 recorded. So, clearly this is my system and on this system there is gravity force that is external force yes or no. I am saying that even if gravity is there just before collision if their velocities are v 1 and v 2 and just after the collision if their velocities are v 1 dash and v 2 dash just before and after I am telling you ok, because gravity is non-impulsive. So, suddenly it cannot change momentum. So, even if gravity is acting I can conserve momentum. So, I can write m 1 v 1 minus m 2 v 2 this is the initial momentum opposite direction momentum is equal to m 1 v 1 dash minus m 2 v 2 dash. So, this is these velocities are v 1 and v 2 just before collision ok and v 1 dash and v 2 dash just after collision ok. This you need to be taking care of because even if gravity is non-impulsive force if you give it some time gravity can change the momentum. For example, you cannot conserve momentum 2 hours before and 2 hours after because gravity has good 4 hours to change the momentum. But if you conserve momentum just before and after collision because gravity is non-impulsive gravity cannot change the momentum suddenly. All of you understood that you can conserve momentum even if there is an external force of spring or gravity understood. Yes sir. You can conserve momentum just before and after ok that you need to be taking care of. But here are these three forces these three forces can be not can be impulsive ok they can be impulsive they can suddenly change the momentum they have a potential to suddenly change the momentum. So, you need to be careful when these three forces are there as an impulsive force ok. So, let us analyze how can we deal with the scenario when there is an impulsive force ok. So, v 2 minus v 2 dash by v 1 minus v 1 dash will be a constant for two given masses. Why? Because sir like if you here if you take m 1 and m 2 out and if you divide you get m 1 by m 2 is v 2 minus v 2 dash by v 1 minus v 1 dash and m 1 by m 2 is a constant. So, that is a coefficient of restitution only right. Ok. In m 1 you can take m 1. So, you are rearranging the terms only so that is fine. Yeah. So, but then that is a good observation. Ok fine let us see how the impulsive forces are been taking care of. So, first we need to quantify what is impulse ok impulse in words we know it is sudden change in momentum. So, we quantify the impulse as if it is just a change in momentum only impulse is change in momentum ok. The SI unit of impulse is meter sorry kg meter per second it is mass into velocity. So, we quantify impulse as change in momentum only and there is a letter associated with it it is called J. So, J is a momentum ok it is sudden change in momentum J is equal to change in momentum. Now, suppose collision is happening between m 1 and m 2 there is an m 2 as well. Now, because of the collision there will be a normal reaction and that will be impulsive sudden force, sudden large amount of force will be there and there will be equal and opposite force. So, this will be f and this is also f ok they are in opposite direction. So, if you write in a vector form. So, it will be f and that will be minus f fine. So, I can say that f is equal to rate of change of momentum of mass 1 like this you can write in the vector form. So, rate of change of momentum which is equal to f. So, change in momentum of the first mass become equal to f d t ok. So, when you integrate again and again integration is coming. So, you need to be comfortable with integration. So, integration of delta p 1 is change in momentum of the first mass this is integral f d t. So, right hand left hand side become change in momentum. So, right hand side should be impulse. So, integral of f d t is my impulse ok fine. So, on this the direction of impulse is that one on that impulse is this way fine and I know that minus j vector impulse is on the second mass this should be equal to change in momentum of the second one ok. So, when I add these two boxed equations I if I added up j and j will get cancelled away. So, I will get change in momentum of the first plus change in momentum of second is 0. Now, this has to be there because there is no external force ok, but then at times there is an impulsive external force. So, when you add the two equations it does not become 0 ok. So, we will see that in numericals how such cases are taken care of. All of you understood about impulse any doubt ok. Let us take a quick numerical on this. I think there will be a little bit of hesitation in terms of solving numericals on this topic let us quickly take a numerical. I am taking slightly tricky question ok. So, so that we can learn multiple things in the same question you just try it out. Suppose a ball is coming and when it is