 It's 60 seconds now. Yeah, check out the channel. It's so many, I am just seeing a boy here. I am just seeing a girl here. I was thinking that. Good evening everyone. Waiting for others to join in. Oh, thank you, Mama. Mama, Mama. That's why I say mute yourself. All right, so how many of you have attempted the homework? I have received not many, but a good number of people have submitted their homework. But what happened to others? UTS, those who have submitted, they don't have UT tomorrow. Doesn't matter whether NFL has UT or you guys have UT. In fact, I would not suggest that UT should affect your preparation at any point in time. So going forward though, we will probably not account for UT. Okay, so this time because the first time UT you might be facing, that is why we have accounted for UT, whether it is NFL or the R&R. But understand this thing that you are competing with everyone in the country. You're not competing in your class only. So whether UT or not, both the schools, NFL and R&R, make sure your assignments are done. Okay, I'm not asking something which cannot be done. I've seen over the years people are doing it irrespective of their UT's or not. Okay, so make sure you are submitting it regularly. I'm not talking about just this time. I'm talking about every week. Every week you have to submit your assignments. Just take a snap and don't take a snap of final answer and send it to me. That means nothing to me. Final answer is written against every sheet also. Make sure you take the snap of the workings and then submit it. Everybody understand this? And probably then only you could see the improvement in your marks in the chemon test. Okay, it's very important. Take it seriously. All right, so today is the last session on this chapter, work power energy. All right, and this is the third session on it. So we are going to complete this chapter today. Let us start. Last time what we did, anybody remember? What did it last time? What was the major focus of the last class? Yeah, you can submit after the class. After the class you take your time. By tomorrow also you can submit. No problem, but do submit. Today, tomorrow, day after it is fine, but submit it. Potential energy. Potential energy was the prime focus last class. Okay, in this class, we are modified form of work energy theorem. Good. So in this class, we will be talking about the power and collisions. Okay, we have done work and energy. All right, the chapter name is work power energy. So we'll be talking about power part of it, which is a very small part. Once you finish power also, we are done with work, energy and power, all three. But then there are some scenarios, like when collision happens, something different we have to take into account. So that is why collisions are dealt separately. Okay, so that is the focus of today's session and we'll be completing this chapter today. Okay, what is going on in school? Anyone? Both in Neffel and R&R. What is going on? Laws of motion. Finished laws of motion. Rotational dynamics, they've started in Neffel. Okay. All right. That's going a little fast. All right, guys. So all of you write down this topic power. Now, till now, our focus was the amount of work done. All right, we never worried about how much time it is taking or how much time it will take to complete a work. We only worried about how much is the work done. So for example, two people are there. Two persons are there. One person, both the persons are climbing the stairs. Same amount of stairs. One person climbed in four minutes. Other person climbs in two hours. The amount of work done is same. Both of them reached from this level to that level. Okay, but then the first one has reached lot faster. Okay, then another practical example like what you are doing, you're preparing for exam. Okay, so suppose you take, let's say two years to prepare for an exam. Someone else is taking 10 years to prepare for the exam. Okay, so you know that preparing for 10 years is useless. Okay, so that's the reason why time factor is very important. We till now never accounted for the time. How much time it will take to do the work. Power is the concept which will take into account of how much time you are taking to do the work. Fine, till now only magnitude or amount of work was the focus. Now the even time factor is coming in. All right, so very plain and simple. The definition of power is rate at which work is done. Write down rate of work done is power. Do you remember any other physical quantity or any other thing in physics which is rate of something? What it is? Do you know anything which is rate of change of something? Acceleration is what? Rate of change of velocity and velocity is what? Rate of change of location or displacement, right? So similarly, you're just like you have written. Acceleration is dv by dt. Velocity is ds by dt. What do you think power will be written as? Power will be dw by dt. Yes, rate of change of momentum is force. There are many things in physics which are rate of change of with respect to time, okay? And power is also one among them. Power is rate of work done, correct? So we know that work done is f dot s, okay? So dw by dt which is power will be equal to f dot ds by dt. Assuming force is not changing with time. This will be the power. So ds by dt is what? What is ds by dt? Velocity, right? So power is represented as dot product of force and velocity, okay? So this is how we represent the power in physics, okay? So just like the work done which is force, dot product of force with a displacement. Similarly, power is dot product of force with velocity, okay? You can also, you know, just like we have done for the work done, you can say it is magnitude of force into magnitude of velocity into cos theta. You can write like this. Or this is also equal to f cos theta into v. Now can you tell me what does this bracket represent? v cos theta is what? v cos theta represents what? Everyone, component of velocity along, along what? Along the force, not along x or y, along the force. Theta is the angle between force and velocity. So velocity cos theta is the component of velocity along the force, okay? This is the component of velocity along the force. And what is this? f cos theta, everyone? Component of correct, component of force along velocity. Just like we have written for the work done, that can be written as force into displacement in the direction of force or displacement into force in the direction of the displacement, okay? Both ways it can be written. Same way for the power, okay? So just like you remember we have done five cases for the work done, right? Same thing is valid for the power also. If angle between force and velocity is 90 degree, what do you think power will be? If angle is 90 degree, power will be 0, right? Cos of 90 will come 0, alright? And can power be negative? Can power be negative? For what angles it will be if it is? For angles greater than 90 degree, it can be negative, right? So for 180 degrees, cos of 180 is how much? Minus 1, okay? So it can be negative if cos theta is negative. Just like work done, work done was negative, right? If theta was more than 90 degree, similarly power can also be negative, right? Okay? So yeah, someone is saying that scalar quantities cannot be negative. That's not correct, okay? If it is a vector quantity, minus sign represents the direction. If it is a scalar quantity, minus sign represents magnitude, okay? So negative work done is lesser than zero work done. Negative sign doesn't represent direction every time. Okay, can you tell me what is SI unit