 Let us recapitulate what we did in our last lecture. We had an idea of looking at the conservation laws which are obeyed in all the inertial frames as per the postulate of special theory of relativity. So, we are looking for a new definition of momentum which could satisfy that particular criterion namely that if momentum is conserved in a given frame it should be conserved in all the frames. When we are looking at that particular aspect, we realized that the conservation of momentum critically depends on the fourth component of what we originally called as momentum 4 vector. Then we try to reinterpret that particular fourth component and invoked that that is to be called as the energy of the particle which was a totally revolutionary and totally different concept of energy which does not have a classical analog. So, this is what we discussed and we said that this particular four vector is now re-termed as energy, momentum energy 4 vector. So, this is what is my recapitulation. We discussed the energy momentum 4 vector. Then in this process, we gave a new revolutionary definition of energy which did not have any classical analog. And of course, as a consequence we arrived at a new definition of momentum. So, we had a new definition of momentum, we had a new definition of energy. Of course, the definition of momentum in the classical limit meaning that the particle speed is much smaller than the speed of light will reduce to the classical definition of momentum. But the same will not be true for the energy definition because energy was a totally different concept altogether. We even said that if a particle is at rest, that particular particle also has some energy which we call as rest mass energy. And the kinetic energy we defined as only the additional amount of energy gained by the particle when the particle starts moving. So, this was totally a revolutionary concept and this has had such a great amount of impact even on a common public that people started almost naming Einstein by E is equal to m c square. E is equal to m c square becomes synonymous with Einstein. People even who may not be understanding physics, they are still aware of E is equal to the glamorous relation of E is equal to m c square. Then we gave the relation between the energy and the momentum of a single particle because once we have to solve the problems, we have to start looking at the relationship between the energy and the momentum and some of the problems we are going to describe today. Then we also discuss a totally new concept which was not existent in the classical mechanics because in classical mechanics a particular particle has to have some finite mass. Here we consider we imagine a particular particle which could have a zero rest mass and that particular particle could also have energy and momentum. The only condition is that in that particular case, the particle must move with the speed of light. So, this was again a new concept which came out from the special theory of relativity of a particle with zero rest mass. Now, today what we will be doing? We will try to work out some of the problems about the collision and specifically we will revisit the earlier problem of the collision which we had started with as a consequence of which we had decided that the conservation of momentum will not be valid in all the frames of reference unless we redefine the momentum. That is the way we had started if you remember how we lead to how we led to new definition of momentum. So, let us look at this particular old example which is here just to remind you that what we had discussed in our earlier case that there was one particular particle M which was moving to the right hand side which we can call as a plus x direction with a speed of 0.6 c. There is another particle which is also of mass M which is moving to the left hand side with the speed of 0.6 c. These two particles collide with each other and eventually get stuck to each other. So, after collision you have a combination of the two masses. If you remember the example which we had given earlier and that time we did not have a new concept of rest mass, we did not have a concept of new energy. At that time we had just talked about one mass because mass was sort of in that sense essentially sort of universal. So, we had said in that particular case that of course, in this particular example when we look at this particular frame of reference then we do find that momentum is conserved because here the initial momentum is 0 and after the collision the final momentum is also 0. Therefore, momentum was conserved. Then we transform my frame of reference and went to the frame of reference of this particular particle. It means we assume another frame of reference in which the particle is at rest or the frame of reference is moving with a speed of 0.6 c along the plus x direction. Then we applied the velocity transformation that we had derived on the basis of Lorentz transformation and try to find out the speeds of these particles in that particular particle's frame of reference which we called as S frame of reference. Then we discussed and we found out that the momentum was not conserved in S. That is where we had arrived at this particular conclusion that momentum need to be redefined. Now, let us look at this particular problem again in the light of the new information that we have got regarding momentum and energy. So, what I would sort of rephrase this particular problem now that we will