about to hit the floor when it is about to hit the floor its velocity is u and this angle is 45 degree or let us keep it theta only this is theta ok. A ball of mass m is coming like this this is angle theta ok coefficient of restitution for this collision is e ok. This is floor of course floor does not move floor will remain fixed coefficient of restitution is e and the coefficient of friction is mu ok. This is mu we need to find out the velocity of this mass immediately after collision just before the collision its velocity is u at an angle theta with the vertical ok. Try doing this all of you just attempt it you may or may not get it, but attempt it so that when I tell you you will able to relate it properly. Shushant says I cannot hear I hope all of you able to hear me yes sir yes sir ok Shushant rejoins rejoins Shushant or now Shushant no Shushant it is let me know once you are done attempting the question it is ok if you not get it sir is it minus e u? minus e u ok. Yes sir and the angle will be the tan theta will be mg by mu mg it will be 1 by mu ok. Who is this name? Sir Tirpan. Tirpan ok others Tirpan that is not correct actually, but then at least you attempted others please attempt quickly ok I will start solving in in couple of seconds all right. So, I will solve it now see this mass when it collides with the floor what are the forces acting on it there will be normal reaction see this is one of its kinds numerical. So, first time when you do this numerical nobody get it right ok, but then you need to pay attention because once you see this solution you should be able to solve all the similar kind of numerical there will be mg force ok and also there will be a frictional force ok which is mu times normal reaction ok. I am assuming that maximum friction force is being exerted ok, but that is one assumption I am making any doubt in this free body diagram? Sir. Ok this is clear now tell me which are the non-impulsive forces is mg impulsive? No normal reaction impulsive what about friction is this impulsive? See if not if ok if normal reaction is impulsive mu times normal reaction is also impulsive simple right normal is if n is infinity mu times n is also infinity ok. So, only impulsive forces are normal reaction and mu times n. So, I can say that integral n dt this is impulse along y direction ok and mu times normal reaction into dt this is friction force this is the impulse along the minus of that is impulse along the x axis because I am assuming this direction to be positive and that upward direction to positive x and y. All of you understood this I am ignoring mg because because mg cannot suddenly change the momentum getting it I am just integrating the time for which it is in contact with the surface it immediately loses the contact any doubt this is clear ok good. So, impulse in the y direction should be equal to change in the momentum along the y direction and impulse in the x direction should be equal to change in the momentum in x direction ball serve, but the ball strikes the floor at an angle yes Anjali that is correct perfectly correct we will take care of that ok we will take components we will take component of just like we have taken component of impulse this is vertical impulse horizontal impulse similarly we will take the component of momentum. So, that is why I am writing momentum along y direction and momentum along x direction clear. So, change in momentum along y direction is what assume that after collision after collision assume that it is going with velocity v and at an angle of pi ok then change in momentum along y direction is mass times v cos pi this is the final momentum along y direction ok and initial momentum along y direction is how much quickly tell me how much is initial momentum no that is wrong minus of mu cos theta is correct my change in the momentum along y direction is minus of minus mu cos theta ok. So, this is the correct equation so, we have got integral of n dt is equal to m v cos pi plus mu cos theta ok this is the first equation and the second equation is minus of integral mu n dt change in momentum in x direction is what final momentum along x direction is m into v into sin of phi ok minus initial momentum along x direction is what tell me quickly m u sin theta m u sin theta ok. Now, how to deal with these two equations how to solve these there is an integral sign also. So, can someone tell me quickly what to how to do now I mean what to do to solve these two equations no one square right ok see here it is this is mu right mu is constant so, I can take it outside the integral are you getting it so, this is same as minus of mu times integral n dt this term ok. So, if I multiply minus mu to this I can equate it to