 I hope all of you understood condition to leave the surface is normal region becoming 0. So, we will write equation the Newton laws of motion second law of motion equation at C and put n equal to 0 that will give me velocity at point C velocity at point B is given. So, I will correlate point B and C with work energy theorem should I solve or should I wait. So, one minute sir. Sir, two minutes. I just draw three by a diagram at point C. Nometer reaction Mg and there is a the intripetal acceleration this is acceleration v square by r. So, at point C equation the laws of motion equation is Mg cos theta minus n this is net force this should be equal to mass and acceleration did you get this equation? Yes. Okay. So, is it B that we are cos inverse of 3 by 4? Yes sir. I doubt that is correct but let us see I hope you are not using a ready made expression anyways. So, normalization is 0. Sir, option C. Option C. Who said this? Tromek. Tromek. Yes that I think if I remember correctly I am for C. So, normalization is 0. So, from here you will get velocity at point C square should be equal to G into r into cos of theta. This is the first relation this is the condition for it to leave the surface okay this should happen otherwise it cannot leave the surface at C. Now I am going to use work energy theorem w is equal to u2 plus k2 minus u1 plus k1 okay. So, w is 0 gravity is doing work but for that I am using potential energy okay and I am going to say that this horizontal line is 0 gravitational potential energy. This represents uG gravitational potential energy to be 0 okay so k1 is half mv0 square so m into v0 square v0 is 0.5 root gr, u1 is 0 right. Now k2 is m into v square where v square is gr cos theta. What about u2? Anyone? u2 is what? To find u2 I need to find how much below it is vertically how much below it is from the 0 potential energy line. So, this distance is how much? Tell me how much is this distance? So r minus r cos theta. This distance is r cos theta okay and the complete distance from here to here is r so that distance is r minus r cos theta so potential energy will be negative right because it is going below the distance it is going below the 0 potential energy line so it will be minus of mg r minus r cos theta fine so just substitute the values v0 is 0.5 root gr v v square is gr cos theta and put everything and I think you will get cos theta to be equal to 1 by 4. Can you confirm it once anyone? Can you solve this? See you get 2 by 3 you will get cos inverse 2 by 3 when it starts from rest you just gently push it forward then I think you will get cos inverse 2 by 3 but if initial velocity is there then angle will be different. All of you understood? Any doubts? Sir what will be lesser if it is going with some velocity? Will it be lesser if it is going with some velocity? You think angle will be lesser? Yes I am saying that if it is already going with some velocity it should quickly fall off. Quickly fall off your argument seems to be correct. Sir it should be 3 by 4. No that argument could be correct but what we have got is the fact okay answer is 1 by 4 only I am wondering about what should I say okay? The answer is 1 by 4 only you cant qualitatively you cant say and arrive at the answer you have to find out what is the answer using equations. No sir I solved these equations only. Oh you got 3 by 4? Sir even I got 3 by 4. Oh then answer is 3 by 4 Shomik you said 1 by 4 can you check? Sir I got 1 by 4. Okay whatever it is this is the equation to solve these are the two equations first equation second equation okay fine let us go to the next question this one this one is very similar to what we have just done okay so do it properly and get the answer. Is it C 3 sin beta equal to 1 by 4? Yes sir even I got 3 sin beta is equal to 2 by 4. Yes sir same sir. Okay fine so it leaves the cylinder at point B right so I will draw the free buried diagram of B this is mg this is normal reaction and expression is v square by r towards the centre so one thing you may have to notice here is that this angle is 90 minus beta it is not beta because beta is given with horizontal okay so you will write mg cos of 90 minus beta which is mg sin beta minus normal reaction is equal to mass time acceleration v square by r so normal reaction 0 when it leaves the cylinder so v square that is velocity at B square is equal to gr sin of beta okay this equation I am assuming all of you got it some of you might have done the silly error taking cos beta instead of sin beta but have you got this relation? Any doubts here? Skanda. Skanda will speak. Doubts am I audible? Oh Skanda typed. Yes sir. So between A and B I will use work energy theorem w is equal to u2 plus k2 minus u1 plus k1 okay w is 0 and let me assume this I am I can assume any horizontal line to be 0 potential energy okay so just for the sake of convenience I am assuming this horizontal line to be 0 gravitational potential energy okay so that is why u2 is 0 because I have assumed that to be 0 k1 is 0 it starts from rest okay k2 is half m v square v square is gr sin beta so this is m into v square so m gr sin beta sin u1 is the initial potential energy initial potential energy is mg into h and