RE-NEET 2026 Laws of Motion & Work Power Energy 🔥Complete Physics Revision

Indiziert: 2026-05-21

[00:00:15]Hi everyone. Good afternoon. Good afternoon. Welcome back my dear friends.

[00:00:20]Quickly join. How are you all?

[00:00:24]A very very good afternoon.

[00:00:31]We are going to today complete Newton laws of motion and work power energy using some of the most important numericals. We are going to complete these two chapters. Quickly let me know in the chat box if I'm clearly audible and visible. Hi, good afternoon Robin.

[00:00:48]Hi Sanjana. Hi Rakshitta. Ma Lakshmi.

[00:00:52]Hello. Good afternoon. Welcome back.

[00:00:57]quickly join.

[00:01:01]How are you all?

[00:01:04]Everything's good.

[00:01:07]All right. So, using these PyQs of I've got a mix of some of the questions are from JW mains, some of the questions are from uh from modules, right? From different books. Let me tell you if you are going to do these many questions with me it will be more than sufficient that means your Newton laws of motion and work energy will be almost over and then you can proceed with mock exams right so please note that we are going to almost 90% of the topics are going to be covered by these important questions other than this you need to revise again you need to revise your notes you need to attempt more mock exam and that will be sufficient for these two topics. All right. All right. Great. So, let's quickly start without any delay. Let's start with the first question. My dear friends, first question is somewhat like this. And see, questions are going to be of you know it. Questions are going to be of slightly difficult level at least some of the questions are going to be difficult. So, do not worry. Attempt it.

[00:02:14]Give your best attempt. After your attempt, I'm going to explain. Right. So the first question says, a modern grand pricks racing car of mass m is traveling on a f flat track in a circular arc of radius capital r with a speed v. If the coefficient of static friction between the tires and the track is mu s, then the magnitude of the negative lift acting downwards on the car is assume forces on four tires are identical and small g is equals to acceleration due to gravity. Right? So everyone knows everyone knows circular turnings of uh of car that means banking of road this is a question related with that right let's attempt it it's a new type of a problem be quick hopefully I'm not covering the question you can see it properly let's begin let's try it while you are trying I'm also give I'm also going to give you some tips right so tip for this question is always draw free body diagram that will help you.

[00:03:18]Come on, be a little quick my dear friends. Let's suppose we have got a car over here. I can also draw a circular track for you.

[00:03:28]Let's suppose this is the circular track.

[00:03:37]And let's say my dear friends, this is the top view of the car. Top view of the car. Fine. Clear? And let's draw the free body diagram of this car. So we are going to draw we are going to have friction over here. We are going to have centrifugal force which is basically mv² by r. Over here we have got friction force over here which is mu into n.

[00:04:04]Right? And then downwards mg is acting. So please note my dear friends that this mg is acting downwards. This mg is acting downwards.

[00:04:20]This mg is acting downwards.

[00:04:25]Right? Let me explain it to you. So let's suppose if if this is a car if this is a car then f then friction force is acting in this direction centrifugal force is acting in this direction mg is acting like this mg is acting like this got it mg is acting like this and a a downward lift is also acting a downward lift is also acting something like this clear so a downward lift is also acting Fine. Let's draw a downward lift.

[00:05:00]Let's call it as L.

[00:05:03]And then we have got a normal reaction which is acting upwards. We have got a normal reaction which is acting upwards.

[00:05:12]Something like this.

[00:05:14]All right. Clear? I've drawn the diagram for you.

[00:05:19]Let me know in the chat box. Hopefully you don't have any doubt in this in this free body diagram. If you have any doubt, let me explain the concept also.

[00:05:26]After solving this question, be a little quick. Let's quickly solve it my dear friends.

[00:05:33]And after this, we are going to talk about circular turnings again.

[00:05:40]Now, let me explain this question to you. First of all, this free body diagram is drawn with respect to the body. That is why always we are we are talking about centrifugal force.

[00:05:52]Centrifugal force is basically a fictitious force. We draw centrifugal force whenever we are talking about whenever we are trying to draw free body diagram with respect to the body with respect to the body. Right? And this is friction force which is actually keeping the car on track. Clear? Now we all know please see here my dear friends. We all know that friction force must be greater than or equal to that means the maximum value that we can write over here is mu is equals to MV² by R. MV² by R. And instead of normal reaction, my dear friends, hopefully you guys can see that normal reaction will be Mg + F L is equals to MV² by R. And if I'm going to rewrite it again, I'm going to write MG plus MU F L is equals to MV² by R.

[00:07:02]And I'm going to write minus mu mg like this. And then f l will become let's take m common.

[00:07:15]So we are going to write v ² by r v ² by mu into r minus g.

[00:07:25]So m * v ² divided by murus g. That means option A is absolutely correct.

[00:07:33]This is your answer.

[00:07:37]This is your answer my dear friends. Do let me know in the chat box if you have any doubt. All right. Please understand.

[00:07:43]By the way, let me tell you what is this negative lift? What is this negative lift? Right?

[00:07:49]Got it.

[00:07:52]What is negative lift? You you have seen you have seen racing cars. Have you do you know about spoilers? Right? You know about spoilers. Let me draw a car for you.

[00:08:11]Here we have got spoilers like this. If you can understand right, we have got spoilers like this.

[00:08:19]Now the these spoilers actually act in exactly the opposite way how the air how the wings of a of an aircraft works.

[00:08:28]Right? Wings of an aircraft craft provides an upward lift. These spoilers provide a downward lift. So when they are talking about they have they themselves have said that it is acting downwards. That means it is acting along mg which means that normal reaction can be written as mg plus L mg + L. All right. Clear sir. Bengali student who English Bengali question paper translation line translate.

[00:08:59]There is a possibility that NTA can commit such blunders right. We we we cannot expect a lot from NTA. I think we should not expect anything from NTA. So it can do such things that it can actually make uh mistakes in translation from one language to other language.

[00:09:18]Other than that I think uh you have got English I think you should stick with the English language because mostly whenever you are going to learn uh from any book from any faculty they are also going to speak in and and at least the questions are going to be in English.

[00:09:32]Okay. So let's quickly move on my dear friends. Hopefully the first question is all good. Let's move to the next problem. Let's see this one. Come on be quick. I want your answers in the chat box. Okay. And if you have any doubt do let me know. Robin your answer is right.

[00:09:50]For the last one in this question they are saying a block of mass m slides down an inclined plane. Inclined plane inclined at an angle of 30° with an acceleration of gx4. The value of coefficient of kinetic friction will be.

[00:10:06]What is the value of coefficient of kinetic friction? My dear friends if you have any doubt let me know. I'm going to explain the concepts also with numericals. Okay. So in this problem my dear friends they are saying you have got an inclined plane and it is inclined at an angle of 30°. It is inclined at an angle of 30°.

[00:10:27]And then they are saying that a block of mass m a block of mass m slides down.

[00:10:37]A block of mass m slides down with an acceleration of with an acceleration of g by4 with an acceleration of gx4.

[00:10:51]And we have to find out the value of mu.

[00:10:53]Very good. Very good. Let's quickly do it. Okay. So the acceleration is g by4.

[00:10:58]Great. Let me draw the free body diagram. And motion on an inclined plane is a is a very very important topic. So let me explain it to you. See let's draw free body diagram of this body. It's very easy. You have got one force which is acting downwards which is mg sin theta.

[00:11:17]You have got another force which is acting in upward direction which is nothing but friction force which is nothing but friction force which is equals to mu into n. You have got a force acting downwards which is mg cos theta which basically means my dear friends you have got a normal reaction acting upward I mean acting along this direction whose value is going to be equals to mg cos theta.

[00:11:46]All right so now after this they are saying that it is accelerating downwards. If it is accelerating downwards that means definitely this force has greater magnitude than this force obviously right and net acceleration will be written by by something like this. My dear friends please see G sin theta minus instead of mu n we are going to write mu g cos theta and this should be actually equals to gx2 as gx4 as they have said.

[00:12:21]So we can cancel out g.

[00:12:24]Instead of sin 30 we we are going to write 1 by 2. We have to find out the value of mu. Instead of cos 30 we are going to write <unk>3x2 and we are we are going to get 1x4. So this will become two. This will become two. Right? And after this I think uh almost we have we have got our answer.

[00:12:45]We are going to get 1x 2 and we are going to get mu uh this is 1 1x 2 - 1 is 1 by 2 and you have got mu * <unk>3 by 2 so 2 and 2 gets cancelled out just a second let me see this two is also cancelelling this two is also getting cancelled out of course so we don't have any two over here which gives our answer as Friends, mu is equals to mu is equals to 1 upon 2<unk>3 1 upon 2<unk>3.

[00:13:22]Please check and let me know in the chat box. Very good. Very very very good.

[00:13:28]Great. Answer B is absolutely right.

[00:13:32]Answer B is absolutely right my dear friends. Now let's move on. Clear? Any doubt? Sorted.

[00:13:42]Moving on to the next one my dear friends let's see this question now let's increase the level of the question by a bit in this question they are saying consider a block kept on an inclined plane inclined at an angle of 45° as shown in the figure if the force required to just push up the incline kindly see to just push up the incline is two times the force required required force required to prevent it from sliding down. Then the coefficient of friction is be a little quick. Let's solve it. Be quick.

[00:14:24]Kindly understand the problem what they are saying. Come on Saksham. Forget about Nava.

[00:14:33]Let's worry about friction. Let's worry about neat. Let's worry about rene.

[00:14:38]Okay. Instead of navia, think about neat.

[00:14:42]Okay. So in this question, the important condition is that we want to find out coefficient of friction such that the force required to push it up the incline is two times the force required just to prevent it from sliding down.

[00:15:00]Okay, great. Let's see. Let's see. First of all, first of all, let's talk about what if it is going to slide down. Okay. Uh sorry, let's talk about what is the force required just to prevent it from sliding down. Listen here. Just to prevent it from sliding down. Clear? So that means you have got mg sin 45 acting downwards.

[00:15:32]That means my dear friends mg by <unk>2 and then you have got a minimum force should act in this direction and by the way you have got also friction force. So friction force is mu n mu n and n is by the way mg by <unk>2.

[00:15:59]Now kindly see mu is always going to be less than one. Clear? That means friends this value let me explain the question.

[00:16:08]This value will always be less than this value. Why? Because mu is less than one.

[00:16:14]Got it? So this force is less than this force. So we will have to apply a minimum force over here. We will have to apply a minimum force over here.

[00:16:25]We'll have to apply a minimum force over here. Clear?

[00:16:32]Now what is the value of this minimum force?

[00:16:36]So this minimum force will be this minimum force will be mg by <unk>2 mg by <unk>2 minus mu mg by <unk>2. So I'm going to take mg by roo<unk>2 common. So I'll get this. This is the minimum force required. This is the minimum force required. Clear? Right? This is the minimum force required. Now next case they are saying that to push it up. Let me draw another diagram for you so that everyone can understand. All right. If you have already got your answer then I'll suggest you to please have some patience. I'm going to quickly move on to the next one. This is our block. Now we want to push it up with a certain force. So let's say this is that force with which we want to push it up. Right?

[00:17:32]And now in this case my dear friends mg sin theta is acting downwards also friction force is acting downwards. So, mg by <unk>2 and now you guys know that this is going to be actually nothing but this is going to be your friction force L and that will be mu mg by <unk>2. Hopefully this makes sense. That means force required to push it upwards is actually equals to mg by <unk>2* because mg by <unk>2 will be common time 1 + mu. And now they are saying that this force is 2 times this force.

