 Okay, good morning. Let us begin with what we started yesterday. So that was the principle of impulse and momentum and before that we discussed work energy theorem and saw that a body typically only in dynamics. For example there was a query about thermodynamics and how does the energy in thermodynamics related with the energy in dynamics and I briefly discussed okay what that means. But in our dynamics course what we discussed is that that there are two types of energies one is the potential energy one is the kinetic energy for conservative fields there is a potential energy but whenever the fields are not conservative for example whenever there is friction then what does the work energy is that the initial kinetic energy plus the work done to move the particle from path 1 to point 2 will be equal to the final kinetic energy of the system. So this was in brief the work energy theorem we also saw that the work energy theorem can be rephrased in the language of conservation of energy if the forces are conservative that T1 plus V1 is equal to T2 plus V2 where V1 can come from gravity when can come from spring energy and if additional electric field example is put in we can also put in the electric field and write down the potential energy but for the time being let us not worry about that our VE as this course is concerned is only based on a potential energy from gravity the potential energy from so we saw that that when a ball for example a cricket ball or a tennis ball okay is hit with either a bat or a racket then in that case for a very brief period a huge amount of force acts on the ball and as a result the momentum changes then we defined what is the concept of impulse okay so the principle of impulse momentum what did it say is that that from Newton's second law the force acting is nothing but rate of change of momentum and so if you do this simple integration then what do we see that I've bar dT is equal to MV2 minus MV1 which is the momentum before the impulse was applied and the momentum after the impulse is over we saw yesterday that impulse typically just means the integral of force over a particular interval impact is a very special case of a impulse in which the duration is very short and within that short period very high forces act and as a result of which this integral even when T2 minus T1 which is delta T can be small but integral f dT can be very large and so there can be an appreciable change in the momentum of the particle note momentum when I say moment change in momentum of the particle I mean in the vectorial sense okay that the vector momentum can change appreciably we saw yesterday that we can also replace this moment this impact with the help of what is called as a mean impact force which is written as f bar and at f bar can be can also be written as integral f dT divided by delta T what is delta T is just T2 minus T1 so this can be said as this is a vector so vector never forget that f average okay so this can be said to be the mean force that acts on it now one thing which we noted that in the cases in the case of impact okay so there is one thing okay so there is a terminology that I have used is impact and impulse okay there was a question also asked on it okay so at least as far as this lecture is concerned the way we are using the word impact is that that the duration of the impact is very small but as we had seen in this yesterday's problem that impulse need not necessarily mean that delta T may be small it only means that even if a force is acting over a constant duration okay for example even if a force acts like this okay T1 T2 and the force may be a variable or it may be a constant like this the impulse is given simply by f dT from T1 to T2 and this is the definition of impulse and it is clearly a vector now we solved this problem yesterday briefly to find out what is the stoppage distance for the car now let us discuss another simple problem that a 120 gram baseball is pitched with a velocity of 24 meters per second okay baseball why example from beer and Johnston 10 American book baseball not cricket so 120 gram baseball is pitched with a velocity of 24 meter per second so what do we know we know the initial momentum of this ball the vectorial direction it comes here after the ball is hit with the bat it has a velocity of 36 meter per second in a direction as shown so you can see that this is 24 in the horizontal direction in words this is 36 at an angle of 40 degrees with respect to horizontal we are also told that if the bat and ball are in contact for 0.015 seconds this determine the average impulse you force exert exerted by the ball during the impact now the speed is given here okay so very simple problem so this initial velocity is given to us so initial momentum is known final velocity is known to us because the mass is also known the final momentum is also known to us now what do we want to do we just simply use the moment the impulse momentum theorem what do we know that the initial momentum plus this impulse okay to take it from 1 to 2 okay that that the impulse starts okay when the ball is hit with velocity v1 and the impulse is over because the contact is lost when the particle again goes along this direction okay with the velocity v2 okay now the x component we can we can rewrite this in terms of components we can rewrite the x equation we can say that our x axis is in the horizontal direction okay wide axis in the vertical direction so in terms of these components what do we have the momentum will be minus mv1 will be the initial momentum okay in the x axis x axis it is only mv1 there is no y component for the initial velocity