of power? Everyone, SI unit of power is joule per second, which can be also written as watts. Okay, have you heard of horsepower? Heard of horsepower? Where have you heard of it? Cars, bikes and all that. So horsepower is a very popular unit. The reason why it is very popular, because earlier times horses were used for the transportation. So it becomes a fair comparison or you can visualize, okay, how much power this car can provide, okay? So horsepower is still getting used and hence we should know how much is one horsepower. Anybody knows how much one horse power is in terms of watts? Anyone? Correct, 746, 746 watts. One horse power is 746 watts. Can you roughly tell one human's power will be how much? This is power of one horse. One horse can do 746 joules per second work. What do you think a human's power will be? Any guesses or if you know anything? No idea? A person's power, do you think horse power is more or humans first of all? Horses will be more, right? Where are you telling human power to be more than 746? So the person's power is roughly 75 watts. This is human. Even if you bring Arnold, then also power won't go beyond let's say 100 watts. So horses are a lot more powerful, at least 10 times more powerful. Their age is of course very less. Their overall age is around 30 years. They don't live more than that, but then the power is tremendous. And that's why humans utilize it earlier for the transportation. Anyways, so that is about the power. It's a very, very small topic. So this is the definition of power. Let's take up a simple numerical and see whether we can solve it. All of you, do this. First step is drawing free bird diagram. If you're not drawing free bird diagram, the high chance you'll get it wrong. Your cables are suspending the elevators, of course. Okay, Rohan got something. Okay, so let me solve it. Everyone please attempt it. It doesn't matter whether you get it right or wrong. If you get it right, doesn't mean that you qualify for J advance. If you get it wrong, doesn't mean that you can't learn it later on. Okay, but then if you do not attempt it, definitely you will be on a losing side. This is the tension. You have 1800 kg. So mg force is there. I'll take GS10 18000 Newton and the frictional force is 4000 Newton. It's going up with velocity V, which is given as 2 meter per second. So acceleration is how much? Acceleration is 0. So forces will be balanced. Net force will be equal to 0. T minus 18000 minus 4000 will be equal to 0. Constant velocity is 0 acceleration. So T is 22000. Okay, so the power supplied by this tension, which is cable, will be positive or negative. What is the angle between velocity and the force? Angle between velocity and force is what? Over here, 0 so I can directly write down T into V because T into V into cos of 0 will come. Cos of 0 is 1. So power is 22000 into 2 watts. That is 44000 watts or 44 kilowatts. In horsepower, it can be written as 44000 divided by 746 horsepower. Understood everyone? Okay, now solve the same question. If it is not moving up, it is, let's say, moving down. Same velocity. My Newton velocity is same, but now it is going down. What is the answer? Will it be positive or negative? First tell me that. Negative. Velocity is down and force is up. So what is the answer? Do it yourself, okay? It is very easy to find out this question from your book, from various sources. Do it yourself, then only you will learn. There are no points in getting the right answer quickly. Okay, good to see that one. Sambhu, Aaryan, Kinshuk, they got it correct. They've got it correct. Those who are in a hurry, they will answer minus 44 kilowatt and that is terribly wrong. Because tension value will change. Not only the velocity is changing its direction. Because velocity is changing its direction, even the friction is changing its direction, isn't it? Friction no longer will be down. It will be up now because it is going down. So new tension T1 plus 4000 minus 1800, sorry 18000 is equal to 0. So T1 will be equal to 14000 Newtons. Okay, so power will be minus of 14000 into 2. That is minus 28 kilowatts. Understood everyone? Anyone has any doubt? Quickly type in. Everybody understood? Yes, we are assuming ideal cases here. Anyone has any doubt? All right. By the way, I forgot to mention at the start, what the main thing is that we are going to reshuffle you guys. As in not just NPS, RNR or NAPL, there will not be any concept of any school now as such. So whatever number of students we have from all the schools, we are going to create a batch. Two batches will be there or can say two type of divisions will be there. One will be the where we'll be focusing on the advanced stuff. And theory will be very less straight away problem solving. Problem based learning will be there. And other will be the mains level. Where in those who are interested in the mains or CT or neat preparation because all these preparations are very similar. Advanced preparation is a little different. And that's how we'll reshuffle. And that will be done by end of this month. So then you will see that there are students from NPSHSR, from DPS East. So we can do all that till the time we are having online classes. So that is what is the plan. I think I've mentioned it last class also. If I'm not again at least. Alright, so now we are getting into collisions. Can you give example of collisions? Anyone? Whatever come in your mind. Tell me what is collision? Two balls collide with each other hitting one particle, the other particle car accident. Hammer and nail. Hammer and naffle. No, nail. When two objects crash into one other. What else? When two objects hit each other with a force and they move back with a force. Bouncing of a ball. Good. So I think all of you have this similar explanation about the collision wherein one object has to hit the other object. So let me just give few examples here and then you tell me which are the examples of collision, which are not the example of collision. Okay. I'll just write down six different kinds of things. The first one, which many of you have said is one ball coming and colliding with the other ball. Okay. This ball come and hit the other ball. All right. The second example. Let's say is this wherein there is a wooden block like this. Okay. Then there is a bullet coming in bullet is coming inside with some velocity and it just goes inside get embedded. It doesn't come out. Okay. Then you have let's say another one wherein a bullet comes in it then pierces it and then come out. It doesn't stay in. Okay. Prakul stop typing. Okay. You can wait. Prakul and Akshath you both. Let me finish first. I don't want your comments right now. Understood. Third is let's say you have a, this is an inclined plane like I guess inclined plane is coming almost every problem now and then. So let's say this is theta. Okay. You have a block of mass M. This is capital M. So this block, you know, both the inclined plane and the mass, both are, both are, you know, free to move. But when this small m is about to move, like when this small m is moving, capital M has also started moving. Okay. Fifth. You have a positive charge. You have another positive charge. It starts moving. The first one started moving towards the other one. And then other one also starts moving. Why the other one starts moving because this positive charge will ripple the other positive charge. Okay. The sixth example. Is this one. You have a bomb. Okay. And it explodes into three pieces. This piece goes this way. This one go that way. This is going that way. There's an explosion here. Okay. So