[00:18:15]Let's quickly do this math. Very simple math my dear friends. I'm going to say that mg by <unk>2 1 + mu is equals to 2 * of mg by <unk>2 into 1 - mu. Mg and mg can get cancelceled out. <unk>2 and <unk>2 can get cancelled out. I have got 1 + mu is equ= to 2 - 2 mu. That means I have got 3 mu is equ= to 1.

[00:18:48]And my dear friends that's it. We have got our answer mu will be equals to kindly see mu will be equals to 1 by 3 1x 3 that means answer is a 0.33.

[00:19:05]Clear? Is everything fine? Be quick. Let me know in the chat box.

[00:19:12]Done and dusted.

[00:19:16]Clear.

[00:19:18]Aushi, Samia, Fzana, Robin, Sanjana, Sakshi, be quick. Everything's done. Can we move on to the next one? This is a nice problem, right? Please understand how to draw how to write down the friction force. How to draw the direction of friction force. Friction force will always resist the tendency of relative motion. What I'm saying?

[00:19:45]Relative motion. Tendency of relative motion. If we if they were saying that it is it is about to slip or we want to prevent it from sliding down, it means that friction should act upwards. It means that friction should act upwards.

[00:20:00]to prevent it from sliding. Right? But still in this case friction is not sufficient enough to prevent it from sliding and that is why we will have to apply a minimum force. So minimum force will be applied in the upward direction.

[00:20:14]So first equation is all done. In the second equation they are saying that we want to actually pull it upwards. So if you if you want to pull it upwards that means relative motion is acting in such a way that block is moving upwards on the inclined plane which means that friction force will act downwards.

[00:20:32]Clear? Done.

[00:20:35]Be quick. Let me know in the chat box.

[00:20:38]Moving on to the next one.

[00:20:40]Moving on to the next one. My dear friends, everyone's attention over here. Next question is on the board. A block of mass 10 kilogram is kept on a rough inclined plane as shown in the figure. A force of 3 Newton is applied on the block. Okay. Okay. A force of 3 Newton is acting on the block. Right. And then they are saying the coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P such that the block does not move downward?

[00:21:16]such that the block does not move downward. So you have to balance the forces. Be quick. Come on. Quickly solve this question. Quickly solve this question. My dear friends, 3 Newton is acting downwards. We can see that a force of 3 Newton is act is acting on the block. Coefficient of static friction is there right and what should be the value of P? Now please understand the how they are framing the question.

[00:21:40]And they are saying what should be the minimum value of force such that the block does not move downward. That means to prevent it from moving downwards.

[00:21:49]Friction force will act in upward direction.

[00:21:52]Friction force will act in upward direction. My dear friends, friction force will act in upward direction.

[00:22:01]Upward direction. Mu mg sin 45° that means this is 60 right this is 60 60 60 <unk>2 this is 60 <unk>2 this is 60<unk>2 this is 3 newton we have got one more force acting downwards which is mg sin theta mg sin sin theta.

[00:22:34]Okay. Now let's just equate these two forces. That means force P + 60 <unk>2 is equals to 3 Newton plus 100 <unk>2 + 100<unk>2 that means P will be equals to 3 + 40<unk>2 okay now find this find the value of this 1.41 41. Find the value of this.

[00:23:08]This is around going to be 32. What do you guys say? Be quick. Huh? So I have written sin 45 now. Sin 45 1x <unk>2 cos 45 1x roo<unk>2. I have done exactly the same thing. Buy kindly see.

[00:23:21]Come on. Come on. Come on. Be quick.

[00:23:27]Be quick. Those is going to be around 32 Newton. Okay.

[00:23:32]This is going to be around 32 nton.

[00:23:37]Please remember <unk>2 is 1 1.414.

[00:23:43]Option B is incorrect. I g option B is incorrect. Option A is correct. If you have any doubt, do let me know.

[00:23:52]All right. So by looking at your responses in the chat box, I'm thinking that you guys are not getting what I've done.

[00:24:01]If that's so please let me know. Is it clear Gorov? Good. Gorov saying yes sir your answer one sir is sufficient.

[00:24:12]Sorted. Okay great. All right Robin. Any doubt? Okay good. Aushi good for Zana Samia 32. Yes 32 is correct answer.

[00:24:21]Let's move on my dear friends. Hopefully everything is clear. See why I'm constantly giving you problems related with friction inclined plane because if you are going to see the pattern of previous year neat exam they have constantly given problems related with friction in inclined plane okay let me explain it again but I'm going to be a little quick see for they are saying that a 3 Newton force is acting okay a force P is acting we want to find out the value of force P fine but they are saying the condition is that find the value of P such that it does not move downward. It does not move downward.

[00:25:00]That means friction force should act upwards. Friction friction force will be mu mg cos theta. Mu mg 10 into 10 cos theta cos 45 1x roo<unk>2. This is 60 by <unk>2. This is p this is 3 newton. This is mg sin theta. Mg sin theta. That means 3 newton plus this is equals to p + this. So p + this is equals to 3 + this. That's it. You have to solve it.

[00:25:25]Okay. Now I'm moving on.

[00:25:30]Casual. Okay. Okay. Fine. Let's move on to the next one. My dear friends, this question says, the time taken by an object to slide down a 45° rough inclined plane is n times as it takes to slide down a perfectly smooth 45°ree inclined plane. The coefficient of kinetic friction between the object and the inclined plane is you have to find out the value of mu. Be quick. Let's solve this question.

[00:26:08]Let's say this is a rough inclined plane.

[00:26:12]Coefficient of friction is mu.

[00:26:15]A block is sliding down.

[00:26:19]Right? A block is sliding down.

[00:26:22]Let's say distance of this uh this inclined plane is s sorry the length and the length of this is also s and then this so kindly see kindly see okay fine this is 45° this is also 45° no problem let's quickly draw the forces this is going to be mg sin 45 this is going to be mg mu mg cos 45 got it so if I'm going to talk about acceleration, right? Let me write it as mg sin 45 and this is going to be mu mg cos 45.

[00:27:04]All right. Next is this is simply mg sin 45 and both of these are being being released. So for this case for this case my dear friends I'm going to write s is equals to half a half a t² for this one I'm going to write s equ= to half a a a a acceleration will be g sin 45 that means in this case acceleration will G sin 45° minus mu cos 45° which basically means I can take obviously you have got g also over here you can take g by <unk>2 common and you'll be left with 1 minus mu so over here you have got g by <unk>2 and you have got 1 minus mu and then they are saying this is taking n times right this is taking the uh they are saying that rough inclined plane is n times as it takes so that was taking t time so this is taking n * so n² t² let's equate these two equations let's equate these two equations and if you have any doubt do let me know do let me know my dear friends so if you are going to equate these two equations we are going to get 1 upon 2 g upon <unk>2 2 1 - mu into n² into t² is equ= to you have got 1 upon 2<unk>2 you have got g and you have got t² my dear friends I can easily cancel out g and 2<unk>2 I can cancel out t² I'll am left with 1 minus mu and they want us to find out the value of mu please remember they want us to find out the value of mu so I'm going to Write 1 upon n² and mu will be actually equals to mu will be actually equals to 1 - 1 upon n² that means d is the correct answer d is the correct answer very good very very very good good everyone any more doubts clear if you remember they have asked a similar question in need 2025 also right and this question is from JW mains 2023.

[00:29:43]I have again and again said the same thing. If you guys are going to do sufficient number of problems from JW mains pyq see these are perfectly of neat level right and they are also slightly new and better ex right so we need to we have already attempted all neat pyqs most of them are easy and we are going to still attempt neat pyqs before neat exam before one week we are going to kill all the neat pyqs but right now what we should do is we should actually increase our level by a bit.

[00:30:18]That's why we are solving questions like these. Now let's move on to the next problem my dear friends. Okay.

[00:30:25]Okay. Now I think uh inclined plane is is fine. Let's start with some questions on momentum. Momentum is also very important. Right. Uh next question is saying Gorov Gorov who are let's let's study my dear friend. 100 balls each of mass m moving with speed v simultaneously strike a wall normally and are reflected back with the same speed in time t seconds. The total force exerted by the balls on the wall is this is a uh this is an easy one. Let's start. I'm going to give you poll for this one. 60 seconds. Let's begin. Be quick my dear friends. Be really really quick. This is an easy one. So I've given it to you. Come on. Come on. Come on. Come on. Come on.

[00:31:18]Be quick.

[00:31:23]Momentum is important.

[00:31:26]Friction is important. Inclined plane is important. All right.

[00:31:38]Okay. Good.

[00:31:40]Good. Good good use Newton's second law. Force can be written as change in linear momentum divided by time taken. Right? And because they are only talking about magnitude, we are also going to find out the magnitude only. You have got 100 balls and each ball has mass m and velocity v and they are reflecting back with the same speed. So the change in momentum for each ball will be 2 mv. So for 100 balls it will be 100 * 2 mv right and divided by time which is t. So it is simply going to be 200 mv divided by t.

[00:32:33]So 200 MV divided by T.

[00:32:37]Right? This is a this was a very easy problem.

[00:32:43]Option B is correct.

[00:32:45]Very good. Great.

[00:32:48]Great my dear friends. So if you guys are able to do this then let's also solve this question. Let's solve this question my dear friends. Be quick. This is also a typical type of a problem.

[00:33:00]They have asked this in in JW mains 2023. A block of mass <unk>3 kilogram is attached to a string whose other end is attached to the wall. An unknown force F is applied so that the string makes an angle 30° with the wall. The tension T in the string is the tension T in the string is let's start 60 seconds is I think sufficient.

[00:33:24]Begin begin.

[00:33:30]Come on.

[00:33:35]I have given you pole.

[00:33:39]In the meantime, let me also draw the uh the forces.

[00:33:44]This is mg, right? So, I can I can actually directly draw mg over here.

[00:33:49]This is tension t.

[00:33:52]So, this is going to be one component of tension. This is going to be another component of tension.

[00:33:59]Kindly see if this is 30°, this is going to be 30°.

[00:34:04]If this is T, this is going to be T cos 30.

[00:34:10]This is going to be T sin 30.

[00:34:14]T cos 30° will be equals to <unk>3 into G. G is 10. So <unk>3 into 10. So first equation is t cos 30 let me save some time t cos 30 is equals to <unk>3 * 10 <unk>3 roo<unk>3 cancel t is coming out to be 20 nton if they are asking us about tension then our answer is over here yes this is and if they are going to talk about force also then that would be t sin 30 t sin 30 that means 20 into 1 by 2 is going to be equals to f that is 10 newton that is 10 newton clear so 20 is the correct answer option a very good robin gorov shib nita aayushi fzana somia good very very good okay now let's increase the level by a bit again. So I'm giving you all types of problem. Please remember whenever you are going to see an easier question on the board, I also want to complete the chapter using all sorts of numericals.

[00:35:33]Now this is a nice question. In this question they are saying please see and if you remember in neat 2025 again they have asked a similar type of a question.

[00:35:45]Hello.

[00:35:47]A root a rooe 34 m long ladder weighing 10 kg leans on a frictionless wall.

[00:35:56]Leans on a frictionless wall. So wall is frictionless.

[00:36:00]Right? And it rest on a 3 m away. It rests on a floor 3 m away from the wall as shown in the figure. Fine.

[00:36:09]FF and FW are the reaction forces of the floor and the wall. Then the ratio of these two is it's a very nice problem.