times average force fx what we had discussed earlier the average impulse force fx into delta t which was a duration of impulse will be the x component of the final momentum okay this is simply the impulse momentum theorem what is that it is mv2 cos 4t cos of this angle so substitute all the values and we will see that the average impulse force was equal to 412.6 Newtons okay now writing down okay the component equation okay in the y direction what do we see that the initial momentum in the y direction is 0 f y is the mean impact force in the y direction into delta t will be equal to the final moment okay the final momentum in the y direction so 0 plus f5 d delta t is equal to mv2 sin 4t solve this and we find out what is the average impact force in the y direction and so ultimately the vectorial impact force is given by this we can just do square root of 413 square plus a 180 5.1 square and this will be the total average impact force that acts on the particle okay so very crude estimate of the impact force because typically the impact forces reaches very high at the peak of the contact okay and then drops down now this is a nice problem so what we have in this problem is as follows a 10 kg package drops from a shoot into a 45 kg cart with a velocity of 3 meter per second so this is a slope this is a cart okay which is used to carry this now what is done is that this particle slides over this and before getting rid of this contact okay before get before losing touch with the shoot what it does is that that it slides off clearly at an angle of 30 degrees the same as the angle of the shoot just when it is leaving okay and then what happens is that when that happens it has a velocity of 3 meters per second and it hits into this 24 kg cart with a velocity of 3 meters per second okay that same velocity at this angle it hits the cart now knowing that the cart is initially at rest and can roll freely okay determine the final velocity of the cart and impulse exerted by the car on the package and third the fraction of the initial energy lost on the impact now the deal is this so we ask ourselves that what are the various forces acting on this system so for example let us take the example of this cart for this cart there are clearly two forces acting one is the weight okay W and other is a normal reaction from the surface because the cart can roll freely okay it is given in the problem we neglect the friction that can be exerted by this surface on the wheels okay because the cart is free to roll freely now the second portion is that we are asked to if you look at this particle okay or this package which we are this package which is traveling at a 3 meter at 3 meters per second then what are the forces acting on it the force that will act on it is 4 due to gravity okay and and that and that's it before the impact happens okay that will be the force that acts on it but now we make a very small assumption okay that by the time the impact happens and this particle settles down so we know from our example what happens is that this will hit okay there will be a lot of vibration set in this car and this and this package will settle down and both of them will start moving with a common velocity okay in this direction so ultimately what happens is that even though gravity force of gravity acts on it okay before the impact also after the impact also okay during the impact because gravitational pull acts in the vertical direction okay we don't write that in the equation of momentum balance that we use for in the horizontal direction so with this metaphor preamble let us see how do we apply the principle of impulse and momentum to the package card now what happens is this that when this hits the card what happens that what are the forces acting on this complete system that this is the initial velocity when it hits okay this will hit it and come to rest in this position now during the duration when the hitting is happening what will happen there will be some impact okay there will be some impulse that will act on it okay in the horizontal direction and in the vertical direction and similarly on the card okay from the ground okay to keep this thing in balance because this cannot move in the vertical direction so there cannot be any change of momentum for this card as a result there can be an impulse that can act from the ground on this but now what we realize is this that as far as the x direction is concerned okay as far as the x direction is concerned what do we see okay we see that the total momentum of this system will be what mp v1 bar okay this is the packet the mass of the packet into the velocity of the packet okay this is the initial momentum of the entire system plus the impulse okay that acts externally on the system because note that between this point and this point there is an impulse but this impulse acting on the package and the impulse acting on the card they cancel out so if we take this entire system as one package okay as one complete system that the package plus the card as one complete system then the the impulses that act here they will cancel out but what do we know is that at mp v1 plus the external impulse acting which is this impulse will be equal to that when this settles down this will be the common velocity for both of them and so mp plus m car okay into v2 will be the final momentum for this complete say a cart and package now what we realize is that that it is very tricky to write down impulse momentum in the vertical direction because this is also unknown but what we realize is that that externally there are no external impulses no external forces acting on this