that what I mean to say is that minus of mu m v cos of phi see I understand this is little too much of mathematics, but then pay attention you will understand it is a like once you do it properly you will understand this lot of such questions are there so, we need to understand this we cannot ignore this have you understood this equation this is my first equation all of you understood yes sir ok. Second equation second equation is the coefficient of restitution equation coefficient of restitution is what E which is velocity of approach magnitude of velocity of sorry velocity of separation divided by magnitude of velocity of approach ok. Now, velocity of separation is what quickly tell me separation velocity is with what velocity the ball is separating v no you I mean why you are taking v cos theta v cos phi v cos phi see separation and approach please write it down separation and approach should be seen along the normal ok please write it down separation and approach should be seen along the normal if you are not moving along normal you are not coming closer or going far you are just moving parallel to the surface so, it does not mean that you are approaching approach will only happen when you move along the normal ok. So, velocity of separation is v cos phi and velocity of this is velocity of separation and velocity of approach we just need to take the magnitude ok. So, that is why I am ignoring the sign otherwise there is a minus sign in the in the formula itself v cos phi divided by u cos theta ok. So, this is my second equation. So, I have two variables phi and v. So, when I solve these two equations I will get the answer now I am not getting into how to solve these two equations it may be difficult because I just made up a question in the actual numerical there will be values of theta and phi given such a way that you will be able to solve it easily, but the concept remains this only ok. Now, please go through it once and let me know if you have any doubts in this question. Sir, could you repeat how you arrived at equation one? Ok, see this is the impulse along the y axis I am ignoring mg because mg is not impulsive it cannot suddenly change the momentum and I know that impulse see here we have derived impulse is integral of f dt ok force dt integral is change in momentum. So, that is what I have written impulse along the y axis is integral of n dt ok which should be equal to change in the momentum along y axis impulse along y axis can only change momentum along y axis it cannot change momentum along x axis and impulse along x axis is minus of mu and dt because I am taking that direction to be positive. So, it becomes minus of mu and dt this should be equal to change in momentum along x axis. Now, look at these right hand sides mu is constant. So, I can take it outside the integral. So, it is minus of mu times integral and dt which is nothing but this itself integral and dt. So, instead of integral and dt I am substituting mv cos 5 plus mv cos theta and I am getting first equation clear? Yes. Any other doubt? So, what are the limits of put for the integration? Ok limit will be such that it should denote time for which the force is being applied ok and we are assuming that the force is applied for a very very small period and limit actually does not matter I have not used the limits to integrate I have just casually written as 0 to tau I do not know what is tau I have never integrated it I just wrote it for for the representation sake. Any other doubt? Yes. Any other doubt quickly tell me? Ok no doubts no doubts fine. So, let us start solving the numericals from the entire chapter the chapter is over now ok when I say chapter is over the theory is over ok. The theory constitute just a 2 to 3 percent of the chapter 97 to 98 percent is only problem practice ok. So, we and you know that we cannot solve all kinds of questions in the class itself. So, 70 to 80 percent of your preparation happens on what you do at home. So, I am just showing you few type of numericals. So, once the class gets over please solve at least 100 questions of work by energy then you can say that I am done. So, if you practice now you will have to only solve 100 questions to be comfortable, but later on when you forget the concepts the concepts are not very fresh then you may have to solve instead of 100 you have to solve 200 questions then ok. So, do yourself a favor and practice in this week itself 100 questions. Anyway solve the question this question which you can see on your screen all of you attempt this one. I have taken simple questions only mains level I have not gone up to advanced level ok. So, this can be useful for your CET mains and NEET all three. Full work power energy is