 okay are you guys able to hear me those who are online okay fine so as all of you know today we are going to revisit the most popular chapter in physics laws of motion okay so at times we do lot of problem practice in this chapter and we tend to ignore the you know finer nuances in this chapter because we at times become complacent with respect to this chapter so I have collected few of the questions which are I am assuming they are something which from which you can learn okay so without further you know wasting too much of time let us start solving questions okay so you can see the first question on your screen right start solving this welcome you have any board practicals tomorrow I hope you are able to clearly see the question if you're not probably the resolution of your video is less so you can increase the resolution as you do for any YouTube video okay anyone able to solve this one here you have to use concept of pseudo force try to solve this problem with respect to the accelerating car itself okay amog has got one answer others so concept of pseudo forces used here fine let me solve this now so you have a mass this is tension along the string will be the tension okay and then you have an mg force fine and along the string there is an acceleration of a and since you are solving in a frame of accelerating reference so you're solving in a accelerating frame of reference so there has to be a pseudo force of value m into a and direction of that force will be opposite to the direction of acceleration fine this is the situation and with vertical this angle is theta fine so you have components of t this is t cos theta and this one is t sin theta so you will get there will be an component of acceleration also so you'll have this as a cos theta and here the acceleration component will be equal to a sin theta fine so now vertical and horizontal we are balancing all the forces t cos theta minus mg net force vertically should be equal to mass mass m acceleration so m a cos theta fine so from here you get t cos theta is equal to mg plus m a cos theta fine so this is your equation number one all right and horizontally you will get t sin theta minus m a is equal to m into a sin theta okay so you will get t sin theta is equal to m into a plus m into a sin theta this is equation number two all right so when you square an ad you'll get t square t square to be equal to m square g square plus m square a square plus here m square a square cos square theta will come and here m square square sin square theta will come so again one m square a square will come right and then here you will get 2 m square a g cos theta plus 2 m square a square sin theta okay so theta is coming in the picture we need to eliminate theta and how you guys have eliminated theta okay so you guys have substituted tan theta as a by g yeah that's correct vector addition of a and g will be along the string have used only this equation have you guys substituted tan theta as a by g have you done that see what will happen is this is a situation right this is a situation so if you balance forces perpendicular to the string like this if you balance okay this angle is theta okay so this is 90 degree and then this is also 90 so this is 90 as theta so this is 91 is theta and this is theta isn't it this is theta yes this is theta this is 91 is theta okay so see you should know what you're doing so you you should know how we arrive at it you may have got the correct answer so I was just thinking how it is than theta is a by g okay so perpendicular to the string there is no acceleration and there is no tension okay so perpendicular to a string I can balance the forces fine so m into a cos theta this direction is the component of m into a and this direction the component of mg is mg cos of 91 theta which is mg sin theta so this is mg sin theta fine so you know mg cos theta minus mg sin theta is equal to zero because there is no acceleration or tension over there so from here you'll get tan theta as a by g okay now we can substitute the value of cos theta here and there to get the value of tension okay if you don't need second equation okay we or you can use any of these two equation to get the same answer understood all of you tan theta is a by g this is the the critical part of this question no doubts right okay so we'll go to the next one now okay Amogh has got the answer others see when you apply a force there'll be a friction between small m and capital M which will be balancing the weight of small m fine so let me solve this now so we have two masses small m and capital M coefficient of friction is given no friction between capital M and ground minimum force applied to hold the small m against capital M so there is no relative motion between small m and capital M fine so if you consider your system to be capital M plus small m if this is your system okay then whatever force you are applying should be equal to small m plus capital M into a right both of them are moving together so you got the acceleration f divided by m plus capital M so with this acceleration if both of the blocks are moving together will move there is no friction so there is no other horizontal force other than capital F okay but now if you just consider your system to be small m okay so if this is your system it will have forces like this this is f this is mg okay and then you will have normal reaction from capital M okay and you'll have a frictional force like this okay and of course it has an acceleration a which you have already found from here okay so when you write down the force balancing you'll get f minus normal reaction this is equal to m into a okay net force equal to mass and acceleration all right and friction should be greater than or equal to in fact since it is balancing mg okay so friction has to be equal to mg all