assume that this particular particle the mass m that I am talking is m naught which is the rest mass of the particle. So, there is one particular particle which has a rest mass m naught which is moving to the right with a speed of 0.6 c. There is another particular particle which has a rest mass m which is moving to the left hand side with a speed of 0.6 c. Let us not talk about this particular thing. Remember this particular picture I have taken just from that old example. In fact, this picture would probably require modification once we have described this particular problem. This is the way we had described it earlier. Now, let us look into the new definitions and try to work out the same thing and try to really convince our self that if momentum is conserved in one frame it has to be conserved in the other frame also which it did not in the earlier case when we did not have the knowledge of new definitions of energy and new definition of momentum. So, initially what we will do? We will look at this particular frame only for where this particular collision has been described. So, I assume that there is a particle with mass rest mass m naught which is moving to the right with a speed of 0.6 c and another particle with mass rest mass m naught moving to the left with a speed of 0.6 c and let us try to write the new values of the momentum that I will obtain from this particular case. Now, if you remember the new definition of momentum was P is equal to gamma u m naught u. So, basically it differs from the classical definition because there is also a gamma u factor and this gamma u as we know is related to the particle velocity which is given by 1 minus u square by c square. So, here we have a collision of two particles and each is of rest mass m naught. So, what I will be doing? First let me try to calculate the momentum of the first particle and then I will calculate the momentum of the second particle as given in this particular frame of reference. So, this is what I have written here. So, this is the momentum of the first particle because it is first particle. So, I have written 1 and this is the particle which is moving to the right with a speed of 0.6 c. Now, this is the expression which again I have just now written which is m naught gamma u times u which is the new definition of momentum. We let me calculate gamma u as we have just now seen. Gamma u is equal to 1 upon under root 1 minus u square by c square and if I substitute the value of u is equal to 0.6 c in this particular expression, we know that this expression would lead to a value of gamma u is equal to 1.25. u is equal to 0.6 c is one of the speeds which leads to a rather simple result or a simple answer for gamma u. So, gamma u will turn out to be 1.25. So, that is what I have written here. This is the value of gamma u which is 1.25. I have slightly reorganized this equation to express momentum in the units of m naught c. So, the speed u is 0.6 c and this is my m naught. So, this becomes 1.25 multiplied by 0.6 m naught c. If I simplify it, this becomes 0.75 m naught c. Now, the second particle is also moving with the same speed of 0.6 c, but it is moving in the minus x direction. So, therefore, this momentum has to be negative. Momentum is a vector quantity. It has to be negative because sin is opposite to it. So, we use exactly the same expression minus m naught gamma u u. Again, gamma u is 1.25 and speed is 0.6 c and the m naught. So, I get same numerical value, but with a different sign. So, second particle's momentum is minus 0.75 m naught c. Of course, the y component and z component, we need not bother because in that particular direction, the momentum is anyway 0. We are assuming that the motion is only along the x direction. So, this is what I have found out that are the initial momentum of the particle number 1 and particle number 2 before collision. And as you can see that magnitude wise, both of them are same. Only their signs are different. So, if I add these two momenta, these two momentum would lead me a zero value, which we had expected even earlier that the initial momentum of the two particles, some of the momentum of the two particles would be zero. So, this is what I have written in the next transparency that p x k where k of course, gets added up for the two particles from 1 to 2 is equal to 0. And this i symbol refers to the fact that this is the initial momentum, which is before the collision. So, clearly this was the initial momentum and final momentum obviously is 0, because if I look back at this particular equation, p is equal to gamma u m naught u. And if u has to be 0, then of course, p has to be 0. And because in this particular frame, it has been given that after the particle gets stuck, they come to a rest. Therefore, this u must be zero in that particular frame. Therefore, p is also zero. So, clearly the momentum after the collision is also zero. Therefore, the momentum is conserved. This is what we had expected that momentum should be conserved in this particular process. However, there is a different situation now. If you remember when we worked out this problem earlier or when we worked out a similar problem in a classical mechanics, traditional classical mechanics, we said that this is an inelastic collision and therefore, we never used to conserve energy at that time. We at that time said it is only momentum, which is conserved. Energy is lost in some non-mechanical processes. In fact, we always meant that