system in the horizontal direction and because of that what we do is that we just break this component so the impulse component in the horizontal direction or in the x direction becomes 0 so mp v1 x what is v1 x v1 x will be just v1 cos 30 will be equal to mass of the package plus mass of the card into v2 just substitute all these components and what we will see is that the combined velocity of this entire assembly which is the package plus the card will be equal to 0.742 now we ask ourselves okay if we apply the same principle just to the package just to determine what is the impulse that is exerted on it from the change of momentum then what we see is that that this is the package this is the vertical component of the impulse force what is this vertical component of the impulse force that between this cart and the package there is an impulse force that act so it has a vertical component it has a horizontal component initial momentum final momentum so what this impulse do that this impulse essentially changes this momentum from the initial to the final you may ask me that we have neglected the weight from the gravity. Because why? Because weight is also acting on this system for a period of delta t. But note as we had noticed earlier, we will see that this force, this force that will come out, this impulse component will be far more than the component of the weight of the packet. So, that is why we neglected, but in principle we should also incorporate the weight of the package. We have said here, we have assumed here that the weight component acts for a very short time and overall weight is much smaller than the impulse components, which is this f x and f y. So, that is why we are neglecting it. Now, what is the x component? The momentum in the horizontal direction plus the impulse in the positive direction acting on it is what? f x delta t will be the final momentum in the x direction just for this particle. But now, by solving this composite system, we have found out what is the velocity of this composite system, which is also the velocity for the package and so, we substitute all the values inside. And what do we see that f x delta t will be 18.56 Newton per second. And similarly, we solve this and what we will see in the y direction when we solve this problem, what do we know that m p v 1 into sin 30 minus y, because that is the y component acting in the downward direction plus f y delta t is the overall impulse. So, the mean impulse in the y direction into delta t is the overall impulse acting on this package in the y direction is equal to 0. So, substitute both these values and we will see that the x component of impulse is acting as expected in the minus x direction and y component acts in the vertical direction. And what we have seen is that this is the total impulse that acts on the particle or that acts on this package. Then the last question is that we want to find out what is the fraction of the energy loss. We know the initial kinetic energy. The initial kinetic energy is given by half m p, p is the velocity of the package v 1 square, substituted it is 45 joules. But when the package hits and complete assembly now moves together the cart and the package they move together with the velocity v 2, which we found earlier from the momentum balance of this composite system in the x direction. Because for this composite system, there was no external impulse acting in the horizontal direction. So, we found out that v 2 and now what do we do? At t 1 or the initial kinetic energy simply this initial velocity is known. Final velocity for the entire assembly is known. So, that kinetic energy is half mass plus of the package plus the mass of the cart into v 2 square and you will see that the final energy is much lower than the initial energy. So, what has happened? So, simply for example, what has happened is that when this package hits the cart you will see that for example, lot of vibrations are set up in the package. Some part of the energy goes into setting up those vibrations. There is some plastic deformation both for the cart and there would also be some local deformation for the package that will also end up in dissipating the energy. Also some part of the energy it gets dissipated into sound and all those dissipation put in together even though the momentum is conserved in the horizontal x direction for the complete system the total energy for the system is clearly not conserved. And also that this part of the momentum is completely destroyed why because this hits it and gets stopped here and when it hits here a majority as we just discussed of the energy gets dissipated in the vibrations that are set up in the cart. Now, the next concept which I want to briefly discuss is the concept of impact of two particles. So, this is something where for example the case we discussed earlier was also case of an impact but here we will specifically discuss that when two particles collide with each other then what happens during that impact. So, let us define a few terminology. So, the way we define impact of these two particles is collision between two bodies which occur during a small time interval and during which the bodies exert large forces on each other. Now, the main terminology is that what is the line of impact? The line of impact is when these two surfaces hit with respect to each other in the previous problem we did not really clearly define what is the line of impact but here we define it why because the surfaces have a clear cut geometry and their normals are well defined you can define tangents and normal on the surface. So, we can clearly define what is called as