there yes Aditya the syllabus was till whatever we have done in work by energy and luckily or unluckily we have done entire chapter now. So, full thing is coming. If something is not given assume it to be ML whatever you want to assume if the answer is independent of that it automatically get cancelled away. Floor friction less yes it is not given here. So, we will assume flow to be frictionless that is a good question. Sir is the answer C 3 4. Ok I will solve it others you are getting any answer will the ball leave the floor with the same angle after collision. Actually that will happen if the collision is elastic, but here collision is not elastic see coefficient of restitution is not 1. Ok. So, this is 45 degrees and this is theta which we do not know let us say mass is m. Ok. Now tell me coefficient of restitution is half given. So, if initial velocity is U and final velocity is V how can you write half as equal to this is velocity of separation divided by velocity of approach right magnitude of that. So, velocity of separation is how much? V cos theta. V cos theta divided by velocity of approach is U cos 45. 45. Right. So, this is the first equation what will be second equation what do you think? Sir U sin 47 is equal to V cos very sin theta. Ok. So, it says no horizontal force. Right. So, you can see that there is no horizontal impulse friction is absent. So, no horizontal impulse this implies that momentum does not change momentum does not change in the horizontal direction. Ok. So, initial momentum initially is initial momentum initial m U sin 45 this is initial momentum this should be equal to somebody's background noises coming please mute it. This is the initial momentum in the horizontal direction just before the collision this should be equal to the final momentum in the horizontal direction just after the collision. So, m V sin theta ok. So, you can see that mass gets cancelled away from here as well. So, V sin theta is equal to U sin 45. So, U root 2 this is my second equation and from the first equation you will get V cos 45 sorry V cos theta Netra you need to mute yourself. Ok. So, V cos theta is equal to U cos 45 into half. So, I will get U divided by 2 root 2. Ok. This is my first equation. Ok. So, how do I get V from first and second equation what should I do now square and add theta is gone. Ok. So, when you square and add you will get V square sin square theta plus cos square theta which is 1. So, this is equal to U square by 2 plus U square by 4 into 2 8. Ok. So, you will get V square is equal to this is how much 8 10 by 16 10 by 16 U square. So, you will have 5 by 8 U square. So, if you multiply half and m both sides you will get half m V square is equal to 5 by 8 times half m U square. So, you can see that the final kinetic energy half m V square is 5 by 8 times the initial kinetic energy. Anyone got option A in this problem anybody was able to solve it. So, it should be 1 minus 5 by 8. Oh yeah. I am sorry about it. I did some silly error here. So, this is the final kinetic energy is 5 by 8. So, loss is 1 minus 5 by 8. So, 3 by 8 sorry about that. Ok. So, loss is 1 minus 5 by 8 times the initial kinetic energy. So, that is 3 8th. Ok. Anybody got 3 8? No one was able to solve it. Ok. But have you understood this? Any doubts here? Ok. Yes. Yes, sir. Ok. Fine. Let us go to the next one. This Akshit Ruchir Parekh they are also here from HSR batch. You also can speak up you cannot you do not need to be muted all the time. Sir, is your option C, sir? Ok. Who is this? Ruchir. Ok. Others? Sir, option C sir 2 by 5 meter per second. Who? Option C 2 by 5 towards the right. Ok. Who is this? Shankin sir. Shankin. Ok. It is hard for me to remember your names looking at your faces. So, remembering your names just by hearing your voice is even more difficult. Sir, option C. Option C. So, three people have got the option C. Anybody else? So, C C. C C. Wait sir. See clearly we have to use conservation of momentum here in the horizontal direction and there are three masses. Their initial momentum is 0, right. Everything was at rest. So, initial momentum is 0. So, total final momentum should also be 0. Ok. I will solve it now. So, initial momentum is 0. So, final momentum should also be 0. So, I have to take one direction to be positive. So, I am taking that direction to be positive. Ok. The right side. So, 0 should be equal to 80 into 1. Ok. Then momentum of 60 kg person is in opposite direction. So, that should be taken negative. So, minus of 60 into 2 and the plank has mass 100. So, I will take plus 100 into v. I do not know which direction the plank will move. So, I am taking at plus 100 into v. If that direction is not correct, v will come out to be negative. So, nothing to be worrying about. 