Hello sir .I am in class 12(2021batch) and want to start pathfinder now and I don't think I would be able to solve the entire book. So can you please suggest me which section( MCQ, build your understanding etc.)of pathfinder is important for jee advance.
first finish your regular assignments, previous year questions, selected IRODOV problems and then if you have time and feel confident enough then only start Pathfinder and you can target doing MCQ of each chapter first.
@@PhysicswithAkashGoyal Sir i have done jee relevant portion of irodov of whatever has been taught till now(in 11th)..I don't really have much time to solve another book so is doing pathfinder absolutely necessary or something? I go to Fiitjee and do 85%of their material.
Why gravitational potential does not came in account as disc move up with velocity v and reach some height h where its final velocity will become zero then there will be maximum extension
Sir in the first method, why haven't you considered the gravitational potential energy of the particle of mass m at the instant when the spring achieves maximum elongation???Shouldn't the potential energy of the particle be included in the equation of conservation of energy?
The system moves on a horizontal plane. And if you are talking about gravitational pe of the two point mass system in the question then it's very much negligible and should not be accounted if there is NO hint in the question
Sir can u plz tell how will the approach change if they were connected with a string and we were asked to find tension in the string during their subsequent motion
in such cases, there has to be non-conservative force provided otherwise if we ignore the non-conservative force effect, then the final velocity would be v0 and T=mv^2/R..
Sir, in first question "kx" will always be acting perpendicular to disc .. so work done by force to change v° will be zero.. so v° should not change? (ie v2=v° always)
Normally to find Maximum extension we equate spring energy at maximum extension with 1/2 * (mu) (Vrel)^2. this will give correct result only if velocities are along the line joining the particles. and at the time of maximum extension Vrel becomes zero. But here Vrel is not zero..... component of relative velocity is zero along line joining the particles but not zero perpendicular to it.
Sir its a blessing to study from you. Advanced questions don't seem tough anymore pls keep giving adv level questions for jee adv 2020
this playlist was the best..cm frame concept cleared thanks to you sir
Fabulous problem and solution💙COM frame analysis is too good.
OK
@@aryanshah17 OK
Bhaiya 💖💖💖
@@vedbrahmbhattiit-j4497 ok
@@siddhant7806ok
lovely stuff
best playlist on whole youtube for mee ........
Hello sir .I am in class 12(2021batch) and want to start pathfinder now and I don't think I would be able to solve the entire book. So can you please suggest me which section( MCQ, build your understanding etc.)of pathfinder is important for jee advance.
first finish your regular assignments, previous year questions, selected IRODOV problems and then if you have time and feel confident enough then only start Pathfinder and you can target doing MCQ of each chapter first.
@@PhysicswithAkashGoyal Sir i have done jee relevant portion of irodov of whatever has been taught till now(in 11th)..I don't really have much time to solve another book so is doing pathfinder absolutely necessary or something? I go to Fiitjee and do 85%of their material.
sir ji in com frame kinetic energy is 1/2(mu)vrel^2 but in reduce mass frame there is additional 1/2Mvcm^2 term
Any method without using angular momentum??
sir Why have you taken velocity along spring to be same for both blocks in the final state?
Why not -1/2 kx² as speing elongated
spring energy is always positive
Sir can we conserve angular momentum about a moving point?
if we are in its frame and we properly apply psuedo forces, then yes.
Why gravitational potential does not came in account as disc move up with velocity v and reach some height h where its final velocity will become zero then there will be maximum extension
Read carefully.
In the first line it is given that system is on a smooth horizontal floor and all the motions are along the surface.
Thanks sir awesome explanation 😀🔥👍🏻loved it 🙏
Sir in the first method, why haven't you considered the gravitational potential energy of the particle of mass m at the instant when the spring achieves maximum elongation???Shouldn't the potential energy of the particle be included in the equation of conservation of energy?
The system moves on a horizontal plane.
And if you are talking about gravitational pe of the two point mass system in the question then it's very much negligible and should not be accounted if there is NO hint in the question
@@PhysicswithAkashGoyal Got it. Thank you sir
@@Pratham._.Kamath even if u take the difference will be zero irrespective of the size of the body
Sir can u plz tell how will the approach change if they were connected with a string and we were asked to find tension in the string during their subsequent motion
in such cases, there has to be non-conservative force provided otherwise if we ignore the non-conservative force effect, then the final velocity would be v0 and T=mv^2/R..
just a centrifugal force equation.
sir plz upload questions realted to thermodynamics
aakash ji why cant we solve it via reduced mass method
because we do not know the final relative velocity ?
please help!!
Sir plz explain why did u add Kcm in kinetic energy term ? We will only use Kwrt Cm only 🙏😀plz clear my doubt sir🙂🙏
Sir, in first question "kx" will always be acting perpendicular to disc .. so work done by force to change v° will be zero.. so v° should not change?
(ie v2=v° always)
?
Spring is initially perpendicular to velocity Not after that... I have explained this in the solution...
@@chiragmaheshwari9769 brother its perpendicular intially and for max elongation it again has to get perpendicular
Sir why can't we use reduced mass directly here??
Normally to find Maximum extension we equate spring energy at maximum extension with 1/2 * (mu) (Vrel)^2. this will give correct result only if velocities are along the line joining the particles. and at the time of maximum extension Vrel becomes zero.
But here Vrel is not zero..... component of relative velocity is zero along line joining the particles but not zero perpendicular to it.
Best analysis from com.. 😍
sir i have a doubt that spring force is always perpendicular to velocity so why is velocity changing?
No it's not " always" perpendicular to Velocity. I have explained this in the video
Ok sir thank you
Very helpful thanks
OK
Thank you very much sir
Sir can u provide a proof for writing kinetic energy in COM frame as K_com+K_reduced mass
Already there in CM frame playlist
Thanks sir
thank you sir
Sir laws of motion also plzzz
Loved it
COM is love ❤ now 😋
very low volume!
Sir q no. 1.267 🙏🙏
revised
Solution quality is soooo good and audio quality is soooo bad 🤣🤣🤣🤣🤣
Balancing Act 😂.
I am thinking to re upload these videos having sound issues ☺️
@@PhysicswithAkashGoyal koi nahi sir..... your doing very nice job thore se earphones to ham le hi skte hain......
Very low voice.
Sorry for that... Plz use headphones
Yes its almost not audible
amazing stuff