 All right, welcome everyone again waiting for a few more to join in people are still joining. So how many of you have come prepared as in done a lot of problem practice on work by energy and then attend attending the session. How many of you have done that only four or five I can see five six did little from module just now I mean it is nothing it is just you wasted your time. If you're doing just before the session for 510 minutes while you're doing it. This drink water and come it is it is worthless. Okay, you need to make efforts. Okay, it is not that you can just come over like that. When I say come prepared, you have to revise everything that you have done on this chapter one year back. Okay, no one revision is worthless. Okay, you could have as well not come prepared. Fine so now you know you are towards the end. All right, so don't fool around with yourself. The do only meaningful things. All right, you have lack of time. I mean, are you busy with something else during this time. Those who haven't come prepared. Do you have anything else that is going on some exams are running or anything like that. Everybody is free. Okay, so if you come unprepared, it is you will get worsely affected by it. Fine, so I am cautioning you. I am warning you I am. I want you to do well. And that is why I'm telling you all of this. Those who listen will do well. Okay. All right, so I mean let us start the today's session, which is on work by energy. All right, so today we are going to complete this. The revision of work by energy. Who is this. Okay, people are still joining so that is why I'm not able to start properly. All right, so this chapter was there in grade 11. Again, if you have joined. Okay, now whoever comes will enter after five minutes. And so this chapter was there in grade 11. I hope all of you remember the, the sweet memories of mechanics and work by energy. All right, and this chapter is chapter in which we are dealing with motion variables, which are basically not a vector quantity, like work power and energy. These are the three things we are focusing on this chapter. And apart from that, we also consider collisions to be part of this chapter. Fine. So let us quickly proceed and continue from there on. So the first thing that I would like to highlight is about the linear momentum. Now, all of you might be well worse with the principle of the linear momentum conservation. All right, so it says that if if there is some sort of collision between two objects or multiple objects. And there is no external force. If external force is zero. Then you can conserve the momentum. All right, there are cases in which external forces zero along the x axis, but it may not be zero along the y axis. Then you can conserve momentum along the x axis. And vice versa. Fine. So basically you need to find a direction along which external forces not there and conserve momentum along that direction. Fine. Now, when I say external force, when two or more objects, they collide with each other, they will exert force on each other. Right. I'm not talking about that force, any other force other than what these objects are applying on each other. Okay, they might be applying spring force to each other, normal force friction force, whatever forces between these two, when they are colliding. That is not external external to the entire system. Fine. So when collision happens, whatever is the momentum before and whatever is the momentum after the collision, they are equal. Right. And this is true. The condition of momentum is true. Like we can say five hours before the collision and five hours after the collision also whatever is a momentum. All right. But if external force is not zero, let's external force along the x axis is not zero. But only non-impulsive forces are there, then can I conserve momentum or not? Everyone, do you remember this? External force is not zero, but non-impulsive forces are there. Right. So we can conserve momentum just before and after. We can conserve the momentum. Why? Because only the impulsive forces can suddenly change the momentum. Impulsive forces such as normal reaction, friction, tension, they are the impulsive forces, they can suddenly change the momentum. Okay. But if collision is happening, even though there is external force, but if it is non-impulsive, then also can conserve momentum. But just before and after collision. Okay, immediately. So all of these things we have already, I'm sure that we have covered it when we were in 11th. Now when the collision happens, there are some special cases that you need to remember. Okay. So if it is an head-on collision, this is a head-on collision. What is a head-on collision? Head-on collision is the collision when the line of velocities is along the common normal. Line of velocities along the common normal. Fine. When head-on collision happens and masses are equal, then velocities get exchanged. Clear to all of you. Okay. And there are various cases of collisions. Okay. And for example, let's say elastic collision, inelastic collision, and partially inelastic collision, perfectly inelastic collision. So there is something called as coefficient of restitution E. Okay. Which is defined as coefficient of restitution E, which is defined as the, you remember that what comes in the numerator, velocity of, correct, correct. It is velocity of separation divided by velocity of approach. Okay. And this equation is something which you can apply not only for the collision between the two point masses, but between the two objects which are extended. For example, this ball can come and hit an inclined plane like this. It can come like that and hit the inclined plane. All right. So how do you write coefficient of restitution formula then? Suppose after hitting the wedge goes like this, you need to consider these velocities of separation and approach along some line, right? Along which line will you find out the separation and approach? Good. Good. So you need to consider the separation and approach along the common normal. Remember these things. So oblique collision, just one variation is there in the way the two objects collide with respect to how their velocities are. One is that the velocity is along the line of the common normal. The other variation is when the velocity is not along the line of common normal. Fine. So for example, let's say this object, it goes like this and hits this object. When this object comes near, let me move this near. When this object comes near and it tries to hit, look at the common normal. Common normal is along this. Along this direction, the normal ratio will be applied, but the velocity is not along this red line. This is called the oblique collision. Just a definition, alright? Nothing, let's say there is no extra principles or there is no extra law regarding oblique collision. Just it is a type of collision, okay? You can still use conservation of momentum the way we are using it and also you can write the coefficient of recitation E. But when you're writing coefficient of recitation formula, you have to find the component of velocities along this red line. Fine. Alright, so remember I'm not teaching you. I'm going very fast because my intention is just to refresh your memories. Are you getting it? So I'm just trying to make you remember those things which you already know. Suppose you do not know something, suppose you do not know about coefficient of recitation and this is the first time you're looking at it. Then I would suggest that after the class would get in touch with me, okay? Then I can help you. Otherwise, if I start teaching the theory, the entire class will get over and will not be able to do any questions. The main purpose is solving problems. Alright, work energy theorem. There are a couple of variations to the work energy theorem. First of all, let us understand what is work. Work is the some change a force can create and the change in physics is displacement. So if there's a displacement in the direction of force, we say that that force has created some change. Alright, so work is done by that force. So that is the reason why work done is represented as the force multiplied by displacement in the direction of the force. Here, we are assuming that the we're assuming forces constant throughout. If force is not constant throughout, then we have to use this expression f dot dr to find out the work done. Alright, and as per the work energy theorem, it says that it says that if you take a mass and multiple forces are supposed acting on it. Okay, you need to find work done by all the forces acting on it one by one. First you do that, add all the work done, total work done on that mass should be equal to change in the kinetic energy. Okay, so work energy theorem relates what is happening right now with whatever will happen later. K2 is final kinetic energy, K1 is initial kinetic energy. So there's a connection between the two points. Alright, so this is plain and simple work energy theorem, but there are some modifications also to it. For example, you may like to define something called as potential energy. Okay, potential energy is nothing but a simpler way of accounting the work done by a special kind of force. Okay, so what we do is that you split the