right and since maximum value of friction is mu times n so mu times n has to be greater than or equal to mg all right so get the value of normalization from here substitute there and you'll get the answer okay so these are the equations any doubts no doubts I'll go to the next question now this one again on friction okay poor we got it others all right I think this is also something which was similar to the previous one with respect to concept involved okay so all you have to do is to balance the forces fine so we have let me draw a diagram over here itself this is mg this is f sine 60 f cos 60 fine this is mu n the maximum frictional force and this is normal reaction okay so this is small m all right so whenever you have you know whenever you have a situation where they are asking you to find maximum or minimum value of friction you know you you can directly write mu into n as your friction force right mu into n is actually the maximum value of friction force so they will try to ask what is the maximum value of f such that it does not move so basically what is happening is that the only variable force is friction right static friction varies from 0 to mu n isn't it so whatever variation happens because of the frictional force only so either friction will be zero or the maximum value that is why the limiting condition happens okay so from here you will get f sine 60 plus mg minus normal reaction is equal to zero all right and horizontally I'm assuming that there is no exhibition right so f cos 60 minus mu times n should be equal to zero fine so these are the two equations and you just have to substitute the values and get the answer all right the fact that you are using mu into and your frictional force you will get the maximum force only because f is equal to mu n by cos 60 so when this mu n would have been just friction force you're using the maximum value which is mu into n so you'll get the maximum value of force also fine let's go to the next question okay there is a variation as in some are saying b some are saying a so let me tell you the final answer the answer is a the answer to this question is a okay amok you have done like this only okay which is 3a yeah okay fine so I think some of you are making silly errors please don't do that silly errors can cost you you know very very dearly I mean you you I think you have already experienced okay five marks can cost you around 10,000 ranks you might have realized that some of the trickiest questions from laws of motion chapter they are from friction only should I solve see so these type of questions you might have already done but here there are three blocks okay so what will happen is that the acceleration of b can be any value okay it can be anything because it is connected with the string so string can pull it off at infinite acceleration also no problem okay but string is not connected with a and c okay so what is moving a and c is frictional force horizontally okay so frictional force is moving them horizontally and the lesser friction is available on a right the c will have higher friction because normal reaction on c will be more between b and c normal reaction is more than between a and b all of you understand this fine so because of that the the limit in maximum acceleration the limit in maximum acceleration will be on a okay if the if all of these three are moving together okay the first chance as in the the most probability of slipping is for a only okay because a has the least friction because it has the least normal reaction okay so let us try to find out what is the maximum possible acceleration for a fine so this is block a fine so it has mg and there is a normal reaction like this and there is I'll put the maximum friction force which is mu into n right normalization is clearly equal to mg okay so if acceleration is a okay so mu mg is the maximum horizontal force this should be equal to m into a all right so a max which this mass can have is mu into g beyond this it is not possible for the block a to accelerate fine so we will assume that this is the acceleration for all three masses okay so if you assume that then let's say this is the acceleration okay and there are three masses that friction between c and the floor is 0 right we are her smooth floor so we have t should be equal to 3m into maximum acceleration which is mu into g okay and similarly block d if suppose has mass m then I can write mg capital mg minus t is equal to capital M into a which is mu into g fine so t is equal to 3m mu g right so I can write capital mg minus 3 mu mg is equal to mu capital M into g all right so from here you will get capital M to be equal to 3 times mu mg divided by g into mu plus 1 fine so you can see that g will get cancelled off g is gone fine just substitute the values and you'll get the answer for capital M fine any doubts 1 minus mu yeah I am sorry there is 1 minus mu over here okay so I would be more careful if I write an exam I don't make silly errors then fine let's go to the next question now this one anyone time here where is Khushal I don't see him I haven't heard of heard from him due to board exams probably he got a pretty decent percentile guys board exam preparation you can start let's say second week of February all right the starting of the second week fine so two blocks m1 m2 are connected by string negligible mass the coefficient between m1 and the horizontal surface is 0.4 when the system is released m1 m2 starts accelerating what additional mass to be placed over mass m1 so that the mass slide with uniform speed okay if it is sliding with uniform speed the acceleration is 0 okay and if acceleration is 0 all the forces are balanced all right so you can consider m1 and small m to be a single mass if acceleration is not there okay and then you can treat it as if it is a big