I actually equated the PM distance ,means the (root R^2w^2+v^2/2 )*v/root 2 *g =PM .PM is calculated from traingle law PM/sin wT =R/sin (180-wT)/2 ,Where T is v/root 2 *g
Block Strategy Revision for JEE Advanced 2023 - th-cam.com/video/WoGmb5b4kQU/w-d-xo.html Best Practice Book of 700+ Advance Illustration for physical situations on Physics Problems. Check at - amzn.to/3qyIK8H Best Chapterwise PYQ Book of JEE Advanced with detailed comprehensive solutions. Check at : amzn.to/3ZXotZs Download academic resources for JEE Mains & Advanced 2023 from PG Telegram - t.me/physicsgalaxyworld For best online courses on JEE & NEET, download Physics Galaxy mobile app. Download link : bit.ly/PhysicsGalaxyApp
Sir, one doubt, should i do my coaching material before or after the Advanced Illustration and PYQs, or should i leave it and focus only on Adv Illustration and PYQs? Please help Sir.....🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
But sir isme to solution galat ho gaya na. Agar boy ko ball catch karni he to uske liye ball ko wapas land bhi to karna padega. Matlab ball ki to resultant veloicity ki range he usko exact circumference pe khatam bhi to hona padega. aisa bhi to ho sakta he ki ball circumference tak pahuchne se pehle he land kar jae. To sir mai finally keh raha tha ki isme ek aur eqaution banegi jo Resultant velocity ko uski projectile ki direction me karegi => 2 * tan⁻¹(v/✓2Rw) = theta
Sir concept matched...could think till this level ...after mains was feeling like I lacked practice but now thanks to you I gained confidence that i do not need to solve many ques...just focus on the concepts and try to do the same question with multiple approaches which would be more beneficial.
I reached upto equating time, missed: Calculating the value of θ using Resultant horizontal velocity: Lesson: Always draw neat and to the scale diagram taaki resultant nikalna naturally aajaye yaad
I was able to think most part about it the mistake i made tht i thought it will catcht the ball at other point of circumference i mean I considered theta 180 degree and GOT wrong thank you sir for this amazing perspective!!
Thanks you so much sir motivation through this question this question clicked me conceptually and iam able to solve instead of low performance in mains
Once sir mentioned the word "3D" the question seemed pretty obvious to me and I got the procedure....I lagged behind in taking theta in terms of velocity of projection and the radius
sir i was able to get the idea of that resultant velocity thing but can't get into PM and time of flight part. I'll try my best to evolve my thinking. Thanks for showing us the approach towards jee adv qs. Edit : PM = range of the projectile
why resultant velocity is not included in time of flight it seems that tangential velocity only affect range and direction of projectile not time of flight
9:02 sir Person is in touch with table which means he doing circular motion also that's why you added linear velocity due to circular motion but the ball is in air and have only projection velocity not Circular velocity which cause ball to move towards centre only, it means person can catch only ball opposite side of daimeter. WHY YOU ADDED CIRCULAR VELOCITY TO BALL ALSO. I MAY WRONG?PLZ EXPLAIN SIR? How ball also has rw velocity?
@@ninja_42 yes, boy do circular motion and have linear velocity with wr but when you throw the ball in air it was not follow you because it has not circular velocity. It has velocity towards centre and can you only catch the ball when you reach the just opposite the daimeter means you travel distance of πr(half circumference ( ball must have range range 2R)
Thank you sir, really a nice concept..... although I didn't solve on notebook and pen, I could figure out the steps of solving....just didn't form the equations.
Sir can't we just simply use concept of relative motion Like at t=0 and t=final the displacement of ball wrt boy should be 0 For ball by vector laws components along x axis we have Rw of ball and -Rw of boy which cancels out In y axis we have vcos45 and deacceleration of -w²r since boy is moving with centripetal acceleration which will be parted to ball in his frame in -y axis In z axis we have vsin45 and g We don't care about the z axis but in y axis the displacement should be 0 so by 2nd equation of motion we can find time of 0 displacement which is √2v/w²r By this time we can calculate theta and other things
Sir I was able to get the second question exactly...till the last expression...just for that angle I calculated it using the angle between the velocity vectors...