these are the six different type of scenarios. You have to now tell me which one are collisions. Okay. You can type in your answers now. No hurry. Okay. Take your time and answer it. So is is fifth is fifth collision or not fifth one. Why? Because they're not physically touching. There's no contact. Can you define contact? What do you mean by a contact? Anyone contact means what forces exerted on each other. Okay. Does contact mean that electron and proton from one hand touching the electron proton on the other hand? No. There has to be some sort of force among each other presence of some normal force or any force. Okay. In fact, contact never happens. Fine. So talking about contact is useless saying that there has to be a contact in a collision is meaningless. So first I will tell you what is the answer. Answer is all are the forms of the collisions. One, two, three, four, five, six. Everything is collision. Okay. Now your English definition of collision is different. Where in only when car crashes on the tree, then only it is a collision. Okay. That is how we understand it in day to day life. All right. In physics, just like we have work done is going to force into displacement and you doing any work like you completing the assignment according to physics work done is not there because force and displacement are not there. Okay. Similarly, in physics in mechanics, the definition of collision is different. Okay. The definition of collision is this one right down. If one object affects the motion of the other object, we say collision has happened. Okay. Plain and simple. Now you look at all the example. One object is affecting the other objects motion. Look at the fourth one is in presence of this small m is making the capital M move heavier. This is small m is faster. The capital M moves. Okay. Over here, sixth one might be a little confusing to you. But sixth one also the how big this particle is and how fast it is splitting outwards will determine what will happen to the other two pieces. Okay. So everything is an example of the collision. Okay. Anyone has any doubt here? Kirtana, have you understood? We will discuss it a little bit. I mean, this is just a start. I'm just telling you that, okay, fine. These are collisions. We haven't yet started discussing about the collision. I'm just telling you that open up your mind. When you talk about collisions, open up your mind and don't restrict your thinking that only, you know, only example number one is collision. Okay. Example number one is the first thing that will come in your mind. Where two objects come and hit each other. Look at example number two, one object goes inside the other object. Example number three, one object goes inside and comes out. Example number four is both the masses are already in contact. Contact is not getting made during the collision. During the collision, they're affecting each other's motion. That's all. Example number five, there's no contact at all. And example number six, originally only one mass is there. Okay. So all our examples of collision. So whenever we discuss anything related to collision, all these six examples are, I mean, everything is valid for these six examples. And these six examples are not the exhaustive examples. There can be many more examples also. Fine. So this is the collision and we are going to start the study of collisions with this concept of write down conservation of momentum. Type in your doubts, if any, can constant motion be called? Yeah, why not? There can be constant motion that are collision, but you will understand a bit more when you hear a little bit more about the collisions. Okay. You did this topic under laws of motion in school. Yeah. In fact, many times this topic is done with system of particles or rigid body motion. That is a next chapter. Okay. But I want to do it in work energy chapter because I think that is the appropriate place. In laws of motion, you might have done about the impulse. Okay. And momentum, but conservation of momentum should be done in work energy chapter only. Anyways, suppose you have two masses. Okay. Suppose you have two masses. We are taking example of the most obvious form of collision, but whatever we do is valid for all six forms. And you will see the way we analyze is valid for everything. Let's say mass M1 is moving with velocity V1. Okay. Mass M2 is moving with, let's say velocity V2. This is M2. Okay. And the collide. During the collision, what will happen is that they will exert force on each other. Right. These two masses, they're colliding. So they will exert force on each other. Okay. So the way it will be, let's say I draw these two again. Now in contact, let's say this one is M1. And this is, sorry. This is the M2. Okay. On which direction they will exert force on each other. Anyone? Suppose they are disc, circular disc. Which direction they will exert force on each other when they collide? What do you think? Along the line joining the centers, right? They will exert normal reaction on each other. Isn't it? So let's say second mass exert force in this way. This, they will exert normal reaction. Okay. They are smooth. The force is the normal reaction along the normal. So this is, let's say F12. Force on M1 due to M2 is F12. What is the force on the second mass? What can you say about the force on the second mass? Yeah, that is a name of the force. F21 is the name of the force. What you can say about F21 equal and opposite. Very good. So F12 and F21, they are equal and opposite to each other. Which law have you used here? Anyone? Newton's third law, right? Correct. So this is the mechanism. All right. Now can I say, can I say that F12 is rate of change of momentum for mass M1? Can I say that during the collision, during the collision, whatever force is there, F12, it will change the momentum of M1. Is it correct? And minus F12, which is this force will be equal to rate, the momentum is changing for the second mass. I can say this. Now if I add them up, I'll get zero on the left-hand side. This is equal to dP1 by dt plus dP2 by dt. Fine. So if I take d by dt common, I'll get P1 plus P2. These are vectors by the way, is equal to zero. What does it mean? Derivative of a variable is zero. What does it mean? Derivative of P1 plus P2 is zero. So P1 plus P2 will be a constant. So P1 plus P2 will be a constant. So initially whatever is the sum of P1 and P2, initial, this is equal to sum of P1 and P2, final. What does it mean? Before the collision, when they have not been interacting among each other, they were not creating any force on each other. Let's say the momentum is P1 plus P2 initially. Same momentum will be after the collision also. Even though momentum of M1 is changed, whatever is the momentum of M1, it is changed. Momentum of M2 can also be changed. After collision, but when you add them up, the sum will not change. Everyone understand this point? Now tell me, yeah, this is called conservation of momentum by the way. Tell me what is the assumption here? Assumption is what? A big assumption we have made here. No heat. Where have we made assumption of heat and all? There can be resistance. There can be resistance. Okay? It doesn't matter because then this force will not be along the normal. There will be tangential force also, but even tangential force will be equal and opposite or not? Tangential force, if there is a friction, this will exert this way, that will exert that way. Even friction if you consider, it will add up to zero. No, no, no. It's not about energy. Have we considered energy anywhere? It's about force only, isn't