Sequence wise lecture of ABJ sir class 11. Complete playlist of class 11 physics for Jee mains and advance . For another subject or class 12 playlist , Check on channel (if it's not there , check after few days).
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Watch Complete YouTube Lectures Distraction-Free Download the Competishun App 👉 https://play.google.com/store/apps/details?id=xyz.penpencil.competishun&pcampaignid=web_share 🌐 Login for Courses & Test Series: 👉 https://i.competishun.com/ *Popular Courses for IIT-JEE:* Pratham | Prakhar | Prakhar Integrated I | Praveen | Pragyaan C3 | Champ | CATS | ABC | PYQ Express UTS | ATS 👉 Explore all: https://competishun.com/courses/ 📞 Support: 8888-0000-21 | 7410-900-901 | 7410-900-908 --------------------- Topics covered in this Chapter Centre of Mass Class 11 lecture by ABJ sir IITD with timestamp: 00:00 - 06:05 COM Problem 1: In this problem, we have a cart of Mass M and a boy of mass m is on the cart at one end (According to the given Diagram) and this system is moving with initial velocity u. Now this boy jumped from the cart with the velocity v wrt cart. So we have to find the velocity of the cart after Jump. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system. 06:15 - 16:46 COM Problem 2: In this problem, we have a cart of Mass M and two boys A and B each of mass m is on the cart at both ends (According to the given Diagram) and this system is initially at rest. Now Boy A jump with velocity u relative to Cart, and after it, boy B jumps with velocity u wrt Cart. So we have to find the final velocity of the cart. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system. 17:16 - 26:28 COM Problem 3 (Part A): In this problem, we have a wedge of mass M (Free to Move) is at rest initially and a block of mass m is moving with velocity u. (Situation will follow the given diagram) We have to find the maximum height achieved by m on the Wedge of Mass M. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system. 26:28 - 38:10 COM Problem 3 (Part B): In this problem, we have a wedge of mass M (Free to Move) is at rest initially and a block of mass m is moving with velocity u. (Situation will follow the given diagram) We have to find maximum velocity of mass M. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system and energy conservation. 38:50 - 54:38 COM Problem 4: In this problem, We have two blocks of Mass m and M connected with a spring of spring constant k, initially at natural length. Initial Velocity of Mass M is u. We have to find the maximum extension in the spring, velocity of centre of mass of the system, initial velocities of blocks wrt COM, the momentum of system wrt COM and maximum extension in terms of reduced mass. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system and energy conservation. 55:00 - 01:02:44 COM Problem 5: In this problem, We have a ball of mass m projected with velocity u at some angle wrt ground. At maximum height, this ball explodes into two equal parts each of mass m/2. If one part comes instantaneously at rest, then Find the range of both parts. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system and energy conservation. 01:02:54 - 01:14:52 COM Problem 6: In this problem, We have two blocks A and B each of Mass m connected with a spring of spring constant k, initially at natural length. This system is at rest initially. Now, A force F is applied on block A. So we have to find acceleration of COM, Velocity of COM at any time t, Momentum of the system at any time t. Also if at any time t, Block A has velocity v, then what is the velocity of block B. Also find the displacement of COM at any time t. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system and energy conservation. 01:15:10 COM Problem 7: In this problem, We have two blocks A and B each of Mass m connected with a spring of spring constant k, initially at natural length. This system is at rest initially. Now, A constant force F is applied on block A. So we have to find maximum extension in the string. To solve this problem, ABJ Sir used the Law of conservation of linear momentum of a system and energy conservation. This is the Fourth lecture of Centre of Mass class 11 for JEE Mains & Advanced. There are nine lectures in one shot for Centre of Mass Class 11, in which ABJ sir covers all topics, concepts, examples, and problems important for JEE Mains and Advanced. This Centre of Mass Class 11 Physics lecture consists of the Law of conservation of the linear momentum and problems related to this. This lecture is for the Science student of class 11 who is preparing for JEE Mains & Advanced or Olympiad. ABJ sir also covered JEE ADVANCED PYQ in this lecture. Prerequisite of the Chapter – There is no prerequisite for this chapter. Basic mathematics and calculus is required to solve good problem and derivations.