total work done in terms of conservative force and non-conservative force. I hope all of you have heard these terms conservative and non-conservative forces. What are they quick? conservative non-conservative forces, what are they? What is conservative force? Right, conservative doesn't depend conservative force. Right, the work done by the conservative force doesn't depend on the path. So if you are at point number one, you are at point number two, you can go like this, you can go like that, or you can go like this. So if it doesn't depend on the path, why should I break my head in finding this integral? Because in this integral, this is dr, right? And there's a dot product, so angle theta will be changing along the path. So why should I find the work done for the conservative force? So that is the thought process of introducing potential energy. So as per the definition of potential energy, work done by conservative force is negative of the change in the potential energy. It becomes a state function, depends only on initial and final points, k2 minus k1. So what you can write it as, work done should be equal to u2, delta u is u2 minus u1. So work done is u2 plus k2 minus u1 plus k1. So please understand that this equation and this equation both are same. But in this equation, this w, you have to find work done by those forces for which you are not accounting the potential energy. For example, you account the potential energy for the gravity, don't find the work done due to gravity, if you're using this equation. If you're finding the work done by gravity, then use this, you're not using potential energy expression. Okay, so I hope all of you know basics about the potential energy, potential energy of the gravity is defined as MGH, right? We assume that the mass is small size, much smaller than Earth, which is how most of the objects are. And we are near to the Earth surface, so Earth for us looks like a flat thing. Okay, so in this chapter we are dealing with such situations, the mass is small, near to the surface, surface is flat. So this H is the height from where this height is, everyone. What is this H is? Height from where? Height from the Earth surface or it can be from anywhere. Height from a reference point or a reference line, there's a difference. Surface, good. It is not even, it is not line, it is not a point, it is a surface. So, but then usually our problems in physics are two dimensional. So we assume that surface is line when I'm looking at it in two dimensional. So in every question, if I have to define the gravitational potential energy, I will draw a horizontal line. And that horizontal line for me becomes a reference for the potential energy. So every time I write potential energy, it is a relative potential energy, relative to that line. So that line is zero potential energy below it, below height H minus MGH potential energy above height H plus MGH. And that horizontal line, you should not change while you're solving the problem. Once you fix it for a mass, it is fixed for the entire problem. Okay, so this is the potential energy due to the gravity, potential energy due to the spring. I'm sure all of you are experts is half kx square where what is the compression or extension? What is the compression or extension? It can be either, it can be either, doesn't matter. So you should just look at how much it is deviated from its natural length. So two centimeter compression, same potential energy, two centimeter extension, same potential energy. You don't need to break your head whether you should take plus or minus and things like that. All right, so enough of the Gyan, which you already know, let's solve a question, solve this. Gayatri, you're able to join? Gayatri is there? Okay, great. So what is the answer? Okay, so can you got it? Anusha, Aditya, others, anybody else? That's it. Okay, so I'm sure many of you are about to get the answer, but I'm creating a pressure just to hurry you up and make you aware that look others have already got the answer. Okay, so those who are in the middle still to get the answer, it is not that others know more than you, it is just that they have practiced more questions. So you also do a little bit of more problem practice without looking at solutions and your speed will come up. Anyways, so I hope many of you have typed in the answer. So can I conserve energy between point A and point C? There are two points, A is bullet is not hitting, bullet has not hit yet. B is bullet has hit and C is the maximum height. So can I conserve, can I use working as a theorem between A and C? Can I do that? Let me take a poll. Can I use working as a theorem between A and C? Alright, I'm ending the poll now. Ending the poll now. This is what people have said. So, you know, working as a theorem is what? Working as a theorem is kind of conservation of mechanical energy. It is kind of that expression. Alright, and here when the bullet hits the mass M2 and it goes inside ballistic pendulum, fast moving projectile fired into the bullet embedded. Okay, so when it goes inside, don't you think that some heat energy will be generated? There's a high chance that heat energy will be generated. So some mechanical energy is lost between this event and event B and working as a theorem is not capable to capture it. Fine. So first, you need to reach from point A to B by not using conservation of energy, but by using something else because mechanical energy is not conserved. You need to reach B after that you can go to point number C. Okay, so first step would be conservation of momentum. So M1, what is the velocity? Let's say velocity is V1A, it is written okay. This is equal to M1 plus M2 into VB. So this is the conservation of momentum. Momentum is conserved. Alright, and then I can use work energy theorem. And many times when you use work energy theorem, W would be zero. Many a times. And usually W will be the work done by the friction. So once you notice that work done is zero, even this tension, you know tension will always be perpendicular to the movement. Work done by tension is also zero. Work done by gravity, you're considering it as a potential energy. So whenever work done is zero, you immediately you can write it as U2 plus K2 becoming equal to U1 plus K1. So I can say that this line is my reference potential energy UG0. So U2 is M1 plus M2 GH. In case of any doubt, you can start typing. Don't wait for me. This is equal to U1 zero plus kinetic energy half M1 plus M2 VB square. So using this, you'll get the value of H. Alright, so I've seen many of you have told that you can use conservation of mechanical energy between A and C, which is not correct. Fine. Alright, let us move forward to this. We are going to have MCQs little later. Right now I'm just giving you some simple questions so that it get it refreshes your memories. Done. Okay, let us solve it now. Anyone about to get the answer? Okay. All right. So a railroad of mass capital M that can roll without friction initially mass small and standing on the car. Then the velocity of the car if the man runs to the left. Relative speed is VREL. Okay. So let's say car is moving with velocity U. Fine. Can I conserve momentum or not in this case? This is smooth. I can conserve momentum. But then to conserve the momentum and my observer is standing on the ground. So he is looking at the absolute velocity of the person and the rail. So the absolute velocity of the person is how much if relative velocity is VREL. In the decision situation, the velocity will be V relative minus U. This is the velocity of the person and velocity of the rail is U. Initially everything was at rest. The total momentum was zero. Zero should be equal to M times V relative minus U plus capital M into U. So what you get U as M into U minus capital M into U. This is equal to small M into V relative. So you get like this M V relative divided by small M minus this. No, no, no, no. I made a mistake. There is a minus sign here. Momentum is a vector quantity. So we can't treat it like a scalar. So minus M U become plus. Clear to all of you. Type in quick. Then we'll move to next. Right. Let us go quickly do this. Whenever a physics question comes, you need to ask yourself what is the condition for that to happen. They're asking find the maximum velocity of the wedge. So what is the condition for maximum velocity of the wedge that if it is not clear, you will not get the answer. So what is the condition that maximum velocity of wedges attain? Okay. Small M should have max velocity then capital M would have max velocity in opposite. What does it mean? I didn't get that. When you think maximum velocity of the wedge would be when the ball is over here somewhere at the bottom. Yes or no. See. Good question. Look at this. What exactly is happening when the ball is coming down ball is applying normal reactions like that. Yes or no. Throughout the motion. It is applying normal reaction like that. So there is a component of normal reaction in the horizontal direction. So as the ball is moving down, it is trying to accelerate the wedge in this direction. And as soon as it reaches bottom, the component. In that direction, the force vanishes. Acceleration becomes zero. Right at the bottom most point, acceleration