overall energy is never sort of disappearing. It is always conserved. But what we meant at that particular time is of mechanical energy, that if you are taking potential energy and kinetic energy, those type of energy is the mechanical part of the energy. They are not conserved because the energy is lost in some non-mechanical form like heat or whatever it is. So, we were never conserving the energy at that particular instant of time. But now, we realize that with the new definition of energy and momentum, the energy must also be conserved in this process. Because remember, the conservation of momentum, universality of conservation of momentum depends critically on the conservation of energy. So, unless energy is also conserved, momentum will not be found to be conserved in other frame. So, in this particular frame also, I must conserve energy, but with a new definition of energy and not the classical definition of energy, which involve only the kinetic energies. So, we conserve the energy. Therefore, I must write the energy of the first particle and energy of the second particle. If I write the energy of the first particle, the expression is E is equal to m c square, which is the well-known expression. And this m, we have said, is gamma u times m naught, where m naught is the rest mass of the particle. So, if I have to write the energy of the particle, which is total relativistic energy of the particle, then what I have to use for E is gamma u m naught c square. So, this is what I have written here for the first particle. E 1 is equal to m naught gamma u c square. Gamma u, we just now discuss, is 1.25. So, I put it 1.25. I get this equal to the energy of the first particle as 1.25 m naught c square. The second particle also moves with the same speed. Therefore, gamma value is again 1.25, though its speed is in negative direction, but the energy is a scalar quantity. Therefore, I will not put a negative sign. It is the same energy, which is 1.25 m naught c square. Therefore, the total initial energy, summation of E k i must be equal to this energy plus this energy, which means 2.5 m naught c square. So, what it says that the initial energy of the system of the particle of the two particle was 2.5 m naught c square. And I now want that after the collision, again this energy should remain same. It means the combined particle, which has got stuck should also have the same energy of 2.5 m naught c square, if this particular particle has to obey the conservation of energy, which we expected to be obeyed according to the new rules of energy and momentum conservation. But if I take the earlier expression, where I have assumed that the two masses just get stuck. And remember, in that particular picture, I had drawn two masses and I have written 2 m at that particular point. Then this 2 m obviously cannot have energy of 2.5 m naught c square, because the particle is at rest. What essentially it means that this is not correct. When the two particle have got stuck to each other, of course, u is 0. So, when I write the total energy E, gamma u is 1, because this is 1, because the particle is at rest. Therefore, this energy must result with a different m naught. It means the rest mass of the particle has changed. And this gamma u m naught c square should be equal to 2.5 m naught c square, which I have written earlier. What it means that after the particle have got stuck, though they have come to rest, but their rest mass energy would have gone up. Because this gamma u being 1, this means m naught is equal to 2.5 times. So, the two particles, once they have got stuck, their mass is no longer 2 m, but it has become 2.5 m naught. Their rest mass has gone up. Why rest mass has gone up? How you would explain on the basis of relativity? You will say that that energy which was lost, it was lost into some non-mechanical form of energy, but as far as E is equal to m c square is concerned, that does not look into that particular aspect, because overall energy must be conserved. And of course, that energy is no longer of the kinetic form. Therefore, u is still 0, gamma u is still equal to 1, but that infest, that particular energy shows itself in the form of increased rest mass, because if the masses have become probably warmer because of the collision, because of the loss of energy, this will eventually result in its increase of rest mass. So, what it shows that the two particles after they have got stuck to each other, their rest mass has gone up. That must be true if momentum and energy both have to be conserved in this particular frame of reference. So, this is what I have written here. Hence, the final energy should also be the same, even though the speed of combined particle is 0. So, gamma u is 1. So, 2.5 m naught c square must be equal to nu m naught c square, because gamma u is equal to 1, which obviously means that m naught is equal to 2.5 m naught, the rest mass of the particle has gone up. So, this is what I have written here. Hence, the rest mass of the combined particle has increased and m naught, the new rest mass of the particle is 2.5 times m naught. So, remember that is what we have said that if the two particles gain certain amount of energy, then this rest mass goes up. If they lose certain amount of energy, if they tend to become cooler, then again the rest mass will go down. That is what is the new way of looking into the energy. So, as we said that there is no difference between energy and mass, in relativity