the line of impact. What is the line of impact is the common normal to the two surfaces of contact during the impact. So, two spheres for example interacting with each other this is the common normal and this is the line of impact. Now, central impact what does that mean central impact means impact for which mass centers of the two bodies lie along the line of the impact that is another thing that we are going to deal with that the two centers essentially lie along the line of the impact. Ultimately direct impact is the impact for which the velocities of the two bodies are directed along the line of impact. So, this is a direct impact that the velocities are also directed along this line of impact or the or also the line connecting the two centers of the small particles whereas in an oblique impact what can happen is that that one particle has a velocity in this direction other has a velocity in this direction even though when they impact with each other there is a well defined normal. So, there is a well defined line of impact and this is also a center impact why because the centers of these two spheres lie on the straight on the line which is also the line of impact but the velocities are not directed along the line of the impact. So, these are the various impact parameters. So, first briefly let us look at what is direct central impact. So, what is direct central impact that the line of the impact and the velocities are in the straight line they are all directed in the same along the same line. So, what we say is that upon impact the body undergo a period of deformation during which the bodies are contact with each other and move with a common velocity. So, these two spheres they come together hit they keep moving moving till the bodies deform to their maximum it is a very small deformation that happen in those contacting spheres and then they start separating from each other and then completely lose contact then they start moving separately again. So, this is the overall impact procedure. So, this is called as a period of restitution. So, one is a period of deformation in which the bodies come they keep going going going deformation happen and the period of restitution is what that they start losing their contact and then they completely lose contact. So, the time during which the deformation the contact is being formed is called as a period of deformation and the time between during which from the maximum deformation to complete loss of contact is called as a period of restitution. So, these are bunch of terminologies if you wish to you can have a look at BJ 10 for example and just go through a terminology. There is bunch of terminology why I am using this because I need to use this when I go to what is called as the coefficient of restitution. Now, clearly in this case if you look at these two particles there will be some impulse acting here some impulse acting on this phase and ultimately what we realize is that the total momentum has to be conserved because there is no external force acting on this other than the impulsive force and the impulsive force okay it cancels off for this particle and this particle why because impulsive force exerted by A on B and by B on A are exactly equal and opposite to each other. So, then the conservation of momentum then says that M A V A plus M B V B is equal to M A V B prime plus M B V B prime okay if velocities along the right direction this is positive is velocities in the left direction this is negative. Now note one thing if I want if I know initial V A I know initial V B I want to find out what is final V A and what is okay this should be M just note there is one change this should be M A V A prime okay there is a mistake here. So, the second relation is required why because we have two unknowns we have velocity as the unknown okay initial velocity is given but the velocity final velocity of particle A and final velocity of particle B are both unknowns and we need an additional relation and that additional relation is provided what is called as the coefficient of restitution and the way we define the coefficient of restitution is as follows. So, look at this sphere okay let us zoom only on sphere A and not on sphere B okay we will zoom on that sphere later. So, when you zoom on sphere A sphere A keeps coming okay with velocity with velocity V A keeps coming coming it hits with the second sphere now what happens when sphere B hits sphere A it starts the both of them start deforming because they want to go into each other okay but because okay because of the the overall rigidity or overall elasticity of the entire material they cannot interpenetrate each other. So, they keep on deforming each other. So, first phase is that that if we zoom only on the sphere okay let us not concentrate on the other sphere then there is a impulse that keeps acting on it keeps acting keep acting till it deforms the second sphere to its maximum and the period between which the contact has just happened and when the maximum deformation has happened for either of the spheres that is called as the okay period of deformation. So, what we realize in this period is that just note the signs. So, M A V A minus this impulse P D T during what duration during the period of deformation okay will be equal to M A into U what is this M A into U that when it keeps hitting the particle the particle or the ball is still moving. So, till that period of deformation happened what is the that initial is when before the contact happens the initial momentum is M A V A till the deformation happens the intermediate momentum it has is some M A times U okay. So, this is the period