of the wedge on the left hand side becomes zero. And then the wedges start decelerating because now normalization will be in the opposite direction. Does it make sense now? Who has it? Is it clear? Okay. So you can see here that you can conserve the momentum from here to there. There is no external force. So let's say the small m, small m was at rest at the top capital M will also address a zero momentum. This is equal to M into you for small m minus capital M into V. And then what can I do theorem? Right. U2 plus K2 is equal to U1 plus K1. There is no external force that is doing work. So the U2 is, let me say that this is my zero potential energy line. The bottom most point where the ball reaches. So U2 is zero. K2 is half M U square plus half capital M V square. U1 is small mg R. R is given plus K1 zero. So all these two equations, you'll get the value of V that is your maximum velocity. Okay. Someone is asking, can't we also say conservation of center of mass as the region? Conservation of center of mass as the region means what? What do you mean by that? Reason for? Since no external force is applied, so what? I didn't get your reason for what? Of course. Yes, yes. It is moving because center of mass should be conserved. See, same thing you can understand by many principles. Fine. So right now we are focusing on momentum conservation. Same thing you can. I mean, suppose I ask you how much the wedge has moved, then you have to do it by using the conservation of center of mass. All right. So let's move to the next concept, impulse. Impulse is also a made up concept like everything in mechanics. They're all made up concepts, but they help us visualize the scenario in a better manner. Okay. So impulse is defined as a large amount of force for a very short interval of time. Impulse by the sense of it can suddenly change the momentum. That is the idea behind it. Impulse suddenly change the momentum. Okay. And if you want to quantify it, what is the value of impulse? The impulse is simply change in the momentum. You can see the entire derivation here. Impulse is defined as simply change in the momentum. Okay. Integral of FDT is the impulse which is change in momentum and mathematically you might have done the, this thing, area under the graph in mathematics, right? So integral FDT is nothing but area under the curve. So if in any of the problems, if you get the, you know, the curve F versus T, whatever kind of curve there is, area under that, you can equate that to change in the momentum. It is change in the momentum. Okay. It is not the final momentum. There might, if the initial momentum is zero, then yes, area is the momentum. But if it is not zero, then you have to determine the area would be change in the momentum. Okay. So let us take a question on it. It may not be visible clearly. Now, you want me to draw the figure for this? I can do that. This is a figure. Solve it now. You have to use impulse equation here. Can you conserve the momentum between these two? If not, what is the reason? The reason is this tension force, which is impulsive force. Okay. Just write the equation and let me know that you're done with the equation. People are still joining in. All right. Done. Someone is done. Others. Hmm. Okay. So those who haven't done much of the impulse related question, look at the solution of this one very carefully. Okay. So when this mass and all are equal masses, right? The block is mass and pan is equal mass, mass and falls, right? So this is the pan on this pan. There will be one impulse force T and one more impulse force when the mass hits it in that direction to impulse force it normal force from this when it comes and hits and tension from the above. Okay. Let's say after hitting it start moving with velocity V immediately after hitting. This is a situation. So the impulse would be integral of N minus T net impulse force DT. This is the impulse. Okay. This is equal to M times V. Is this equation clear? Everyone type in. Is this equation is clear or not? Okay. Then this one, small M, small M when it hits, you can see here I'm ignoring gravity. You can see in the first equation I'm ignoring gravity. Why gravity is ignored when we write the impulse a fourth because gravity is not impulsive force. Impulsive forces are usually very, very large. I can just give you a sense of it. Suppose you're standing on the ground. Normal reaction is just MG. It is non-impulsive. Then suppose you're jumping from 10 meters. 10 meters is a big height. Okay. Let's lower it from five meter. If you jump, then also you can break your bones. Why? Because now it is not the gravity. Normal is not equal to gravity. Normal is way higher. So then during the collision, impulsive force act, normal reaction suddenly raises its magnitude. And that's why we ignore the gravity kind of non-impulsive forces. So on this small M, on this small M, then you have the normal reaction in opposite direction. That is the impulse force on the small M. Always take care of the direction. When we wrote the impulse equation for the pan, I'm considering downward positive. That is where normalization is positive. Tension I'm taking negative. And M into V is also positive. So there is a sense of direction in all the equations. Okay. For small M, I'm writing NDT. So for small M, I'm assuming upward to be positive. Okay. This is equal to M. Final velocity of small M is this way, which is V. Initial velocity was in the same way, U. So both are negative. So final velocity minus V, minus of initial velocity. This is the equation. Second equation is clear. This is for the pan. This is for the particle. Okay. Now for the mass. Mass is experiencing only T. Upward. And what is its velocity? What is the velocity of the mass? M immediately after impulse. Is a constraint. It is constrained to move up with velocity V. If it is going down with velocity V. Okay. So integral of TDT is equal to M into V. All right. So nobody is interested in doing the integral part. Okay. So what you can do, you can subtract second and third equation. You can get integral of NDT N minus TDT to be equal to U minus V minus M into V. So instead of this, now on the first equation, instead of integral of N minus TDT, you can write this and you can solve it now. M into U and you open the bracket and become minus of two M into V. This is equal to M into V. How many of you getting V is equal to U by three? U by three is the answer. Okay. Slightly, you can say off the track type of question because when you hear this chapter work by energy, you feel that is very welcoming and all the simple questions are there in the chapter. So what is my task? My task is to make you aware that there is a line in the room. So you should be aware that there are some tricks which are hidden in every chapter. Okay. So can't we just equate change in the momentum? Once you see the answer, you have all kinds of ideas. Getting it, but you can't conserve momentum because impulsive forces is there. Yeah, you may get the correct answer also. Luckily, that's it. You can, why can't we conserve right before and after collision because tension is impulsive force, right? T is the impulsive force. How can you conserve momentum for the plan? You cannot. Someone is asking in the example jumping on the floor. What is the expression for total normal force? See what happens when you jump from here, you hit the floor. You hit the floor. So which you suddenly come to rest. So acceleration is in which direction? Tell me your, your velocity is like this. Suddenly it is stopped. Acceleration is upwards. Acceleration upwards and it is a huge acceleration because you had at least five or six meter per second velocity. Suddenly it comes to rest. Acceleration is huge. So then you write n minus mg is m into a and normal action is mg plus a and a is way larger compared to G. So G is ignored. Understood. So this, do this question on restitution. I'll give you options for this. Maybe you guys are feeling that where are the options? Here are the options. Find their velocities. No, I'm asking you find the velocity of velocity of eight keys. It's a head on collision. Just to make it like simple. What is the answer? Well, there's one who has joined us. Okay. Let's have a poll for this first mcq. We should have a poll. Anyone about to get the answer? Anybody else? I can. I already have many answers. Okay, I launched the poll now. All right. So ending the poll. This is what you guys have said. Nobody wants to take a as an option. Okay. Let's see. Let us see. Ball of 4 kg moving with 12 collides with a mass of 8 kg moving with four in the same direction. It is, it always helps, you know, once you draw the diagram. This 4 kg 12 meter per second. This is 4 meter per second. Four into 12 plus eight into four is initial momentum. After the collision, the losses are V1 and V2. This will be able to four into V1 plus eight into V2. Then coefficient of restitution E. This is point five. That is equal to velocity separate separation. What is the velocity separation? Look at the scenario. Tell me what it is. Separation separation. Yeah. No separation is V2 minus V1. Separation is after the collision. Okay. Approaches what? Approach with what velocity are approaching 12 minus four. So you