0 0 0. So, 10 of v will be equal to 12 minus 8 that is 4. So, v will come out to be 4 by 10 meter per second which is 2 by 5 meter per second. So, option C is correct. Ok. A simple question on conservation of momentum. Ok. Any doubts? I am going to the next question now. That is one. First tell me when the maximum compression will happen. What is the condition for maximum compression to happen? M1 will compress. M1 will compress the spring, but it will keep on compressing. So, when, what is the, what will be the condition for which the compression in the spring is maximum? At that position, 0. Correct. So, both will move in the same velocity, both will move in the same velocity. So, that is the condition for the maximum compression. Ok. The velocity of M1 will decrease and of M2 will increase. Till the velocity of M1 is more than velocity of M2, spring will keep on compressing. As soon as the velocity of M1 equals to the velocity of M2, the spring will stop compressing and that is the maximum compression. So, you can conserve the momentum and you can conserve the energy also. You can use work energy theorem. Anyone? Anjali says it is 5 others. Ok. I will solve it now. How should I wait? Quickly tell me. Tirpan, should I wait? Yes, I will wait. Ok. Meanwhile, I will just write down the equations that will be used here. First equation is conservation of momentum. So, between the points before the collision, and when the compression in the spring is maximum. So, initial momentum is mass M1 which is 2 kg is moving with 10 meter per second. So, 2 into 10 is the initial momentum of M1 plus initial momentum of M2 is 5 kg of M2, 5 into 3. This should be equal to, when the both of them starts moving with the same velocity. Let's say velocity is V. So, 2 into V plus 5 into V. So, 2 plus 5 into V. All of you got this one? This relation? 12 plus 15, 27 by 7 meter per second. This is the velocity with which both of them will move at the time when the compression in the spring is maximum. Now, I can use work energy theorem. Ok. I can use work energy theorem here. W is equal to U2 plus K2 minus U1 plus K1. Ok. K1 is half into 2 into 10 square, half MV square for mass 1 plus half into V square, M is sorry, M is 5. 5 meter per second. 20 plus 15, 37. 35 by 7. 35 by 7. Ok. So, this is 5 meter per second. 5 meter per second. Right. So, K1 is this. U1 is what? Initial potential energy is 0. Ok. K2 is equal to half into total mass, which is M1 plus M2, 5 plus 2, 7. 7 into 5 square. So, as if it is moving with the same velocity, so I directly write both of them together. Or you can write half into 2 into 5 square plus half into 5 into 5 square. So, this will only come. U2 is equal to Kx square. W is 0. So, when you solve it, the value of x is the answer. Can you quickly get the answer? How much it is? It is in this entire chapter, you just have two things to take care of. Condition of momentum and work energy theorem. Nothing else you have to do. There should not be any confusion. Previous equation. Previous equation. Are you talking about the momentum equation? Anjali, that is momentum equation you want to see. The first one is conservation of momentum. Ok. So, I am conserving the momentum before the collision and when the compression in the spring is maximum. So, I have got the velocity when the compression in the spring is maximum, which comes out to be 5 meter per second. Condition for maximum compression is that both the mass move with the same velocity. Just one second. And the work energy theorem. In the work energy theorem, I am applying the work energy theorem again between the same two points. Ok. So, initial kinetic energy, you know how I have written initial kinetic energy, which is half m 1 v 1 square plus half m 2 v 2 square. So, that is half 2 into 10 square plus half 5 into 3 square. Ok. And initial potential energy is 0 because spring is not compressed and also everything is happening on the horizontal line. So, even the gravitational potential energy is 0. Fine. And finally, spring get compressed. So, the potential energy is half k x square. Ok. And the kinetic energy is the final kinetic energy, which is half m v square plus half m 2 v 2 square. So, that is what half 7 into 5 square is. And work done is 0 because there is no other force other than for which you have considered potential energy, which is doing work. Is it clear now? All of you, let us go to the next question now. This one. Sir, what is the previous answer, 21 centimeter? See, answer I did not store, but exactly how what we have done. It is just solving the equations now remaining. Ok. Focus on this question now. I will send you the final answers later. You can answer