mass and energy are related by fundamental constant c square. A system has more energy, it has more mass. If system has a lower energy, it has a lower mass. Mass and energy are always related. In fact, as we have said earlier that you can express energy in terms of masses or mass in terms of energy. So, many times when we say mass of electron, many times people will tell it in the unit of MEV or mass of the proton in terms of the MEV or GEV, while mega electron volt, giga electron volt, these are the units of energy and not of mass. But then still we can express them because the two are related with a fundamental constant c square. Now, let us look at the S frame of reference because that is the way we have started. From S frame we concluded that the rest mass of the particular particle has gone up after the two particles have combined with each other. Now, remember this rest mass is a force killer. So, this rest mass is not going to change when I go to S frame of reference. So, I must use the same rest mass of the particle even after I change my frame of reference from S to S prime. But now, let us recalculate the momenta in S frame of reference. When I try to recalculate the momenta, of course, the values of momenta individually will be different. But eventually, momentum must be conserved and energy must be conserved because I expect the conservation principles to be universal. So, let us go to S frame of reference. So, let us look at the velocity of the first particle just before the collision. Remember, we have two particles in S frame one going this way, another going this way. Now, I go to a frame of reference of this particular particle. It means I go to a frame in which v is 0.6 c. Now, looking at this particular frame of reference, I go to this particular frame of reference with v is equal to 0.6 c. Calculate the speed of this particular particle which obviously should be 0. Calculate the speed of this particular particle. Find out the total initial momentum. Then, again I look at that particular particle which has been obtained after the collision. Find out the speed of that particular particle. Calculate the momentum of that particular particle and try to see whether I get the conservation of momentum and energy. That is what I want to do. So, I go to this particular frame of reference S which has v is equal to 0.6 c. First, I look at this particle itself for which u will be 0.6 c. I apply velocity transformation. So, this is what I have applied here. So, u 1 x is equal to u x minus v. So, u x is 0.6 c. v is 0.6 c. 1 minus u x into v divided by c square. Both u x and v are 0.6 c. c square gets divided. So, I get 1 minus 0.36. c square gets cancelled rather and because numerator is 0. So, u 1 x prime is 0 which is expected because once you go to the frame of that particular particle, the speed of that particular particle must be 0. If you are in a frame, you do not notice its own speed with respect to anything. According to you, the speed of the frame is 0. So, obviously, 1 x prime is 0. The momentum then of course is 0 because we have just now discussed if the speed of the particle is 0, the momentum is 0 even with the new definition of momentum. So, there is no problem over that. So, momentum of the first particle is indeed 0, but let us look at the moment of the second particle that will not be 0. So, let us find out the speed of the second particle. We use exactly the same expression. I write u prime to x. Now, for second particle, the particle was moving to the left with a speed of minus with a speed of 0.6 c. So, I write u x is minus 0.6 c minus v which is same 0.6 c because I am going to the same frame of reference of the first particle divided by 1 minus u into v x because of negative sign. This will become plus and this expression will remain as 0.36 because both u x and v are 0.6 c each c square will cancel. So, you get the speed as minus 1.2 divided by 1.36 c. This calculation we have done earlier also. When we had done this particular example earlier, only thing at that time we had used the old definition of momentum. Now, I will be using the new definition of momentum. So, using this particular u, I must calculate now gamma u because the new definition of momentum not only involves the speed and mass but also involves gamma u. So, let us calculate gamma u. This is being little longer, little big. You will find out that expression little bit more complicated but eventually the result is simple. So, that is what I have written. The moment of the second particle before collision is thus. This is the gamma u value 1. So, not the numerator here but 1 divided by under root 1 minus u square by c square. The speed was 1.2 divided by 1.36 c. So, c square and c square cancels. So, numerator you have remained with just 1 upon under root 1 minus 1.2 divided by 1.36 square. Let me just rewrite here, just to make it clear. So, u hat p 2 x prime must be equal to gamma u m naught u. Now, this gamma u I can write as 1 upon under root 1 minus u square by c square. This was 1.2 divided by 1.36 c divided by, of course, square divided by c square will cancel. So, what will remain is this is square. This is m naught and u will become 1.2 divided by 1.36 c. So, what I have done in this transparency? I have brought this particular thing up into the numerator to express this in again the units of m naught c. So, this is what I have written here. p prime 2 x is equal to minus 1.2 divided by 1.36 divided by