have these two equations equation number one equation number two. Right. Some of you are messaging that they want to change their answer. You can do that. Who can stop you here. But in the exam when you write. Once you make a silly error. You make a silly error. You can't correct it because you don't know in the exam that you have made a silly error. Option D is the correct answer. Fine. All right. Let us move ahead. And something oblique. Let's just write the equation for this and move ahead. Okay. There are many small concepts in the chapter. I don't want to miss any. So we are not finding the final answer for this. Just write the equations and let me know. Once you're done, then I'll write the equations. Then. Rashmika is done. Others. All right. One is 2 kg other is 4 kg same radius. And this is going with four meter per second. And this is going with. Two meter per second and that collide. And this is equal to one third. Find the final velocity along the line of impact. So now let us say. They are line of impact velocities. After the collision. This way. So I would assume that let's say is the. Four kg mass. Two kg mass. The four kg mass has. Velocity of. V2. Along the line of impact. And V1 perpendicular to it. Okay. And the two kg one has. Let's say. V2 dash along the line. And perpendicular to it is V1 dash. Fine. So. First thing I can do is conserve the momentum. Along the. Along the green line. So this angle, what is this angle? Everyone. How much is this angle theta? Everyone. Find out what is that angle. Five by four. Is it. This distance from here to here. Is to our, and this is our. This angle theta is such. That. Sign theta is half. So theta is. 30 degrees. Five by six. Okay. So when we conserve the momentum. Two into four. Into cause of 30 degrees. Minus. Four into two. This is also 30 degrees only. Cause of 30 degrees. This is the momentum. The. Green line. This should be equal to. I'm considering downward to be positive. Right. So two into V2. Plus four into. Two into V2 dash. This is the first equation. How many forgot this equation correct? Okay. Only one. Fine. The momentum is along this line. The initial momentum is two into four. Sign of 30 degrees. Minus. Four into two. Sign of 30 degrees. This question is. At least slightly above level compared to J means. But then it is good to. Know all these concepts. Okay. This is equal to. A perpendicular to that. I have assumed. That doesn't positive. Right. So. Minus of two into V1 dash. Minus of four into V1. Two equations of momentum. I have taken along the normal and perpendicular to the normal. Because. They're asking velocity along the normal only. So I have to consider that way. And coefficient of restitution. Let us see how many. If you got that correct. That is more important. Just. Let me create a space. E equation. E. Which is one third. Velocity of separation is what. What is velocity of separation? What you will write in this case. V2 minus. V2 dash. And velocity of approach. And velocity of approach. And velocity of approach. And velocity of approach. With what velocity they are approaching each other. For cause 30. Plus. To cause 30. All right. See this component. Coming towards that component also coming towards. They're both the both components along the normal. They come towards each other. So they will add up right there approaching each other. Fine. All right. Let's move ahead. You are making all kinds of errors possible. All the possible errors you are making. Not leaving any kind of error. Yeah, only three questions. And you'll get the answer. Fine. Next topic is the circular motion. Circular motion. When it is done. It can be uniform motion. Non uniform motion. All right. So. Uniform motion is a situation in a circle motion. In which. The speed is constant. Fine. In non uniform motion. Speed is not constant. Speed varies. So how you tackle these situations. In circle motion. We tend to use the angular variables. For example. Let's say a uniform circle motion like this. Velocity is constant. Right. So what is changing is the. Angle. It has reached here. An angle theta is made. So if we track theta. We will be able to find out everything about the. Particle. So that is the reason why we write. Let's say if it has traveled. A distance of. S. So you know by the definition of the angle. S by R is equal to theta. S is equal to. R theta. And. Rate at which the distance is not displacement is distance. Rate at which distance is. Changing is the speed. Which is equal to R d theta by dt. So if you track the angle theta. You can find out the speed easily. So it will be R d theta by dt is omega. Speed is R omega. Okay. And. We also have learned. That. Whenever circle motion is there. There must be a radial acceleration. Which is equal to U square by R. So U is R omega. So it becomes. Omega square R. Radial acceleration. All right. So tangential is not there in the uniform. Circular motion. All right. In a non uniform circle motion. At every moment. Whatever is available or whatever the speed. The radial acceleration would be. Speed is square divided by R. Okay. At the same time. There would be tangential acceleration. Tangential acceleration is. The rate at which speed is changing. Not the velocity speed. Okay. Take the magnitude of the velocity and differentiate that. You will get the tangential. Acceleration the rate at which magnitude of the velocity is changing is the tangential acceleration. This is equal to R. D omega by dt. D omega by dt. Is alpha. R alpha. Okay. So. Non uniform acceleration. Two exceptions are non uniform motion. In a circular. There will be two acceleration. Like that. Now. We are. Using angle theta. To understand the circle motion. But the equations of motion. Which we have learned. They were in terms of displacement. Velocity and acceleration. If you are tracking the angular variables. There should be. Equations. On the. Angular variables. So. That you can derive easily. By defining. Omega as. D theta by dt. Alpha is. D omega by dt. Now this is the definition. They will be always correct. Just like velocity is always great of change of displacement. Acceleration is always great of change of velocity. No matter. Acceleration is uniform or non uniform. Similarly this. This is the definition of alpha and omega. Alpha is also equal to D omega by. D theta. Okay. When you solve the differential equation. For. Constant. Angular. Acceleration alpha. You will get. Omega equals to omega naught plus alpha t. Okay. Omega square is equal to omega naught square plus two alpha delta theta. And delta theta is equal to omega naught t plus half alpha t square. I'm sure all of you have seen these equations. Many times in grade 11. Is there anyone who hasn't seen these equations. No one. Good. I have seen questions in. The. The complete exams. Where. They will ask you to find. The. Angular. Velocity. Suppose in this case. Let's say there is a rod. Okay. Rod of length L. It is given. And. This is the situation. The two points. Point A and point B. They are moving. In the given way. B is moving like this. Is going like that. This is V2. This is V1. Theta two. Theta one. Okay. First tell me what is the relation between V1 and V2. Is there any constraint relation between V1 and V2. If rod is of fixed length. Everyone. Correct velocity along the road should be same. Good. So V1 cost. Theta one should be equal to V2 cost. Theta two. If we do cause it. I do is more than V1 cost. Theta one. The length will become stretched. If it is less length will get shrink. Both are not happening. This is that. Now. If. If you talk about the perpendicular component velocity. They need not be equal. Okay. If you talk about the perpendicular component of the velocities. They may be different. If they are equal. The rod just moves straight up. Without rotating. Okay. If. The company components are not equal. If let's say V2 sin theta two. Is not equal to V1 sin theta one. Then the rod starts rotating. If you talk about the perpendicular component. The rod should rotate. Let's say V2 sin theta two is more. So the rod after sometime becomes like this. After sometime to rotate like that. Right. So how you find omega. To find the omega. You the best way is. You stand here at one of the corners. You can stand anywhere actually. Doesn't matter. But if you stand on one of the corners. It would be better because you already know this corner. Velocity. So you will be seeing the relative velocity. Of the other corner. With respect to you. So you will see what velocity V2 sin theta two. Minus V1. Sin theta one. Okay. So this corner will try to describe a circle. Keeping you as a center. Because whatever it does. The lens should be fixed. So it will become a server motion. And V2 sin theta two. Minus V1 sin theta one. If this is the tangential speed, they should be equal to omega times radius, which is L. So omega is. This. Okay. I've seen question in which they asked you to find omega. Okay. You may be wondering that. What this has to do with. Power and energy. But then. Somewhere you have to study circular motion. Right. So little bit of it. We have studied in motion in 2D. And a little bit of it. We study here. And if you're writing an exam, which is completely exams. When you look at the