Actually, for the boy to catch the ball, both of their x and y coordinates at time t must be identical. However, the boy's velocity in the negative y direction is always decreasing while the ball's vertical component of velocity is constant. This implies that the displacement in the negative y direction of the ball will forever be greater than that of the boy and hence, they can never be at the same position at time t which is not 0. I have mathematically analysed the result ashish sir derived in this video and can confidently say that the derived equation has no solution(v) for all real values of w and r which are constants in our equation.
But I think that when the ball will actually complete its range then automatically x and y coordinates will match to that of the person at the same time
You are correct When we look at in in only dimension (eyes pointing towards +ve x axis)we observe that always the ball will be ahead of the boy along -ve y. So they can never coincide
You are correct. I have not analysed it mathematically but I was able to think the same. I was searching such type of comment and found yours. Thankyou.
Exactly. I was just so confused how it could be even possible without having the boy _throw_ the ball in the +ve y direction. Without a throw in the +ve y direction the ball will always go farther than the boy in the y direction, hence never meeting at the same time and point in space after t=0. Thanks for writing the comment, I do not feel left out.
I thought entirely right...for one moment i calculated l as vh^2/g then realised upward speed is vsin45 but horizontal isnt vcos45 anymore so projection angle I think changed then went again to basics range=xspeed*timeofflight
Sir mene another method se keya h batna ki sahi h ya nahi Method: First let projectile towards the centre axis is X and downwards from top view is Y To mene Range = v²/g jo x coordinate ho gya or y= (2)½ wrv/g ho gya Ab ye point circle per hoga to centre se distance radius ke equal hogi so waha se equation bani Plz sir batna koi concpet missing to nahi h??
Sir you told in your previous videos that we need to follow finger technique and read questions and their solutions but you said in this video that doing pyqs and solutions will not help so sir can you give a little clarity on that. I am sorry if my doubt is a little silly I'm just not able to understand what's the right way to go about it.
@@omprakashnarayan1327 Yeah I understand that but suppose if I am using finger technique with the 700+ advanced illustration book then am I going right and if not which material am I supposed to refer to for reading with finger technique and learning.
But @Physics Galaxy sir isme to solution galat ho gaya na. Agar boy ko ball catch karni he to uske liye ball ko wapas land bhi to karna padega. Matlab ball ki to resultant veloicity ki range he usko exact circumference pe khatam bhi to hona padega. aisa bhi to ho sakta he ki ball circumference tak pahuchne se pehle he land kar jae. To sir mai finally keh raha tha ki isme ek aur eqaution banegi jo Resultant velocity ko uski projectile ki direction me karegi => 2 * tan⁻¹(v/✓2Rw) = theta
I was literally thinking the veryyy same thing like how in the earth it is possible to... I was writing this👈 but suddenly a thought came to my mind It's about frame... The thing that we are missing, you must try to look it from a distant position and you'll nottice that the ball will have RW velocity too... It's easy if uknow some part of inertial and non inertial frame...
@@Khokalaa Fuck U I hate increasing my comp. But okayyyyyy!! Imagine if you're traveling in train it will appear to you that you are at rest but you'll be seeing trees speeding or in simple words the world outside will see you as they are in rest with respect to earth and you in motion with respect to earth... Similarly if the projectile is at P and P rotating then P will also have the very same angular velocity... If you're saying that only person has that RW velocity and not the projectile then they won't be moving together and the projectile will be left behind...
if block A of 2kg is kept on block B of 4kg And block B is on a inclined plane making an angle of 30 degree from the horizontal if coefficient of friction between block A and Block B is u1=0.3 and the coefficient of friction between block B and Plane is u2=0.2, and a external force F along the incline is acting on block B(the external force F is the minimum force along the incline needed to make sure that block A does not slip over Block B). find the direction of friction force acting between block A and block B and also find magnitude of F
TQ SIR ..., MAINE ATTEMPT KARTE WAKTH CURCULAR KA DYAAN NA DENE KU MISTAKE KARDI THI .... SO , MADE THE BOY TO COVER SEMI-CIRCLE IN SAME TIME AS THE PROJECTILE WOULD LAND COVERING THE RANGE OF 2R ... LATER , ANALYSED MY MISTAKE... , ACHA LAGA BY VISUALIZING... !! TQ SIR , ♥️💪🔥
Sir, your question is wrong ❌❌❌❌as there will be no solution. Because the ball has wr velocity in downward direction at the same time boy has wr velocity but on circular path so at any given time the downward displacement of ball will be greater than boy, so he would never be able to catch the ball. A correction can be made if he throws the ball by aiming at a point which is above the center O as then the downward velocity would decrease and the boy can catch the ball.