mass of m plus m1 so downward force will be m plus m1 into g this is your normal reaction okay and you will have your friction force as it is sliding okay even though accession is 0 the velocity is there so clearly it is sliding and hence maximum friction force will be applied okay the limiting friction has to do with the thing that it is sliding or not it is sliding even though accession is 0 fine so mu into n will be friction had it been static then friction would have been between 0 to maximum value this is tension t accession is 0 so we know that from here normal reaction will be equal to m plus m1 into g fine and t should be also equal to mu times n all right and for m2 if you see which is hanging vertically you will have m2g and this is tension okay so we will get t minus m2g is equal to 0 so t is equal to m2g fine so from these two equations I will get mu times n which is mu times m plus m1 into g to be equal to m2 into g right we need to find the value of m so you can cut this and m will be equal to m2 by mu minus m1 fine so just substitute the values from here and get the answer the correct answer for this question is B B for Bombay right let's go to the next one this one all of you okay I think this is something which is straightforward you have a force mass is there displacement is there you need to find the speed of the ship okay so you can get the value of acceleration as force by mass fine and you can use v square equal to u square plus 2 as where s is equal to 3 meters and u is equal to 0 so you can get the value of velocity okay fine so I think this was straightforward we'll move on to the next one this one anyone okay a more got it others guys little quick fine I'll solve it now so this is from your school experiment directly so if this thing happens you know that coefficient of friction is tan of phi okay so I hope you know this okay so this coefficient of friction the distance traveled by the block if thrown up the inclined with an initial speed of u0 is what so we have block thrown up with velocity u0 okay but you can see that downward forces over here okay there will be two forces there will be mg sin theta okay and there will be frictional force as well mu into n okay where normal reaction is what normal direction is mg cos theta fine so total force in this way downward direction the net force is mg sin theta plus mu times mg cos theta mu is tan theta so it will become mg sin theta only again I am saying theta instead of phi but it's okay right you can understand so 2 mg sin theta is the force right so acceleration is force by mass so that is 2g sin theta along the incline is the acceleration okay so what we can use is v square equal to u square plus 2as because it is about velocity and distance right so 2as we will be using okay so v is 0 initial velocity is u square and it is deceleration 2g 2g into 2 sin theta into displacement so displacement is u square by 4g sin theta so that's our option is correct fine we'll go to the next question in case you have any doubts please type in quickly this one okay others okay fine let me solve it given object takes n times as much time to slide down as it as it takes to slide down or to go to go up to go up coefficient friction we need to find out okay so when when the object is going up the acceleration we just found out right acceleration will be equal to see I will be you know skipping a lot of steps going forward because I think that now you guys are ready I am not teaching you a concept so I can skip these steps assuming that you will take care of the small small nuances fine so accession will be actually g sin theta plus mu into g cos theta fine this will be actually opposite to the velocity so this is deceleration okay and when it is coming down the the acceleration will be g sin theta minus of mu g cos of friction will be acting you know opposite of g sin theta when it is coming down and when the object is going up friction will be acting in the same direction as the g sin theta so that is why two different acceleration going up and coming down all right and good thing is a sin 45 is equal to cos 45 so you can simply write it as 1 plus mu times g sin 45 and this will be equal to 1 minus mu times g into sin 45 so these are the two accelerations okay total distance is same let's say L is a distance fine so you can it's about time right so time it takes to travel a distance of L okay so like that you can now going forward you can solve it fine so you have this kind of situation let's say distance is this distance is L okay no no no no I think what I have assumed was a situation where you know it's yeah I thought this is some other question okay I was thinking that there can be a situation where the object is thrown from here it reaches here or object is released from here or just thrown down and it is reaching here the alum the same incline but of course that's not the case over here so there is no friction in the next case so it is just g sin theta which is g sin 45 degrees okay yeah it is going sliding down only both the cases okay so you have to use s equal to ut plus half 80 square only so t will come out to be under root of 2s by acceleration okay so you can get the value of t1 yes Lalitha t1 is 2L divided by 1 plus mu g sin 45 and t2 is under root of 2L divided by g sin 45 okay so what is given as it is sliding down it is sliding down what is given as t1 is two times of t2 fine so acceleration will be g sin theta minus mu g cos theta something weird happened in this question okay this is what will be there okay any doubts sorry to confuse you in this particular question any doubts guys on this one this one the solving too many questions also has repercussion like I have solved so many varieties of question that I tend to think whenever I look at a question I tend to think that