 Next question. It's a rod of mass m and length l. There is a bullet that comes from here. The bullet of small m comes from this side. It hits this rod, comes out from the other side. Should I simplify this or it is okay? Simplification the bullet doesn't come out. It gets embedded inside. So it's like that. Okay. Bullet becomes a part of the rod. It doesn't come out. Alright. It got embedded inside the bullet. Embedded inside the rod of mass m and length l. Okay. This rod along with the bullet swings by an angle of theta max. It gets shown by an angle theta only. You need to find the velocity of the bullet. So where does it get? This is l by 4. So it's small and insignificant in comparison to capital l. No. It was that good move. No. It will move. It will still move. Velocity could be high. Anyways, do it now. I have just one question. Can I use consumption of energy between this point and that point? Yes. No. This collision, loss of energy will happen. No sir. Once it's collided. Once it collided, after that, this point or that point you can use. But not between this point and that point. Because of collision, some energy will be lost. It is an inelastic collision. Can I use conservation of linear momentum just before and after collision? Yes. I cannot use because there is a force generated at the hinge. So you just find i and n. What I can use? I can use conservation of linear momentum or not? No. No because there is an access force from this. What I can use then? Conservation of angular momentum. Angular momentum I can use about which axis? Fixed axis. About this axis. Reason? There is no torque about that. Just before and after collision. Here I use conservation of angular momentum to find the angular loss just after the collision. And from that point onwards I use conservation of energy. Can I do it? Like you ordered me. What is the initial angular momentum? This axis. What is the angular momentum of the bullet? mv. m into u into 4L by 5 perpendicular distance. Yes or no? This should be equal to i omega. Initial angular momentum should be equal to the final angular momentum. Conservation of angular momentum. What is the moment of inertia of the fixed axis? Capital ML square by 3 plus small m into... How many of you got this equation? I got all of them but the answer is like... Have you got this one? I am asking only that. And anybody else got this equation? I understood how to write this. Moment of inertia of fixed axis into omega is the total angular momentum of the rigid body. Now rigid body has bullet also a part of it. That is why we have to add its moment of inertia. So from here we get omega. And once you get omega forget about the fact that collision has happened. Now it is conservation of mechanical energy. So which you can write like this. W is equal to u2. Can I assume this to be my zero potential energy? Potential energy initially. See initial critical energy is what? i into omega square where i is this. u1 is what? You can write down the potential energy of the bullet and the rod separately. So you can say for the rod it is minus of mg l by 2 plus for the bullet minus of mg into 4l by 5. Understood? And k2 is what? k2 is zero. U2 I have to write now. U2 is the final potential energy. For the capital it will be this distance. Which will be what? l by 2 cos theta. So m into g into by 2 cos theta plus minus, minus potential is negative. mg into 4l by 5. Please understand how to solve this rather than just noting down. Is it clear? So guess what? The chapter is over. And now I can give you the mixed question. Multiple concepts together. See whenever you solve problem, you know almost every time if you solve properly you will feel that oh this concept was not done, this little thing was not told. So like that there can be many different things. And these small small things are your own learning when you solve problems. Do not rely only on the theory which is taught in a class. I spent around a month teaching this chapter still. There are a lot of small small little things which you will understand when you solve problems. And it's not that we are not solving problems. We are solving problems. But in the size of a chapter is as big as combining entire kinematics 1D, 2D, laws of motion and walk-by energy. All these four chapters together, the size of that is the size of this single chapter. So we can't do each and every variety of the problem in the class. Even if we spend one month like what we have done. But then it is up to you to practice that. All right. Let's start. We'll start with the simple ones and then go to the complicated ones rather than directly jump into it. All right. So now we can take up questions from whichever topic we like. And you can also start attempting questions anywhere from the chapter. All right. All right. So let's take this question first. There's a rod of mass m and length l lying on a frictionless surface. Start asking. Mass m coming with velocity u hits the rod at a distance l by 4 away from the center. When it hits the rod, after it hits the rod, it comes back with the same velocity u by 2. You need to find out after it comes back with u by 2. Straight back. You need to find vcm and omega. vcm the rod and omega. Use conservation of linear momentum. Again, use the frictionless surface or internal surface or internal external forces. 0. Initial momentum is m into u. This should be equal to final momentum minus m into u by 2 plus capital M into vcm. So vcm will come out to be equal to how much? vm u by 2 by m. This is your vcm. Straight forward, conservation of linear momentum. Sometimes, whenever we learn something new, we forget that we can apply whatever we have learned earlier. So this we haven't learned in this chapter. This we already know. Conservation of linear momentum. Now, can I use conservation of angular momentum? About which axis? About center of mass. About any axis I can use. But center of mass would be the easiest one to apply because about center of mass, angular momentum is Icm omega. I can literally use about any axis. About any axis torque is 0. So what will be the initial angular momentum of the system? m into u into l by 4. Yes or no? This will be equal to what? Minus of m into u by 2. Now, angular momentum has changed for the particle after collision. Then what plus Icm which is m l square by into omega. How much omega will come from here? 9 m u by 2 m l. How did you get 9? Yes, 9 m u by 2 m l. No, so 2 capital m l. 2 capital m l. How many of you got VCM? Omega, they are not getting it? I didn't even get u by 2 backwards. Yeah, u by 2 backwards. In HCR momentum, it comes to rest. Okay? Find out the point which will be at rest immediately after collision. Point which will be at rest immediately after collision. Similar to what we have done sometime back. It's here. How the rod will rotate after collision? It's l by 3. Rod will rotate like this and it will move forward like that. So, it has to be above only. Below what will happen? Omega into l is also this way and v is also this way. So, is it l by 3 above? l by 3 above. Yes, sir. Let's see. Omega into d this way at the velocity and this way it will be VCM. So, total velocity will be VCM minus omega d. So, d will be VCM by omega. Yes, sir. VCM plus and omega is this. So, it will come out to be l by 3. So, at a distance l by 3 from the center above, that point will be at rest immediately after collision. Okay? Any doubts? Is there always a point at rest? It's not necessary. So, how do we know there will be one here? Because we have found out l by 3. If suppose d comes out to be more than l by 2, then that is not on the rod. If you try the same thing below the rod on the bottom side, you get a minus thing right from the center. Below is physically not possible. Both omega into d is also the same way as v. Obviously, we get a distance as minus which means etc. Mathematically, you will get minus. But people say that minus of 1 in terms of distance has no physical. It's an imaginary number itself, minus 1. It doesn't have a physical significance.