[00:36:21]Let's start. Be quick. Let's begin my dear friends. Be quick.

[00:36:31]Quick quick.

[00:36:39]See, let me tell you one thing.

[00:36:44]If you guys are giving me answers in in 5 seconds, in 10 seconds, then there can be two or three possibilities, right?

[00:36:51]There can be two or three possibilities.

[00:36:54]First is that you have done this problem earlier and you remember the answer. You know that C is the correct answer.

[00:37:01]Second is you're constantly you're seeing the question you're opening the next tab you're googling it you're seeing that option C is correct and then you're giving the answer in any which way why you are wasting your time this is the time where you should practice all of a lot of questions even if you remember that C is the correct answer for this one I don't want to know the correct answer I want you to attempt it I am here so that you can actually practice practice some of the questions with me. If you guys are simply using chat GPT, you're using Google, you're using uh or you are simply remembering uh the answer from your memory, then I mean there are lots of things on YouTube, there are lots of things on on uh on internet why you are wasting your time over here. I don't want to know the correct answer. I want you to solve these questions with me. Okay? So, please solve please solve. This is not a problem that you can actually solve in 5 seconds.

[00:38:11]Let's say this is our this is our rod.

[00:38:17]Let's say my dear friends this is our rod.

[00:38:22]Okay.

[00:38:34]Let me now draw the pre-body diagram for this.

[00:38:39]Let's start with the usual suspects like mg. Okay. MG is acting downward.

[00:38:47]MG is acting downward.

[00:38:52]Done.

[00:38:53]Reaction by wall is act will act in this direction.

[00:39:00]There is no friction over here. There is no friction on this wall. This wall is smooth. My dear friends, this wall is smooth.

[00:39:12]So there is no friction. If there was friction, understand? If there was friction, then the tendency was to slip down. Then friction would have acted in upward direction. But right now there is no friction. Over here you have got a normal reaction by by floor. So this is FF, right?

[00:39:36]They have not talked about whether this floor is rough or smooth. Sorry. Whether this Yeah. Whether this floor is rough or smooth.

[00:39:45]Now it makes no sense if you are going to say sir this floor is smooth. Why?

[00:39:50]Because then this force cannot be balanced. Clear? So this floor will have to be rough and this is going to be friction force. This is going to be friction force and I'm saying this is going to be rough.

[00:40:10]This is the free body diagram. All right. This is the free body diagram.

[00:40:15]clear now fine very very good if you are not using Google I'm not I'm not saying you are not you're using Google but in any which case you cannot find the answer for this question under 5-second right if you if you can solve the question under 5second then I should I should be the one who should learn from you guys not the other way around now please see first of uh we are going to do two things. this these types of questions these type and let me tell you one more thing we are not we are using these questions to to actually complete our chapter right when we are going to be one week uh when there is going to be one week time for your neat exam at that point of time I'm not going to do anything I'm going to simply give you questions and questions right now let's discuss everything related with these problems let's not jump to the next question as soon as we get to our answer. Okay. So my dear friends in these types of problem the concept is that the rod is in equilibrium. These types of rods are in equilibrium. These types of rod are in equilibrium.

[00:41:32]And equilibrium basically means two things. Equilibrium basically means two things. First is net force will be equals to zero. Summation of all the forces along x y and z axis are going to be zero. Second uh net torque is going to be zero. Net torque is going to be zero. So let's first use this equation.

[00:41:55]Uh upward direction force is FF.

[00:41:58]Downward is only mg. That means F F is equals to mg.

[00:42:06]This is FW.

[00:42:10]This is F. And by the way, let me also write this thing as capital, right?

[00:42:14]because they have used capital. So this is FW. So this is going to be F W equals to friction force F.

[00:42:24]Now let's do again one thing. Please remember that this whole this complete length is roo<unk> 34 m. Root 34 m. All right. And now let's talk about this is about summation of net forces is equals to zero. Now let's write summation of net torque. So we are going to find out torque about this point. We are going to find net torque about this point O. So net torque about 0 is zero. Uh yeah it will be zero. Right? So this distance is going to be <unk>34 by 2. And if this is mg this is going to be mg.

[00:43:06]This is going to be mg mg cos theta. And by the way this we can easily find out what this angle is going to be. Uh if you want if you want us to find out cos theta cos theta will be simply base divided by <unk>34.

[00:43:26]Right? So we are going to write first is mg cos theta multiplied by <unk>34 divided by 2 and then the opposite torque will be applied by this and this is going to be something like this my dear friends and please know whatever this angle is we are going to have same this angle so this is also theta right so this is going to be fw sin theta so This should be equals to f w sin theta.

[00:44:05]Clear? Right? Hopefully everything is sorted.

[00:44:11]Now the most important thing in these types of problem is after this you can find out fw. You can find out FW right because see sin theta will be uh this side will be under root of 34 minus 3 that means this is going to be under root of 31 this is going to be under root of 31 right so you can use this equation and you can write down mg mg is uh this was 10 kg so 100 into cos cos theta. So cos theta is 3x <unk>34 3x <unk>34 * <unk>34 by 2 is equals to f_sub_w and time sin theta and sin theta is <unk>3 sin theta is going to be my dear friends perpendicular upon hypotenus that means roo<unk>31 divided by <unk>34 so this gets cancels out and this is why I was saying that please do not give your answer so quickly. Right? Okay. Got it?

[00:45:21]So this is what I was trying to say. Now please see from here you can get to know FW. So FW is known, right? FF is known.

[00:45:32]Hopefully this makes sense. If you can know FW, you can also know FF, right?

[00:45:38]And please understand they are they are talking about FW and FF. Now I'll have to use some space over there. See FF is actually a different kind of a force. F F is actually a different kind of a force. Right? Let me tell you uh let's do one thing my dear friends. Instead of writing it as FF now let's write it as normal reaction N for a moment. Let's write it as normal reaction N.

[00:46:08]uh hopefully everywhere else is all sorted right and f will be actually under root of normal reaction squared plus plus f².

[00:46:23]So the ratio will be ratio will be my dear friends fw divided by ff fw divided by ff will be I think we have already found out fw.

[00:46:39]FW is going to be uh we have found out FW and FF is this FW divided by under root of N² + F².

[00:46:53]This is how you are going to get to the correct answer my dear friends. And then yes correct answer is I believe for this one. Let me see what is the correct answer for this one.

[00:47:24]The correct answer is I believe A. The correct answer is I believe A. Yes.

[00:47:30]But you're going to have lots of calculation in these types of problem.

[00:47:34]So even if you have if you know the correct answer, if you have solved it previously in some mock exam, do not skip it. Solve it again. Solve these types of questions again. Solve your fw from here. Substitute in this equation and then you are going to get to the correct answer. Got it? Hopefully this makes sense my dear friends. Quickly let me know in the chat box if you have any doubt. And you guys were you guys have given me answer of this question in in less than in less than 5 seconds. Be quick. Find the correct answer. Moving on to the next one. My dear friends be quick. Clear.

[00:48:11]Two blocks A and B of masses 1 kg and 3 kg are kept on the table as shown in the figure. This question is also nice.

[00:48:19]Okay. Okay. Let me explain the previous question again. Somia, do you have any doubt in the previous question? You can you can ask your exact doubt dear. You can ask your write down your exact doubt. Then I'm going to come back to this question. Ask your exact doubt dear. I'll come back. I'll come back.

[00:48:38]In this question they are saying that we have got two blocks A and B. The coefficient of friction between A and B is 0.2. So coefficient of friction between A and B between A and B is 0.2 and coefficient of friction between B and surface B and surface is 0.2. So both of them are same. The maximum force F that can be applied on B horizontally so that the block A does not slide over block B is come on be quick be quick free body diagram I'll explain the free body diagram I'll explain the free body diagram okay no problem let me explain the free body diagram again see free body diagram is very simple you have got floor like this. You have got wall like this. And you have got a rod somewhat like this. You have got a rod somewhat like this. Here mg will act downwards.

[00:49:46]No doubt about it. There is a tendency of this rod to slip in this direction.

[00:49:52]So friction force will act in this direction. This is a normal reaction by the floor.

[00:49:58]by the by the floor. So that that means that because of these two forces there will be a there will be a reaction force. Got it? There will be a reaction force somewhat like this. Please understand this concept. This is this is the concept that we have done in in in the in friction the the this is the reaction force. These are simply the two components of those reaction forces.

[00:50:23]Clear? And you another reaction force will be drawn something like this.

[00:50:28]All right. So this is mg.

[00:50:31]This is normal reaction. This is reaction by wall. This is friction force. And this is your F floor.

[00:50:42]Hopefully now this makes sense. Clear?

[00:50:45]Your doubt is sorted. Can I move on to the next question?

[00:50:51]Can I move on to the next question? Is it all fine?

[00:50:57]Can I move on to the next one?

[00:51:02]Come on.

[00:51:05]Two blocks A and B. Now, now let's solve this particular problem. They are saying the maximum force F that can be applied on B horizontally so that the block A does not slide over the block B.

[00:51:19]Hopefully now the last question is all sorted. You have to find out the net force F so that the you have to find out the maximum force F so that the block A does not slide over.

[00:51:32]Come on, let's start. Be quick.

[00:51:46]Be quick.

[00:52:25]in questions like this. Hold on, let me explain it to you. Whenever you have got two block problems, right?

[00:52:35]First of all, understand what is the maximum acceleration of A?

[00:52:42]What is the maximum acceleration of A?

[00:52:44]What could be the maximum acceleration of A? Maximum acceleration of A will be in in one way. When block B will move in the forward direction, friction force on block A will act in the forward direction. Now think about it. If you have any doubt, Somia, Robin, if any student has any doubt, do let me know. Friction force will act in the forward direction and this is going to be the maximum value will be re uh mu that is 0.2 mass is a mass is 1 kilogram. So that means 10. So that means maximum force can be 2 Newton which means that maximum acceleration of a block can be can be 2 / 1 that means 2 m/s squared.

[00:53:33]Please understand this means that the maximum acceleration with which block A can move is 2 m/s squared.

[00:53:43]You're going to say how sir? For a simple reason because see on block A there are no other forces other than friction. No other forces other than friction. So if block A has to move if block A has to move it will move because of friction force and maximum friction force can be mu mg that means 2 newton.

[00:54:04]So maximum acceleration will be 2 Newton divided by mass of A that is 2 m/s squared.

[00:54:10]This means that this combined system, this combined system of a + b should move with a constant acceleration of 2 m/s squared and not more than this because if it will move more than this then friction force cannot supply that much force. It cannot provide that much acceleration and it will start to slide.

[00:54:35]Got it? Now for a + b please understand we are going to draw a friction force in backward direction and this is going to be mu which is 0.2 into mass. Mass will be 1 kilg and 3 kg that is 4 kg multiplied by 10. So this is going to be 8 newton. So what is this force? So that means f - 8 is equals to total mass time total acceleration. That means F should be equals to 16 Newton.

[00:55:09]F should be equals to 16 Newton. That means option A is correct. Clear? Be quick. Let me know in the chat box if you have any doubt friends.

[00:55:23]Be quick friends. Be quick.

[00:55:26]If you have any doubt, do let me know.

[00:55:35]Clear.

[00:55:45]Can we move on to the next one?

[00:55:58]All right. Great.

[00:56:02]So this was combined bodies. Moving on to the next one.

[00:56:10]A body of mass 1,000 kilogram is moving horizontally with a velocity of 6 m/s.

[00:56:16]If 200 kilogram extra mass is added, the final velocity will be conservation of linear momentum.