where is h? This is h. How much is this length? This length is r cos alpha this is r cos alpha and this one is sin of beta yes or no? Is it understandable? Yes sir. This is r cos alpha minus r sin beta okay Ananya have you understood what is written here mg h is written here h is r cos alpha minus r sin beta okay there is no Ananya here no Ananya h is here Ananya Ramchandra three Ananyas we have oh that is why these guys are keeping quiet huh so I should tell the full name Ananya Ramchandra have you understood Ananya ar Aniruddha have you understood fine so when you simplify this you will get the answer okay so what answer you will be getting you will get three sin beta three sin beta is equal to 2 cos alpha so you will get c okay when you simplify this let us move to the next question this one you can solve one by one 13th first you answer 13th question hmm anyone first you need to find out what is the minimum velocity required for it to reach the point d for an object to just to reach point d normal reaction at d should be 0 right it loses contact just loses contact at d the normal reaction becoming 0 there is the condition for minimum velocity once you find that velocity that into 2 by root 5 is the initial velocity anybody got the minimum velocity? so root 24 g r 5 5 the whole root 24 g r by 5 root 5 g r is the answer for the minimum velocity at a okay so I will quickly do this I will do a lot of practice okay this chapter I mean you can't just attend the lecture and hope to do well in the numerical that will that has never happened and it will never happen okay you have to back it up with lot of problem practice at home otherwise everything is going waste so at the top most point this is normal reaction then there is mg also and acceleration is let's say v square by radius is let's say capital R so v square by R so at the top most point n plus mg net force is equal to mass times acceleration mv square by R so normal reaction is 0 at the top most point if it just leaves the surface so for normal reaction 0 v square should be equal to g times R okay this is the velocity at point d okay now between a and d I am going to use work energy theorem w is u 2 plus I don't know we have done this many many times okay so can we just use the motion in a vertical circle formula root under root 5 g l that is for the pendulum that's for pendulum right this is not a pendulum although the final answer looks like a pendulum one only because it turns out to be similar equation but it's not a pendulum that vertical circle motion we have done for the pendulum where tension is there here normal reaction is there anyway so w is 0 k 1 is half m u square where use a velocity at point a u 1 is 0 k 2 is half m v square so it will come out to be mg R by 2 and u 2 is mg into this distance that height which is 2 R okay so this is mg h so mg into 2 R from here you will get a velocity to be equal to root over 5 g R this is the minimum velocity required for the object to reach point d all of you understood this any doubt on this one minimum velocity to reach point d is root of 5 g R okay understood okay so this is the minimum velocity required for it to reach point d so currently the velocity is 2 by root 5 times the minimum velocity so velocity of projection right now is 2 by root 5 times of root 5 g R that is 2 root g R this is the velocity okay so at what angle the block gets separated from the track okay let's say that this is the track it starts from here let's say it separates there at an angle of theta okay so this is point a and c so if it separates at point theta then again you can write down the laws of equation relation this is mg this is normal reaction and expression is v square by R okay so if you write the laws of motion equation if it had to be find max velocity what we have to do at the maximum velocity is infinite this is a minimum velocity after this whatever velocity you give it will cross d there is no limit up to max for the maximum velocity minimum velocity there is a limit so at point c if you write the Newton's laws of motion equation you will get mg cos so this is theta so this will be 9 to n theta so you'll have mg sin theta there mg cos of 9 to 1 theta so mg sin theta minus normal reaction is equal to m velocity square divided by R so v square should be equal to g R sin theta okay where it loses the contact now I am going to use the work energy theorem between a and c now okay so w is 0 okay w is 0 and u2 is what u2 is this height this height is R plus R sin theta okay so I'll write mg R plus R sin theta so mg R 1 plus sin theta plus half m v square at point c v square is this so it will be mg R sin theta this is u2 plus k2 okay u1 is 0 and k1 is half m into this velocity square this one okay so this will be half into 4 into g R okay so once you solve this equation what answer you'll get can anyone tell me I got c so c c even m will come so again cancel out mg R mg R mg R from everywhere so I'll get 1 plus 2 sin theta 1 plus 2 sin theta minus is equal to 0 right so sin theta is half wait what's the error we have done can you identify what's the error we have done oh there will be half here this is v square so half m v square is the