question. You don't think that which chapter belongs to. Same question can belong to multiple chapters. So it is all mixed up jumbled up. Don't need to worry about it. But will we get same result if you stand at the center of the road? Of course. Yes. When you stand in the center of the road. You must know. What is this velocity? Let's say this velocity was. X. Now you will not observe V2 sin theta two minus V1 sin theta one. You'll observe V2 sin theta two minus X. Okay. You'll get the same answer. You can read yourself and message me if you don't get the same answer. The hint is the rod is a rigid body. Omega is same everywhere. So you can write the center points velocity in terms of omega. And then I mean it'll be little lengthy. But still you can try it. Do this. Everyone. Should I give you the options for this? Yes, I can. Option number A. Is. Omega. Seek. Squire. Theta. D. This is omega. Okay. Whatever is given to us. Two. Omega. Seek. Squire. Theta. Seek. Three. Omega. Seek. Squire. Theta. Four. Omega. Seek. So some of you got the answer. And Sharon you're guessing. He's saying yes. So I am not here to. He'll make you learn how to guess. Okay. If you have to guess, you don't need to attend the classes. We are here to learn. Do it properly. Okay. So let us do it now. It's not a spot light s rotates in a hundred plane with an angle loss of Omega spotlight P moves. Along the floor and it's in three meter find the velocity of thought. P. So whatever is the velocity of a spot P. That is equal to rate of change of this distance or not. All of you agree that Dx by DT is velocity of point P. Everyone. Right. is that and you can write x as 3 tan theta, 3 times tan theta it is. So, when you differentiate dx by dt is equal to 3 c square theta d theta by dt, d theta by dt is omega. So, 3 omega c square theta option c. Okay. I hope it is clear. Let us move forward. So, I'll circulate this PDF after the class. I'm skipping some questions because of the lack of time this PDF has more than 70 slides I think. So, we can't do everything. Okay. So, sometimes I have seen that in the question they will ask you to find radius of curvature. All right. Now, I know that you have that differential equation form of finding the radius of curvature, but they don't expect you to use that. So, you don't need to memorize it. Fine. Radius of curvature is simply the if you, if you find out the perpendicular component of the acceleration, whatever is going on. If you find the perpendicular component of the acceleration, perpendicular to the velocity at that moment, whatever is a perpendicular component of the acceleration, that should be equal to speed squared divided by radius of curvature. That's how the if it is moving in a circle, centripetal attrition is u square by r. That is how it is defined. So, if radius of curvature is r, no matter whether it is moving in a circle or not. If you find out the perpendicular component of this direction, if you find out the acceleration, this acceleration should be speed squared divided by radius of curvature. Okay. How many of you have seen question in which you need to find radius of curvature of the projectile motion at the top of the projectile? How many of you seen that? Okay. If you have not seen that, you haven't done a lot of problem practice. I mean, you have, you have not done enough problem practice because it is seen regularly nowadays in almost every text book. Okay. Banking of roads. Banking of road is again, nothing new in terms of the concept. There is no new concept here. Same concept laws of motion is applied to find out what is a minimum velocity with which a vehicle should be moving so that it doesn't skid. All right. So, when you bank the roads, then the maximum possible velocity becomes larger. Okay. If you don't bank it, then theta is 0, the maximum possible velocity is root of mu g r. But if you bank it, you'll get this. Okay. And I think there's a minus sign in the denominator. Not very sure though. Should be cos theta minus mu sin theta. But anyways, just verify this in case you are relying on this. I think there should be a minus in the denominator. Okay. So, this is the banking of roads. Don't worry. There will not be a direct question on it. But this is a good situation to analyze the movement and understand how you can apply your concepts. Considerative force, non-conservative force, we already discussed. Okay. There are these kind of questions that are asked. Try solving this. I'll give you options for which part the answer is. The answer is for the second part. Okay. Some of you got something. Let's try to solve. Force is k y square i cap plus x square k x square j cap. Right. So, work, this force is not constant. So, work done is integral of f dot d r. d r is what? How can I write d r in terms of i cap and j cap, everyone? d r would be d x i cap plus d y j cap. Okay. So, the work done would be equal to integral of k. So, k y square d x, when you take a dot product, plus x square d y. Now, what to do? Everyone, what to do? So, you need to get some relation between x and y. Otherwise, you will not be able to solve this. Fine. So, how will you get it? You are going from 0 to a a in a straight line like this. Fine. This is what? y is equal to x line. So, throughout the motion, y is equal to x. Throughout the motion, if y is equal to x, even d y is equal to d x. Fine. So, this is equal to integral of k, y is equal to x, x square d x plus k x square d y is d x. So, x goes from 0 to a, x goes from 0 to a. So, we will get the answer as d. Clear to all of you? Type in, quick, quick. How it is clear? How it is clear? I mean, do you think it will come out to be this? Easily, we can induce silly error. Easily. You are very, you can say, you want to get the answer quickly. If that is the mindset that I should quickly get the answer somehow. I don't have time. I'll get the answers by doing something. Then you'll make a silly error every time. That is how these exams are designed. x square d x integral is what? x cube by 3. So, there is a factor of 3. First question, first part, consider it as homework and try doing it. Very similar way. You have to do it. First, find the work done from here to here, then find the work done from there to there and add them up. Try it yourself. Let me know. Then we see some questions related to the equilibrium. Yes, first part was homework. Type of equilibrium on the basis of stability. So, equilibrium means that the net force is 0. And there can be many kinds of equilibrium. They are called stable equilibrium, unstable equilibrium and neutral equilibrium. For example, this one, if we take these two examples, let's say a ball is here at rest. This ball and that ball both are in equilibrium. But if you push it a little bit, it will not going to come back. It will going to go away. But this one will come back. So, a stable equilibrium will do oscillations. Unstable will not do oscillations. We just go away. Fine. But both are stable. Sorry, both are in equilibrium. This one is unstable. This is stable. And if you keep a ball like this here, this is the neutral. Fine. How will you check whether it is a stable, unstable or neutral? See, the fact is that if you move this way, the force has to be in the opposite direction. Force should get created in the opposite direction. Then it is stable equilibrium. Now, how will you check that? We will check that by finding out if df by dx. This should be positive or negative for stable, everyone. df by dx. You changed dx little bit. It should be negative. It should be negative. And do you guys remember force is equal to derivative of x like this? Do you remember this? So, double derivative of the potential energy should be less than 0. Or double derivative of potential energy. If it is greater than 0, you say it is a stable equilibrium. And if double derivative of potential energy is less than 0, you say it is unstable equilibrium. Many times in the numerical potential energy is given as a function of x or y. So, you have to do all of this circus to find that out. Okay. Anyone has any doubts? Quickly type in. No doubts. Let us take a question on it. Double derivative, if it becomes 0, neutral. The f is not changing with x. Find out. This one. All right. Others? Force is equal to minus of du by dx. This kind of derivative, I hope you know, when you do this kind of partial derivative, you are considering other variables to be constant. Like for example, if y would have been there, y is treated as a constant when you differentiate like this. This is equal to minus of 2ax plus b. When this is equal to 0, it is an equilibrium situation. So, x is equal to b by 2a. This is a point of equilibrium. Now, whether this is stable or not, we can find it out by taking a derivative of the force. df by dx comes out to be minus of 2a, less than 0. So, it is a stable equilibrium. Okay. Or if you are very mathematically driven, you do not want to, you know, think, so you can, as a matter of fact, just find out the double derivative of potential energy. This should be greater than 0. Okay. You can memorize it like this. All right. So, it is a stable equilibrium like this. Can any situation, there can be more than one stable, unstable? Yeah, why not? There can be