[00:56:24]Conservation of linear momentum.

[00:56:30]Somebody's saying hubli per set available sir need coaching car hubli per set available you I think you are trying to ask me about the mock exam in hubli most probably yes most probably yes you can check that in uh you can check that on PW website okay or you can call the vidya pete center for that sir friction is backward due to floor yes s because Block was moving in this direction. So, friction force will act in this direction.

[00:57:03]Block was moving in this direction.

[00:57:04]Therefore, friction force will act in always in in in in a such a way that it will resist the tendency of relative motion. Clear? Now, let's solve this problem. This is a simple problem of conservation of linear momentum. The question is saying that a block a body a body of mass 1,000 kg was moving with was moving with with a velocity of 6 m/s and another block of 200 g 200 kilogram was kept on it.

[00:57:42]200 kg was kept on it. So we are going to simply say initial momentum is 1,000 * 6 and finally it will become 1200 * v. In today's class we are going to solve many complex problems of conservation of linear momentum and work energy theorem because we have to complete work power energy also in today's class. So don't worry this is a very easy problem. Okay. So this gets cancels out. This is 2. This is five.

[00:58:09]Answer is going to be 5 m/s.

[00:58:13]answer is going to be 5 m/s.

[00:58:18]See if you guys are asking any doubt please note that I'm going to definitely explain that. I will explain.

[00:58:28]All right. Moving on to the next question. An object is thrown vertically upwards at its maximum height. Which of the following quantity becomes zero? At its maximum height. If a object is thrown horizontally upwards, If an object is thrown horizontally upwards, then at the maximum height, its velocity will become zero and its momentum will become zero. There will be potential energy, there will be acceleration, there will be force. Force will act mg.

[00:59:01]Acceleration will be G downwards and potential energy will be mgh, mgh at any height. Right? Any doubt from the last one? Let's see this question. This is a again a nice problem. Nice problem my dear friends. A balloon has mass of 10 g in air. Okay, it has 10 g of mass in air. The air escapes the balloon at a velocity of 4.5 cm/s.

[00:59:31]If the balloon shrinks in 5 second completely then the average force acting on the balloon will be average force acting on the balloon will be in dine in dine my dear friends. All right. Dine basically means dine basically means g cm square cm/s squared.

[01:00:01]Be quick. Solve this question.

[01:00:05]Let's begin. Let's begin, my dear friends.

[01:00:10]Let's begin.

[01:00:15]You have got a balloon of mass 10 kg.

[01:00:18]You have average. You have to find out the average force.

[01:00:26]Come on.

[01:00:33]Quick be quick then I'm then we are going to move on to the next one.

[01:01:30]Come on, you have to use See, whenever they are going to talk about average force, whenever they are going to talk about average force, please remember force will be written as DP by DT.

[01:01:46]DP by DT.

[01:01:48]And then they are saying that you have got a 10 g You have got a balloon like something like this, right? And let's say that air is escaping from this balloon.

[01:02:04]Air is escaping from this balloon such that its initial mass was 10 g and air is escaping at a speed of 4.5 cm/s and it completely shrinks. That means its mass that mass is of air right in air it mass goes to zero in 5 seconds.

[01:02:32]in 5 second. Right? So in this dp by dt we are going to write d mv by dt and we are going to get we are going to get v * dm by dt because they are talking about average I'm going to simply write this thing is given 4.5 and dm by dt is 10 g of mass is changing in 5 seconds. So that means this is going to be 2 and answer will be 9 g cm/s squared or 9 d. Answer B is correct my dear friends. Please note answer B is absolutely correct.

[01:03:20]Clear?

[01:03:22]Let me know in the chat box. This is a different problem because over here they have many times I have said this that in these types of problems where you need to use v dm by dt in in in concepts like rocket propulsion they actually give this value as in question they somehow give this value as gram per g per second or kilogram/s right in this question they have not mentioned anything like that clear.

[01:03:55]Yes. Yes. S please understand this is a nice problem. If let's suppose in a in a problem they are saying that mass is decreasing at a rate of at a rate of 2 g per second 2 g per second then instantly it will it will actually click to you that okay definitely we'll have to use this formula.

[01:04:16]But over here they are saying that initially it is 10 g and then it becomes it shrinks in 5 seconds. It shrinks in 5 second right? That means you and they are talking about average. So we instead of dm by dt we are going to write it as d m by delt and for d m we are going to simply write 10 g. Change in mass is 10 g in 5 second. That means 2 g per second is the rate of change in mass. Clear?

[01:04:44]Can we move on my dear friends? Be quick.

[01:04:49]Let me know in the chat box if you have any doubt.

[01:04:55]If you have any doubt do let me know.

[01:04:57]Moving on to the next problem.

[01:05:00]Moving on to the next problem. Answer my question. Answer my question.

[01:05:06]Everything's clear.

[01:05:08]In this question they are saying a bag is gently dropped on a conveyor belt. So you have got a conveyor belt like this.

[01:05:32]right? You have got a conveyor belt somewhat like this. Fine. And it is moving at a speed of 2 m/s. It is moving at a speed of 2 m/s.

[01:05:44]the coefficient of friction between the conveyor belt and the bag. So let's suppose we have got a bag over here.

[01:05:54]We have got a bag over here and then they are saying the coefficient of friction between the conveyor belt and bag is 0.4. Initially the black bag slips on the belt before it stops due to friction. The distance traveled by the bag on the belt during during slipping motion is nice problem. Let's start. Be quick.

[01:06:22]Be quick. Let's solve this question. You have to find out. You have to find out the distance traveled. The distance traveled by the bag on the belt during this slipping motion. during this slipping motion. All right, kindly start. And when it is gently dropped, okay, so if it is gently dropped, that means at this point initial velocity is zero.

[01:06:51]Initial velocity is 0 m/s.

[01:06:54]Initial velocity is 0 m/ second. Be quick. Come on. I'm going to start poll for this uh 90 seconds. Let's start.

[01:07:17]Come on.

[01:08:38]See what do you mean by slipping?

[01:08:40]Slipping basically means that you have kept this block on the surface. Now friction force is acting on it. Right? And friction force is acting this surface is going in this this surface. My dear friends, this surface is moving in forward direction.

[01:08:57]That means with respect to this surface, this block is moving in backward direction. Which means that friction force on this block will act in forward direction.

[01:09:08]Friction force on this block will act in forward direction. And if friction force will act in forward direction, we are going to write it as mu * mg. And I believe mu is 0.4 and mass is mass is my dear friends we don't require mass you know what we don't require mass let's find out acceleration so acceleration will be simply mu into g so mu is 0.4 4 * 10 that is 4 m/s squared 4 m/s squared right please understand now we are going to use the first equation v = to u + a t let's keep v v = to 2 m/s v = to 2 m/s u is zero acceleration is 4 and multiplied by t so t will be actually 1x 2 m uh sorry 1x2 seconds. So in 1x2 second in 1x2 second my dear friends this block will acquire a velocity of 2 m/s.

[01:10:18]In 1x2 second the block will acquire a velocity of 2 m/s and after that time there will be no relative motion. After this time, after 1x2 second, after 1x2 second, the block will acquire a velocity of 2 m/s and relative motion will cease. Relative motion will seize and if there is no relative motion, then no friction is required and then there will be no slipping. Clear? So, we have to find out this displacement. So s is equals to u t +/ a acceleration is 4 and t².

[01:10:58]So this gets cancels out and answer is is going to be 0.5 m. So 0.5 m.

[01:11:05]Let me know if you have any doubt.

[01:11:08]Be quick.

[01:11:10]Done. Any more doubt?

[01:11:16]B is absolutely correct. Okay, some of you have selected option A as well. Very good. Robin, Sara, Shib, Fzana, Vikas, Sonia. Good. Let me know if any student has any doubt in this question. We can use V² minus Oh, so I'm so sorry. You can use V square minus U square equals to 2 A S also. You can use that. No problem. You can use that.

[01:11:42]Absolutely. You can use that. Clear?

[01:11:45]They were not talking about time. You can use V square also. Any more doubt clear? Can can we move on to the next problem now?

[01:11:54]All right. Now let's see. Hopefully we have solved all the other questions of NLM.

[01:12:01]Moving on to the next one.

[01:12:04]This question please see.

[01:12:08]Come on let's begin. I'm going to explain this question to you. So don't worry, pay attention. This question is saying a small ball of mass m is thrown upwards. So let's assume that this is ground my dear friends. You have got a small ball of mass m. We have thrown it upwards right with some velocity u.

[01:12:32]And the ball experiences a resistive force a resistive force mv² resistive force m kv². Right? So see we know that we have got we have got acceleration due to gravity but along with acceleration due to gravity we have got one more force. So this is mg and we have got now one more force that is m kv² right where v is the speed where v is the speed. So my dear friends, now the net acceleration will become the net acceleration will become it is acting in downward direction. It will become G + K into V² G + K into V² G + K into V². Right? So they are saying that what is the maximum height attained by the ball? We we cannot use equations of motion over here. Equations of motions are not valid.

[01:13:40]Equations of motions are not valid in these types of problem because acceleration is not constant.

[01:13:49]Acceleration is variable. Acceleration is changing with speed.

[01:13:54]So whenever you cannot use equations of motion please come back to old school methods that means differentiation and integration right that means my dear friends differentiation and integration simply write acceleration equals to dv by dt but hold on they are actually not talking about time they are saying find out the maximum height attained by the ball so max for maximum height instead of writing dv by dt I'll write it as v dv by dh v dv by dh a very important formula we use this ver this uh this form of acceleration whenever we want to relate velocity and displacement along with acceleration this is a very very important formula a equals to v dv by dh usually you you write it as v dv by dx Right? Right now I'm not talking about X. I'm talking about H. That's why I have written V DV by DH. Now this means that I'm going to write V DV by DH is equals to * of G + K into V². Now if you are new to this type of problem, let me explain how to solve. After this step, my dear friends, we actually we separate the variables. We separate the similar variables to two same sides. That means I'm going to write V dv divided by I need some space. I'm going to write it over here. We are going to write V dv is uh divided by g + k v² and equals to minus dh minus dh we are going to write something like this okay and after this we are going to integrate We are going to integrate it right or what you can actually also do is you are going to then say that let's suppose then when height was zero initial velocity was u and when height is h velocity will become zero after this integration you'll get to your answer quickly write down in the chat box if you have got any doubt be quick my dear friends any doubt till here after this you can you can assume g + kv² to some some constant if you if you remember how to solve these types of problem then that's that's best but if no then you can assume it to be some constant let's say we can we can use it as some constant P or let's use it as C right and let's differentiate it with respect to DV right so if you're going to differentiate it with respect to DV this will become zero this will become 2 KV and this will become D c by dv. So dv will be equals to d c div / 2 k v or you can directly use natural log there is also an identity okay if you don't remember that identity then I'm explaining it to you so you have got v dv divided by i g + kv² is c and instead of dv instead of dv my dear friends we are writing d C divided by 2 K V. So V and V gets cancels out and over here you have got integration of DH and from 0 to H and this is from 0 to 0 to uh sorry from U to 0.

[01:17:56]Now integration of 1 upon C with respect to C is going to be natural log. So natural log answer is going to be correct. So either it's going to be B or C. Quickly let me know in the chat box if you have got any doubt my dear friends. Be quick. According to me answer C will be correct because we have got an extra two over here. We have got an extra two over here. Quickly write down in the chat box if you have got any doubt. Everything's clear. This is a nice problem. This is a nice problem.