it? I'm waiting for the answer, correct answer. The bodies can start moving in any direction as long as momentum, some of their momentum is same as the before. Doesn't matter. Mass does not change. No, we haven't made that assumption. P1, inside P1 and P2 mass is there, no? No one. No damage. No, there can be damage. Doesn't matter. Constant velocity, not required. Velocity can change. In fact, velocity is changing. After collision, don't you think their velocity will change? Bodies need not be rigid. It can be compressible. Where have we assumed that bodies have to be rigid? Whatever we have assumed. In fact, when collision happens, there is a small amount of deformation that happens. I mean, in the first case of collision. Okay. The assumption is that, yes, isolated system. Correct. So, Lavannaya got it correct. So, there is no external force. No external force to the system. Are you getting it? This is very important. Only force that is present is water force that's exerting on each other. Right? From outside, someone should not exert force. Otherwise, what will happen is that over here, from force outside, plus force between them will be equal to change in momentum. When you add it up, left-hand side will not be zero if outside force is present. Okay? So, this you need to understand. You can apply conservation of momentum right down. Only when there is no external force to the system. The masses when they collide, they can exert force on each other. Okay? So, let me go back to this slide over here. You can look at exam number four. Example number four. One mass is exerting force on the other mass. Continuously. Okay? So, that is fine. From outside, they should not be any force. Is there a force from outside, by the way? Example number four. Can you tell me? Is there any force from outside? Other than what they're exerting on each other. Small m and capital M. Gravity. Gravity is one. Is there any other force from outside? Friction. Friction between small m and capital M, Raghav. Normal reaction. Very good. This could be the external force. Even the friction between the surface and capital M is an external force. Friction, yes. Between floor and capital M is an external force. But if friction is there between small m and capital M, that is fine. That is internal. Okay? So, I hope little bit clarity is there. We are going to analyze it a little bit more. Write down. So, we have written conservation of momentum to be P1 plus P2 to be constant. P1 can change. P2 can change. But P1 plus P2 cannot change. Okay? Let's say P1 is Px1. ICAP plus PxPy1. What is Px? Anyone? Px is what? Component of momentum along the x-axis. Very good. Plus Px2 ICAP plus Py2 jCAP. This will be constant. So, you can write it as Px1 plus Px2 ICAP plus Py1 Py2 jCAP. This should be constant. If this is constant, can I say this has to be constant and this also has to be constant? Some, even if the sum is constant, then also component along z-axis, component about y-axis has no implication on component along the x-axis. So, they have to be individually constant. What I'm trying to say is that momentum along x-axis should be constant and momentum along y-axis should be constant individually. Independently, they have to be constant if entire momentum is constant. Okay? Now, can you tell me if net external force along x-axis is not equal to zero? Is not equal to zero? Can I say Px1 plus Py... Is this constant? Is this equation number one valid? If external force is present, equation number one is valid or not? No, it's not valid. What about equation number two? Is equation number two valid? Okay. Yes, equation number two is valid. Whatever happens in the x-axis has no implication on whatever happened along the y-axis. So, you will understand you now know the power of vectors, isn't it? If external force along the y-axis is not equal to zero, but of course, external force along x-axis should be equal to zero. Over here, we are assuming that external force along y-axis is zero when we write this. So, if this is along y-axis, net force is not zero, but along x-axis it is zero. I can say Px1 plus Px2 is constant. Anyone has any doubt till now? Type in. Quickly. What do you think momentum along x-axis will be? Mass into mass into what? Velocity along x-axis. Okay, so you guys know it, right? Component of velocity along x-axis, if you multiply with mass, you will get momentum along the x-axis. Okay, so you know how to write it. So, what is initial momentum and final momentum? How you write it? Mass into initial velocity is initial momentum. Mass into final velocity is final momentum. Okay, got it? So, let's solve few questions now on whatever we have done. Any doubts? Alright guys, so you have you have a mass. This is smooth. You have a small m coming with velocity v1. It goes inside capital M then comes out with velocity v2. Mass remains same. So, v1, v2 are given. You need to find out what is the velocity of the capital M after this has happened. Do you see, can I say horizontally there is no external force in the horizontal direction? There is no external force. Can I say that? So, can I conserve momentum horizontally? Yes, the condition is there. Horizontally I can conserve momentum. So, whatever is the momentum for both the masses, both initially will be the momentum of both the masses finally. So, write down the expression and tell me the answer. Momentum is what? m into v1 plus what is the moment of capital M? Initial capital of momentum is 0. This should be equal to what is the final momentum of small m? This for capital M it is capital M into v. So, from here you will get the value of v to be equal to m times v1 minus v2 This comes from the conservation of the momentum itself directly. Everybody understood this? This example everybody understood? Type in quickly. Let's proceed further. Have a boat. It is placed on a surface of water almost no, I mean zero friction you can assume. Ideal case. It is there on the water zero friction. There are two persons. This person has everything was at rest. Everything was at rest. This person suddenly started moving with v1. That person started moving this way with v2. You need to find out with what velocity the boat start moving. Let's say velocity is v. The person's masses are m1 and m2's boat masses capital M. What is your what is your system? Are you taking all three masses in your system? Are you taking all three or only two? All three you have to take. All three. For collision there can be n number of masses colliding. So you are taking all three and then you have to see a direction in which net external force is zero. Which direction it is? Horizontal. Or if you have taken x axis to be horizontal then x axis because x axis could be vertical also. So horizontal direction is the direction in which net external force is zero. So momentum will be conserved. What is the initial momentum of everything? Zero. Zero should be equal to m1 v1 then minus sorry zero is initial momentum that I am equating to final momentum now. So I have taken this direction positive right hand side. So m1 v1 is positive quantity minus m2 v2 negative quantity plus capital M into v. So from here you will get v to be equal to m2 v2 minus m1 v1 divided by capital M. These are the some of the straight forward examples which you can have with whatever you have studied till now. So I hope things are very clear now at least the basics are clear. Now we can we can get into the details now. Shouldn't we consider m1 m2 also for boat mass as they are in the boat. Yes we have considered m1 m2. Isn't it? Haven't we considered