one by one. Thirteenth question you can answer first. Thiroption D. Ok. Other? Thiroption D. Those who think they have got 13th move on to the next one. Others, what is the answer for 13? Sir, 14 is A. I am, see, you can keep on solving 14 and 15. I am asking about 13 only. So, before, till I ask 14, please answer, please do not answer 14th one before I ask. Others, did you get the 13th question? Ok. So, for the 13th question, we need to apply the work energy theorem. W is equal to U2 plus K2 minus U1 plus K1. Ok. So, W is 0. K1 is 0. And let us say ground is my 0 potential energy. So, U2 is 0. U1 is MGH. And K2 is half MV square. So, V is equal to under root 2GH, which is none of these. Ok, wait, wait, wait. I am sorry. B is movable, right? Even B can move. Ok. I have assumed that B is fixed. Ok. So, that is not correct. I think most of you might have assumed B to be fixed. Those who got B answer, have you assumed wedge B to be fixed? That is wrong then. Ok. That is not correct. B can move. Block A is placed on wedge B. Block and two wedges are same mass M. So, why mass is given? Because even wedge can move. Ok. So, when this is coming down, wedge is sliding backwards like that. Understood? So, please solve thirteenth question again. All of you. The hint is, let me give you a hint. When the mass comes to the bottom most point, its velocity is horizontal and the velocity of this mass is also horizontal. Ok. Horizontally, there is no net force on these wedge and mass together. So, you can conserve the momentum in the horizontal direction. And then, you can use work energy theorem. Same two things you have to use. So, then the only answer remains the same. No, no. Please attempt thirteenth again. Wedge B and C are not fixed together Abhishek. B is a independent wedge and C is independent wedge. Ok. So, let us first write conservation of momentum, conservation of momentum equation. Initial momentum along the horizontal direction is 0. This should be equal to final momentum. So, let us say velocity of the mass M is V1 and velocity of the wedge. Velocity of the wedge is V2. Velocity of the mass is V1. Ok. So, 0 should be equal to M into V1 minus wedge's mass is also M. So, M into V2. So, from here you get V1 is equal to V2, conservation of momentum. Clear all of you? How we write conservation of momentum? Ok. Now, conservation of energy equation. Conservation of energy is this. Ok. So, the value of, I will write it again actually. See the value of K1 is 0. Ok. U1 is MGH. K2 is half M into V1 square plus half M into V2 square. Ok. So, you can write it as M into V1 square plus half M into V2 square. Now, since V1 is equal to V2, these two will add up to give you M into V1 square only. And V2 is 0. U2 is 0. W is 0. So, when you solve it, you will get V1 is equal to root of GH. So, option A is correct. Please go through this solution again and let me know if you have any doubts for the 13th question. First, I have conserved the momentum in the horizontal direction. Ok. That is little counterintuitive because you never conserve momentum in such kind of scenarios. But these scenarios do exist. Ok. Good to know that because you do not want to know this at the time of exam. So, you should know it beforehand. So, first is conservation of momentum. Second is conservation of energy. And you do not have anything else to use in this entire chapter. Is it clear to all of you how we have done 13th? Yes, sir. They are shunkin, right? No, sir. Netra. Netra. Oh, sorry. Ok. So, no doubts. So, let us solve 14th now. Question number 14. A question number 14 also can be solved in a very similar manner. Sir, when the block A comes to rest on C, that time C can be moving towards the right, sir. Ok. What is the condition for block A to reach the maximum height on C? Tell me. Potential energy is 0. Quantity energy is 0. Quantity energy is 0? Quantity energy of A. No. It could be moving with the block C, no? Block C is going horizontally. So, it could be moving with it. The condition for the maximum height is that velocity in vertical direction becomes 0. We have done in projectile motion also. We keep on using it. Ok. So, velocity along y direction is 0. So, only velocity remain should be along x axis. That is the condition. Getting it? So, when you conserve momentum in the horizontal direction, the horizontal component of the velocity which you use in the conservation of momentum is actually the total velocity itself at the maximum height. Is it clear? Ok. I will explain it again. Suppose this is the maximum height. Suppose this is the maximum height up to which mass can move. It cannot move above this point. Ok. Then, only velocity possible should be horizontal. Vertical direction velocity, this