okay this is I just read one or two sentences I like okay fine this is that question which I have solved already so that creates confusion and I'm solving with you only so I it is not that I have read these questions so these question have taken from some new book okay so I want to solve these fresh with you so that I can solve I solve like a student with you so that I can think exactly the way you will be thinking while solving this question for the first time 9 B others okay fine I'll solve it now so excitation of the block B we need to find out assuming the surfaces and the pulleys to be smooth okay so you can see that on block A this is the force okay and there is a pulley to have tensions both sides okay this is mass 2m alright so let us say that excitation is a 1 for block A okay so you can write F minus 2t is equal to 2m into a1 fine so this is your first equation and for block B block B will have one tension by the way it is just one string so one string entire single string since it is so everywhere tension be same fine so horizontally only these three tension forces will be there so 3t let's say that excitation is a2 for B so 3t is mass of B is 4m so 4m into a2 fine this is your equation number 2 alright now the problem is that there are three variables P a1 and a2 so you need to get the constraint relation okay now if you look at the string it is like this string goes like this takes a turn fine then goes like this takes a turn fine it is like this this point is fixed let's say this is point 1 okay this is 2 3 4 5 okay this length is l1 is l2 and this is let's say l3 fine so I can say that l1 plus l2 plus l3 is a constant fine so if you differentiate it and then again differentiate it you will get a excitation of length l1 excitation of length l1 plus excitation of length l2 plus excitation of length l3 to be 0 fine because you have differentiated it now excitation of length l1 is what a1 is this excitation is a point 1 excitation which is connected to B so this point 1 excitation is a2 point 2 is accelerating this way with a1 fine so excitation of l1 which is the you know how fast is the excitation of increase in length of l1 it is given as a1 minus a2 fine this is excitation of length l1 with this excitation l1 is increasing plus excitation of length l2 now l2 will follow a similar pattern as l1 okay because you can see that 4 point moves with the pulley and pulley is connected with mass B so 4 excitation is same as 1 similarly 3 excitation is same as 2 so this is a1 minus a2 fine plus excitation of length l3 okay now this point is fixed this point doesn't move but that point moves this way with excitation of a2 so excitation of the length l3 it is actually deceleration because length is decreasing so it is minus of a2 so this is equal to 0 fine so you get a relation between the excitation so you have 2 a1 minus of 3 a2 to be equal to 0 right so this is your third third equation so when you solve these three questions you will get the answer all right let me tell you the answer for this one the answer is d 3f by 17 these are your like the level is same as your mains or it is above the mains which you have written this one you can solve this question using the concept of shm very easily okay viscist and purvik has answered that is correct it's shm q right or is it system as a whole q oscillates without slipping slipping of q will happen on p so basically both the blocks are oscillating together okay but anyways let's solve it now the angular frequency of oscillation will be under root k by total mass total mass is 2 m m plus m fine so this is omega right the excitation in shm is minus of omega square x all right now you have to understand that when the q when this q accelerates okay that is caused by uh you know friction as well as the spring force fine so let's say this is the spring force okay so this is um k times x fine and you have acceleration this way which is omega square x fine and this is mass m all right and all this uh all this is actually provided by uh friction this excitation is provided by friction as well as the spring force so let us say friction is fr okay any doubts till now in fact rather than looking at the upper block let's look at the lower block because in the lower block only friction is applied okay so let's say this is the frictional force fr and there is an acceleration of omega square x fine so if mass is m then the friction should be equal to m into omega square x right we need to find maximum value of friction that will happen when x is the amplitude so m into omega square is k divided by 2 m all right into uh the maximum value of x is a this is a all right so k a by 2 is the maximum value of friction on the lower block and which is same as the friction value on the upper block okay all right now if you want to solve using the upper block then you have to do it like this that um let's say you need to find maximum value of friction okay so let's say friction is this way okay the friction will be maximum when acceleration is in the direction of friction okay you can uh you know write the force equation itself and realize fr minus kx fr minus let's say plus minus kx is m omega square x okay so fr is omega square x plus minus of kx okay so now omega is under root k by 2 m so you will end the value of x let's put it as a so you will get k a by 2 plus minus k into a but then you know here it will never happen that acceleration is opposite of the force fine acceleration will always be because of the force so it will not happen that friction and kx will be in the same direction okay and acceleration is in the opposite direction so acceleration has to be uh in fact acceleration has to be in the direction of this spring force itself because the spring force is this x is from the mean position so uh mean position must