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bhaiya ek doubt h ki agr ball b pe hum sirf vertical velocity hi manenge mtlb ki ball A drag krke ball B ke position m ajaye , mene yeh soch k solve kiya toh ENERGY CONSERVE , FORCE BALANCE yeh toh same agye h prr MOMENTUM IN X-DIRECTION nhi a parha h smjh . aap yeh bta do mera assumption vertical velocity ka galat h ?
u is not the velocity at any time of B, it is the initial velocity given to B. So, as B moves, its direction changes due to the tangential force applied by string... Hope this helps..
@@jeesimplified-subject and even if we consider the tension force.. the string remains taut all along so no distance covered against force.. so work wont be done either way
Got the ans root 3gl ! but i was sceptical as the length wasn't given. My method was that i considered an angle x where the centripetal force is equal to tension second eq made was cons of energy, similar to in que. Then centripetal force eq had cosx in it so found that angle to be around pi\6. Then got the. Ans
When B will start lifting, the angle between string and ground changes and Tension also changes as B decelerates and hence, Horizontal force on A = Tcosθ also changes so, it'll be a double variable force.
exact same method.. and i did it be myself(not the kind of person who doesnt know anything but rocks the comment section saying 5 min mein ho gaya).. yeh waala genuinely 5-10 mins mein ho gaya and mostly because of intuition.. by the way you should have given the length of the string in the problem statement atleast should have mentioned it.. because there is always a possibility that the answer could be taken out without the length of the string and then to verify our answer we have to look at your answer and if our answer is wrong and we see your answer it erases the whole point of the question.. so please mention all the parameters in the question beforehand
Friction hota to energy conservation wali equation mein friction ka work done ko consider krna hota , also momentum in x direction will not be conseerve
Since we have velocity in tangential direction, centripetal force(tension) and no forces like friction, why not take it in circular motion? Also when the circular motion becomes invalid, we know it must have happened due to movement of a. So here we have the favourable situation.
It shud remain taut always cuz the COM_x (locus of COM will be a vertical line) cannot be touched, so the balls will never come towards each other, else the ball A wud have already lost contact. (he cud have mentioned these details in the video)
and by this u'll get that velocities of balls shudn't have a component along the line joining AB hence u may take a line passing thru A (perpendicular to the plane) as the instantaneous axis of rotation.
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Just do T=mv²/L=mg and use energy conservation. Solve in 15 seconds with this method
Nice... Did the same...
Explain your physics behind the first equation !
Same problem but with given Data came in our Test[PW]
11th
bhaiya ek doubt h ki agr ball b pe hum sirf vertical velocity hi manenge mtlb ki ball A drag krke ball B ke position m ajaye , mene yeh soch k solve kiya toh ENERGY CONSERVE , FORCE BALANCE yeh toh same agye h prr MOMENTUM IN X-DIRECTION nhi a parha h smjh . aap yeh bta do mera assumption vertical velocity ka galat h ?
u is not the velocity at any time of B, it is the initial velocity given to B. So, as B moves, its direction changes due to the tangential force applied by string...
Hope this helps..
How do you justify the tension here is a non conservative force . Tension is not always conservative.
Tension is an internal force here
@@jeesimplified-subject and even if we consider the tension force.. the string remains taut all along so no distance covered against force.. so work wont be done either way
Got the ans root 3gl ! but i was sceptical as the length wasn't given. My method was that i considered an angle x where the centripetal force is equal to tension second eq made was cons of energy, similar to in que. Then centripetal force eq had cosx in it so found that angle to be around pi\6. Then got the. Ans
When B will start lifting, the angle between string and ground changes and Tension also changes as B decelerates and hence, Horizontal force on A = Tcosθ also changes so, it'll be a double variable force.
exact same method.. and i did it be myself(not the kind of person who doesnt know anything but rocks the comment section saying 5 min mein ho gaya).. yeh waala genuinely 5-10 mins mein ho gaya and mostly because of intuition.. by the way you should have given the length of the string in the problem statement atleast should have mentioned it.. because there is always a possibility that the answer could be taken out without the length of the string and then to verify our answer we have to look at your answer and if our answer is wrong and we see your answer it erases the whole point of the question.. so please mention all the parameters in the question beforehand
IDK about being an educator but you are definitely going to be a champ entrepreneur
I have solved this question when you have posted and I got it right
Good, keep grinding
tq so much bro
Bhaiya agar ground pe friction hota to kya ise mai rolling ke sath compare kar pata?
Friction hota to energy conservation wali equation mein friction ka work done ko consider krna hota , also momentum in x direction will not be conseerve
Yes bhaiyaa❤
0:57 bhaiya ye irodov ka sawaal hai mene bana rkha hai
Bhaiya pathfinder ke solution k lie koi series start krdo
Not a series , but for a structured videos of such level questions you can join Set of 60
Observe.adapt.excel.
Why B is moving in circular path and A as its centre , since A is not fixed ?
Circular motion is with respect to each other not with respect to ground
Since we have velocity in tangential direction, centripetal force(tension) and no forces like friction, why not take it in circular motion? Also when the circular motion becomes invalid, we know it must have happened due to movement of a. So here we have the favourable situation.
i used T= mu^2/r-2mg+3mgcostheta and put T=mg.but got root 6gl
Did you understand your mistake then ?
Because A is not fixed
My approach was similar but I stuck on one thing. Length of string is not given in question so how can we consider it in final answer?
Basically usko as parameter consider krle uss form mein answer nikal skte ho
Bhaiya how would you find the net workdone by tension over here
Dono balls pr individually Work Energy Theorem apply kro, you will work done by Tension
lekin bhaiya agr question he smjh na ayya ...mtlb uski language aur data to kya kareee?????
plss reply
Dm me on telegram
bhaiya but why will block A have a velocity? we should consider it to be still.
Then momentum in x direction will not conserve
No string length given and mentioned that the string will remain taut throughout the motion or not??
It shud remain taut always cuz the COM_x (locus of COM will be a vertical line) cannot be touched, so the balls will never come towards each other, else the ball A wud have already lost contact. (he cud have mentioned these details in the video)
and by this u'll get that velocities of balls shudn't have a component along the line joining AB hence u may take a line passing thru A (perpendicular to the plane) as the instantaneous axis of rotation.
How is U' = 0.
Couldnt understand
There is no non conservative ya koi external force jiska koi work hai, so mechanical energy will conserve
Bro helpful video thanks❤️❤️
Bhaiya set of 60 mei live sessions one on one hote hai?
Not officially, but you can always connect with me and Pratham.
Answer in 2 root(gl)?
Nope, you made a mistake here
Thanks bro
1st