many like this. Unstable, stable, unstable, stable, unstable. And I can draw one neutral also. This is neutral. Yeah. Okay. Somebody asking, is there double derivative positive or stable? Double derivative of what? Double derivative of what? Double derivative of potential energy is positive. Okay. And derivative of the force should be negative. Force is equal to minus of du by dx. Negative of the derivative of potential energy is the force. So, that is why one is positive, other one is negative. Now, let us get into the objective type of question, MCQ. I think I have now all questions. I'm done with the pointers which I have to highlight. I think I have told you a few things which you might have forgotten or you do not know. So, now we will be solving questions from anywhere in this chapter and sometimes you may learn some new theory from the question itself. So, do it. So, here I have not told you, power is what? Power is rate of work done. This is equal to also f dot v because work done is f dot dr. So, when you do dw by dt, it becomes f dot dr by dt. So, f dot v it becomes just a basic theory. Okay. Many of you got the answer. Let us see. Body of mass m at rate uniformly from rest to velocity v1 in time t1 instantaneous power delivered to the body as a function. Okay. So, here I mean it is roughly we are assuming that acceleration is constant. So, acceleration is v1 by t1. This is actually an average acceleration but we are assuming that it is a constant. So, the force is mass m's acceleration and v1 by t1. So, the power is force times velocity. This is the v1 in time t1 power. Yeah, this is v1 only but I have found out the average power. Sorry about that. Okay. I have found the power at that instant. Let us say at any moment the power is m into v1 divided by t1 multiplied by velocity after time t. So, velocity after time t is initial loss is 0 plus at v1 by t1 into t. So, you will get m v1 square by t1 square into t. Okay. Option one. But then yes, it should be written uniformly. Okay. I assume it is uniformly which was correct. Good. Everyone, just one. Do this. Everyone. Yeah, you may get dimension analysis by using that. You may get the answer during the practice but in the exam, trust me, you will not get the answer by using dimension analysis because they are smarter than you at least. All of us even me. So, they will definitely give at least two or three options which have the same dimensions and one of them will be correct. Okay. Here also I can see many of you got the answer. Is it very straightforward? Car of mass m has engine which can deliver power t. Minimum time rest to speed v. Okay. So, power is dw by dt. Okay. Work done basically if I am using the maximum possible power p. Right. For a minimum time t, I have to use maximum power possible throughout the motion. Power is a constant then. So, work done is p into t. Right. Work done becomes p into t. And what else we can write here? Minimum minus p is v. So, work done is the change in the kinetic energy. W is changing the kinetic energy half m v square. This should be equal to p into t. Right. So, t would be equal to m v square by 2p. So, option number one. Okay. I hope those who got the answer, they have done it properly without dimensional analysis because trust me in the exam there will be one option like this, other option like that. So, all this, this, this will have the same dimension. You can't use, you can't rely on dimensional analysis to solve problems. Do this, everyone. Anyone close to the answer? Hmm. All right. Many of you got some answer. Let me proceed. So, if I look at the graph, the two things should strike me. The area and the slope. Okay. You should start thinking about the area and the slope. There is something which has to do with the area of the slope. Here it is f and t. We know that the area of f and t is what? What is the area of f t curve? Change in the momentum. Right. So, initial velocity is 5 meter per second. Body acted upon a time-dependent force like this. So, the area, you have to consider the sign of the area also. Downward area is negative because force is negative. So, 20 into 5 plus half of this base is 5, height is 20. And this is also same, right? So, into 2. And this is also same. So, in fact, into 3. How much is this area total? Anybody got this? 100, 50 into 3, 150, 250. I told that consider a sign and then I myself not considering a sign. Minus. Sorry. Sorry. So, 100 minus 150, minus 50. All of you got it? Minus 50. Sometime I make silly errors. Pardon me. So, this area is 20 into 5 plus half of 5 into 20 minus of half 5 into 20 into 2, right? 100, 150 minus 100, 50 plus 50. This is equal to change in the momentum. Final velocity is let us say v. So, 2 into v minus 5. Okay? So, 25 is equal to v minus 5. Final velocity is 30 meters per second. You have to find work done. And sometime, you know, they will not write 20 here. They may ask you what is the work done in 15 seconds. So, we are in so much hurry in getting the answer. We do not even read till what time they are asking. So, we will do it like this, which is completely wrong. And then there will be an option sitting there. It will make you believe that you are right. But then you are not. So, carefully read the question. They are asking 20 seconds, right? So, that is all fine. But they may ask you 15 seconds. So, then you have to take area till here only. Anyways, so, work done is what? Work done according to the work as it is your own. Work done on a body means work done by all the forces, including gravity and friction and spring and everything is equal to k2 minus k1. That is half m into final velocity square minus half m into integer velocity square. How much you get this? 875 is the answer. Okay. All right. Let us proceed. I think you might have seen this question somewhere, but still do it. Okay. Archer got it. Prabhu got it. Shall we discuss now? Okay. Let us discuss. Power is constant. Power is rate of work done. Okay. Since power is constant, the work done is power into t. So, power into t is change in the kinetic energy. It starts from rest. So, finite kinetic energy is half m v square initial is zero. This is what it is. Okay. And what we are trying to find out? We are trying to find out displacement s as a function of v. Okay. Till now, displacement has not come in the equation. And we need to bring it up. Power is equal to force into velocity, which is mass and acceleration into v. If you substitute it over there, v, this is a constant. Right? This is a constant. So, acceleration, I need to bring s in the equation. So, I can write it as v dv by ds into v. This is a constant, which is power. So, m v square dv is equal to power into ds. I do not need to do this. So, when you integrate this, zero to s, zero to v, power is constant, so comes out of the integral. So, m v cube by three is equal to p into s. You can see that s is proportional to v cube. Option number three. Okay. I hope it is clear. Let me go to the next question. We can write f dot s here. What do you mean by that? No, you are assuming force is constant. Power is constant. Right? Power is constant. Velocity can change. Force can change. Multiplication between velocity and force is a constant, not force. This one, you want to do it? We will come back to this. We have done enough of power now. Let us do something else. Power is a small topic. Do this. If time permits, we will again come back to those questions which we have skipped. All right. Many of you got the answer. These are some of the questions which you should not get the wrong answer. It is moved slowly. So, at every moment, velocity is almost zero. Okay. So, the total work done on the mass is zero. So, work done by the force plus work done by the gravity should be equal to zero. Chaining and energy is zero. Work done by force is w f. Work done by the gravity is minus m g h because gravity forces down. Displacement in that direction is opposite to it. Zero. So, work done by the forces plus m g h. Okay. Let us take five minutes break. We will meet after that. All right. So, let us continue. We will move forward. This one quick. Everyone, this is very simple. All of you should, okay, let me put a pole. Hmm. I am ending the pole now. Only spring force is there. All right. So, you can see the pole result. Those who have marked A and B, you have not come prepared today at all. Answer is D. Okay. Okay. I have to share the results. This is the result. Fine. Many of you have picked A and B. A is the majority. Answer is D. So, we did, we have discussed, right? The work done by the spring force is half k x1 square minus half k x2 square. You remember this? Depending on x1 and x2, if the spring is already extended and you move this way. So, spring is doing positive work. It is pulling and moving also this way. If a spring is in natural length, you are moving that way. So, spring is doing negative work. Spring force is this way, but pull is that way. Okay. So, depends on initial and final position, it may be positive or negative. This is the formula for the work done. Okay. These kind of questions, when you see in the exam, your eyes will light up. You will be like, oh, very simple. A theoretical question, I don't need to do, I don't need to write the