be on the left hand side so since it is shm the acceleration has to be towards the mean position omega square x okay so friction uh in fact it will be kx minus friction should be equal to m omega square x okay so friction will be equal to you know it is always good to solve this question by using the lower block but anyway i'm telling you how you can solve using the upper block as well this kx minus m omega square x so when you substitute k a minus m omega square x becomes k a by 2 you will get the value k a by 2 only okay so like this you can solve this particular question okay fine let's go to the next question now this one diagram is not clear actually um no this is not clear let's go to the next one this one so you have to solve this question in the frame of elevator so better to use pseudo force okay resist is saying c okay let me solve this now um the ascending acceleration is g by 4 the coefficient of friction between m or m2 is there the floor is smooth point your forces applied acceleration of the bar and the plank you need to find so simply whenever you draw a diagram in in accelerating frame or reference just add a pseudo force everything else remains the same so one pseudo force m2 into g by 4 will act okay and then m2 g will act and from here normal reaction will be there okay there is no acceleration vertically since you are solving with respect to the accelerating frame of reference all right and there might be um there might be a frictional force like this okay says m2 and because of the friction force there can be acceleration let's say a2 fine so I can write the force balance over here so n uh minus m2 g plus sorry m2 g minus m2 into g by 4 is equal to 0 fine so you get the normal reaction and horizontally fr is equal to m2 into a2 okay and maximum value of friction is mu n right so mu n should be greater than or equal to m2 into a2 this is your first equation and in the second equation you will have free bar diagram for m1 have m1 g by 4 downwards this is equal to m1 g there will be a normal reaction let's say n1 okay and there will be a force like this and the pair of this friction force like that tell me am I missing any force am I missing a force in m1's free bar diagram quickly tell me okay so there is a force normal force the pair of this force okay so at times we miss forces don't do that okay although it might not have affected your answer in this question but what my point is that you have missed right so if there would have been a friction between m1 and the ground the answer would have been wrong all right so let's now balance the forces vertically we have uh n plus m1 g by 4 plus m1 g minus n1 is equal to 0 fine and horizontally we have f minus frictional force to be equal to m1 into a1 so let's say its expression is a1 fine so frictional value the maximum friction value is this all right so you can actually first assume that both the blocks are moving together okay if you assume both the blocks are moving together then whatever force you are providing that force okay should be equal to m1 plus m2 into acceleration assuming both are moving together fine both have the same acceleration if they are moving with same acceleration that acceleration should be equal to f divided by m1 plus m2 okay so the value of f and m1 m2 are given there is no friction there is a friction between m1 and m2 but that is internal force right so this is the acceleration that must be there if they're moving together all right and if the maximum value of friction if mu into n okay is less than m2 into this acceleration then friction cannot support this acceleration for m2 getting the point so this is not possible then all right so a2 will be equal to mu n by m2 then okay and then you can get the value of a1 also you can substitute the value of friction to be mu into n from here you'll get the value of a1 okay so first you need to check whether they are moving together or not if they are moving together acceleration should be equal to this okay and if this acceleration doesn't suit this equation or this in equation then they are not moving together all of you understood this one this is important any doubts this one okay should I solve fine looks like not getting answer okay let me solve it now this is mg and this is t okay let's say acceleration is a for this it is simple right it is just mg minus t let's say this is acceleration a3 okay mg minus t is equal to m into a3 all right for block a you have t and you have like this this is mass m okay so 3t equals m into a1 fine assuming this is acceleration a1 fine and let's say this has acceleration a2 all right so that is for a block a for block b this is mass m m b is a2 okay so for block b we have 4 times t to be equal to m into a2 fine so all of you got these three equations yes or no now let us try to find out the constraint relation okay the string is like this it's like this so I just give you a hint try to get the constraint relation 1 2 3 4 5 6 7 8 take it as 9 all right this is l1 l2 l3 l4 and l5 all right so first step is to write the length l1 l2 l3 l4 plus l5 is constant okay now try doing the same thing which we have done earlier you want me to solve double differentiated fine so acceleration of a1 sorry excision of this what should I write instead of this quickly tell me a l1 will be equal to what no no no no no no there you guys have not understood this concept probably see a l1 when I say I am asking you acceleration with which this length is expanding okay both a1 and a2 they are decreasing the length okay so excision of length l1 will be negative of a1 plus a2 this is a l1 what will be a l2 a l2 is also same okay so minus of a1 plus a2 a l3 that is also same a1 plus a2 fine what about a l4 tell me quickly a l4 is what what is l4 l4 is a length between 7 and 8 where is 7 7 is this