of deformation. Now what happens is that the maximum deformation of one sphere has happened and now the sphere okay start to go back to the original shape and that particular period as we just discussed is called as a period of restitution in that what happens the shape which was kind of deformed slowly starts going back to some final shape. Now if the impact is perfectly elastic the sphere will completely recover its original shape but many times for example there is a permanent deformation that happens in the sphere and so after the contact is lost you will see that there is a rest there is a impulse which acts in this restitution period impact slowly slowly slowly the impact starts decreasing but the overall impact during that period of restitution is given by integral R okay R for restitution dt but the final momentum of that is what that this momentum in the intermediate period plus this impulse okay which is the in the period of restitution which will be the final momentum which this body A will have after the contact is lost so just during the contact deformation deformation that is the period of deformation starts slowly the recovery which is the period of restitution lose contact and whatever velocity you get now is the final momentum and the coefficient of restitution is defined in this way okay this is a highly empirical definition so you may ask me what is the basis behind it there is a very simple basis behind it is that that if this that if the if the collision is perfectly elastic okay that's a simple logic behind it the collision is perfectly elastic then what happens then the then the whatever deformation you get during the period of deformation okay you recover back deformation fully okay everything happens reversibly like the spring the spring skids getting compressed and when the particle starts to come out the spring again come back to their original position so that is a perfectly elastic response deformation and restitution are exactly equivalent to each other okay now in that case okay what we also expect is that for example in the case of spring the overall forces that keep acting during deformation of the spring or also the overall forces okay and the overall time that acts when the particle for example loses contact with the spring so taking that example is in mind we say that this rdt is the restituting impulse PDT is the deforming impulse and these two impulse for example the ratio of them we call as a coefficient of restitution now if this ratio is equal to 1 then what do we realize is that that if this ratio is 1 then there is a perfect the restitution deformation are exactly equivalent to each other and so this will become 1 and if that there is no restitution at all you hit it the deformation stays then this impulse becomes 0 only deformation no elastic rebound at all this is perfectly plastic deformation in that case this coefficient of restitution becomes 0 so it's a highly empirical concept okay it is not exact but in engineering mechanics and for many many engineering problems okay very simple engineering problems okay this to the first order or as a first cut is a very decent thing to do for this collision impact problems now we just take the ratio of this we can immediately see that this will become u minus va prime va minus u the same thing can be done for now particle B and what you will realize is that that coefficient of restitution will again be equal to vb prime minus u and u minus vb you may ask that why I have used the same u think about it that when the particles keep deforming ultimately when the maximum deformation has happened both of them okay will keep moving okay in the same direction with the same momentum why because if they don't then it means that there is still some relative velocity between them and the deformation can keep happening so this u will be the same okay at the end of the period of deformation for both particle A and particle B you just divide one with the other and what will we get is that vb prime minus va prime okay both directions of vb are along the positive x direction so vb prime minus va prime will be equal to e times va minus vb that is the additional equation we have in in addition to what is the initial equation for momentum balance or momentum conservation and you will see that this we have two equations to unknowns and we can find out what are the final velocities of both A and B if initial velocities are given now for a perfectly elastic impact is equal to 1 and you can also show that when e is equal to 1 then the kinetic energy is also conserved there is absolutely no dissipation and in a perfectly plastic impact e is equal to 0 and you will see that there is significant of amount of dissipation means the initial kinetic energy of the system will not be the same it will be lower than the finite kinetic energy of the system now we will not go into the details of that but the same thing can be done for oblique impact problems oblique impact means the particles don't impact along the line of impact but the impact like this but it is still a central impact problem what is central impact again that there is a well-defined normal and impulse forces act only along the line of impact so in that case what we can do is that we can write the impulse momentum equation okay we can write the impulse momentum equation in the x and y direction first we realize is that because there is no impulse in this tangential direction okay the the velocities of these particles in the tangential direction will not change at all okay that v a tangential will be equal to v prime a tangential okay even after the impact