m1 m2? Pachita. No it is not included. It's not included in it. Body move by initial momentum is zero. Initially everything is what was at rest I said. Later on it started moving. Kiran why you joined so early? One hour late you came. Sorry sir. No sorry tell me the reason. Sir I didn't the time go sir. I was doing something else. Something else what exactly? Everybody is interested tell me what you are doing listening to you. Next time you are not allowed like this. Understood Kiran. Right guys so let us proceed. We are going to discuss more about the collisions. Right now there can be there can be many types of collision. You can classify the collision now. So suppose I have to study little bit more about the collision. I will have to first classify and then study it. So with respect to classification there can be two broad based category that can have for the collisions. Have you heard of elastic collision in elastic collision? Have you heard of these terms? I mean that is better. You know you knowing something will be not good actually. Now I can tell you everything from start and you will understand everything. Otherwise you will be like oh two years back someone said told me something else. What happens in laws of motion? Right head on collision you heard. Okay. Yeah even that is there. Right down elastic collision. One type of collision could be elastic and there can be in elastic collision as well. These are just name. Okay. So when a bullet hits the Superman the elastic collision happens. The bullet will bounce off. You know in every collision do you think energy some of the energy gets converted into heat and sound or not? What do you think? It gets converted into heat and sound. Light also. There can be many different form the energy is converted. So if you just talk about before and after the collision if you just talk about before and after the collision immediately before and after potential energy won't change much. Potential energy will be constant. Suddenly high difference will not be there. When I talk about the energy before the collision I mean kinetic energy. Okay. Immediately before and after collision if there is energy loss okay in terms of heat and sound that energy loss is because some kinetic energy earlier in terms of kinetic energy some part of it got converted into heat and sound immediately after collision. Okay. So with respect to energy only these are the two classifications. Elastic collision is an ideal case. Okay. It is an ideal case which never happens but this is one of the extremes we have to discuss. In elastic collision no kinetic energy is converted into heat and sound or any kind of thing. Kinetic energy before the collision is equal to kinetic energy after the collision in the case of elastic collision. Okay. So write down momentum is conserved there are near to the practical collision have you played billiards pool snooker have you played that that is near elastic collision. Marbles also you can say write down no kinetic energy you know if if I if I drop a marble on the floor and if it is an elastic collision no kinetic energy is lost immediately after the collision so velocity just before the collision is equal to velocity just after the collision that will keep on bouncing till infinite time till eternity it will keep on bouncing up and down if kinetic energy is not converted into heat okay so ideal cases don't exist no kinetic energy I mean till till I guess two or three years back we used to say that even getting 100% marks in J mains is an ideal scenario that doesn't exist but one of the person got 360 out of 360 in J mains couple of years back so I can't take that example now when you throw it up and down you're giving it energy so kinetic energy no kinetic energy is converted to heat heat sound so we can conserve we can conserve kinetic energy just before and after collision okay this we can do this is the ideal scenario ideal scenario elastic collision ideal scenario we can assume on paper I can tell you that elastic collision is happening okay so many questions are on elastic collisions then in elastic collision do you think momentum will be conserved or not everyone momentum is conserved here also momentum is conserved but some energy some kinetic energy got converted to these things heat light sound okay because of that we cannot conserve kinetic energy just before and after the collision okay so these are the scenarios elastic collision is a specific scenario in elastic collision is everything other than elastic okay now there can be a specific scenario of in elastic collision also what that can be the elastic collision is when total energy initially is equal total kinetic energy initially is equal to total kinetic energy finally entire kinetic energy is conserved is perfectly elastic collision what could be the other extreme what could be the perfectly in elastic collision what do you think the other extreme of it all kinetic energy is consumed but that is doesn't happen all kinetic energy if you say that entire kinetic energy water was there earlier becomes zero the momentum won't be conserved then to make sure momentum is conserved okay so I can understand first thing that will come in your mind is that the other extreme has to be when the final kinetic energy becomes zero but that may happen may not happen depending on momentum is conserved or not so you can write the perfectly in elastic collision is a case maximum max loss of initial kinetic energy happens one more thing you can keep it as a point here between between two masses perfectly in elastic collision happens when both start moving together after collision or both of them could stop after the collision so basically they should not go away from each other after collision they should just get stuck they should just get stuck on each other a car hitting a tree is an in elastic collision okay can you point out any in elastic collision in this case this one over here any in elastic collision which example example number two correct okay and it is not that this thing is coming to rest if this is coming with velocity u this entire thing could start moving with v but whatever the velocity of the block is the velocity of bullet also okay both are moving together after the collision that is perfectly in elastic collision anyone has any doubt quickly type in no doubts okay have you heard of conservation of momentum before like in class ninth we should be already knowing a lot of things problems on recoils you have done okay alright so let us continue our discussion we had two classification elastic and in elastic elastic is one extreme perfectly in elastic is the the other extreme okay now you can classify the type of collision into the way collision is happening also for example I can have you heard of head on collision head on collision have you heard heard of it okay I will talk about example here first write down head on collision in the head on collision also there can be elastic and inelastic collisions okay it's like you know let me take an example let's say you have set of 108 elements you can classify them with respect to which are liquid which are gas which are solid then you will be like let me find out which are metals inside the metal also you will have some of them could be solid and few of them could be liquid so similarly we first classified the collisions into broad categories based on energies conserved or not now we are classifying the collision depending on how collision is happening okay now how collision is happening let's say I am talking about head on collision