and 8 is this when you consider the length between 7 and 8 which point is moving point 7 is moving 8 is static 8 is not moving okay so acceleration of l4 is minus of a2 minus of a2 okay acceleration of l5 is what tell me excision of l5 is what a3 a2 plus a3 this is equal to zero fine so so you will get three times I mean now you can solve okay you have these equations you have equation one equation two equation three just simplify this equation four and then you will be able to solve it fine we will go to the next question now any doubts four week I have full video on constraint equation relation okay so you can go to centrum academy channel or just remind me I will forward you that constraint motion relation video okay I recommend all of you watch it because at times the question becomes tricky because of the constraint relation nothing else do this one I'll forward it right away while you are solving this I'll call you in some time I'll call you in some time in a class I'm forwarding you guys video on constraint motion okay anybody got the answer over here okay should I solve okay I'll solve it now you're getting a others what you're getting fine so we have a smooth ring of mass I'm slide on a fixed horizontal rod string passes over a fixed pulley m by 2 at an instant angle 60 degree initial action of the ring is what is asked okay so just draw the free bar diagram it's a simple question so let us draw the free bar diagram of m by 2 first so we have m by 2 g this is t okay let's say acceleration is a 1 fine so for for m by 2 we can write m by 2 g minus t is equal to m by 2 into a 1 fine so this is your equation number one okay now for the ring ring is getting pulled by tension t but that tension is making angle 60 degree with the horizontal okay so this is tension okay this tension makes 60 degree with the horizontal okay so all right so basically you know the ring is constrained to move horizontally only so acceleration will be horizontal but horizontally the component of tension will be just t cos 60 all right vertically also there will be a component of tension which will be t sign of 60 okay which will be balanced by normal reaction between rod and the ring okay there could have been friction between the ring and the rod okay so if friction is fr uh fr or coefficient friction is mu then you can write t cos 60 minus mu times n you know horizontally but there is no friction so horizontally there is no other force other than t cos 60 so t cos 60 net force is equal to m into a 2 let us say okay a 2 is the rings acceleration forward this is your second equation fine and you can see that there are three variables t a 1 and a 2 all right so one thing you should remember for an in extensible string okay in extensible string along the string acceleration should be same fine so along the string if you take component of the ring along the string it will be a 2 cos 60 fine this acceleration which is along the string should be equal to a 1 which is the acceleration of this point along the string okay so this is your third equation which is your constraint one fine now tell me quickly which option you are getting right fine let us go to the next question this one let's do one thing we'll have a break those who don't want to take a break can attempt this question right now it is 601 we will meet at 615 p.m okay so let us start open that link so looks like some of you have solved both the questions let's take 16 first then we'll go to 15 okay so block is resting on a hundred plate in xy plane coefficient of friction is mu the plate begins to move with velocity in x direction what will be the time after which blocks start sliding on the plate okay so your velocity is this so acceleration is derivative of velocity it will be 2bt fine and the acceleration on the block is provided by the friction which is mu times n okay so if normal direction is upward and this is mg so normal direction will be equal to mg all right so the force forward direction is mu m into g this should be equal to m into a okay so acceleration should be equal to mu into g so this is the max possible acceleration so when this is equal to 2bt that point onwards if acceleration increases okay then the block will not match that acceleration block will keep on having this frustration only but the belt is moving with more acceleration okay belt or the plane is moving with more acceleration so it will slip okay so the time will be equal to mu g divided by 2b okay so mu yeah d okay and let's go to the previous question so massive wooden plate known unknown mass capital M remains an equilibrium in vacuum where n bullets are fired per second the mass of each bullet is this strikes a plate center velocity v coefficient of restitution is given the mass m is what okay so if coefficient of restitution is given and this this block of mass capital M is at rest right so after the collision you know initial velocity with which it hits is v after the collision the bullet is actually coming down with e into v okay that is a coefficient restitution is given it is not an inelastic collision fine so for one bullet uh one bullet let's say mass m the change in momentum is m into v into 1 plus e okay initial momentum is m into v the final momentum is minus m into e into v so changes m into v into 1 plus e okay let us say that number of bullets are capital N so this is the total momentum p okay the force it will generate is rate of change of momentum fine so we have m into v to 1 plus e times dn by dt fine and dn by dt is small n okay so the force is small n 1 plus e times m into v this force must be equal to m into g fine so capital M will become equal to m into v 1 plus e times n divided by g fine so like this you have to solve this particular question so here option c is correct you are getting c okay fine now let's