[01:18:27]Pay attention. They can ask this question in neat 2028 2026 renit exam if they want to increase the level of the question. After this all you need to do is you need to integrate. Clear?

[01:18:41]Done.

[01:18:46]Clear my dear friends.

[01:18:49]If you have got any doubt you can ask me. After this you are going to integrate it. You will get to the correct answer.

[01:18:56]Moving on to the next one now.

[01:19:00]Come on. Now let's solve this question.

[01:19:04]Be quick. In this question, they are saying a block of mass 25 kilogram is pulled along a horizontal surface by a force at an angle of 45° with the horizontal. The friction coefficient between the block and the surface is 0.25. The block travels at a uniform velocity. The work done by the applied force during the displacement of 5 meter of the block is come on let's solve this question be quick a block of mass 25 g is pulled along a horizontal surface at an angle of 45° with the horizontal.

[01:19:43]So we have started work per energy. Now you have got a block like this.

[01:19:53]And there is a force acting on this block at an angle of 45°.

[01:20:02]At an angle of 45° and this force is we don't know what this force is, right?

[01:20:14]We don't know what this force is. Let's say this is F and this is 25 kg.

[01:20:24]And we have got also friction. Friction coefficient is 0.25.

[01:20:28]Done. So let's draw all the other forces. There will be mg, there will be mg, there will be f cos 45, there will be f sin 45, there will be friction force. Right? So this is mg.

[01:20:43]This is f <unk>2.

[01:20:46]This is also f<unk>2.

[01:20:49]And this is mu * of normal reaction. And normal reaction is mg - f roo<unk>2.

[01:21:01]The block travels at a uniform velocity.

[01:21:03]This is very important. This is very important. When they are saying that the block travels with uniform velocity, it means that net force will be zero.

[01:21:13]It means that net force will be zero. It means that net force will be zero. And if net force is zero, we are going to equate these two forces. That means mu mg minus mu f by <unk>2 is equals to f by <unk>2. We are going to equate these two forces my dear friends. Clear? So I think we know each and every value. uh mu is is 1 1x4 mg is 250 minus 1x4 f by <unk>2 or what I can do is I can write this this as f_sub_x <unk>2 1 + 1 by 4 so you can actually uh write four four you can cancel out this four with this and we can find out the value of f from here. Okay, we can get and this is by the way this is five, right? So this is going to be 50. So f will be 50 <unk>2. f is going to be 50 <unk>2.

[01:22:28]f is going to be 50 <unk>2. The work done in a displacement of 5 m. So displacement is in this direction. So work done will be f <unk>2. So that means 50 <unk>2 / <unk>2 * 5 m. So answer will be 250 jew. Answer will be ah maybe we have taken the value of g as uh 10. If we'll take 2 9.8 I think we'll get to the correct answer. We'll get to the correct answer my dear friends.

[01:23:04]Let's check it again.

[01:23:06]Let's check it again.

[01:23:15]Let me see. Let me see if we are going to You guys tell me if you are getting the correct answer 250 after taking 10.

[01:23:24]See, we have taken if you are going to take that as uh 9.8 then we'll get to the correct answer. You'll write down in the chat box. Be quick. Be be quick. Let me know for this question friends.

[01:24:05]Come on. Come on. Come on. Let me know.

[01:24:07]Correct. Now clear.

[01:24:11]This is This is right. We have taken 10.

[01:24:15]They have they must have taken it as 9.8. No problem. Everything else is fine. If you have any doubt, you can ask me. Else I'll move on. Done.

[01:24:25]Done. Done. Done. Done. Done. All right, moving on to the next one.

[01:24:34]A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 m/s. Here they are saying it as take g as 9.8. So we will take 9.8. Clear?

[01:24:47]Next they are saying the coefficient of friction between the surface and the block is 0.5. The distance covered by the block before coming to rest is all right.

[01:24:56]Clear? So it is sliding on a surface with an initial velocity of 9.8 and coefficient of friction between the block and this is thine. Let's quickly solve it. So you have got a block like this and friction force will act in the backward direction. It has got certain initial velocity and initial velocity is of 9.8 m/s.

[01:25:21]9.8 8 m/s and friction force will be by the way I have to actually find out this is mu m into g and acceleration due to this will be simply mu into g so mu is 0.5 and g is 9.8 8. So they are saying the distance covered right. So let's use v² = to u ² + 2 a s. So v is 0. U is u is my dear friends 9.8 2 acceleration is 0.5 into 9.8 into s 9.8 gets cancels out 2 * 0.5 is 1 s comes out to be 9.8 m.

[01:26:11]B option is correct.

[01:26:15]Done. Clear.

[01:26:18]Any doubt in this question? Be quick.

[01:26:21]Be quick.

[01:26:25]I think we have completed almost uh 20 25 questions till now and these are nice level problems.

[01:26:36]Yes, B is absolutely correct. If you have any doubt do let me know else I'll move on to the next one my dear friends all right let's move on you can also discuss your strategies if you have any doubt okay if you're thinking that uh if you have any problem in in preparing physics for your He neat exam you can ask me uh you are you are asking about sir daily life timing. So see we uh the timing of live classes are being posted on this batch on the community section uh I think one day before the live class. So you can actually please come on this channel and daily we have got either chemistry, biology or physics class. Right. So today we have got physics class. Next class will be on Saturday for physics.

[01:27:39]Right? So you can actually catch the timing. You can get to know the timings and the uh day uh about the classes on the community section of this of this channel. Moving on to the next one.

[01:27:53]Come on.

[01:27:58]A uniform metal chain of mass m and length L. Okay. So we haven't done any pulley problem till now. So let's do one. A uniform metal chain of mass M and length L passes over a massless and a frictionless pulley. It is released from rest with a part of its length L is hanging on one side and rest of its length L minus L is hanging on the other side of the pulley at a certain point of time when small L is equals to capital L divided by X. The acceleration of the chain is g by 2. The value of xs the value of xs. All right, let's do this problem my dear friends.

[01:28:43]You have got a chain which has got a total mass of m.

[01:28:50]Let me explain this problem to you.

[01:28:54]Total mass is m.

[01:28:57]Total length is l. So mass per unit length will be m divided by L. They have said that at a certain point when L when small L is equals to L by X when this is when this is L by X when this is L by X right fine so this length will be L - L by X mass of this is going to be kindly see mass of this will be small M will be L byX * M by L that means it will be M byX so this is this will be my dear friends M byX * G m by X m by X * G mass over here I'm find mass over here will be m by l multiplied by x l - l upon x I can cancel out this l so I'll get mx * x -1 over here I'm getting mg by x right clear obviously this is heavier so this will go down so net acceleration will be written something like this.

[01:30:35]Net acceleration will be written something like this.

[01:30:40]This force that is m / x -1 * g minus m g / x whole divided by total mass that is m. m gets cancels out and a net is coming out to be they have they have said that a net is g by 2. So g by 2 is equals to g by x * x -1 - g by x. So g can be also canceled out. You can also cancel out g. So this will become x this will become 2x -2 and minus1. So uh this will become 3x and this will become 3.

[01:31:33]So x is coming out to be one.

[01:31:36]Ah just a second.

[01:31:41]Just a second. See my dear friends if we have done any mistake. Be quick. Be really really quick.

[01:32:03]Have we done any mistake?

[01:32:11]Okay, just a second.

[01:32:13]You don't have to select it as L by X.

[01:32:16]Yeah.

[01:32:18]Okay. Fine. Okay. Okay. Okay. Okay. Just a second. Just a second. Just a second.

[01:32:23]Just a second. No problem. No problem, my dear friends. No problem. Hold on.

[01:32:28]Hold on. Hold on. Hold on. Hold on.

[01:32:31]Actually, they have this is we have to find out the value of x. My bad. We have to find out the value of x. I have chosen I have chosen the fraction to be x.

[01:32:42]Concept was absolutely correct. But we can't select x. We can't select x, my dear friends. We cannot select x.

[01:32:52]So what we are going to do is when L is equals to L by X right okay so let's do one thing let's write this mass as M by L * * X let's do something like this let's say this is actually equals to X and this will be L minus X so this force will be this force will be m by l * l - x * g concept is exactly the same there is no change in concept concept still a net will be equals to the net pulling force which is m by l * l - x * g minus m by L * X into G whole divided by total mass M. Now A is given by G by 2. A net is given by GX2 and this will become M by L. L - X * G - M by L X * G all / M. M can be canceled out. G can be cancelled out.

[01:34:18]Right? G can be canceled out and I can multiply this L2 to this side and I can multiply this two to this side. So this will become 2 L - 2X - 2X.

[01:34:31]Now please see kindly see see my dear friends now now everything is clear. See initially I have said I have selected let's say this is L by X that is wrong.

[01:34:42]Let's suppose that this is X part. This is X part. So this will automatically become L minus X. If this is X length and total length is L.

[01:34:53]Kindly see I'm explaining it to you.

[01:34:55]Total length is L and I am saying this length is X. So this will become L minus X. Right? This will become L minus X.

[01:35:04]Downward total downward force will be mass of this part of the length. So this is L minus X. Unit mass is m by l. So m by l multiplied by length. Similarly m by l mult*lied by length. Clear? Solve that we'll get to our answer. I believe we'll get to our answer.

[01:35:27]Uh this will become minus l and this will become - 4x and we want to find out we want to find out x. So x will be lx4.

[01:35:42]x will be lx4. Clear? Now let me know if you have any doubt. We have got our answer my dear friends. Answer is correct. Four is absolutely correct.

[01:35:54]This is a very very nice problem. You should practice questions like these.

[01:35:58]This is based on a very simple concept of pulley.

[01:36:04]And you all know these types of questions my dear friend. Let's suppose we have got a pulley. It is attached over here. We have got two blocks.

[01:36:12]Something like this. Clear? We have got two blocks. All of these ropes strings are massless.

[01:36:19]Let's say this is m_sub_2. This is m_sub_1. So we we simply write down the net acceleration as net force divided by total mass divided by total mass. So net force is m_sub_2g. Let's suppose m_sub_2 is greater than m1. m_sub_2 is greater than m1. M_sub_2 g minus m1 g divided by m1 + m_sub_2. So exactly same thing I have done over here. Please see a net is fn divided by m total. A net is fnet that means the mass of this part of the chain acting downwards. So mass is m / l time length of this part time g. So this is m_sub_2 g. This is m1 g. So m_sub_2g minus m1 g divided by total mass m.

[01:37:12]Total mass m. And they have already told you what is a net. A net is g by 2.

[01:37:18]After that you can get to get to the answer. Clear. Quickly write down in the chat box if if you have liked the explanation if everything is good write down in the chat box. Be quick and then I'm going to move on.

[01:37:33]Then we will move on to the next one.

[01:37:37]Be quick.

[01:37:41]Any more doubt?

[01:37:45]Any more doubt my dear friends?

[01:37:49]Clear. Clear. Clear. Clear.

[01:37:54]All right.

[01:37:55]Then we can move on to the next question.

[01:37:59]See these questions are very very nice problems. You are not going to see there are very less channels where you can solve questions of this standard for your reit exam. Right? Stop solving only neat PYQs for your renit exam because you might get a little shock in your final renit paper. NTA can can actually set a slightly difficult exam. Right? So this is this is the level of question that we should actually go for. Moving on to the next one.

[01:38:35]A block of metal weighing 2 kilogram is resting on a frictionless plane. Fine.