equation. But these are the questions in which maximum people will get it wrong. Done. Anyone? All right. Let us solve this. Consider system, coefficient of friction is mu. System is released from rest. Find the work done by the friction. Now, in this case, friction is doing positive or negative work. Okay. All options are negative. So, we cannot eliminate that. Speed of the block is this. When the speed is this much, it is released from rest. So, equation is what? Equation is 2 mg minus mu mg divided by 3m. How much it is? Anybody calculated this acceleration? Five. Okay, you get five. So, distance traveled is half 80 square, half, no, no, no, not that way. Well, our final velocity is given. So, we will use this to get s. So, 10 square is equal to 0 plus 2 5s. So, s is 10 meters. Okay. The work done by the friction is minus of mu mg into 10. Okay. So, 0.5, whatever it is, what do you get when you do it like this? Option four, right? All right. So, that is the answer. You can see your kinematics is mixed. Not only kinematics, laws of motion, kinematics, work by energy, all the chapters are there. Okay. So, most of the questions in the exams will be like that. Do this. This looks interesting. But this one is, I think, question from laws of motion. But anyhow, do it. I do not know whether it is laws of motion or whatever it is. Try getting the answer. All right. Let us see. What is the maximum compression of the spring? If the lowest spring is shifted rightward with expression A, how many of you have picked this option? Option three. Option three is not correct. Okay. Why? Because condition for maximum compression is what? Maximum compression, if you have equated forces, then that is not correct. Maximum compression is when the relative velocity of M on the plan should become 0. That is the condition. Okay. So, yes, it is work energy chapters question, not laws of motion. So, what we will do is that what we'll do, we'll apply the work energy principle while standing on the lower block. But when you stand on the lower block, there will be a pseudo force acting on the small M with a force of M into A. So, on this block, I can apply work energy principle. So, work done on block M in my reference frame should be equal to change in the kind of energy. In my reference frame, it was at rest. And when the final compression, the maximum compression has reached, then also with respect to me, it should be at rest. Other way, we'll keep on moving backwards. Right? So, work done on the mass M by M A forces M A into X. And work done by the spring is minus of half K X square. This should be equal to 0. So, answer is 2. Is it clear to all of you? I've been quick. Is it clear? I can explain it to you in another manner. Okay. Let me explain it to you in another manner. So, when you stand on the lower block, when you stand on the lower block, the upper block, the small M will do oscillation or not. Will it do SHM or not? Everyone, will it go like this? Let's start performing SHM because pseudo-force is applied. Right? So, when you equate pseudo-force with K X, what is that? What is that? Is that a mean position? It's a mean position. So, however much further it was initially at mean position, that much further inside it also will go. So, extreme positions are on the both sides of the mean position. So, at the mean position, the compression is M A by K. So, it will go further by M A by K this way. Clear now? Yeah, you have to change the frame of reference and understand. Every time those who are in a hurry to get the answer will make this mistake of equating forces for maximum compression. Okay? I am sort of warning you that if you are in a hurry, you will not do well in exams like J. You should be calm and composed. Okay? Relaxed. Fine? And if you get a simple question, don't run over it. Give it full respect, try to get the correct answer, then move ahead. Difficult question, you may not get the answer. So, simple question, if it is there, make sure you take the right option. Do it. All right. Shall we discuss now? Arnav and Kinshuk have answered others. 30 seconds. Okay? Poll. Okay, I will put a poll. In 30 seconds, I will put a poll. Here comes the poll, ending the poll. This is what you guys have done. You know, what kind of question will affect your marks in a worst possible manner? The questions which everybody gets right, but you got it wrong. Getting it? So, those simple questions in which you make silly error will hurt you the most because 10,000,000 people will get it right and you got it wrong. So, 10,000,000 people are ahead of you immediately. Difficult question, if you get it wrong, you might be behind, let's say, 5,000-6,000 people. But simple question, if you get it wrong, you are behind by lakhs. Okay? But at the same time, if you get difficult question right, you will be ahead by 10 lakhs. So, it is a tricky scenario, but making a silly mistake in a simple question is a crime and you will pay for it for rest of the life. Okay? Don't do that. Kinetic energy is half m v square. Now, this is what k versus x. So, I told you that whenever there is a graph, it is all about area or slope. So, if you see a, b, c, d, the difference is in terms of slope or something like that. So, you tend to find the slope of the kinetic energy in terms of x. dk by dx is the slope of k versus x graph, which is half du by dx. U du by dx is what? This thing is what? Acceleration, m times a, which is, so wherever the slope of k versus x is highest, acceleration is highest. Where it is at d, it is highest almost like a vertical line. Okay? a, oh, you say a as in acceleration, not option. Now, this is the kind of question which you have seen so many times. Right? This is, this you can do it. If you have done basic problem practice, you might have solved already 3, 4 times that kind of, do this. Okay, let's do it. x is equal to 50t minus 5t square. You need to find work done on the particles or work done by all the forces. So, I mean, if you, if you correlate with x is equal to ut plus half 80 square, you'll clearly understand that acceleration is minus 10 meter per second square or you can do a double derivative of this x. Differentiate once, you get velocity, differentiate again, you get acceleration, which is minus 10. So, the force is, it's a constant force, mass is 2 kg. So, 2 into 10 minus 20 newtons. Work done on the particle in first 5 seconds, what will be the x? x is 15 to 5 minus 5 into 5 square. So, 250 minus 125, 125 meters. So, x is positive, force is negative. Work done will be negative. So, minus 20 into 125, option number 2. So, here, someone is asking, here does negative direction of force only means in the negative x axis? See, you have taken x coordinate and this is half of a, and this is what? This is a vector equation. You're treating like a scalar equation, but it's a vector equation. So, minus sign means what? When x is positive is negative. So, there in opposite, the acceleration is in opposite direction of x. So, acceleration is in the direction of force. So, there in opposite direction. Clear? What if x is negative? If x is negative, then force is in the direction of x. You can see, see what is happening is, this is the kind of thing that is happening. Initially, the object is let's say moving it 5 meter per second hypothetically, but the acceleration is in this way. So, what will happen? It will move up to certain distance then turn around and then start going in that direction. So, till it turns around, the force and displacement are in the same direction. But once it turns around, then the change happens. Never call it deceleration. Deceleration is a made up concept to make sure that standard 9th gets understand little bit about the kinematics. But in reality, deceleration will ultimately leads acceleration only. Initially, it will decrease the velocity and then start increasing the velocity in the opposite direction. This one? Okay. How many of you got the answer? Is this force is given if the potential energy is asked? So, you know that negative of the work done by the force is change in the potential energy. Good thing is that it is assumed that at x equal to 0, potential energy is 0. So, Ax dx, so Ax square by 2 comes out to be potential energy. When you put the limit, u1 is 0. Yes, option 2. How do you option 4? Option 2, it comes out to be Ax square by 2. Do something different. This one? Power. Where is power here? It is potential energy. There is no power here. Do this. Forget about the option. Solve it properly. I will tell you all the shortcuts at the end. Right now, focus on learning. Okay. Shall we discuss now? Nishchal, Aditya, Thanush. They got it. No one else. All right. Let us discuss. So, what will happen? This block will try to track the hemisphere and it will leave the hemisphere at certain angle. Let's say it leaves it here. Do you remember what was that angle? What is this angle? Theta. Oh, I have to take attendance. Okay. Yes. Good that you remember that, but we will derive it. Let's say here when it leaves it, its velocity is V. After leaving, it will become a projectile and travel like that. Okay. So, let's say before leaving, it makes an angle theta. So, this distance is r minus r cos theta. So, we can use the conservation of the mechanical energy. The work done is zero by any external force. Gravity is doing work, but we are considering potential energy for that. Okay. So, let's say this red line is, let's say my zero potential energy line. So, u2 is zero, k2 is half mv square, u1 is mg, r1 minus cos theta, k1 is zero. Okay. And if it is leaving at that point, then what is the condition for that? If it leaves, the condition for leaving any surface is normal reaction and friction becoming zero. Okay. So, normal reaction, you can equate along that direction, n minus mg cos theta, where there is an acceleration also. So, mg cos theta, acceleration is towards the center, minus n is equal to mass times acceleration. Acceleration is centripetal acceleration towards the center, v square by r. So, if it leaves the surface, normal reaction should become zero. So, when you put it as zero, you can substitute the value of mv square over here, and you will get the value of cos theta to be equal to 2 by 3. 