go to the next one 17 so when the block is about to topple the point of application of normal reaction will shift earlier it was applied through the center of the base now it will be applied through the corner fine so let us try to solve this one uh we have this plank when it is about to be toppled so it will topple like this okay so the normal reaction will shift at this edge and you will have mg force like this okay and uh you will have a frictional force like that let's call it as fr fine so this is for this small m for the capital M this is a situation okay and both of them are moving together so both of them should have same acceleration okay any doubts till now anything okay so for this capital M I can write f minus frictional force is equal to m into a okay and for the small m I can write friction to be equal to m into small a fine so acceleration if they are moving together should be equal to f divided by m plus m all right so friction should be equal to m f divided by m plus capital M fine so the capital F value will become a m plus capital M divided by small m into frictional force frictional force the maximum value of friction force I'll put here because I want to maximize capital F so mu times the normal reaction which is mg fine so f will be equal to mu times m plus capital M into g fine so this is one of the limiting force okay but it may happen that before even this force reaches okay this body could topple okay then it can it should not even rotate as well fine so this is one limiting force okay which will be equal to what one third of m plus small m is 12 into g let's say 10 so 40 okay so this is around 40 Newton's fine so answer can be 30 also okay because the object could topple before even reaching 40 so it should not topple as well okay so we need to find the condition for it to not to topple okay so what I'll do is that I will balance the torque about the center of mass when it is about to topple understood so I can say that the torque due to friction is fr into h by 2 fine this should be equal to the torque due to normalization which is n into h by 8 fine so friction should be equal to um n by 4 n by 4 where n is mg so one fourth of m into g so friction cannot exceed this value if friction exceeds this value then the object will topple fine right earlier we had taken friction to be one third of mg but friction cannot be more than this all right so the value of force from this equation from this equation the value of force is uh equal to m into a which is m divided by f divided by m plus m plus frictional force which is one fourth of m into g can anyone solve this and get the value of f is it coming out to be 30 quickly solve you understood right what I'm trying to do over here all of you understood it's not a point object it is 30 so the answer is 30 answer is not 40 no doubts let's go to the next question now this is I think the usual scenario we'll skip it anyone okay now I will solve it there go you have to draw the diagram carefully okay draw it neatly like this and then the bead is over here okay let's say bead is over here it will experience a force mg okay what does it experience it'll experience a normal reaction okay normal reaction will be that way but the good part about normal reaction and mg is that you have the knowledge of the direction in which they will be applied okay so normal reaction will be normal to the surface and there is no friction right um no friction bead is released from y equal to 4a tangential acceleration when it reaches the position y equal to a okay now if you if you see what is asked is tangential acceleration right so you need to at least draw the tangent okay so this is the tangent okay along this direction I need to balance the forces or write the Newton's this thing Newton's law of equation okay so if let's say the slope of the tangent is this is theta slope of the tangent is tan theta okay so this angle will be 97 is theta what mg is making with the tangent right so at along the tangent mg sin theta will be equal to m into a and it out right so a will be equal to g sin of theta now the problem is how to find theta but good thing is that you know that dy by dx is the slope of the tangent which is tan of theta fine so y is equal to x square by 4a from here dy by dx is equal to x by 2a and this is tan of theta all right now when uh y is equal to a when y is equal to a uh x will be equal to 2a fine so tan of theta become equal to 1 so theta becomes 45 degrees all right so acceleration you know is g sin theta so sin of 45 is root 2 so g by root 2 answer is b all of you understood okay let's go to the next question dy by dx was x by 2a at y equal to a the value of x is 2a the equation of curve is x square is equal to 4a y so you can see that so put the value of x at y equal to a which is 2a 2a by 2a is 1 so tan of theta is 1 so theta is 45 yeah have you guys seen the pattern that almost every one of you and I mean whoever you talk to in Bangalore all the students not regarding to just symptom students invariably their chemistry percentile is low it doesn't matter whether you are in symptom or not by the way I'm not saying that okay what I'm saying is in Bangalore students chemistry percentile is less what does it mean that some like that many students are ahead of you and of course since almost everyone in Bangalore has lesser percentile in chemistry so they they are non-Bangalorean students okay and who are they they are studying in places like Kota Jaipur Udaipur Hyderabad fine they they know that it's a chemistry that's a game changer so they they put a lot of extra effort so even though you know in maths and physics they cannot beat a very easily a talented person but they they take it as a challenge that okay fine they'll compensate that with chemistry okay so all of them would have got