[01:38:41]It is struck by a jet releasing water at a rate of 1 kg m/s. So this is what I've been telling you that they are going whenever you are going to see a unit like this of 1 kg m per 1 kg per second that is mass rate that is mass rate dm by dt. Then the initial acceleration of the block is okay fine at a rate of 10 m/s. So this is very easy. So you have simply have to write F is equals to V * DM by DT.

[01:39:15]So V is equals to 10 m/s and D M by DT is equals to 1. So answer is 10 and acceleration will be 10 divided by mass that is 2 which is 5 m/s squared.

[01:39:35]So one difficult one easy one difficult one easy right.

[01:39:41]Come on let's move to the next one.

[01:39:45]Again you have to use the concept of a net. A net will be net force divided by total mass. A net will be net force divided by total mass.

[01:39:56]A net is going to be net force divided by total mass. So let's draw the forces over here. You have got force acting in this direction, force acting in this direction. You have got tension but I don't have any problem with that. Right?

[01:40:09]This is if this is T then this is T.

[01:40:12]This is M g sin 30. Sin 30 is 1 by 2.

[01:40:19]This is m g uh g is 10 into sin 60 sin 60 is <unk>3x2 friction is is not there frictionless surface right so a net will be kindly see this is 5 m this is 5 newton this is of course 20 * <unk>3 so this is greater so we are going to write 20 <unk>3 3 - 5 divided by total mass. Total mass is 1 + 4 that is five. Cancel cancel four.

[01:41:00]So answer will be 4 <unk>3 - 1 4<unk>3 - 1 4<unk>3 -1 and friction is not there.

[01:41:17]Friction is not there my dear friends right so 4 <unk>3 minus 1 that means option D is correct clear any more doubts Robin any doubt be quick let me know in the chat box mg sin 60 mg sin 30 right and we don't have any friction so simply write this force minus this force divided by total mass.

[01:41:48]Moving on to the next one.

[01:41:52]Let's solve this question.

[01:41:55]Another problem.

[01:41:57]Clear? Good.

[01:41:59]A balloon of mass m is ascending with an acceleration a.

[01:42:05]You have got a balloon and it is moving up with an acceleration of a.

[01:42:16]It is moving up with an acceleration of a.

[01:42:20]Fine. If it is moving up with an acceleration of a then let's figure something some things out. See it has got total mass m definitely mg is going to act downward.

[01:42:34]mg is there right?

[01:42:38]Mg is acting downward and still if it is going up with an acceleration of a that that means there must be some force which is providing this upward acceleration and that is f. So initially we can write that f min - mg is equals to m into a. That means f is equals to m * of g + a.

[01:43:07]This is the driving force because of which the balloon is accelerating upwards. Clear? And then they are saying what fraction of its mass should be dropped.

[01:43:22]What fraction of mass should be dropped.

[01:43:25]So let's suppose mass is being dropped.

[01:43:30]mass is being dropped. So if mass is being dropped now the new masses of the balloon is m minus m. Capital M minus small M. Capital M minus small M. This is the new mass. In that case, the acceleration becomes In that case the acceleration becomes 3A. The acceleration becomes 3A. You have not changed the driving force. So driving force will remain same F and downward force is changed minus M - M * G and equals to M - M * 3 * of A. But what they want us to find uh what fraction? They want us to find out small M. They want us to find out small M.

[01:44:26]Right? Clear? So let's substitute the value of this capital f over here. So we are going to get m g + m a - m g + m g is equals to 3 a m - 3 a m. So mg and mg gets cancels out.

[01:44:50]And my dear friends please note that I want to find out small m right. So bring this small m over here. So 3 a m + m g is equals to 3 a m - m a. So you'll be left with m * 3 a + g is equals to that will be 2 a m 2 a m. So small m will be 2 a m divided by g + 3 a.

[01:45:26]Nice problem.

[01:45:29]Nice nice nice problem.

[01:45:31]Clear. Quickly write down in the chat box. Be quick.

[01:45:37]See these types of concepts are new. It can be new for you right because if if you have solved neat pyq then write down in the chat box if you have solved questions like these Robin Henry everyone gorov saksham everything's clear with this question please remember that this is the driving force this is because of the engine of the hot air balloon right engine of of that thing which because of which it is accelerating so we are not changing anything in that bind force is not over here but we are going to ignore things like bind force as long as as long as they have not mentioned it. See they have not talked about volume of of this.

[01:46:20]They have not talked about the air around it. They have not talked about the density of those things. So we are not going to assume bind force. Right now this is your common sense that okay if you are going to drop something down that will not change the driving force but that will change the mg force because of that net acceleration will increase in upward direction. This is a very nice problem. Okay. Moving on to the next question. You guys can see this is the beauty of these live lectures. We could have recorded these and we could have simply uploaded it. Somebody has asked me about bind force. This is a very good doubt that what about bind force? We don't have to assume it. Why?

[01:47:00]Because they have not mentioned anything about that. Clear? Moving on to the next one.

[01:47:06]Moving on to the next one, my dear friends. Let's move.

[01:47:12]This is the next problem.

[01:47:14]Come on. Let's solve. As shown in the figure, a 70 kilogram garden roller is pushed with a force of 200 Newton at an angle of 30° with the horizontal normal reaction. So this is a normal question.

[01:47:30]Nothing too difficult about this. Be quick. Nothing too difficult about this my dear friends. Solve this question.

[01:47:52]Be quick.

[01:48:02]Shall we move on?

[01:48:06]This is a force. We can also draw a force like this. This is acting at an angle of 30°.

[01:48:15]This is 30°.

[01:48:20]And this force is of 200 Newton.

[01:48:24]This is 20 200 sin 30.

[01:48:33]This is 200 cos 30. I don't have to do we we don't have to do anything with that. Right? We don't have to do anything with that. Along with this you also have got mg.

[01:48:45]So plus mg. So mg is 700 nton.

[01:48:50]Clear? So normal reaction will be normal reaction will be simply the sum of these two things.

[01:49:00]So that means uh 100 plus 700 that is 800. So this is an easy question.

[01:49:09]800 Newton is the correct answer.

[01:49:13]800 Newton is the correct answer. Clear.

[01:49:18]Any doubt?

[01:49:24]Any doubt my dear friends? Be quick.

[01:49:29]Clear. Somebody has said B. So please kindly check.

[01:49:36]All right. Moving on to the next one.

[01:49:42]Moving on to the next one.

[01:49:48]Let's see this question.

[01:49:51]Let's see this one.

[01:49:54]Question is saying, let's read it together. It's a it's an easy one. Don't worry. Consider two blocks A and B of masses 10 kg and 5 kg kept on a frictionless table.

[01:50:07]table is frictionless.

[01:50:10]The block A moves with a constant speed of 3 m/s towards block B kept at rest. A spring with spring constant of 3,000 Newton per meter is attached with the block B as shown in the figure. After the collision suppose the blocks move blocks A and B move along sorry A and B along with the spring in constant compression state move together then the compression in the spring is okay. We have to find out the compression.

[01:50:43]We have to find out the compression.

[01:50:51]Can I draw the diagram over here because I don't have much space.

[01:50:56]Let me explain this question to you.

[01:51:00]Okay, I I'll give you some time then then I then I'll explain. Solve solve my dear friends.

[01:51:08]Block is coming now. See I have made a copy of this. So do not get worried. I'm removing this question from here right and I'm solving first of all please see that it is moving with some velocity V and that velocity is 3 m/s.

[01:51:46]This is 3 m/s.

[01:51:51]All right.

[01:51:53]It will move. It will it will start to compress this this spring. Now, please understand what's going to happen when it will start to compress the spring.

[01:52:04]Right? When it will start to compress the spring, there will be a point.

[01:52:09]There will be a point when something like this will happen.

[01:52:15]You have got a spring over here and you have got a block and you have got another block over here and it will start to compress it. It will start to compress it and the spring will get compressed and spring will apply forces along this direction.

[01:52:38]Please understand I'm explaining you the mechanism. Okay, spring forces are acting like this. So because of the these spring forces, the velocity of this block B is increasing. The velocity of this block A is decreasing.

[01:52:56]Initially this is zero. Initially this is 3 m/s.

[01:53:02]So when spring will get compressed the spring will actually apply force in in this direction in opposite direction.

[01:53:09]Right? And when it will apply force in opposite direction this spring force will cause retardation for block A. So its speed will start to decrease from 3 m/s and it will start and this spring will force will start to increase the velocity of B at a particular point at a particular sorry at a particular point. My dear friends what will happen is that they will start to move with a constant velo they will start to move with a common velocity.

[01:53:42]So all of them are going to move with a common velocity.

[01:53:46]At that point this compression is maximum.

[01:53:50]You have to remember this. At this point this compression is maximum.

[01:53:56]All right. At this point this compression is maximum. So initially this will start to decrease. This will start to increase. At a point they will become same and at that point this will become maximum. Now how to find out?

[01:54:09]First we are going to find out that common velocity. So we are going to start with this. Let's say this is our final state and let's say this is our initial state.

[01:54:20]So first thing first let's apply conservation of linear momentum. Let's apply conservation of linear momentum.

[01:54:28]So we are going to write mass 10 into 3 that's it. This is at rest equals to mass 10 + 5 15 multiplied by common velocity. This will become five. This will become two. That means common velocity comes out to be 2 m/s.

[01:54:52]Common velocity comes out to be 2 m/s.

[01:54:56]So they are moving with 2 m/s of velocity.

[01:55:01]Now again let's apply one more law.

[01:55:05]One more conservation that is conservation of mechanical energy. Next is conservation of mechanical energy.

[01:55:16]This is the initial initial point. This is the final point. We are going to write initially we have got only kinetic energy that is half mass into v² is equals to then we have got half mass mass of both of these blocks that means 10 and five I believe so half m v² + half k x² and k is 3,000 3,000 * x² half can be cancelceled out right this will become uh 90 this will become 60 this will become this is 3,000 * x² so x² will become 30 divided by 3,000 so this will become 1 upon 100 that means x will become 0.1 m 0.1 m is correct answer clear done be quick let me know in the chat box 0.1 m is correct answer this is your entire problem entire question C is the correct answer C is the correct answer along with this question I have explained each and every concept related with this concept initially it will strike Right? Spring force will act in either directions.

[01:56:51]Sorry, spring force will act in both directions and because of those spring forces, velocities will change at a point they will achieve a common velocity. This is our final state. My dear friends, whenever you have got a quotion like this, this is going to be your final state. And now you don't have to look for any intermediate state. You just pick the initial state when the block was not colliding, right? and apply conservation of linear momentum and conservation of mechanical energy these two things. All right, clear moving on to the next one. If you don't have any doubt.

[01:57:27]Okay, let's move on to the next question.

[01:57:35]This is our next question.

[01:57:39]This question is saying a particle experiences a variable force f = to 4x icap + 3 y² jcap in a horizontal xy plane.

[01:57:51]Assume distance in meters and force in Newton. Okay, I'm also starting poll for this 60 seconds. Let's start. And then they are saying particle moves from point particle moves from point 1 comma 2 to 2a 3 in the xy plane. In the xy plane then kinetic energy changes by concept is let me tell you the concept.

[01:58:22]Concept is work energy theorem.

[01:58:27]Work energy theorem. Work energy theorem says work energy theorem says that work done by net force is equals to change in kinetic energy.

[01:58:46]Work done by net force is equals to change in kinetic energy. Right? So they are talking about change in kinetic energy. So let's find out the net work done. Let's find out the net work done.