you could look at this question in another way from the person's perspective and think of the disc to be stationary. then the person looks to throw the ball not at the centre but somewhere at the circumference to another person. and the v(h) as sir wrote it would be the only velocity and not a resultant one. the angle of inclination of the projectile would be same (45 degrees).
sir i was able to visualise the direction of velocity of the boy and the ball but failed to equate the time of flight. watching this video gained me motivation to try harder, thank you sir.
Sir should we not take the centripetal acceleration the ball experiences in consideration to find the range You multiplied the resultant velocity with time simply
i was able to grab the theory of the question but now i am curious as this is an interesting question. now i am this train of thought : if we throw the ball towards the centre with enough velocity and the moment we throw the ball, the horizontal disc starts to accelerate, would it be possible to catch the ball diametrically. i solved theta for 180 degrees in this equation and as far as my math took me, it was not possible( for constant omega).
Hmm i guess you can if the ball was thrown just before disc started rotating and then disc had enough linear velocity for boy to reach and catch on time.. But if it was thrown exactly whe disc started rotating, you migh get what you want if the ball is thrown with rw in the opposite direction of rotation, making a required angle in that direction to cancel out velocity component due to rotating disc
@@snehapeter4686 hmm i get what you are trying to say. If you throw the ball to catch it diametrically while the disc is rotating. the trajectory of the ball might me similar to how the sun rotates the earth wrt us. not a perfectly vertical parabola, a slighty titlted one.
Sir maine angles ka use Kiya sirf or mera xota soln aaya Velocity vector ka angle v/√2 se @ mana or tan @ nikala Phir circle ke centre pr angle π -2@ substend hua jab ball catch hui or same time me boy ne π-2@ angle ko cover kia So π-2@=wt => t=√2v/g tan@=√2Rw/v 2(π/2 -2@)=√2vw/g 2arccot√2Rw/v=√2vw/g Finally v= √2 Rw tan(w v/√2g)
Sir I was able to solve the question correctly, just a silly mistake, even I had taken it as cos instead of sign. But I was able to solve it completely thank you sir!
Sir my approach is different i am equating displacement in x and z direction and qeta equal to wt and finding t from z direction dispalacement AM I CORRECT
Sir, can we solve the second question in the following way? As triangle PMO is isosceles, angle OPM = 90 - theta/2 So, tan (90 - theta/2) = Rw√2 / v Where theta = w × time of flight of ball. Solving this, I got : tan (wv/g√2) = v/Rw√2
Sir correct me if i am wrong but if we take that man rotated by an angle theta and then write his x and y direction displacement in terms of Rcostheta and Rsintheta then correlating these displacements with the displacement of our projectile as final coordinate will be same , we can then use sin2theta + cos2theta = 1 And find the value of v in proper terms of w , R and g .
Exactly my method We equate x and y displacements of ball and man and then get t from time of flight But here problem is we have more equation(3) than variable(2-time of flight and v) Hence it is never possible to catch the ball
Can this be done by keeping the disc in rest and applying the velocity and relative acceleration. As in this case we will get velocity only along y and x axis and acceleration in -x axis ( the centripetal acceleration in reverse direction applied to the ball). In this way it will simply become standard projectile question but with horizontal component of acceleration too.
why resultant velocity is not included in time of flight it seems that tangential velocity only affect range and direction of projectile not time of flight🤔🤔🤔🤔🤔🤔🤔
Can anyone tell that advanced illustration uploaded on channel and 700+ jee advanced illustration book, are both having the same questions amd illustrations?