kinetic energy this is 1 by 2 so sin theta plus sin theta by 2 is 3 by 2 sin theta so sin theta will come out to be 2 by 3 so option c is correct see if you see that the complication is with respect to solving the equations because ultimately how you use the equation is the same doesn't matter which question you are solving same thing we are doing again and again just that equations are getting formed in a different manner so that that is what it is making it slightly tricky so it's not the physics which is making it difficult it is the calculation or mathematics which is making it tricky so that that problem will go away when you solve your own lot of questions okay so other two questions 14 and 15 I'll keep it as homework alright so let's go to the next one so here is the question guys all of you suppose you have a pendulum of mass m okay this is a pendulum of mass m and length is l okay you have given it a velocity u which is equal to under root of 3 gl okay this is the velocity with which this is thrown like that initial velocity you need to find out the angle the string will rotate the string may go like that before it becomes big before it becomes slack okay so you need to find out what angle it makes with the horizontal before it becomes slack all of you understood the question right yes okay solve it see it is not root 5 gl had it been root 5 gl it will swing to a full circle but it is less than root 5 gl so it will not swing to the full circle so is it sign inverse of 1 by 3 other sir I got sign inverse of half okay I got sign inverse 1 by 3 yes sir I also made a mistake itself 1 by 3 so if it the string if it becomes slack at the top most point tension will become 0 right so this is mg this is mg this is tension there is 91 a theta okay so I can say that t plus mg cos of 91 a theta which is mg sin theta this is the force towards the center this should be equal to okay this should be equal to the mass time acceleration so this is m v square by L now if if string is slack at that point if a string is slack tension is 0 okay so that is why v square should be equal to g into L into sin theta or sin inverse of 1 by 3 okay let's say this is point number 1 and this is point number 3 so now I am applying work energy theorem between these two points w is equal to u1 plus k1 right u2 plus k2 minus u1 plus k1 okay w is 0 k1 is half m u2 so u2 is 3gl is k1 u1 I am assuming this line to be 0 potential energy this horizontal line I am assuming 0 potential energy so u1 because of our assumption is 0 k2 is half m into v square so half m into gl sin theta okay u2 is what anyone mgl plus l sin theta because I have to see vertically how much above it is so this distance is l sin theta because it is 90 degree plus l fine so this is the equation now I can cancel out mgl ln l gone so I will have 1 plus sin theta plus sin theta by 2 minus 3 by 2 is equal to 0 okay this is what I will get right yes is what I will get so I will have the 3 sin theta by 2 minus 1 by 2 is equal to 0 so from here I will get sin theta as 1 by 3 so theta is sin inverse 1 by 3 I hope all of you have understood this no doubt right understood okay now there are few questions which are which connects one chapter to the other chapter okay so let us solve one such question which connects work by energy chapter with the projectile motion okay there are so many such questions which connect multiple chapters together okay so what happens is a bomb is thrown a harmless bomb it is thrown at 37 degree from the horizontal with a speed of 100 meter per second okay at the highest point when it reaches the highest point of its trajectory the bomb explodes at the highest point and the ratio of mass there will be two parts that comes out the ratio is 1 is to 3 mass of one part is one third of the mass of the other part okay the smaller part the velocity of smaller part becomes 0 immediately okay and the smaller part because of that falls vertically down like this okay you need to find out what happens to the larger part larger part will go forward and let's say it reaches here so you need to find out this distance the range of the larger part is what all of you understood the question it explodes at the highest point that is what I said right at the highest point it explodes okay hello solve it now this is also one of its kinds numerical you may not get it right but at least try it we need to find the distance cover the third part second part yes from the initial point of projection what is the distance so we know that the second part is projected horizontally after it plus we don't know that that is not given can I conserve momentum when explosion happens yes or no can I conserve momentum yes sir I can conserve momentum this is a collision only it is a collision I can conserve momentum so at the highest point the momentum is only along the horizontal direction yes or no horizontal direction is the momentum before the explosion after explosion the momentum of the smaller part becomes 0 okay so that again that is like a hint I have given you now try it so what sign 37 oh it's given you should remember by the way it will not be given in the exam remember 37 degree is a 345 triangle Pythagorean