inside the head on collision also there can be elastic collision inelastic collision okay so write down head on collision first I will tell you what it is with examples which you know pretty well now all of you have played Karam right all of you somebody who hasn't played Karam any unfortunate in chemistry you have played Karam anyone who hasn't played Karam Karam board everybody played good yeah now even if somebody hasn't played will not say that I haven't played okay so suppose this is a coin this is a coin Karam and you have a striker this is a striker suppose you know many a times what we want is coin to go in this direction right this direction we want coin to go so what we do is that we hit the striker along this line only whichever direction I want the coin to go I hit in that direction this is head on has it ever happened that you want the coin let's say this coin which is there but you are hitting it your striker is moving in this direction okay but you don't want the coin to go in that direction you want the coin to go in that direction has it happened right cut you take a cut isn't it notice that I was just taking the example of the Karam board in example number one you want the coin to go in the direction of the velocity of your striker in example number two you want the coin to go in some other direction other than your velocity of the striker okay the first example is head on okay second one is oblique both of the cases only point contact is there it's not the entire striker is touching this only this point is touching over here probably this point is touching only one point touches okay head on and oblique same thing happens in the pool, snooker, billiards also sometime you want the the balls to go in the direction of your striker okay so you have a head on collision most of the time in fact you take a cut you don't want it to go in the direction of your striker's movement you take an oblique collision now exactly what happens in this scenario in the head on scenario we will talk about oblique little later okay in the head on in the head on collision force due to collision force due to collision is the force which they are applying on each other force due to collision is along the line of velocities okay it's not necessary that this coin is at rest this coin can also be moving but it has to move along that line only and when this striker hits the coin they exert force also in the direction of the velocity so that is the reason why both before and after the collision their velocities are along that line only whereas in this case when the coin hits this so when the striker hits the coin the force is in this direction and this ball will experience force in that direction along the common normal so the common normal is not along the line of velocity so that is why you will have velocities in the different directions everybody clear about these examples quickly type in have you understood what is oblique and what is head on this should be very clear because further discussion in on this only and numericals will be on these things only oblique will discuss the will discuss oblique right now focus is head on okay head on you understood okay so let's go ahead with the understanding of the head on collision right now so first case in the head on collision is elastic head on collision so you have let's say two masses one is going with u1 velocity other is going with u2 velocity their masses are m1 and m2 okay collision has happened and after collision their velocities are changed okay now this guy is moving with v1 that mass is moving with v2 their masses are again m1 m2 only m1 m2 okay it is elastic head on collision before and after the line of velocities are horizontal only okay it is happening on let's say on a table frictionless table so you need to write down the equations here right it what are the two equation conservation of momentum and conservation of energy right it let me know once you are done conservation of momentum written what will be the equation initial momentum is m1 u1 plus m2 u2 this should be equal to m1 v1 plus m2 v2 this is conservation of momentum now conservation of energy have you written this equation energy equation have you written initial energy is half m1 u1 square plus half m2 u2 square this is equal to half m1 v1 square this is numerical okay this is take it as a numerical this is not theory theory is over long back theory was I can tell you till where theory it was first two slides are the theory everything is numerical okay so half m2 v2 square everybody understood these two equations type in quickly now now I want to find out v1 and v2 in terms of everything else okay so if we have to find out v1 and v2 in terms of everything else you will first you will get worried because there is a square term it is difficult to directly get into solving these two equations to get v1 and v2 okay so we will do some sort of trick here all of you look at it equation number one first I will modify one I will take m1 on the left hand side m1 u1 minus v1 is equal to m2 v2 minus u2 okay this is a modified form of equation number one equation number two half I will remove then I will take m1 this side I will get u1 square minus v1 square which is u1 minus v1 into u1 plus v1 this is equal to m2 v2 square minus u2 square which is v2 minus u2 v2 plus u2 now what comes in your mind when you look at these two equations anything what is the first thing that will come in your mind most obvious thing you have to think in a free manner as if you are solving puzzle okay forget that this is mathematics or physics if you look at it what should it come in your mind divide divide it if you divide it m1 u1 minus u1 this you get cancelled you will get u1 plus v1 is equal to v2 plus u2 or u1 minus u2 is equal to v2 minus v1 okay so you have got another equation which you can say equation number 3 now solving 1 and 2 isn't it same as solving 1 and 3 what do you think solving 1 and 3 will give you the same answer as solving 1 and 2 or not quickly type it solving 1 and 3 yields same thing because 3 is coming from first 2 only now solving 1 and 2 is easy or solving 1 and 3 is easy what do you think 2 and 3 is easy 1 and 3 why solving 1 and 3 is easy because both 1 and 3 are linear equations okay they are linear equations now before solving it before going ahead tell me what is u1 minus u2 something which you know already something which you already know it is that what is u1 minus u2 are you able to identify that no no no something which we have been using continuously in motion in ah good Weber got it Madhumati got it v1 minus v2 is velocity of approach isn't it what is v2 minus v1 what it is v2 minus v1 is velocity separation okay so see it has come here also that velocity separation and approach concept so for elastic collision you first write it down before solving it this is true for every elastic collision for elastic collision velocity of approach before before the collision is equal to velocity of separation after the collision along the common normal not only head on it is valid for the oblique collision also but you have to take component along the common normal approach and separation happens along the common normal we will discuss it when we talk about the oblique collision right now for head on it is straight forward velocity of approach is equal to velocity separation okay so very important aspect of the elastic collision okay alright now please solve equation number 1 and equation number 3 and tell me the final answer okay all of you solve equation number 1 and 3 tell me