[01:58:57]Got it? Because change in kinetic energy is equals to net work done. So net work done will be simply integration of f f_sub_x do dx from 1 to 2 plus f_sub_y dy from 2 to 3. And then when you are going to integrate 4x 4x with respect to x² you're going to get 4x² by 2 1 2 when when you're going to different 3 y square you will get 3 y ^2 / 3 sorry 3 y cub divided by 3 from 2 to 3 this will become two this three will get cancels out square of 2 is 4 uh 4 - 1 + 3 and this is this is I believe 27 - 8 come on let me know the answer what answer we are getting from this one so whatever this three is also getting cancels out so we don't have three over here. Right? So this is six, this is six and this is 19. So that means 25 joule is the correct answer.

[02:00:31]25 jou is the correct answer. Clear. Any more doubt? Be quick. 25 jou is the correct answer. My dear friends, can we move on to the next one now?

[02:00:47]Let me know.

[02:00:57]All right, let's see this question. Conservation of mechanical energy.

[02:01:07]Question is saying a simple pendulum of length 1 m. A simple pendulum of length 1 m has a wooden bob of mass 1 kg. has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10 ^ minus2 kg. 10 ^ -2 kg moving with a speed of 2 into 10 ^ -2 m/s.

[02:01:32]The bullet gets embedded into the bob.

[02:01:34]The height to which the bob rises before its swinging back is all right. So you have got a pendulum and a block is attached to this to this uh to this string right a wooden bob of mass 1 kg. So this is of 1 kg and then a bullet is coming and bullet's mass is 10 ^ -2 kilg and its speed is 2 into 10 ^ 2 m/s and it gets embedded in the bob.

[02:02:26]This means that they are saying that it's an inelastic collision. It's an inelastic collision. So the first thing will be conservation of linear momentum.

[02:02:35]Conservation of linear momentum.

[02:02:39]Which means that which means that we are going to simply say mass of the bullet 10 ^ minus2 multiplied by v 10 ^2 is equals to mass of block plus mass of bullet time velocity. So this gets cancels out. My dear friends, I can I can write 1 + 10 ^ minus 2 can be written as V can be written as 1. So this will be 2 m/s.

[02:03:12]So we are assuming that it is moving with 2 m/s. Right? So this thing will move with 2 m/s.

[02:03:22]And then what height will this achieve?

[02:03:25]Let's find that out.

[02:03:29]Let's say this height is h. So we are going to simply say half m. I have ignored the mass of the bullet because it is 0.01 kilogram. I have ignored 0.01 kilg in front of 1 kilg.

[02:03:48]Right? So half m into v² and v is by the way 2. So square of 2 is equals to m g into h. So this gets cancels out. So h will be equals to 2 / 10 or 0.2 m. So this will be nearly around 0.2 m. Nearly around 0.2 m. My dear friends, if you're going to assume the mass also you you must get to the around same answer. must get around same answer.

[02:04:30]Please see.

[02:04:34]Please see my dear friends. Any more doubt?

[02:04:40]Clear.

[02:04:43]Clear.

[02:04:56]Can we move on to the next one?

[02:05:10]All right.

[02:05:13]Great.

[02:05:15]Okay.

[02:05:23]This is a new problem, a new type of a problem. Okay, new type of a problem. A wedge of mass capital m equals to 4m lies on a frictionless plane. A particle of mass small m approaches the wedge with a speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by see wedge is a different thing. Okay. Whenever you are going to see a wedge, please note that this can move.

[02:06:01]So we have got a situation somewhat like this. Kindly read the question. By that time I'm drawing diagram for you.

[02:06:10]You have got a wedge like this.

[02:06:16]You have got a wedge somewhat like this. Surfaces are frictionless. No doubt about it. But what's going to happen when this block will actually will strike this wedge. Right? This is moving with a speed of v. This is a this is moving with speed v. It has got mass m. It has got mass 4 m. So this block this wedge will also start to move with some velocity.

[02:06:53]It will also it will also acquire some velocity and this at that point let's suppose it reaches at a certain height and we want to find out that height. So first of all we'll find out that velocity. So we are going to use again conservation of linear momentum.

[02:07:15]And we are going to write m into v is equals to m + 4 multiplied by vdash that means or vw velocity of wedge. So velocity of wedge will become m will certainly gets cancels out. It will become V by 5. Vx 5.

[02:07:42]That means it will acquire a velocity of V by 5 over because of this. Vx 5 because of this. And it will keep on moving. Right? And now between this point and this point let's apply conservation of mechanical energy.

[02:08:00]So we are going to write in initial energy of this uh system is half m v² is equals to then you have got half 4 m and m is 5 m and then v² by 25 and plus mgh m can get cancelled out 2 two and I'll write two over here this will become five so this will become V² - V ² by 5 is equ= to 2 G into H and I think instead of G also uh take fine it's fine. So we have got this will become 4 V ² by 5 is equ= to 2 G into H.

[02:08:47]So this will become 2 V ² by 5. So h will become h will become 2 v ² by 5g 2 v ² by 5g 2 v ² by 5g option B is correct. Let me know my dear friends you like this question. This is again a new type of a problem. You have not seen these questions in your neat pyq. You might not have seen these questions in your notes also. So kindly note you know conservation of linear momentum you know conservation of mechanical energy but sometimes we don't solve problems related with wedges is there any doubt be quick let me know in the chat box let me know in the chat box any doubt Henry Robin J any more doubt here.

[02:09:58]Please ask if you have doubt. Hush, please ask me.

[02:10:03]Please ask if you have any doubt. Clear.

[02:10:18]I mean there is always somebody in the chat. I mean uh there are different people but there's always one student who's always asking about some girl how are you whenever I see chat so nice consistency let's move on to the next one now let's move on to the next question my dear Now I don't have any problem with that but I I I can't really understand why you guys actually choose my chat box for all of these things. I mean see J Singh any romantic girl.

[02:11:19]Okay let's come back to mechanics.

[02:11:23]uh a uniform cable of mass capital M and length capital L is placed on a horizontal surface such that it is one such that its 1 nth part is hanging below the edge of the surface. This is a uh typical traditional type of a problem, right? I mean general question, standard question I I can say. So we have got a table and then you have got a chain and it's 1/8 part is hanging. Kindly see that its 1/8 part it's one nth part is hanging right.

[02:12:14]So mass per unit length is m by l. Mass per unit length is m by l. So this much length will have mass l by n * m by l. That means simply m by n m by n. And please understand one more thing this the total length is l. So this length will be this length will be l by n. So the center of gravity will be at this is going to be mdash into g. The center of gravity will be at this point. The center of gravity will be at this point.

[02:13:08]This length will be l by 2n.

[02:13:14]Half of L byn half of L byn right L by 2N clear half of L byN right so we are going to simply write down the work done by mg force let's write down the magnitude of that so that will be mg mdash into G * L by 2N because if you're going to if you want to lift this chain above if if you want to lift this chain upward word then you want to actually move this center of gravity from this point to this point. Clear? We instead of assuming that we have got this entire chain we are assuming that this entire mass mdash is at this point at the center of mass of this this part right so we we want to lift this mdash mass from here to here so this length will be half of this length this length is l byn so this length will be lx2n so that means mdash g * lx 2n l by 2 n this is sorry this is n so mdash is m by l. mdash is m by n g l / 2 n. So it would be mg l / 2 n² mgl / 2 n² mg l / 2 n² All right. mg l / 2 n² be quick. Let me know if you have any doubt.

[02:14:51]Clear?

[02:14:54]All right.

[02:15:01]Can we move on to the next one?

[02:15:11]Next question is saying next question is saying a ball having a kinetic energy K is projected at an angle of 60° from the horizontal. What will be the kinetic energy? What will be the kinetic energy of the ball at the highest point? So again this is an easier question. If you're going to project a particle from ground and its velocity is let's suppose its velocity is u and this is 60°. So at the highest point only the horizontal component will remain that is u cos 60 that means u by 2. So at initially you have got half m u² and finally we have got half m u² by 4. That means this is 1/4 of kinetic energy.

[02:16:07]D is correct. D is correct.

[02:16:11]Clear. D is absolutely correct.

[02:16:15]All right, D is absolutely correct.

[02:16:22]Moving on to the next one. Vertical circular motion. Come on, let's solve this question. I'm opening the poll for this question. Oh, you guys, can you guys help me?

[02:16:36]I think I have given you poll for I don't know maybe for the veg one I think before that I think it's see let's check yes correct correct correct correct let's start this question 60 seconds is sufficient for this problem if you if you know the If you know the results for vertical circular motion, if you know the vertical circular motions result, then 60 seconds is more than sufficient.

[02:17:15]Solve.

[02:17:27]A bob of mass m is suspended by a light string of length L. It is imparted by a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the topmost position B. The ratio of kinetic energies of A and B. Ratio of kinetic energy A and B.

[02:18:04]See minimum energy that we can actually give this particle so that it completes the vertical circular motion is roo<unk> 5g gl and if you're going to give this root 5g l it will automatically become root l and kinetic energy will become The ratio of kinetic energy will become VA² that means 5 G L divided by GL L that means 5 is to 1 5 is to 1 clear 5 is to 1 is the correct answer option T some of you have also said 2 is to 5.

[02:19:01]How come? Any doubt in this question?

[02:19:04]Please let me know if you have any doubt.

[02:19:09]5 is to 1 is correct.

[02:19:13]Many of you have selected option B.

[02:19:17]Please note this is the minimum velocity. If you're going to give roo<unk> 5 GL then the particle will complete the circle for this velocity.

[02:19:24]This becomes root3 GL and this becomes root GL right inside the root there is a there's a uh we actually decrease the number by two values right so from five it becomes three from three it becomes one all right let's start this problem let's start this problem if the kinetic energy of a moving body if the kinetic energy Poll is on. Please attempt the poll. If the kinetic energy of the moving body becomes four times If the kinetic energy of the moving body becomes four times its initial kinetic energy, then the percentage change then the percentage change in the in its momentum is first of all please know that we can write down instead of momentum we can write down under root 2 m.

[02:20:24]We can write down 2 MK. And now they are saying that kinetic energy is becoming is becoming four times.

[02:20:35]Kinetic energy is becoming four times.

[02:20:37]Right? If kinetic energy is becoming four times, the momentum will become 2* of initial momentum. Please understand 2 m and this will become 4 K and that will become 2 * of initial momentum. And if you want to find out the increase in momentum this is not 200% but this is 100%. Whenever you have got doubt simply do it by this delta P upon P in percentage is going to be final momentum minus initial momentum divided by initial momentum multiplied by 100.

[02:21:13]So this is 100%age increase here. Easy.

[02:21:17]This is going to be 100%age of increase.

[02:21:20]Option A.

[02:21:22]Very good.

[02:21:27]Let's solve this question.

[02:21:30]Easy, difficult, easy, difficult. Right.

[02:21:33]Let's solve this problem.

[02:21:35]A particle is dropped from height h= to 5 r h= to 5r and moves as shown in the figure. Calculate the force with which it push the track when it reaches at point C. When it reaches the point C of circular track. When it reaches the point C of circular track.

[02:21:58]This is point A. This is point B. This is point C. This is point D.

[02:22:03]Let's start quick.

[02:22:24]Come on friends. First of all, we have to use conservation of mechanical energy, right? So, let's go step by step.

[02:22:37]We are going to first take we apply conservation of mechanical energy between this point and this point.

[02:22:49]Right.