2 by 3 is what you will get. Then, after that, it becomes a projectile motion. It's as simple as this. It's like this, where this height is r sine theta. It's like height of the tower is r sine theta, and the velocity is this way. So, the projectile will travel like this. This velocity is v, this is theta, that is theta. So, this angle will be 90 minus theta, with that will be theta. So, this angle is theta. Okay? Wait. You need to just find r sine theta. You need to find the horizontal distance between the point where it leaves, and the point where it is, right? You have to find that distance from here to here, that distance, range, you need to find out. Am I correct? The body starts slipping the horizontal. Oh, sorry, sorry, sorry, sorry. Yeah, I did not read it completely. You just need to find that distance. Correct. You just need to find that distance. What I was trying to find was this distance. It leaves here, then I was trying to find that distance, which is, you know, which is, and I was wondering why this is an objective question. So, that distance is actually r sine theta, cos theta is 2 by 3. Sine theta is root over 3 square minus 2 square divided by 3, that is sine theta into r. So, option number one. Okay? It cannot be 3, right? Because that distance can't be more than radius. It cannot be 2. It can be either 1 or 4. What if wedge is not fixed, then it will become little lengthy, that's all. So, you want to try it, try it after the class, and we can discuss it then. Do this. Anyone close? Okay, let us solve it. See, will this block, block B, B will ever leave the floor if the spring is compressed. It cannot, it can only leave when the spring is extended. And if it leaves, the normal reaction should become zero. So, let's say it leaves though, there will be extension in the spring. So, extension is let's say x not, no x on the channel. So, kx should be upwards, then only it can leave. Okay? And mg is downwards. There will be normal reaction. So, kx plus normal reaction minus mg should be equal to zero. Condition for it leaving the ground is n becoming zero, and x becoming equal to mg by k. And this should be extension. Okay? Now, let's go back. So, at present, if let's say f is zero, if f is zero, spring is compressed by mg by k. Nothing will happen. Okay? So, it has to move up by mg by k. Okay? Then go up by another mg by k. Do you all understand? If it moves up by mg by k, natural lens will be there. Then it is to further go up by mg by k so that this lifts off. Is this clear to all of you? Type in. Okay? Now, as of now, as of now, the, this mass is in its main position. Net force is zero, mg and spring force are balanced. But if I push it down from its main position by a distance of two mg by k, will it be able to reach two mg by k above also? Will it be able to reach here? Yes or no? Yes or no? This at present, it is at the main position, considering the oscillations. Okay? So, if you want to further compress it by two mg by k, you need to have a two mg force. Type in. Is this clear? Now, you can do that energy equation and all those things, it will become little complicated. But this way, you can get the answer quickly. Okay? But how is this minimum force? Minimum, see, if you have more force than that, it will go beyond two mg by k up. So, your extension, the spring will be more than mg by k. Fine? Don't worry. I am going to put this recording at this location. You can understand from there. You can watch the recording also. If you want, I mean, you should watch it once again. Those questions which you found that something completely new, watch it again because you can't just learn immediately. But unfortunately, we have to move ahead. We can't just hard only on one question. Do this. Can the answer be more than one? Can the fraction be more than one? Of course, no. So, you can remove these two. Okay. So, let's move ahead. A neutron traveling with the same some velocity collides with a nucleus of mass number A. So, the mass of neutron is M. Then mass of the atom is A times M, fraction of total energy retained by the neutron. So, we are assuming that the atom is at rest. Neutron is traveling with velocity U MAM. After colliding M into U plus A M into 0. So, initial velocity is 0. M into V1 plus A M into V1. It is an elastic collision. So, velocity of after the collision, this is V2 and this one was moving with V1. So, E which is equal to 1, velocity separation V2 minus V1 divided by velocity of approach. So, you have two equations. V2 minus V1 is equal to U and V1 plus A V2. That is also U. Okay. So, using these two, you will get V1 and V2. So, the energy of the neutron would be half M into V1 square divided by U square. This is the answer. Okay. And it comes out to be option one. Okay. So, I think this was direct only. Don't make mistakes in these kind of questions. Where you just need to have conservation of momentum and conservation of energy. All right. Should not make a mistake in that. Let me give you one question, my own. Better to use E. Yes. If you use the energy, then you get a quadratic equation. Right. I think we discussed it many a times. Why you want quadratic equation? Okay. Arnav is saying if mass is much greater than neutron, doesn't the nucleus remain address and neutron bounce back with double the velocity after collision? Not double the velocity with the same velocity. That is correct. If A is very large, but that is not happening here. We are not assuming that the way the options are put over here. We are not assuming that. Fine. That is an assumption. It or it should be written that A is very large. Like that it should be clearly mentioned. All right. Everyone. This is a liquid of density row. Then there is a cube density sigma. Sigma is less than row by two. So, what you're doing is you are pushing this block. You're pushing this block down like this. Side length is A pushing it down so that the top surface coincide with the level of the water and then leaving it. You need to find out, find the velocity of the block as it comes out completely. Last thing for today. Up to what height H it will be already submerged. Should I do it now? Hmm. There will be a buoyant force. The volume submerged A square X into G. Gravity force sigma A cube G not X H. They both are equal. It is balanced. So, row A square H G should be equal to sigma A cube G. So, H is equal to but why we need to find that. All right. So, now when I'm pushing it down, let's say I have pushed it down completely and at this moment it is X outside. I need to find out when this X becomes A so that it is completely outside. So, right now it is X outside. The net unbalanced force which is in upward direction because it is moving upwards right now is buoyant force. Let's say this is X buoyant force which is row A square A minus H. You have to take the volume that is submerged. G minus sigma A cube G. Here X will be there. A minus X G. This is MG force. This one is MG force and this is the buoyant force. Net force upward is this. This is equal to mass times acceleration. Acceleration you can write it as V dV by dx. All of you getting it? What I'm saying here? Type in. Are you able to understand? So, V dV sigma A. Now, I have to use integral here. Row A, row minus sigma G A integral of dx minus rho G X dx. Just rearrange the terms. X goes from 0 to A. 0 to A. The loss goes from 0 to V. Just spend a little bit more time with this question. You will learn something new. Because till now we never considered that even flutes can do the work. But the buoyant force is just like another force. So, even buoyant force can do the work. So, that is what we have done here just to demonstrate that. So, I would say that's it from my side. We are done with the basic 1% of not even 1% of revision of the work for energy. It takes a lot of time to completely understand the chapter. And I hope you guys have done that hard work. And we'll meet next time and we'll take the next chapter then. All right. Bye for now. You can't consider it a spring. Okay. It looks like a spring kind of force, right? Proportional to X. Don't do that. Shortcut would lead you to the wrong answer. Bye for now. Yes, it will perform SHM. It will. We have done this type of simple harmonic motion last year itself. Bye for now.