why clearly because there is in this assumption of central impact the impulse acts only along the line line of impact there is no impulse component in the direction perpendicular to the line of impact or the t direction and so if you resolve the components of the momentum in that direction you will see that the impulse momentum theorem will tell you that there the velocity the momentum in the vertical or the tangential direction remains the same and as a result this momentum remain the same but we can apply the principle of conservation of momentum along the line of impact and again we can also apply this principle of restitution okay for the component of velocity or the component of momentum along the line of impact okay and then solve these problems why because these two components are found out the only components that are unknown here is that at the final components along this n direction of a and b but we have one equation which is the conservation of momentum and other equation which is this conservation of which is this the coefficient of restitution equation and two equation two unknowns we can solve for the the velocity after the impact both for particle a and for particle b okay so we won't go into the details of the exact same thing okay if you if this particle is hitting an oblique block you can have a look in Beren Johnston okay about this now typically in problems involving energy and momentum okay these are some simple tips that we can follow that for example if you ask a question that we release this mass a from the top okay of this pin point okay this this bob is connected to the hip to the hinge by a string we take it upstairs release it and then you want to know for example what is the final condition of this entire system then what do we do to find out what is the velocity the velocity is zero here to find out what is the velocity here we apply conservation of energy we come here then at this point okay we apply conservation of momentum we apply the coefficient of restitution and from that we can find out that what are the final tangential velocities here and here and after we know that then we can again use conservation of energy to find out what is the final part of this two balls or of these two two particles now we will solve one simple problem okay before taking questions so what is given to us is that that a ball is to thrown against a friction less vertical wall now why do I say friction less we say friction less because okay let me draw a quick figure here so this is one thing which I want to emphasize in this impact momentum problems what we had said okay I am drawing this very big for exaggeration exagerative purposes this is the line of impact suppose this is the initial velocity of particle a this is the initial velocity of particle b mass is ma mass is mb vb now this is the line of impact this is the normal these are the centers okay now what we say is that when the impact happens the assumption is that that this is the impulse that acts on this particle a and by Newton's third law and equal and opposite impulse act on particle b and when we say we are neglecting friction in principle there can also be an impulse in the tangential direction and equal and opposite impulse acting on this in the tangential direction so in these kind of problems central impact problems this component we say that there is no friction between these two okay that's an idealization which is strictly not true but as a first cut if that if that if the dimensions of the ball are small okay so whatever the angular momentum components that will be involved are very small compared to the linear momentum components then we can neglect that so what we say is that this component is neglected and that's what we are doing here we say that there is no friction between them why because these components are essentially equivalent to the friction forces these are the normal forces the tangential forces are friction forces and we are neglecting them and what we say here is that this is the initial velocity this ball hits this wall no friction so the only impulse acts along this direction this is the final momentum and we are asked to find out that knowing that the coefficient of resolution is 0.9 determine the magnitude and the velocity of the ball as it rebounds from the wall now what do we do we resolve the component of velocity of the ball in the in the tangential direction and in the normal direction now note one thing that what do we know okay we know that vn will simply be equal to v cos 30 that is my initial velocity and vt or the tangential velocity will be just v sin 30 now straight away we can apply the principle of coefficient of restitution we say that this wall okay and it's perfectly fine to say that this wall has infinite mass so it will now have no velocity it has no velocity before the impact and after the impact also it will have no velocity so we directly apply the coefficient of restitution okay with the 0 wall and what do we see that 0 minus vn prime will be equal to e vn minus 0 just substitute all the values we will find out that the the final velocity in the normal direction will just be equal to this now how do we find out what is the velocity in the tangential direction we realize that because there is no friction between the ball and the wall there is no there is no impulse force acting on this in this direction and as a result or in the tangential direction the velocity or the momentum of the particle is conserved in this direction and so the velocity in the tangential direction also remains the same so with this simple problem okay we also saw how to incorporate