triplet triangle okay sir should I do it so is it 1,204.8 meter 1,204.8 okay others you've taken G as 9.8 1,000 right what about others 1,600 okay let me solve it now let's say mass of the bomb is m okay so clearly one of the when it explodes one of the particle will have m by 4 mass and other one will have 3 m by 4 right some of the masses should be m only and the ratio should be 1 is to 3 so m by 4 and 3 m by 4 will be the masses so initial momentum just before the explosion it had only horizontal component of the velocity so initial momentum is m into 100 cos 37 any doubt here at the top most point this is the momentum any doubt the momentum of m by 4 is 0 so m by 4 into 0 plus 3 m by 4 into velocity along x direction for the 3 m by 4 okay because I am conserving momentum along x direction right now so I will take velocity along x axis fine so this is anyway 0 m and m goes away so you will get vx velocity along x axis for 3 m by 4 is 400 by 3 into cos of 37 which is 4 by 5 okay so this is 80 320 by 3 meter per second this you've got this one anybody okay so this is the velocity along x axis what will the velocity along y axis for the 3 m by 4 mass 0 because initial momentum along y axis is 0 so along y axis initial momentum is 0 and momentum of m by 4 mass is also 0 so when you write it as 3 by 4 m into velocity along y direction even v y will come out to be 0 okay so there is no velocity after the explosion in the y direction for the 3 m by 4 mass so it is like a horizontal projectile projected like this and this height is what height you can find out from the maximum height formula so maximum height is given as u square sine what is the formula for maximum height u square sine square theta by 2g 2g 2g so this will give me height okay now it is like a horizontal projectile now it is projected with velocity of 320 by 3 meter per second with a height of h so this distance we have done this type of numerical so many times right so time of flight is under root 2h by g so this distance let us say this is r1 okay so r1 will be velocity into time velocity along x axis is unchanged so this velocity into root over 2h by g this is r1 r1 is what this distance and that distance is how much that distance is the half of the range of the initial projectile so that distance the range is u square sine 2 theta by g so let us say this is r2 r2 will be u square sine 2 theta by 2g half of it so answer is r2 plus r1 go through it once and let me know if you have any doubts quickly sir can you just repeat the range this part the range formula is u square sine 2 theta by g now had it not exploded it would have come and hit here this would have been total range so from here to here it did not explode so this distance would be half of the range total range is u square sine 2 theta by g so r2 is half of that so by 2g and r1 we have got from here so total range is r1 plus r2 okay so this is one of the kinds and you will learn a few concepts when you study this type new new type of numericals you have to get exposed to so whenever you solve problems you know do not just solve same kind of numericals again and again so suppose you just take a numerical in which a for horizontal force is applied and you have to find the variation now you cannot solve numericals where horizontal force has different different values all that are same kind of numerical so when you practice make sure you practice different different varieties of the numerical for example this one is another question that we are going to solve now please draw this diagram sir what was the answer for the previous one the answer was equal to it was one thousand one hundred twenty I think where it went I got it you got it one one two zero is the answer one one two zero okay fine so here is the next question all of you please draw this this is the last question for today so I hope all of you get it correct try to do it systematically you will get it correct it is it can be done easily okay do it systematically this is m m and spring constant k a light is spring or spring constant k is kept compressed between the two blocks small m and capital m okay this one is capital m the second one is capital m it is kept on a smooth horizontal surface this surface is smooth okay when released the blocks acquire velocities in opposite direction it was initially compressed after that of course they will acquire velocities in the opposite direction they will move away from each other spring will push both together okay the spring loses contact when the with the blocks when it acquires natural length as soon as the spring comes to the natural length spring detaches itself from the masses fine the spring gets detached when it comes to natural length so if initial compression if initial initial compression of the spring is given if initial compression of the spring is x okay you need to find final speeds of small m and capital m okay so velocity of small m is v1 and let us say v2 you need to find what is the value of v1 and what is the value of v2 have you understood the question any doubt on the question quickly ask the hint is again same momentum and work in a theorem nothing else simple sir the velocity of small m