Hy sir, Sir mane is question mai ball ki xy plane mai velocity ke do components leiya or point p ka net displacement liya in time T or fir ball ke displacement se equate karke mera expression for v is (2w2R2 +2Rg)’0.5 aagya plz tell is it right
sirrrr but previously you had told that we should follow d PYQ analysis book by finger reading method but in this video you are telling that just doing that is not usefull please tell properly which method to follow thank you
Question is based on a flawed logic that i realised while reading, I will attach another person's comment to convey: Actually, for the boy to catch the ball, both of their x and y coordinates at time t must be identical. However, the boy's velocity in the negative y direction is always decreasing while the ball's vertical component of velocity is constant. This implies that the displacement in the negative y direction of the ball will forever be greater than that of the boy and hence, they can never be at the same position at time t which is not 0. I have mathematically analysed the result ashish sir derived in this video and can confidently say that the derived equation has no solution(v) for all real values of w and r which are constants in our equation.
Sir i am prapring with my fill energy sir par advance ka form fee hi 2900 hai so price dekh k sir but mere 87.5 bane they pahle attempt me 2 month ke gap me major familiar issue me padh nhi paye drop ka v soch rhe bcz sir form feee v dal de and result v na bane ( my syllabus in physic 85 % and chem 30 and in math 40 cokplete hai and phy chem wale chapter ke accha grip hai unme sir kya kru Mai
I have a doubt and any sort of assistance and constructive criticism will be appreciated. How is it even possible that the boy catches the thrown ball? Considering the plane of the board as a plane of a compass, as the ball is thrown, due to the translational velocity which has towards the south, it moves south a certain distance during this flight. The southward velocity of the ball remains constant. in the same time interval, P has to move a greater distance than the ball(in the south direction) whilst moving with the same speed as the ball to reach the point where the ball falls.. in such a situation, how is it even possible that P and the ball reach that point at the same time??
Sir, can we add another constraint, Since the ball falls on the disc the final coordinates(assuming P as the origin) must lie on the circle - ( x - R)²+ y² = R²? Since this doesn't take into consideration that the boy catches the ball back, is this another condition the velocity should satisfy?
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i only got 90percentile in mains but still i was able to solve this question on my own ..feeling confident now .thank you sir
Same bro
same 87%ile but got the expression
😂😂😂
Bro ghar me question ban ja rhe the exam me same question nhi ban rhe the 🥲
@JEE
Sir I am glad to tell you that because of your mentoring and your tips I've been able to make it to IIT MADRAS in JEE Adv. 2022
Kese Kiya aapne
To phir se kyu aya time pass karne😂😂😂😂😂😂
@@Saurabh.S898vai ye sab juthe ha😂
@@Saurabh.S898 matlab? Kya tum ye khena chahte ho ki iitians ek min bhi waste nhi karte din ka?
@@user-xg1jv4rw7e faltu ke thing me to nehi karte
Was able to solve both the questions, feeling 💪💪💪
I actually equated the PM distance ,means the (root R^2w^2+v^2/2 )*v/root 2 *g =PM .PM is calculated from traingle law PM/sin wT =R/sin (180-wT)/2 ,Where T is v/root 2 *g
Block Strategy Revision for JEE Advanced 2023 - th-cam.com/video/WoGmb5b4kQU/w-d-xo.html
Best Practice Book of 700+ Advance Illustration for physical situations on Physics Problems. Check at - amzn.to/3qyIK8H
Best Chapterwise PYQ Book of JEE Advanced with detailed comprehensive solutions. Check at : amzn.to/3ZXotZs
Download academic resources for JEE Mains & Advanced 2023 from PG Telegram - t.me/physicsgalaxyworld
For best online courses on JEE & NEET, download Physics Galaxy mobile app. Download link : bit.ly/PhysicsGalaxyApp
Sir, one doubt, should i do my coaching material before or after the Advanced Illustration and PYQs, or should i leave it and focus only on Adv Illustration and PYQs?
Please help Sir.....🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Sir can we do it with shm
By taking two projection in x, y axis
But sir isme to solution galat ho gaya na.
Agar boy ko ball catch karni he to uske liye ball ko wapas land bhi to karna padega.