coefficient of restitution there will be one or two problems which will deal with in the tutorial okay so there are some questions let us take those questions first question is by center 1 2 2 7 in sample problem 13.12 why impulse 1 2 is 0 okay so note one thing that this impulse 1 2 is not 0 okay so let me discuss this briefly so what is happening here is as follows if we look at the cart okay and this package during the impact process separately so first let's zoom on to the package what will we see we will see that this package has an impulse acting on it integral fx okay it has a component in the x direction it can also have a component in the y direction which is integral f y dt its initial before the impact started its initial momentum was mv px after the impact was done the final momentum was okay mp v2 all in the x direction okay now if you write down this entire thing even on the card what we will see is that that the cart will also have an impulse like this the equal and opposite impulse okay integral f y dt y from Newton's third law will act on the card and this another impulse equal and opposite to this will also act on the card now why should there be a impulse coming from the ground because if there is no impulse in the y direction coming from the ground what we will see is that the cart will have a momentum change in the vertical direction so there should be an equal and opposite moment impulse from the ground which can balance the card because the cart is free to is free to roll in the horizontal direction what we say is that that this impulse in the horizontal direction does not exist so for particle this is the impulse diagram for the cart which is this is the impulse diagram now if you put them together okay you put this and this together what will happen is that this internal impulse is plus minus cancel each other this plus minus cancel each other and what we are left with is the figure that is shown here so the internal impulses between the package and the card they cancel each other so we don't take that into account but note that this external impulse in the y direction also remain but what we have realized here is that that is external impulse as a component only in the y direction okay this has a component only in the y direction okay and not in the x direction so if we break down this vector equation in x and y components then along the x direction there is no external impulse so that's why we have taken that to be equal to 0 it is not that there is no impulse but along the x direction for the complete system there is no impulse there is a second question from center 1 0 6 0 when potential energy okay is equal to kinetic energy so this question okay let me answer it in one way so I can answer a very simple question like this we use the example of our below at pendulum we take this pendulum start here okay start here and what do we see that with respect to this reference line okay let us take that this length is l this length is l so from this datum the potential energy v 1 is equal to what v 1 will be equal to simply m g l now kinetic energy initially is equal to 0 and when the ball come here okay downwards what happens with respect to this datum v 2 is equal to 0 and kinetic energy t 2 will be equal to half m v square and in this case if we if you apply principle of conservation of energy assuming that there is no dissipation from any sources and this is equal to v 1 plus t 1 is equal to v 2 plus t 2 and in this case the entire component of potential energy will get converted to the kinetic energy so I hope this is okay then there are two different questions one is by center 1 1 5 7 they have asked please discuss the concept of force acting in the elevator or lift okay so I presume okay that this question is about the brief discussion we had yesterday about the concept question one so let us briefly discuss this let us say that we have an elevator which can either accelerate up okay all down now let us take a person who is standing inside this lift okay it may be us or anyone else now this lift is going up with acceleration a because this person has contact with the ground what happens his acceleration also has to be a now if we draw the free body diagram of this person okay then what are the forces acting on it the forces possibly it can act on the person okay so one force clearly is the weight mg second force will be a force okay normal reaction coming from the contact between the person and the elevator there can also be a sideways friction force f now we apply Newton's law on this this this entire system okay what do we see what we see is that now in principle you may say that there may also be a couple and so on okay so let us not worry about that for the time being the person is stable so its center of mass is directly below the reaction okay so let us not worry about the moment that acts from the force now writing equilibrium of equation in the y direction okay say this is x and this is y so writing equation of equilibrium in the y direction what do we see okay the sum of forces in the y direction is equal to n minus mg but what will this be equal to if the lift is moving up this will be equal to ma and then what do we see we see that the normal reaction will be more than the weight of the person now if the acceleration is negative or the acceleration is downwards then this will be equal to mg minus ma okay and so apparently the person will feel that the weight has little bit decreased as far as the contact with the ground is concerned okay so I hope this address is the question about the forces that act on the person doing the elevator