the final answer I want v1 and v2 in terms of everything else it will take 1 or 2 minutes but get the answer it is totally worth it and let me tell you this is your derivation of something which is there in your school this will be useful in your school exam also okay either you learn it here quickly or you have to spend lot of time later on to learn anyone close to the answer v1 and v2 kinshok are you close to the answer not able to solve 2 linear equations no one arjit what is your answer type it no kinshok you will not get that okay I will write the answer if you solve correctly you will get v1 and v2 these the value of v2 just a minute you solve it I will take some time to write it let's see whether you get it I am sure some of you are not solving it waiting for copying down doesn't help everything is written in the book this is this is that arjit you got this only for v2 madhumati you got it arjit got it okay good v1 will be 2 m2 u2 minus m2 minus m1 this is u2 by the way u1 this v1 and v2 u2 u2 alright these are v1 and v2 so when you solve equation number 1 and equation number 3 1 and 3 you are going to get v1 and v2 in terms of known quantities which are this okay now look at these expressions and tell me if m1 equal to m2 if m1 equals to m2 need not be equal okay we are considering a special case if m1 is equal to m2 what will be v2 equals to look at it tell me substitute m1 equal to m2 what you get will be equal to u1 or not what was u1 what was it initial velocity of initial velocity of m1 that becomes final velocity of m2 so velocity of m1 became velocity of m2 after the collision and what about v1 so velocities get exchanged elastic collision elastic head on collision by the way elastic head on collision between 2 equal masses what if u2 is 0 what if u2 was 0 initially if m2 was at rest this object this is this is suppose at rest before the collision then what will happen after collision if m2 was at rest before the collision m1 will be at rest after collision are you able to understand this everyone type it quickly have you understood this if m2 is at rest before the collision m1 will be at rest after collision and this is exactly what happens over here this ball comes and hit this ball near elastic collision happens this was at rest before the collision after collision this will come at rest this will try to move again elastic collision with this this will become at rest elastic collision elastic collision then this will start moving ok so this is what happens in the newton's pendulum and same thing happens in the billiards also those who have played billiards have you seen when you hit the striker head on collision after collision the striker comes to rest and the other ball starts moving with the same velocity because it is near elastic collision has it happened in billiards or snooker that happens many a times that is the near elastic collision so balls of billiards and snookers they are made very very very very specially they are made of you should not you can't put lontan's ball in the billiards table and start playing with it heavy and very rigid they are heavy and very rigid so they will create near elastic situation this is what perfectly elastic collision where in after the collisions velocity get exchanged if masses are equal and kinetic edges conserved now let's talk about perfectly perfectly in elastic collision have they discussed it in school when they thought collision what are we are doing because it will come in work per energy chapter little bit of intro is there in laws of motion they will again talk about it probably in work energy chapter so right on perfectly in elastic head on collision it's a situation like this where in a mass m1 okay it hits the other mass m2 if it is perfectly in elastic what will happen anyone after collision what will happen they will start moving together they start moving together we so you have to do these two things numerical take it like a numerical okay this is not theory again I am repeating find out find out final velocity of the collision and loss in kinetic energy final velocity you will get it easily consumption of momentum initial momentum is m1u what is the momentum of m2 before collision 0 this should be equal to m1 plus m2 into v right so v will be equal to m1 divided by m1 plus m2 times v use do this second part this I think most of you have got it if somebody did not get it please type in any doubts get the loss in kinetic energy loss in kinetic energy should be equal to initial kinetic energy minus final kinetic energy should it be always positive or it can be negative also can it be always positive or it can be negative what is the minimum value of loss can it be negative okay can never be negative minimum value is 0 that happens in the elastic collision when no loss in the initial energy happens energy cannot be created on its own suddenly final kinetic energy will not come out of nowhere in this case however when a bomb explodes then it can be negative final kinetic energy is there but initial it was 0 because chemical energy got converted into kinetic energy anyways initial kinetic energy is half m1 u2 final is half m1 plus m2 this okay so the loss in the kinetic energy will come out to be half of m1 m2 divided by m1 plus m2 times u2 so it's a positive quantity you are getting everybody understood alright so we have learned about the head on collision which are elastic which is one extreme and perfectly inelastic that is other extreme but then these are the two extreme scenarios there are many scenarios in between wherein it will be it can be written for example for example if I tell you that 20% of energy I don't suppose this kind of thing is there 20% of kinetic energy is lost in collision so you just have to write down the equation you don't have to do anything else just write down the equations let's say this is the situation inelastic collision happens but it is not perfectly inelastic so after the collision they need not move together so this is m1 m2 let's say velocity is u1 after the collision let's say this is u2 after the collision their velocities are v1 and v2 I just want you to write down the equations don't have to solve it what is given is it is inelastic collision wherein 20% of kinetic energy is lost during the collision let me know once you are done momentum will be conserved so m1 u1 everybody could have got this how to write the energy equation how much kinetic energy got converted to final kinetic energy how much initial kinetic energy got converted to final kinetic energy 80% or 0.8 times so 0.8 times the initial kinetic energy m1 u1 square plus half of m2 u2 square this become equal to the final kinetic energy anyone has any doubt quickly type it this is inelastic collision only but partially inelastic collision everyone understood this let's proceed which part last part see 20% of energy is lost isn't it 20% of initial energy is able to convert itself to final energy 20% is lost that is a 0.8 times of initial is equal to final understood right on coefficient of restitution coefficient of restitution is a concept wherein we account for the inelastic collisions which are not perfectly inelastic okay many a times this variable is used to tell a situation related to the collision what is this variable let's try to understand it we'll have a break in some time let me finish this coefficient of restitution amount of break will be 15 minutes only that can happen anytime isn't it write down coefficient of restitution so in case of elastic collision the velocity of separation was equal to velocity of