direction velocity along y direction should be equal to 0. This is what I am saying. Velocity along x direction can be there and which will be equal to velocity of the wedge itself. Wedge is also moving horizontally. So, horizontally it will move with the wedge, but vertically it will stop moving if that is a maximum height. Is it clear? Ok. So, basically what I am saying is that it is its horizontal component of velocity is its total velocity. Ok. Now, I can conserve momentum in the horizontal direction. From this point, when it is at the bottom most location to the point where it is at maximum height, I can conserve the momentum. How can I conserve momentum? Initial momentum is m into v1 which is nothing but m into root gh which we have found out. This should be equal to the final momentum m into velocity along the x direction for the m plus wedge's velocity in the x direction which is also vx only. Ok. So, both the velocities are same along x direction for the wedge as well as for mass m because both are going together. Ok. So, at the highest point, velocity along x axis is root gh by 2. Ok. This is the velocity along x axis and luckily this velocity along x axis is the total velocity of small m block as well as for the wedge. Wedge is anyway moving along horizontal direction only. It cannot move vertically. Even for the mass also this is the total velocity. So, when you write kinetic energy, you do not need to worry about y component of velocity. When you write half mv square that v could be vx itself because vy is 0. Ok. So, I can use work energy theorem. W is equal to u2 plus k2 minus u1 plus k1 over here and w is 0. When I write u2 to be equal to mgh, I will get the value of h. That is what in the question they are asking. u1 is 0. It is it has started from ground. k1 is half m into v1 square and k2 both block and the mass block both block and the wedge are moving together. So, this is 2 m into vx square. So, when you solve it, you will get the value of h. Can anyone solve and tell me what is the final answer? Anyone? No, Ronak. That does not come out as the answer. Option a would be correct if both the wedges are at rest. What happens is that some of the kinetic energy is gained by both the wedges. So, they take away some energy. So, it will not again climb back to the same height. It is h by 4. Yes, h by 4. If you solve these equations, you will get h by 4. Okay? H by 4 is the answer. Tell me. Sir, we cannot write work energy theorem only for the block in solid, sir. No, you cannot. Just for the block technique. You cannot because the normal reaction is doing work on the block. So, that component will come in now. Okay? Okay. Because perpendicular direction also small m is moving along the wedge because wedge is moving. So, you do not want to find out the work done by normal reaction. So, what you do is that you take both of them together. So, then normal reaction becomes a conservative internal force. So, work done by normal reaction is 0. Okay? So, that is what. And you can see also that some energy is taken away by the wedge. So, as if the block has donated some energy to the wedge. So, you cannot conserve energy for mass m only because some energy it has given to the wedge. Okay? Let us go to the next question. We will leave 15th one for the homework. Do this. 9th. We know that work done is integral of f dot dr. If f and dr are in the same or in the exactly opposite direction integral f dot dr is simply equal to f into dr integral or minus f into dr integral. Then this is simply represented as area under. So, area of force and displacement curve is the work done. Remember that. What is the answer? 20 joules. 20 joules. Others? Sudha 20, Ruchira 20, Ruchira also saying 20, Ranganath 20. Okay. I will quickly attempt this one. 20. Okay. Work done in displacing the body from x equal to 1 meter to x equal to 5 meters. So, when I say area under the curve, I mean to say area between the x axis, which is your displacement axis and the curve. Okay? So, I am talking about this area. So, from 1 to 5, you have this area, this one, this one, this and that one. This work done is negative. So, this area you have to subtract. Other areas you need to add. So, I can say that this one block as in one square is of the area of 1 into 5. 5 is the area of one block. These are the three blocks. So, 15 minus 5 becomes 10. Okay? And this is 15 minus 5, 10 and then this is approximately 5 only. This area plus that area, you can say this is one square area which is 5. So, 15 plus 5, 20 minus 5. 