[02:22:51]So at this point it has got total energy is mgh and instead of h I'm going to write mg into 5 r and this is equals to half m into v a² m can be canceled out. This will become 10. That means VA² will become 10 GR.

[02:23:20]VA will become 10 GR. VA this is VA.

[02:23:25]This will become 10 GR. Now we are going to apply conservation of linear conservation of mechanical energy between point A and point C.

[02:23:35]Right? And we are going to write at this point we have got total energy is kindinetic energy/ m va squared that means 10 gr and at that point let's suppose it has got velocity v c we want to find that out so half m vc² plus potential energy will be this is actually uh they are saying that radius of track is r. So this is going to become 2 r. So this will be m gg into 2 into r. So m gets cancels out.

[02:24:17]Right?

[02:24:19]I can also cancel this two this two and I'll write two over here as well. And I please remember I want to find out vc².

[02:24:26]So vc² will be equals to 10 gr and this is 4 gr. So this will become 6 gr.

[02:24:34]This will become 6 GR.

[02:24:38]Hopefully everything is clear till this point. Right?

[02:24:42]This is 10 GR. That was 4 GR. This is 6GR. So VC is clear. At this point, my dear friends, let's draw a free body diagram. You have got mg acting downward.

[02:24:55]You have got mg acting downward.

[02:24:58]You have got MV² by R acting in upward direction, right? And you have got normal reaction in upward direction. I mean the force with which it is pushing the track. So this is mg.

[02:25:14]This is mg. This is m vc² by r.

[02:25:20]Right? Clear? So the normal reaction will be because that's what they are saying.

[02:25:29]Calculate the force with which it pushed the track. So it pushes the track with mg minus mc² by r. So it's going to be mg minus m vc² is 6 g divided by r. R and R gets cancels out and obviously my my dear friends a this one is mg will be less right as you can see so this is going to be 5 mg a option is absolutely correct a option is absolutely correct let me know in the chat box be quick nice problem so we have to apply conservation of mechanical energy from this point to this point then we have to apply the conservation of mechanical energy from this point to this point Clear? Is everything clear?

[02:26:21]Fine.

[02:26:23]Be quick. Let me know in the chat box.

[02:26:30]Any more doubt?

[02:26:33]Any more doubt?

[02:26:36]Option A is absolutely correct.

[02:26:56]clear. Nice problem.

[02:26:58]Please understand if it if it is actually moving along the track obviously it will it will apply the force in this direction. Now if let's suppose net force is is more in this direction. If mg is more then it will fall from track. Clearly, so obviously MV MV² by R MVC squared by R will be more than MG, right?

[02:27:19]And normal reaction will act in normal reaction will be applied by the by the track on the uh on this on the car in downward direction. Clear? Sorted.

[02:27:32]Please understand the dynamics clearly.

[02:27:34]Right?

[02:27:36]All right, my dear friends. Very very good. Let's move on. Let's move to the next question.

[02:27:52]Let's move to the next problem.

[02:27:55]A block of mass m is released from rest at the top of a frictionless incline of height h and theta, an angle theta. It then slides onto a horizontal surface and collides with a spring of constant K. Find the maximum compression in the spring. You have to find the maximum compression in the spring.

[02:28:18]Let's begin. I'm going to give you poll for this question. 90 seconds. Let's start.

[02:28:31]Let's start.

[02:28:33]A block of mass m is released from rest at the top of a frictionless. See all of these questions are for work energy theorem. So now let me know in the chat box these many questions at the very start I have said key after this class your Newton laws of motion and your work energy will be almost over. After this you only have to solve some questions from your mock exams and what you are going to observe is those questions will be definitely of lesser difficulty level than these questions. Right? Let me know. We are going to solve some couple of questions more.

[02:29:29]Again, you don't have to think about the intermediate stages, right? All you can do is this is your initial point. This is the initial state of the system and then they are saying that we have to find out the maximum compression. At maximum compression of course it will come at rest. So this is your final state.

[02:29:51]So simply write down initial energy is mgh.

[02:29:55]This is h. So initial energy is m g h. And that will be simply equals to/ kx².

[02:30:06]And this is your maximum compression.

[02:30:08]That's it. So maximum compression will be 2 mgh divided by k and under root of this. Option B is correct. Very easy question. As compared to the previous problem, this is very easy. Very good.

[02:30:30]Some of you have said a but a is not correct. Please see because of this two you are going to get 2 mgh right.

[02:30:43]Let's solve this problem.

[02:30:46]Under the action of a force, a 2 kilogram body moves such that its position x as a function of time t is given by x equals to t² by 3 where x is in meter and t is in second. The work done by the force in first 2 seconds is all right in first 2 seconds. Again they are talking about work done and when see I have told you this again and again that whenever they are going to talk about work done simply write that net work done can be written as change in kinetic energy. Is that correct? And they have also said that x is equals to t² by 3. So that means v can be written as dx by dt.

[02:31:35]So this will become 2tx3.

[02:31:40]Now 2tx3 basically means that initial velocity will be 0 and final velocity will be after 2 seconds it will be 2 * 2 / 3 that means 4x3 4x 3 m/s 4x3 m/s. So change in kinetic energy will be half m into v² 16 by 9. So 16x 9 jou is the correct answer.

[02:32:16]16 by 9 jou is the correct answer. D is correct. All right. D is absolutely correct.

[02:32:24]All right. Let me know in the chat box if you have any doubt.

[02:32:33]Moving on to the next one. A body which is moving with 10 m/s is sliding up or on a rough inclined plane having an inclination of 30°. Find the height up to which it can go if coefficient of friction of the inclined surface is 0.1.

[02:32:51]Come on, let's do this problem. My dear friends, you have got a surface having an inclination of 30° and then they are saying that a body is moving up with a speed of 10 m/s and coefficient of friction given for this surface is 0.1.

[02:33:20]Find out the net acceleration. Right?

[02:33:22]Find out the net acceleration.

[02:33:26]Mg sin theta will act downwards and friction force will also act downwards. So g sin 30 and mu g cos 30 is that correct friends? That means net acceleration will be mu or I can take g common right 0.1 * 10 * <unk>3x2 + 10 * 1 by 2 this gets cancels out and you can actually write something like this if you want and then you can use v² is equ= to u ² + 2 a into s. v is 0. U is 10. That means 100. This is 2 * 10 + <unk>3 divided by 2. This is s 2 and 2 gets cancels out. And let's please find out 100 divided by 10 + 1.7. Right? So 10 100 divided is 10. So it is slightly less than 10. Find out the correct answer for this one. My dear friends, be quick. find out the correct answer for this one.

[02:34:48]Uh this is actually H Please find whatever the answer is.

[02:35:18]Find the height up to which it is going.

[02:35:19]Acha. Okay. I'm sorry. They don't want to find out the distance. I'm sorry.

[02:35:28]They don't want us to find out the distance. They want us to find out the height.

[02:35:33]Yeah. So if they want us to find out the height then why we are doing this?

[02:35:37]Sorry, sorry, sorry, sorry, sorry. I have read the question incorrectly. They are saying find the height up to which it can go. Find the height up to which it can go. Okay, clear. Fine. We can use this s for for for that also. No problem. So what we are going to do is a body is moving with 10 m/s is a body which is moving with is sliding up on a rough inclined plane having an inclination of 30° fine. If the coefficient of friction is 0.1 fine fine fine so what they are trying to say is that let's let's suppose the block has reached over here and v is actually equals to zero then they want us to find out this height right so use work energy theorem after finding out this s use work energy theorem work done by mg plus work done by friction force is equals to change in kinetic energy right so you can Actually we can actually find out this distance. We can find out this distance my dear friends.

[02:36:42]Let's say this distance is d right this distance is d. So this is going to be minus mg * h. This is going to minus mu mg cos 30 * d and this is going to be -/ m v².

[02:37:05]After this you'll get to your answer and this S is over here. This this S is actually equals to D. Clear? Write down in the chat box sorted. Now you'll be able to do they were talking about height my dear friends. They were not talking about this distance. They were talking about height. So you'll have to apply work energy theorem again. Right?

[02:37:28]Let me tell you work energy theorem is basically net work done is equals to change in kinetic energy. Now this includes each and every force. Work done by conservative force, work done by external force, work done by non-conservative force, work done by pseudo force. All of these is equals to change in kinetic energy. And sometimes we can write instead of this we can also write down it as minus times of change in potential energy right hopefully this is clear sorted all right so in this question they are saying that an object of mass 5 kg falls from rest through a vertical distance of 20 m.

[02:38:24]So they are saying that an object of 5 kg falls from a distance of 20 m right and then fine and then the work done by air resistances and it reaches the ground with a velocity of 10 m/s.

[02:38:58]Again see I'll again tell you the same thing. Net work done is equals to change in kinetic energy.

[02:39:07]That means work done by mg work done by air resistance is equals to final energy is half m v² initial was zero. Work done by mg will be mg multiplied by 20 right it's positive work done work done by air resistance will be negative and that's what we need to find out if you're going to put a plus sign you will automatically get negative so half mass s5 into 100 this will become 50 this will become 250 and this will again become 50 isn't it this will become 50 so this will become 1,000 So air will become -,000 + 250. So it will it it has to come negative. Sorted my dear friends. Any more doubt? Clear?

[02:40:06]All right.

[02:40:09]So I think we have finished sufficient amount of questions.

[02:40:14]Now in next class we are going to talk about a I've got one more problem. This is this was from one of the mock exams, right? Although we can solve these questions in SHM also. But let's have a look at this. Let's have a quick glance.

[02:40:32]Uh a uniform spring of force constant K is cut into three equal parts.

[02:40:39]Is cut into three equal pieces.

[02:40:43]That means you have got a spring and then you are cutting it in three equal pieces.

[02:40:52]Three equal pieces.

[02:40:55]So this is lx3 lx3 lx3.

[02:41:02]Please remember that when you are going to cut them their spring constant will become three times 3k 3k and 3k.

[02:41:13]Note this point and if you want to understand how this thing works please remember that if you have got three different springs of spring constant 3k 3k and 3k and if you are going to join them in series if you're going to join them in series you are going to get this one big spring right and the spring constant will become 1 upon 3k 1 upon 3k 1 upon 3k I'm talking about k equivalent so this is how you can remember whenever you have got a spring just Cut if you and if they are saying it to cut it in equal pieces spring constant will increase and then they are saying then they are saying that these are connected side by side in parallel and this entire combination is connected in series with the third piece that means you have got one spring like this you have got another spring like this and you have got third spring like this so this is 3k this is 3k and this is 3k Now series combination sorry parall combination this means that this is 6k and this is 3k. So finally the k equivalent will be 6k ult*lied by 3k divided by 9k. K and k gets cancels out. This becomes three. This becomes 2.

[02:42:30]That means answer will be 2k. Answer is going to be 2k my dear friends. Class timings.

[02:42:38]for class timings. Let me uh I have already told you that please watch this channel regularly. You are going to see the uh see on the community sections when we are going to have our next class. All right. Clear?

[02:42:53]Done? Everything's okay with this problem.

[02:42:57]Quickly let me know in the chat box.

[02:43:01]So after this now we are going to start rotational motion and center of mass in our next class. These two chapters are over. Uh I am pretty confident if you're going to solve these levels of questions, it will be good enough for you. After this, you can complete your neat pyqs, you can you'll be good enough for giving your regular mock exams. All right, clear. So, I'm going to see you in next class with center of mass and rotational motion. Okay, bye-bye. Bye. Bye. Thank you so much. Bye-bye. Bye. Take care.