Matlab ball ki to resultant veloicity ki range he usko exact circumference pe khatam bhi to hona padega.
aisa bhi to ho sakta he ki ball circumference tak pahuchne se pehle he land kar jae.
To sir mai finally keh raha tha ki isme ek aur eqaution banegi jo Resultant velocity ko uski projectile ki direction me karegi
=> 2 * tan⁻¹(v/✓2Rw) = theta
Here we can also use cosine rule to derive the expression
Cos(theta)= (2r^2 -l^2)÷2r^2
L=range
Theta=wt w angular velocity
t time of flight
wonderful
sir it was extremely helpful in getting to know how actually we need to develop otbt thinking. please keep posting these videos.
Sir concept matched...could think till this level ...after mains was feeling like I lacked practice but now thanks to you I gained confidence that i do not need to solve many ques...just focus on the concepts and try to do the same question with multiple approaches which would be more beneficial.
Whats your jee mains score
Hogya iit?
Hogya iit?
I feel lucky to have a mentor like you,
At the same time i feel jealous that it is available on youtube and it is increasing competition day by day.
I reached upto equating time, missed: Calculating the value of θ using Resultant horizontal velocity:
Lesson: Always draw neat and to the scale diagram taaki resultant nikalna naturally aajaye yaad
I was able to think most part about it
the mistake i made tht i thought it will catcht the ball at other point of circumference i mean I considered theta 180 degree and GOT wrong
thank you sir for this amazing perspective!!
Thanks you so much sir motivation through this question this question clicked me conceptually and iam able to solve instead of low performance in mains
Keep calm and cool
Once sir mentioned the word "3D" the question seemed pretty obvious to me and I got the procedure....I lagged behind in taking theta in terms of velocity of projection and the radius
Har baar ashish sir ko dekh ke ek happiness ki feeling aati hai 😊
Feeling good sir i catched the thought for the problem when you gave the time all Because of you sir
sir i was able to get the idea of that resultant velocity thing but can't get into PM and time of flight part. I'll try my best to evolve my thinking. Thanks for showing us the approach towards jee adv qs.
Edit : PM = range of the projectile
why resultant velocity is not included in time of flight it seems that tangential velocity only affect range and direction of projectile not time of flight
9:02 sir Person is in touch with table which means he doing circular motion also that's why you added linear velocity due to circular motion but the ball is in air and have only projection velocity not Circular velocity which cause ball to move towards centre only, it means person can catch only ball opposite side of daimeter. WHY YOU ADDED CIRCULAR VELOCITY TO BALL ALSO.
I MAY WRONG?PLZ EXPLAIN SIR?
How ball also has rw velocity?
Yes, sir table is doing circular motion not the ball...why you added linear velocity due to circular motion in ball.?
Ig the disc was already rotating before the boy even threw the ball, so just imagine yourself instead of the boy and it actually does make sense
@@ninja_42 yes, boy do circular motion and have linear velocity with wr but when you throw the ball in air it was not follow you because it has not circular velocity. It has velocity towards centre and can you only catch the ball when you reach the just opposite the daimeter means you travel distance of πr(half circumference ( ball must have range range 2R)
Yes sir plz reply 😢
@@ninja_42 do you understand what I am asking? What is my doubt?😢
Thanks for all your efforts sir. I will definately get into iit this year ❤️
how?
you studied from where
Kaha ho bhai
@@Abnik62 lpu🤣
Thank you sir, really a nice concept..... although I didn't solve on notebook and pen, I could figure out the steps of solving....just didn't form the equations.
Sir can't we just simply use concept of relative motion
Like at t=0 and t=final the displacement of ball wrt boy should be 0
For ball
by vector laws components along x axis we have Rw of ball and -Rw of boy which cancels out
In y axis we have vcos45 and deacceleration of -w²r since boy is moving with centripetal acceleration which will be parted to ball in his frame in -y axis
In z axis we have vsin45 and g
We don't care about the z axis but in y axis the displacement should be 0 so by 2nd equation of motion we can find time of 0 displacement which is √2v/w²r
By this time we can calculate theta and other things
Sir I was able to get the second question exactly...till the last expression...just for that angle I calculated it using the angle between the velocity vectors...