under root 1 minus 1.2 divided by 1.36 whole square m naught c. If you calculate this 1 upon under root this particular gamma u value that turns out to be equal to 2.125. This multiplied by 1.2 divided by 1.36, eventually you can calculate this will give me a value of minus 1.875 m naught c. You remember this is minus because the speed of the second particle was negative. If you just look, this is speed is negative because if you are sitting on the first particle, the second particle moves in minus x direction. So, we now realize that once we go to the s frame, s prime frame of reference, I am sorry, then the momentum of the first particle is 0, but not of the second particle and therefore, the initial momentum is not 0. The initial momentum which is the sum of the momentum of the two particle is equal to minus 1.875 m naught c. So, what we find that the total value of the momentum has changed once the frame has changed, which is obvious, which happens even in the classical mechanics that once you change the frame of reference, you do find that the speeds will change and therefore, momentum will change. So, in s prime, this is the net momentum. Now, if whatever I am saying is correct, then with the new definitions of momentum, this should also be the momentum of the final particle which is the combined particle, but for that combined particle, I must use now a different rest mass, which is 2.5 m naught and not 2 m naught. So, let us look at the velocity of the combined particle after the collision. In s frame, which was the original frame in which we described the problem, this particular particle after the collision was at rest. So, it means u x was 0 and we are going to a frame of reference with v is equal to 0.6 c. So, therefore, this will be 0 minus 0.6 c divided by 1 minus u x v divided by c square 0 multiplied by 0.6 c divided by c square. This term becomes 0. So, denominator becomes 1 and this is just equal to minus 0.6 c. Again, this particular result I had obtained even earlier when we had discussed this problem for the first time. This is the final speed. That is why I have written f here. Now, I have to calculate the moment of the combined particle. When I write the moment of the combined particle, then in that particular case, I have to use this new speed and calculate its gamma, then multiplied by m naught, then multiplied by u. Now, this is 0.6 c and 0.6 c we have just now discussed gives me a very clean value of gamma u which is 1.25. So, gamma u is 1.25. The speed of the particle we have just now calculated is minus 0.6 c. This is what we have just now calculated here. It is minus 0.6 c. So, this is minus 0.6 c and m naught. But this m naught is now the new m naught because this m naught is a scalar. Once I change, it is a four scalar. Once I change from one frame to another frame, this m naught is not going to change. So, for this m naught, I must use 2.5 m naught. Once I use this as 2.5 m naught and substitute these values and multiply, I indeed get minus 1.875 m naught c. So, you realize the trick here is to put m naught is equal to 2.5 m naught, which essentially comes according to relativity from the conservation of energy. And as we have discussed that conservation of energy is must in order to see conservation of momentum. So, we just now saw that now that problem is solved in S frame of reference also the momentum is concerned. Let us look at the energy whether in S frame of reference even the energy is concerned. It should be everything that we are saying is correct. So, now let us check whether energy is also conserved in this particular frame. We go to S frame of reference and calculate the energy of the particle. How do we calculate the energy of the particle? Let us take the energy of the first particle. When we take the energy of the first particle, this particular particle was at rest. So, we have just now written e is equal to gamma u m naught c square. Now, once I go to the frame of reference of the first particle, then of course, gamma u is 1, because the speed of the particle itself is 0. Therefore, this e is just equal to m naught c square. So, the energy of the first particle is equal to e is equal to just m naught c square. For the second particle, the speed is not 0, but it is minus 1.6 divided by 1.36 c. So, using that particular value, I must calculate gamma, which I have actually done earlier. So, that gamma u multiplied by that m naught, which is the rest mass of the second particle multiplied by c square, that will give me the energy of the second particle. So, this is what I have written here in the transparency. First particle is just clean m naught c square, because this particle is at rest in S frame of reference. It is the second particle, which is moving in S frame of reference and it is moving with the speed of 1.2 divided by 1.36 c. c square c square cancels out. So, gamma is this much, which we have just now calculated. In fact, we have not calculated, but we have just now told that this particular gamma u turns out to be 2.125, of course, multiplied by m naught c square. So, the total initial energy becomes 3.125 m naught c square. So, this is the total initial energy of the system. Let us look at the final energy. As far as final energy is concerned, you have only one single particle now with a rest mass of capital M naught. And that particular mass is not at rest in S frame of reference, but moves with a speed of minus 0.6 c, which we have just now