better percentile in chemistry and not so good percentile in physics and maths those who are you know non-Bangaloreans or those who have been to like in north who are very very focused about the competitive exams so they they just force themselves to study chemistry only so that they get ahead so chemistry is a subject which cannot be ignored you know it it's a like I have spoken to few students and it is sad that almost everyone has this problem of low percentage in chemistry and at the end of the day you have no one else to blame other than yourself okay that should not happen again fine so here um okay this is the situation we need to find the acceleration of A with respect to ground fine so if you look at A's acceleration fine so let's say A is the outcome of B and C okay A is the outcome of B and C now then what will happen is that it has to move along the wedge with acceleration C okay and relative to the wedge A is the relative acceleration expression of small A um A is the expression of A with respect to C okay fine so A is the relative acceleration this is the relative acceleration with respect to wedge okay so the acceleration along the wedge okay if if you just add them add the acceleration of the block A A with the acceleration of the wedge you'll get the total acceleration because small A is relative to the wedge so you add the acceleration the wedge onto it you'll get the net acceleration understood so let us try to do that so let's take this as coordinate this is x coordinate and this is my y coordinate fine so summation of acceleration along x direction is A minus C cos of theta and acceleration along y direction is minus of C sin theta fine so the total acceleration of A is root over this square plus that square fine so this will be equal to A square plus C square minus 2 A C cos of theta so this is the acceleration of A under root of this this is the acceleration of A with respect to ground and none of the option much as that is it value of A is the acceleration of block A with respect to C that this should be correct it is independent of B it has to be independent of B it should only depend on A and C see B B has nothing to do with acceleration of B see acceleration of block B cannot see B cannot push A and even if like suppose string is taught the outcome is A okay outcome is A what is acceleration acceleration of A is relative acceleration of A with respect to wedge so this is actually a vector A minus vector C this is a relative acceleration actual acceleration of A minus acceleration of wedge is the relative acceleration of A with respect to C so the actual acceleration of A is equal to the relative acceleration plus acceleration of the wedge right so B is coming nowhere in the equation so it's a vector addition and this has to be correct so or there is some misprint in the question okay anyways this is I think wrong question we'll move ahead this one whose expression will be zero is there a block whose expression will be zero tell me that first tell me that is there a block whose expression will be zero there is there is a block expression of C will be zero see the string was anywhere not applying any force on C who is applying force on C the spring is applying force on C and there's a gravity force fine so the force on C okay was equal to zero and it is the spring force which was balancing mg fine so cutting of the string will not suddenly decrease or increase the spring force so for a split of second or some few seconds the force on C will be balanced only it will be zero only because spring will take time to again you know calibrate itself it will remain extended the same amount immediately after cutting the string so the force on C will be balanced cutting of string will affect the net force on A and B immediately it will not affect net force on C immediately fine so C should be zero so it cannot be this okay block C has to be zero so there is a high chance that answer is between B and D okay now can you solve it quickly it's just a free by diagram so from C you can get kx is equal to mg by the way this spring force is something else right so this is let's say kx1 it may not be same amount of extension for block B we have this is mg so for block B we have mg plus kx is equal to t okay now t is suddenly gone for B getting it and kx was mg so 2 mg force is downwards this has to be equal to mass into acceleration so acceleration of B will become 2g this is acceleration of B fine so this cannot be correct this cannot be correct so option B is correct all of you understood acceleration of A let's try to see whether we can get the value of that because it depends on extension in the spring anyways let's try so if you look at this system all three masses put together fine then you have kx1 balancing 3m into g okay so for A we have mg plus t to be equal to kx1 okay now t suddenly gone so net force will be equal to mg minus kx1 this has to be equal to m into A so mg kx1 minus mg is equal to m into A kx1 is more than that fine so kx1 1 minus mg will be equal to m into A fine so kx1 is 3 mg so 3 mg minus mg 2 mg is equal to m into A so acceleration of A is 2g but A will move up with this acceleration and B will come down with 2g so both will have same acceleration one goes up the other comes down understood okay so now just I think one minute is left so we can't solve a question but then what I have is 10 questions are there still left so if you're serious about j mains or j advance because these are above j mains level question like one or two are mains level question the others are above mains level so before going to sleep today try solving all the remaining questions and take a snap of your solution and send me across okay fine so that's it for today we will meet again according to whatever schedule we have circulated bye for now