Actually, for the boy to catch the ball, both of their x and y coordinates at time t must be identical. However, the boy's velocity in the negative y direction is always decreasing while the ball's vertical component of velocity is constant. This implies that the displacement in the negative y direction of the ball will forever be greater than that of the boy and hence, they can never be at the same position at time t which is not 0. I have mathematically analysed the result ashish sir derived in this video and can confidently say that the derived equation has no solution(v) for all real values of w and r which are constants in our equation.
But I think that when the ball will actually complete its range then automatically x and y coordinates will match to that of the person at the same time
You are correct
When we look at in in only dimension (eyes pointing towards +ve x axis)we observe that always the ball will be ahead of the boy along -ve y.
So they can never coincide
You are correct. I have not analysed it mathematically but I was able to think the same. I was searching such type of comment and found yours. Thankyou.
Exactly. I was just so confused how it could be even possible without having the boy _throw_ the ball in the +ve y direction. Without a throw in the +ve y direction the ball will always go farther than the boy in the y direction, hence never meeting at the same time and point in space after t=0. Thanks for writing the comment, I do not feel left out.
@@darshan5044 actually ball ke pass Rw velocity hogi na in -ve direction kyunki initially ball bhi ghum rahi thi
I thought entirely right...for one moment i calculated l as vh^2/g then realised upward speed is vsin45 but horizontal isnt vcos45 anymore so projection angle I think changed then went again to basics range=xspeed*timeofflight
Sir i solved the whole question by myself thanks sirr feeling confident now
Please also guide something about chemistry and maths also. It would be a huge help.
Follow block strategy video for fusion revision for jee advanced 2023
Sir mene another method se keya h batna ki sahi h ya nahi
Method:
First let projectile towards the centre axis is X and downwards from top view is Y
To mene Range = v²/g jo x coordinate ho gya or y= (2)½ wrv/g ho gya
Ab ye point circle per hoga to centre se distance radius ke equal hogi so waha se equation bani
Plz sir batna koi concpet missing to nahi h??
Sir same doubt
Thank you sir
You are the only trust for us!!❤
Sir centrifugal force bhi to lag rahi h?????
Sir you told in your previous videos that we need to follow finger technique and read questions and their solutions but you said in this video that doing pyqs and solutions will not help so sir can you give a little clarity on that.
I am sorry if my doubt is a little silly I'm just not able to understand what's the right way to go about it.
Try to extract concepts from pyq and try another ways to do same questions 🙂
@@omprakashnarayan1327 Yeah I understand that but suppose if I am using finger technique with the 700+ advanced illustration book then am I going right and if not which material am I supposed to refer to for reading with finger technique and learning.
But
@Physics Galaxy
sir isme to solution galat ho gaya na.
Agar boy ko ball catch karni he to uske liye ball ko wapas land bhi to karna padega.
Matlab ball ki to resultant veloicity ki range he usko exact circumference pe khatam bhi to hona padega.
aisa bhi to ho sakta he ki ball circumference tak pahuchne se pehle he land kar jae.
To sir mai finally keh raha tha ki isme ek aur eqaution banegi jo Resultant velocity ko uski projectile ki direction me karegi
=> 2 * tan⁻¹(v/✓2Rw) = theta
In the last question, taking angle OPM as theta , we can get something more definitive, since tantheta = root2 wr/v , then angle POM is -tan 2theta
I was literally thinking the veryyy same thing like how in the earth it is possible to... I was writing this👈 but suddenly a thought came to my mind
It's about frame... The thing that we are missing, you must try to look it from a distant position and you'll nottice that the ball will have RW velocity too... It's easy if uknow some part of inertial and non inertial frame...
How ball has rw velocity? Plz explain?🥲
@@Khokalaa
Fuck U I hate increasing my comp. But okayyyyyy!!