seen. So, when I calculate gamma u for this particular particle, I should use 1 divided by under root 1 minus this speed of the particle, which is 0.6 c and c square, of course, cancels out. So, I get this gamma u, which I know is 1.25. So, for this gamma u, I will put 1.25. This m naught we have just now seen is a 4 scalar must be equal to 2.5 m naught, small m naught times c square, you indeed get that the total energy of the system is 3.125 m naught c square. So, we do find that energy is also conserved. So, we work out at the same problem and try to relook at the energy and momentum conservation. And we do find that energy and momentum both are conserved in S as well as S frame of reference. Let us look at one another simple example, which is somewhat similar example, but still have chosen to describe this particular example of another inelastic collision, because this is the type of problem in which we will use energy momentum conservation, which I had not used in the first problem. The first problem was more to illustrate that the new definition of energy and momentum that we have now obtained will lead to universality of energy and momentum conservation. Here, I want to give an example, which is more like a problem in which we have been asked to calculate certain specific things and for which I may require translating between energy and momentum, knowing energy and momentum, calculate speed, calculate momentum, blah, blah, blah, all those things. So, let me describe this particular example. So, there is a particular particle of rest mass m naught and it has a kinetic energy of 6 m naught c square, what has been given remember is kinetic energy. This we should be careful when we are talking of kinetic energy and when we are talking of the total energy. So, of course, it has a very high energy, very high kinetic energy, because this is 6 times its rest mass energy. So, this particular particle strikes and sticks to an identical particle at rest. So, now we have been given the problem in a frame in which one particle is at rest and another particle which is coming and hitting it. So, one particle is at rest, another particle comes and hits it. So, this is the way we have described this particular problem. Now, we say identical particle, it means its rest mass must also be imagined, because then only its particle, this particle is identical particle. So, these two particles come and collide, one particle is remain stationary, another particle comes and hits here. Let me first, sorry, complete the problem. What is the rest mass and the speed of the resultant particle? What has been asked in this particular problem is what is the rest mass and the speed of the resultant particle? These two particles strike and get stuck to each other, just like in the problem that we have described earlier. Energy momentum conservation, not just momentum conservation, both momentum and energy conservation. Let us write initial energy. When we write the initial energy, there is one particular particle which was only at rest. So, that has energy of m naught c square. There is another particle which had a kinetic energy of 6 m naught c square. Obviously, that particle would have also had a rest mass energy of m naught c square. So, therefore, if I take the total energy of that particular particle, that will be 6 m naught c square plus m naught c square will be equal to 7 m naught c square. So, this will be for the particle which is moving. This was its kinetic energy. This was its rest mass energy. Remember the expression that we had written earlier, that K is equal to m c square, K is equal to m c square minus m naught c square. So, the total energy is sum of kinetic energy and rest mass energy. So, this is what I have written here. This is the total energy of the particle which is moving. This is the total energy of the particle which is stationary and this adds up to 8 m naught c square. So, total initial energy of the system is 8 m naught c square. Now, I have to calculate the momentum. Of course, momentum of the particle which is at rest is 0. There is no problem because that particular particle is at rest. So, u is 0. Therefore, momentum has to be 0. I had to calculate the momentum of the second particle. But what has been given is only the kinetic energy of the particle. Therefore, I must use an appropriate expression to find out from this energy the momentum value and for that we are using the expression which is well known, which we have described earlier. E square is equal to p square c square plus m naught square c to the power 4. Now, for the particle which is moving, this E was 7 m naught c square as we have just now discussed. This is m naught square c to the power 4. Substuting this particular expression, I can find out what may be the value of momentum. So, this is an expression which helps me for a single particle to convert energy into momentum or vice versa. This is what I have written here. So, p square c square, I have written as E square minus m naught square c to the power 4. This slightly simplified this equation taken the under root and divided by c square. So, I will be getting this as under root E square, which is 7 m naught c square. 