approach okay so the value of coefficient of restitution is given as velocity of separation it's just a hypothetical you can say a variable which is created which is equal to velocity of separation divided by velocity of approach okay now tell me the value of coefficient of restitution in case of elastic collision for elastic collision what is E E is equal to one clearly numerator is equal to denominator isn't it why there is a mod sign because velocity is a vector I don't want to divide with a vector and I don't want to include signs and everything just the way it is like I have created it I can create it any which way just like that elastic collision is coefficient of restitution E is equal to one what about perfectly in elastic collision what is E all of you answer think of it and answer everyone two bodies are colliding perfectly in elastic the other extreme what with the velocity of separation if it is perfectly in elastic both are moving together velocity of separation will be zero right so numerator becomes zero so no matter what is the denominator E will be zero so one extreme has E is equal to one other extreme has equal to zero so for everything else E has to be between zero and one you can say everything is included in this alright so many a times in your numericals the value of E is given okay E can be given as let's say 0.6 or I can tell you E is 0.7 like that so if I give you the value of E you have to write down the equation related to it you have to find out what is the velocity of separation what is the velocity of approach divide it and equate that to whatever is the value of E okay and again I am writing here that velocity of separation and approach to be taken on the common normal so this concept of coefficient of restitution is valid not just for the head on collision but for any type of collision okay oblique collision any type of collision you can think of E you can have in the question and specially when when an object collides with infinite mass object we can't write conservation of momentum or energy equations what is the example of this when an object collides with infinite mass object yes that's correct you have to take component along the normal to find velocity of separation and approach infinite give me one day to day life example where in a mass collides with infinite mass ball hitting the ground right ball hitting to the ground or ball hitting to your floor okay during those cases you can't conserve energy or you can't conserve momentum mass of earth into velocity of earth when you conserve the momentum so during those cases when ball hits the floor or wall or ball hits the earth all these cases coefficient of restitution will be given to you to find out what will happen after the collision okay everything everyone is clear about coefficient of restitution everything is clear we'll take one numerical before the break then clear right we'll take a numerical on coefficient of restitution now alright don't hold your doubts to yourself please ask the doubts okay doubts are the ones where in maximum of learning happens and doubt has to be specific it cannot be that okay this can you do it again or something like that it has to be very specific you know we can say coefficient of restitution can it be this if that scenario happens what will be that so something like this try to create doubts it will be better for you okay even if you don't have doubts try to think in a critical manner doubt will automatically come alright there is a ball which is coming and hitting the floor from a height of H let's say this is the height H okay coefficient of restitution is E E is the coefficient of restitution between the ball and the floor okay so you need to find out what will be the velocity of the ball immediately after the collision first find out all of you immediately after the collision what will be the velocity of the ball of course it has to be upwards okay how much the ball has no initial velocity ball is just dropped off did we meet after your J advance test sorry came on test advanced test did I talk about it so I have seen more and more of you coming there yours is the biggest batch toppers are see some names there very few names I can see in top 10 alright so velocity yes velocity just before hitting is root over 2gh right you can use v square equal to u square plus 2as okay this is the velocity just before hitting E is what velocity of separation which is what you want to find out divided by velocity of approach which is what root 2gh velocity of separation divided by approach is E so velocity of separation is E times velocity of approach now you have to tell me up to what height it goes how much is this distance up to which the ball bounces off h1 is what if you answer immediately your answer is wrong immediately how root 2gh between 1 and 2 you use v square equal to u square plus 2gs 2as little late or you joined since a start only so v square equal to 0 plus 2gh so v is root 2gh now then if you joined after the bridge program you must be must have solved a lot of questions on kinematics you have discussed it alright anybody got this h1 very good Rohan got it, Arjun got it again you have to use this equation only v square equal to u square plus 2as but now you are using between this point and the point over here what is the final velocity at the maximum height over here 0 initial velocity is this velocity square that is E square into 2gh expression is what what is a minus g so minus of 2g h1 so from here you get h1 is equal to E square h so this is after first collision after first collision now can you tell me after second collision how much height it will go up to h2 will be what after second collision how much you don't have to do it again understand the pattern here now E square h1 now E to the power 4 times h then h3 is E square of this which is E to the power 6 h like that ok so suppose you have to find out the total distance it travels before coming to rest can you create a series all of you you understand what I am trying to say here have you understood the question just create a series write down 5 6 terms and show that you have to add these terms to get the total distance traveled initially it has to travel distance of h then what should I write plus don't you think it will be 2 times E square h it goes up and then comes down same distance then goes up comes down the same distance goes up comes down the same distance goes up comes down the same distance 2 E square h plus 2 E to the power 4 h 2 E to the power 6 h this you have to add till infinity ok this will be the total distance traveled before coming to rest you have not done how to add the infinite series so I will not get into that but then I have just created a series so till here you must understand everybody understood type in everyone type have you understood if there is a doubt type your doubts clearly quickly still there are so many things about the collision your ncrt book has very small thing about the collision but we had to study so many details of it that is why it is taking a lot of time because I am not teaching you like a theory I am teaching you like the way you will see in numericals ok so using numericals we are trying to understand the collisions so that is why it is taking time anyways we will take a break now 6 19 right now after 15 minutes what will be the time 6 34 very good alright go and take your break have something and come back hello am I audible are you able to hear me alright so let us start as in continue we were studying collisions and till now we had studied the head on