means more quate. It means your ball bearing will flow uniformly through the liquid. It won't experience hitting the sides of the container. It won't have a turbulent experience. So. This equation only applies if we have a laminar flowing fluid, a fluid that allows our ball bearing to flow in laminar conditions. So the drag force is equal to six pi viscosity times the radius. So if we're doing an experiment to prove Stookes law, we drop a smooth ball bearing through a viscous liquid. If this ball bearing hits the sides or if it is going to be affected by anything. That could give us turbulence. This relationship here will not be. When you go swimming, does your school, Jackson have a swimming pool? Oh, I actually, I don't know, as there's a swimming pool. I haven't. I haven't going swimming, not gone swimming. When I go swimming, I like to swim in the middle lane because the side lanes I go slower, but in the middle lane I go really quickly. So it's the same thing. The ball bearing, if it's flowing near the sides of the container, might experience a bit more friction. Okay, I hope you can swim. Can you swim? I'm not sure about that because I'm not sure whether my tonyeah, whether my tongue is injured or not. I'm not sure about that. Okay. But if I but I'm maybe 80%, 80% sure I can go to swimming, okay? When I swim, I like to swim in the center of the swimming pool, not at the sides. There's less turbulence there. Okay, let's read this question, Jackson. A force was applied across the ends of an iron bar. The following stress strain graph was obtained. Stress strain, the shaded area represents what. Strength and stress earth are strength and strength, and the area area is not the gradient is spring constant, but the area is. Strain times. So that could be. So Yeah. So if we multiply these two quantities together, f over a times, change in length over length. I'm multiplying this by this. The area is that times that. So the force over the area times the change in length over the length. So that's equal to force times distance. Force times distance. What's that equal to? What quantity is equal to force times, distance for moment? Maybe not no more times distant, maybe spring constant. So it's equal to the work done. Work is force times distance and area, which is length times length times length. What is area times distance? It's like length times length times length. What quantity do we get there, Jackson? Volume W over W volume. So which witches are I choose first one? Where did the two come from? The area. I have Yeah okay, the answer is b. Yeah. So I have. A half times work times volume. I mean, I think mathematically, you're right, it should be a half work over volume that the shaded area represents. So it's just work over volume and it would be multiplied by a half if we were working it out mathematically. But work over volume is the correct answer. Okay. The graph shows the effect of temperature on the viscosity of butter. So viscosity in Pascal, in killer Pascal, seconds against temperature. A student wants to spread butter on some bread. Explain why it is easier to use butter at room temperature than straight from a fridge. So there's two Marks, so we need to stay two points. Tiwhat butter, you know what butter is in England and not butter? Yeah. Toast and butter and jam. Yeah do you does your family keep butter in the fridge or in the kitchen open? Yeah. They keep butter in the fridge and use it for cooking. Yes. Why is it easier to use butter at room temperature to put on bread or scones than straight from the fridge? Easier to use butter. Because in the fridge, the temperature of fridge is 62. Maybe maybe that is not practical experience. So we can see that if we straight from the fridge, the viscosity is higher than the room temperature. So why is that difficult? It will not spread on the bread. He's y Yeah so fridge temperature. Is about four degrees in the so it's about four degrees centigrade. So the viscosity is high with butter straight from fridge, but at room temperature about 15 to 20 degrees, viscosity is lower and it spreads more easily. Yeah so less vissity means it will easy to use Yeah okay. That's room temperature. Okay. Please, I don't know why I have this twice, read the following article, then answer the questions that follow. I think we leave this. The graph shows a typical stress strain curve from for wood and steel, steel stress, so restrain wood stress over restrain. Discuss how the the use of steel rather than wood has made the construction of faster and taller roller coasters possible. Would you like to read this quickly? The fastest, tallest and longest dive coaster on which amusement park thrill seekers can experience free fall is set to open next year, next summer at Cedar Point in sandasty, Ohio. Valal Reis, designed to take riders up to 66m peak, from which they plummeted vertically with an acceleration g and feel weightless. So you're taken up and then you plumb it. The advent of steel frame roller coasters in 1959 made taller structures possible, whereas height remains one of the best ways to attain intense speeds. A coaster can also be shot from its starting point via electromagnetic propulsion or a catapult. Cars on these launched coasters have the potential to go from zero to 130 km per hour in 2s. Although coasters can definitely go faster, they're limited by the acceleration whose higher speeds would require roller coasters reach the peak speeds in a matter of seconds. This is achieved. The achieved acceleration is what causes g forces, which allow riders to feel an increase or decreased sense of their MaaS. These g forces can be dangerous, but they are also well understood by physicists. So roller coasters are built according to strict standards that keep them well within safe levels. Posters are only permitted to accelerate up to six g. So why do you think it's better to make roller coasters from steel compared to wood? Pand wood. So firstly, they have the different material. And maybe they have the different stiffness. Yeah, good. If you think stress is compression force, so they were able to make taller steel framed roller coasters, became taller structures, so you can go up to 66m. You go from zero to 130 km an hour in 2s. So that's a very high acceleration, high g force. So steel has a greater Young modulus than wood. The breaking stress of steel is much higher than the breaking stress of wood. Wood can snap, so wood will break much sooner at much lower forces per unit area compared to steel. So steel can withstand much greater forces per unit area compared with wood. Okay, let's try this question. The force of two newtons acthree by point 3m and stored on a force of two newtons appliapplithis energy. 呃。Good, slow and that I think that is we can calculate the spring constant and we use energy equal to a half K X. So ok equal to two divided divide 0.3 that is. Et p? 6.7, a half fa, half fx or a half kx squared two times 0.3 times that 0.5, 0.3, that's the answer. 0.3, choose. No point three. Yeah. I think we should do some more of these questions. Is that what youlike to do tomorrow? Or shall we do something different? I want to know more about this part. Okay. So we'll continue with these questions tomorrow. That's fine, Jackson. So enjoy your Sunday, and I'll talk to you tomorrow. Bye bye. I miss.