7 m naught c sorry, not E square, but 7 m naught c because c square I have already divided 7 m naught c minus m naught square c to the power c square because c square has already been divided. So, this becomes 49 minus 1, which is 48. So, I will get the momentum of this particular particle as under root 48 m naught c. So, the momentum of the first particle is 0. The momentum of the second particle is under root 48 m naught c. Hence, after the collision, we have just now discussed that the two particles collide and that has been given in the problem and they get stuck to each other. If they get stuck to each other, what remains is only one single particle and that single particle because of conservation of momentum must have the same momentum as the initial moment of the two particles, which is under root 48 m naught c and it must also have a energy which is equal to 8 m naught c square. So, all you know that the new particle which has been formed as a resultant of the combination of these two particles will have now a energy equal to 8 m naught c square and must have a momentum of under root 48 m naught c. Using again the same expression, I can find out what will be the value of m naught capital m naught, the new mass, new rest mass of the particle which we have obtained as a result of getting stuck, the two particles getting stuck. So, I use the same expression e square is equal to p square c square plus m naught square c to the power 4. e we just now said is 8 m naught c square. So, I squared it. So, this becomes 64 m naught square c to the power 4. So, this is my e square term plus p square c square. My p was under root 48 m naught c, I square it. So, I get 48 m naught c square, m naught square c square and this was p square c square. So, there is another c square which will make it m naught square c to the power 4. Remember here, this is the old m naught because this momenta and energy have been written terms of the mass of the original particles, rest mass of the original particle. But now you have got a new particle as a combination of these two particles, which may have a different rest mass which I am writing as capital m naught. So, e square is equal to p square c square plus capital m naught square c to the power 4, which must be the mass of the new particle. I just substitute it here, I just solve it, 64 minus 48 and you will get m naught square and you take under root of that, you will get m naught is equal to 4 m naught. So, in this particular case, you will find that the rest mass of the new particle that you obtain is 4 times the rest mass of the original particles with which this particular particle was formed. So, it has become the rest mass is now 4 times that rest mass. Next question is to find out the speed. Once we have obtained the mass, we have found out rest mass and we have found, we know its momentum and energy, I can always find out its speed. That is simple. Let us see how there, I have given two different methods, but one can choose whatever is convenient. So, as far as the new particle is concerned, we have to find out the final speed. We have said e is equal to gamma u m naught c square, that is what we have written. This particular particle obviously is not at rest, that is why we have to find out the speed. This particular thing must be equal to gamma m naught c square, which must be equal to, because 4 times m naught was its capital m naught. What we have written is that m naught was equal to 4 small m naught. So, if energy e is equal to 8 m naught c square, this I can write this as 2 times 4 m naught into c square. This 4 m naught is capital m naught. So, I can write this as 2 capital m naught c square and this being equal to gamma u m naught c square, it means gamma u is equal to 2. That is what it will tell that gamma u is equal to 2. So, this is what I have written here in this transparency, that e is equal to gamma u m naught c square. This 8 m naught can be written as 2 times capital m naught. So, this is 2 m naught capital m naught c square, giving me gamma u is equal to 2. Now, either I can use the momentum expression here, which I know is gamma u times m naught times u to calculate this u, which I have done here. I know the momentum of the particle, combined particle is under root 48 m naught c and this is equal to gamma u m naught is equal to 8 times m naught, because this capital m naught is 4 small m naught. I can calculate u. This turns out to be equal to under root 3 divided by 2 c. There is other way that you have already known gamma u. So, use direct expression. So, this is an alternate way of doing it. Look at gamma u value directly. Gamma u we have just now discussed is 1 upon under root 1 minus u square by c square and we have seen that this is equal to 2. So, I take inverse of this particular thing and take square of this thing. So, I get 1 minus u square divided by c square, which is equal to 1 upon 4 and if you just solve, you get the same result u is equal to under root 3 by 2 c. So, the new particle, the combined particle would now be traveling with a speed of under root 3 by 2 c as far as this particular problem is concerned. So, what we have seen in this particular problem that how one can convert from energy to momentum or from speed. Now, we just get a practice of using these particular expressions the way we would like to use them to convert from energy to momentum or vice versa or to convert into speed. So, actually at the end of this particular lecture, I would just like to summarize that in this particular lecture, we have just given two examples of what we classically call as completely inelastic collision and we saw that both these examples, how energy and momentum both have to be conserved together. Thank you.