🔥2D, NEWTON LAWS OF MOTION, CIRCULAR MOTION , WORK POWER ENERGY In ONE SHOT | Part 2 || WBJEE 2026

Hinzugefügt: 2026-05-10

[00:00:00]Perfect.

[00:00:10]More.

[00:00:20]Hello. Hello. Hello. Hello everyone.

[00:00:22]Hello.

[00:00:38]Hello everyone.

[00:00:53]on that message.

[00:01:12]Good evening.

[00:01:16]Take speech.

[00:01:46]All is motion. Okay.

[00:02:16]solution.

[00:02:19]Okay.

[00:02:42]So 2D motion 3D motion 3D motion here.

[00:03:10]Please watch.

[00:03:24]So 2D motion 2D coordinate system mainly this is X and this is Y. Okay. So XY system it origin. So this is referred to as the basic cart system.

[00:04:21]So if I mark a point here, so the point must be t some x y. Okay. So this is referred as 2D plane. This is a plane.

[00:04:39]So that is y So here you see what you see that if I can write acceleration if I have to write acceleration that is acelation throughout throughout the class but throughout the journey So acceleration can be written as the dt of d. Okay. This is also a vector quantity. This is also a vector quantity. The first order derivative which also can be written as v vector dot dot also written as ddt. DTP of some function can be as acceleration vector APA.

[00:06:15]Clear? We can write.

[00:06:28]So acceleration AC cm is equals Acceleration by a x i + a yse.

[00:07:02]Acceleration Xapelation clear Hello.

[00:07:45]Please interact with me.

[00:07:55]Acceleration is a plus 2D representation of acceleration.

[00:08:01]2D motion. Sorry d V X I + V Y X by DT dx by DT * I plus DY by DT * of this is velocity 90% App is not. App is not working.

[00:09:24]So if I will try to draw axis like this is the yaxis this is the xaxis particular point at a particular time instant. P is the position of this particle.

[00:10:10]dx dx dx by dt.

[00:10:44]So this is can be written as dt by dx2 by dx² x d2 dx² okay what is 3D motion xy Is it clear? This is the xaxis. This is the yaxis. This is the zis aisle.

[00:11:27]person.

[00:11:37]Hello. Hello.

[00:11:39]All okay.

[00:11:54]Hey, sound that sound.

[00:13:10]Hey, select your hand.

[00:13:43]So if I velocity dt.

[00:13:51]So if I differentiate this term, so that becomes bx bz by dt.

[00:14:30]dx by dt i + dy + dy by dt plus dz by dt. So what we find that is acceleration.

[00:14:40]Acceleration can be written as X dot I Y dot J + Z do. So what is actually already X d X² with respect to time? Yeah.

[00:15:07]Okay.

[00:15:10]Double dot mining.

[00:15:35]full syllabus cover with the quality of content and having every topic.

[00:15:54]J advanc.

[00:16:09]So this can be written as particle.

[00:16:19]projectile motion.

[00:16:33]So projectile motion. What is motion?

[00:17:03]Max is called the Max is a good question, good quality of question, then good quality of question.

[00:17:38]Good quality of questions are there.

[00:17:47]So see this is if I take this as the point P XY Max flight P O Ascend O.

[00:18:29]Clear.

[00:18:43]So Hello.

[00:19:24]Velocity remains along the xaxis.

[00:19:27]Velocity remains Actually gravitational.

[00:19:45]Very good.

[00:20:00]Mix mechanics.

[00:20:42]equation of equation of It is the combination of two independent motion horizontal motion here.

[00:21:33]Sorry.

[00:21:35]Equation of motion intu cos theta ux u x u cos theta = u cos theta into theta Y at least 50 A= G Yus GT G² GT.

[00:22:41]Y = U sin theta into tus² y² = squareg height at time T I t = U sin theta into tus² Y = X² Angle Next special conditions special If theta becomes hy = -² 90°.

[00:25:13]This one at theta= to 45° at theta= 45° the range will Okay.

[00:25:43]Maximum because because 45° 45 advanc Advanc 90 max.