will be x y 2 root k y m have you assumed both the masses equal no sir that is wrong answer one is a small m other is capital m don't you think velocity of small m depends on how big or small the capital m is the capital m is huge small m will have all the kinetic energy if capital m is very less small m will have lesser velocity spring is pushing both the blocks with the same force both sides so smaller block will accelerate more bigger block will accelerate less spring has fixed amount of energy to give so like that you can think and if you are getting answer independent of one of the masses that is wrong should I do it somebody said something should I solve okay I will solve now first is conservation conservation of momentum tell me what is the initial momentum initial momentum R M n what it is Vibhavashudevan what is initial momentum tell me all bodies are at rest so initial momentum is 0 this should be equal to the final momentum after spring comes with natural let's say velocity is V1 and V2 so this will be equal to m into V1 minus capital M into V2 okay this is the first equation which is conservation of momentum okay now I am going to use work energy theorem now I don't know whether you have seen it but I am like bored of doing same thing again and again because every problem even the situation is different but when you solve it you will see that all of them look same okay if you if you are not able to visualize it to be same then probably you may have to do some practice after some practice you will see that you will feel every question is same k1 is 0 both the masses at rest u1 is half kx square initial compression is there right u2 is 0 and k2 both the masses have kinetic energy now k2 will be equal to half small m into V1 square plus half capital M into V2 square and W is 0 because spring is the only force that is doing work and for that you have considered the potential energy right this is my second equation fine so when you solve these two equations you will get V1 to be equal to root over k times capital M divided by small m capital M plus small m times x okay this is V1 and V2 you will get it as kmx divided by capital M into small m plus capital M see all you have to do is substitute the value of V2 in terms of V1 from here V2 is equal to small m by capital M times V1 so when you put it here you will get the value of V2 once you get V2 again use this equation to get V1 all of you understood yes sir yes sir yes sir okay so today we have exhaustively covered the chapter work by energy and again I am telling you if you don't do practice you will remember my words these two chapters work energy work by energy by the way power is something which we haven't covered it's a very very small topic just two minutes will take so what I was saying was that work by energy and laws of motion these are extremely important topics so please do practice for your own sake practice at least hundred question of work by energy then you don't have to worry about chapter anymore okay I quickly cover the power part of the chapter okay it's extremely small it takes five minutes please write down power power is nothing but rate at which work is done rate of work done see it is like you know two people doing the same amount of work but one person doing faster so the time factor comes into the play so time is a very very important factor for example if you are hungry and food is getting cooked you don't want food to be cooked in two days of time the amount of energy use will be same but if your cook get if your food get cooked in let's say half an hour the rate at which energy supplied to the food is much faster right so time is a very very important factor another example could be a very strong person doing a work and a very weak person is also doing the same amount of physical work I'm talking about so the weaker one will take a lot of time but amount of work then will be same okay so that is why we need to also take into account the time factor when we calculate the work done and power is one of the physical quantity which takes into account the factor of time so it's in the definition is rate of work done the power is simply derivative of work done dw by dt okay now work done can be written as f dot dr the entire dw can be written as f dot dr f dot dr divided by dt dw is f dot dr divided by dt okay now this can be rearranged as f dot dr by dt now what is dr by dt its velocity right so dot product of force with the velocity is the work done fine and the SI unit SI unit of work is SI unit of power sorry is watts okay so there are direct question on power there will not be any complicated question there will be a force given and the velocity of object is given so all you have to do is take a dot product of force and velocity will get the power or at times w is given as a function of time for example w is given as 2 t square so power is derivative of w it will be 4 into t okay so questions like these you will see for the power there will not be any very very difficult questions on power okay so that's it guys this chapter is over and next topic will start from the book 2 okay fine so we will meet next week more sorry bye bye for now thank you sir