15 is the answer. No, sir. It is 20 only. Why? Both the triangles are 10 each. From 1 to 5, x from 1 to 5. So, yeah, it is half. What? Triangles together are 10. Dreaming from 1 to 5, not from 0 to 5. No, no, it is, I think I am correct here. See, these two triangles, this triangle and that triangle. This triangle occupies only, this triangle occupies less than 5 or not. This triangle is less than 5. Okay? This is less than 5 and this will add up to it. You can add this smaller part over here and say that this plus this is full square which is 5. Okay? So, this entire thing is 5. 5 plus 5 plus 5 plus 5, 20 minus 5, 15. So, 15 is the answer. Two triangles are not 10. Two squares are 10. Okay? So, when you add two squares area, it is 10. Not the two triangles area is 10. Okay? And you cannot find entire area. What they are asking you? Area or the work done from 1 meter to 10 meters. So, you need to find just area from 1 to 5. Okay? Is it clear? Yes, sir. Okay? So, be careful. This one. It is a simple application of work energy theorem. Okay? Just write down work energy theorem. W is equal to U2 plus K2 all that and substitute the values. Okay? Step number one, write this and just put the values. So, 11th answer is option D. Okay? You continue going, continue solving the next one. I do not know the final answer. We will talk about it. Others, have you attempted 11th? Do you know work done by the spring formula? Do you remember that? Work done by the spring is, anyone say it? Half K x square. That is wrong. Half K x2, x1 square, sorry, minus half K x2 square. This is a work done by the spring from x1 compression or extension to x2 compression or extension. Okay? Now, the natural length of the spring is 2R. Okay? And the other string is, okay? This is 1.5 times R. Now, how much spring is extended initially? Can you find out x1? How much it is? R by 2. R by 2. How it is R by 2? This distance is what? The hypotenuse. This is how much? This is under root of 2R square plus 3R by 2 whole square. So, this will be 5R by 2. Natural length is 2R. So, extension is 2R. 8R by 2 minus 2R. So, natural length is 2R. So, extension is 5R by 2 minus 2R. This will be R by 2. Okay? So, and x2 is 0 because it comes to natural length. So, x2 becomes 0. The work done by the spring is half K which is 4Mg by R into R by 2. Okay? So, you will get as Mg R. Sir, into R by 2 whole square. Sir, R square by 4. Oh, yeah. So, Mg R by 2. So, you will get option A. Yes, Aditya, that is correct. Good that you remember the formula. Now, solve the 12th question. 12th question can be solved by applying work energy theorem. This is a work energy theorem. Okay? Others? So, I got D. D. D, sir. D, okay. Others? Yes, sir. I got A, sir. Okay. Let me solve it now. So, when I apply work energy theorem, I am, remember that I am, I am taking work done by the spring as the potential energy. So, when I apply this equation, W is 0. Okay? Sommick don't sing. You want to sing? Let's have Sommick. Sommick, please sing now. Everybody, listen to Sommick. Sommick, come on. Sir, that bothered me, sir. Sorry, sir. Okay. Fine. So, W is 0. So, U2. U2 is the potential energy, right? So, there are two kinds of potential energies here. We have potential energy due to gravitation and potential energy due to spring. So, let's say that this is the horizontal line for which gravitational potential energy is 0. So, finally, the gravitation potential energy is 0, but the spring potential energy is there. So, U2 is half k into x square, which is, we have already found out the value of half kx square, which is mgR by 2. So, I will directly write that. I will not calculate it again. So, mgR by 2 is U2 plus k2, that is what I have to find out. What is the velocity? So, half mv square minus U1 is what? When it is here, at this location, the spring potential energy is at this location, at this location, even spring is relaxed. So, spring and gravitation, both potential energies are 0 in U2. But initially, there is a spring potential energy, which is half kx square, which is mgR by 2, as well as gravitation potential energy, which is mg into 3R by 2. So, both the potential energies are initially itself. Finally, both potential energies are 0 plus k1, k1 is 0. Okay. Any doubt in this expression, all of you? Have you understood how we got this? Understood? Yes. Okay. See, so I will simplify this further. I will get g into 2R. When you add these two, g into 2R is equal to v square by 2. So, from here, you will get v is equal to 2 under root gR. So, option A is correct. I hope it is clear. Let us go to the next question. Or let us, should we take a small break? You guys want to take a break? Can take a 10 minutes break? Okay. Only one person wants break. Yes, sir. Okay. So, we will take a break for 10 minutes. Right now, it is 10.44. We will meet at 10.54. So, this is your break time. Go and eat your breakfast. Eat after 10 minutes.