[00:26:39]Next time to reach the maximum height t= u sin theta by g = y compent velos P U sin theta U sin theta GT U sin theta is nothing but U sin theta Y here.

[00:27:32]Hey share please.

[00:28:00]science based on logic that is the py sin of theta sin theta minus gtity theta is nothing but the u important.

[00:28:58]Everyone clear or not?

[00:29:01]Yes. Very good.

[00:29:03]That next one say What is the project uh if the projectile speed u from height h above the term if projected if the particle is projected with a speed of u from height hle T= U sin theta U sin theta +<unk> U² sin² theta + g / g clear sir what is the name name of what is the name of I don't know what is the name of the I don't know what cos clear everyone Next range dep and compliment angles the range must be equal and time of flight must be equal.

[00:31:00]Clear? Next.

[00:31:02]At the highest point u= to infinity horizontal and dy=0 u at any two point at any two point on the same level the speed is same by energy at any two points on the same level. What is mean by this? This is a good concept. Like if I draw this projectile somehow symmetry.

[00:31:50]Is it all of you?

[00:31:59]Next question. Next. Next. Next.

[00:32:15]Everyone very bad.

[00:32:47]Open.

[00:33:00]Sorry.

[00:33:16]equations.

[00:33:25]this topic = sin theta U sin theta plus roo<unk> under u² sin² theta + 2 gh / Sorry for names. You got u cos.

[00:34:16]watching.

[00:34:20]Please next minimum speed. The minimum speed to reach a point xy from trajectory equation. Minimum U when theta theta is equals to U² Y capital by GX² / 4 Y toal power by 2. Is it clear? Minimum speed to reach a point XY from Next.

[00:35:27]Give the range for for given range r and u sin 2 theta sin 2 r² theta is become sin inverse of r g by u² hit on hit target at x Y x tan theta x capital g x² x tan²ad for real solution Real solution g x² g into gx² + 2 y² u² = hello everyone isn't everyone Next. Maxiveness.

[00:37:05]If second projectile is made from the height of highest point with the same speed at an angle, then the minimum range Max= 45.

[00:37:45]Zoom the portion. Zoom.

[00:37:58]most of next. Next present.

[00:38:33]Those who were present in the previous class, please mark your attendance.

[00:38:39]Wait, please mark your attendance. Those who are president, please please request Is there any problem here? Hello. Hello.

[00:39:36]Hello.

[00:39:52]cos 5 and y = r sin by d rs Clear velocity component cos R I s and watch I do sin - r cos here actual speed is equals to R² R² + this R5 R square clear acceleration what is the acceleration a C acceleration rus Ra square.

[00:41:30]Next question.

[00:41:35]Equation of motion in case. Equation of motion gravity vector vector vector = * sin rs of sin r + cos.

[00:42:05]Sorry.

[00:42:20]This is the gravity component.

[00:42:26]And you finally got the last RUS R dot sorry R dot 2 R dot 5 dot equal = - cos here at t =0 r =0 at a condition t = = r = = = 5 = 90°= clear and R dot become an angle.

[00:43:52]add sorry this ang= u² sin theta alpha sin theta + alpha / g cos + square. Is it clear to all of you?

[00:44:27]Is it clear?

[00:44:32]Okay.

[00:44:42]For maximum for maximum rate for max rate for maximum rates theta becomes 45° 45 that theta velocity.

[00:45:38]Okay.

[00:45:42]Only the horizontal remains only.

[00:45:46]Okay.

[00:45:52]at the highest at the highest point only the horizontal energy remains same. If any of them is square= to/2 g = Z become zero. G becomes to Z.

[00:47:16]Next.

[00:47:20]Excellent.

[00:47:38]Clear.

[00:47:45]Clear everyone.

[00:48:04]This is the approach.

[00:48:07]This is the relative velocity.

[00:48:14]Yes.

[00:48:20]This is P1. This is P. Yes.

[00:48:33]U velocity velocity velocity Just Vlog So nothing of a respect with respect to velocity of respect and VA is equals to VA plus VC.

[00:49:56]Same concept.

[00:50:05]concept.

[00:50:07]Next closest question.

[00:50:40]VW is nothing but the velocity of load in steel water with respect to and velocity. VWR is the velocity of water with respect to the river bank. VBR velocity of respect to the ground. Then we can write VB R is equals to VW plus V. Clearly for the motion along the line the angle is sin inverse of theta equals to sin inverse of PWR by P.

[00:51:24]Clear. Okay.

[00:51:29]Sorry. This is diamet.

[00:52:08]velocity r= rn notation Expect rain appears to fall vertically to the manicity.

[00:52:48]Two particles of bear. Relative position vector remains constant. Relative part actually approach.

[00:53:27]formula that is V R V sorry V A B cross V A B R V A R A D closest we are here we Uh only projectile motion and same gravity change.

[00:54:44]So A1 2 vector A1 same next condition it compen as R* of E Rap R plus theta theta dot into E power theta I cross Vlog.

[00:56:14]The first question.

[00:56:20]So please do it fast.

[00:56:47]Omega is constant. Which of the following particle is moving to this?

[00:57:07]I do velocity bar cos omega t i + b sin omega t r omega minus sin Omega t omega cosmelocity is this where omega is constant. Which of the followant is total energy of the particle?

[00:57:59]V² + cos square 1 omega omega is constant m² m omega² m² by omega half square cosme B Omega square the trajectory of the particle Next question.

[00:59:35]Next question.

[00:59:47]Interact live.

[01:00:23]A body of mass mically sorry at an angle theta then the horizontal magnitude of the angular momentum at = v Rizal velocity of theta and the magnitude of the structure = k into ru sin theta by U option C. Next.

[01:01:19]Next. Next.

[01:01:44]Fighting horizontal. Sorry.

[01:01:57]Second question.

[01:02:24]Everyone please The relative velocity of B with respect to A.

[01:02:51]Okay.

[01:02:56]60 opposite.

[01:03:10]Relative velocity of B. Relative velocity of B with respect to A with respect to velocity. Check for data types of Next attention is developed in this with both mass and the velocity speed moves with a speed of virtual is what is the gravitational acceleration this is the gravitational Next T cos theta= MV² by R cos theta = M sin theta = MV² by vertical and the blocks of the screen.

[01:05:02]square x² + cos theta.

[01:05:21]Next question.

[01:05:36]Okay.

[01:06:03]solution.

[01:06:15]Good.

[01:06:23]That what is this question? Projectile to projectile 45° plus 45.

[01:06:34]The ratio of the time of flight t= 2 u sin 45 t1 and it is 2 u sin 45 - ala correct t1 t2 is sin of 45 + 14 Sin alpha cos theta sin alpha cos theta + cos alpha sin theta / sin alpha cos theta - cosine here for cos alpha + sin alpha cos - two projectiles of same initial velocity Yes, two projectiles are projected with the same initial velocities at an angle 14 with respect to the horizontal ratio of their science. Yes.

[01:08:28]used by actually x and Equals of the projectile at the function of P4 = = 43.6 63 - 4.9 then the angle in the by the time by dx can be written as dy by dx into BTS.

[01:09:37]Solve solution for clear Sorry.

[01:09:56]Fine. Please present. Please not in English.

[01:10:01]Please note next.

[01:10:30]Sorry question.

[01:10:51]Sorry.

[01:11:10]And I don't need sorry that first second law f is equals DP by DP which is also nothing momentum T= into dy den or A star Gus A. Clear condition.

[01:13:11]Fore is the exhaust velocity. Next question.

[01:13:33]Next.

[01:13:42]They question the questions. question.

[01:14:11]First question on this screen.

[01:14:29]50 plus 30 C force required to push to a body of a rough surface uh is not while source Good.

[01:15:11]Basically, sorry.

[01:15:42]We have to solve the solution.

[01:16:13]Who's the next 9 p.m.

[01:17:01]Quick proper questions here.

[01:17:17]And of 15 plus 15 plus students watching student Next question.

[01:18:01]Next question.

[01:18:37]Huh?

[01:18:54]9 p.m. or the class complete mechanics.

[01:19:31]Complete mechanics are fluids.

[01:19:42]properties and thank you. Take care.

[01:20:18]I think both concept.

[01:20:36]Thank you.