Imagine if you're traveling in train it will appear to you that you are at rest but you'll be seeing trees speeding or in simple words the world outside will see you as they are in rest with respect to earth and you in motion with respect to earth... Similarly if the projectile is at P and P rotating then P will also have the very same angular velocity... If you're saying that only person has that RW velocity and not the projectile then they won't be moving together and the projectile will be left behind...
if block A of 2kg is kept on block B of 4kg And block B is on a inclined plane making an angle of 30 degree from the horizontal if coefficient of friction between block A and Block B is u1=0.3 and the coefficient of friction between block B and Plane is u2=0.2, and a external force F along the incline is acting on block B(the external force F is the minimum force along the incline needed to make sure that block A does not slip over Block B). find the direction of friction force acting between block A and block B and also find magnitude of F
I was able to thing about equating time of both motions but i was not doing projectile motion vector analysis!
Sir why acceleration due to centrifugal force is not considered in the 2nd case??
thank you sir i actually solved the question on my own and feel great about myself for it!!
TQ SIR ...,
MAINE ATTEMPT KARTE WAKTH CURCULAR KA DYAAN NA DENE KU MISTAKE KARDI THI .... SO , MADE THE BOY TO COVER SEMI-CIRCLE IN SAME TIME AS THE PROJECTILE WOULD LAND COVERING THE RANGE OF 2R ...
LATER , ANALYSED MY MISTAKE... , ACHA LAGA BY VISUALIZING... !!
TQ SIR , ♥️💪🔥
Sir honestly I will tell.
I could only think till Vh (net resultant) of Horizontal component Vcosθ and Rω. 😭😭🥺. Couldn't think beyond it.
try to think more and more on framing new such questions from PYQs... it will help in building your thinking...
@@physicsgalaxyworld Thank you sir ☺️.
@@physicsgalaxyworld Sir yesterday I have ordered PYQ analysis by Physics Galaxy
Sir, your question is wrong ❌❌❌❌as there will be no solution. Because the ball has wr velocity in downward direction at the same time boy has wr velocity but on circular path so at any given time the downward displacement of ball will be greater than boy, so he would never be able to catch the ball.
A correction can be made if he throws the ball by aiming at a point which is above the center O as then the downward velocity would decrease and the boy can catch the ball.
Sir I am able to find the exact expression for velocity, it can be done by using pythagoras theorem, darwing perpendicular from M on OP
you could look at this question in another way from the person's perspective and think of the disc to be stationary. then the person looks to throw the ball not at the centre but somewhere at the circumference to another person. and the v(h) as sir wrote it would be the only velocity and not a resultant one. the angle of inclination of the projectile would be same (45 degrees).
No. In person's frame what you've written is not correct because ball will experience Coriolis force in frame of disc... think again on this !
@@physicsgalaxyworld noted sir. thank you for correcting my mistake.
V=root(2Rg-2(R^2)(w^2))
If we take:
Rsin(wt)=Rwt
Rcos(wt)=R-tvcos(45°)
Square and add to get condition on v using t=2vsing(45°)/g
sir i was able to visualise the direction of velocity of the boy and the ball but failed to equate the time of flight. watching this video gained me motivation to try harder, thank you sir.
please make block strategy for JEE 2024
Sir should we not take the centripetal acceleration the ball experiences in consideration to find the range
You multiplied the resultant velocity with time simply
Exactly Same Approach by me sir. Literally same thought process while attempting.
sir, i got the full question.feeling confident now...
Sir , in the expression finding question the parameter R was not mentioned , but your expession has R parameter in it ?
Sir please bring more such kind of question solving videos this was really very helpfull
Sir visualization ek dam perfect tha mera same yahi kiya tha magar expression tak nhi pauch paya
Sir i was able to form the same expression sir. Thank you sir🙏🏻❤️
Thank You So Much Sir🙏❤️
Solved on my own,
Derive the same expression in the reference frame of boy.
i was able to grab the theory of the question but now i am curious as this is an interesting question. now i am this train of thought :
if we throw the ball towards the centre with enough velocity and the moment we throw the ball, the horizontal disc starts to accelerate, would it be possible to catch the ball diametrically. i solved theta for 180 degrees in this equation and as far as my math took me, it was not possible( for constant omega).
same bro I took theta 180 degree (got wrong)
then reconsidered to some small theta !!
Hmm i guess you can if the ball was thrown just before disc started rotating and then disc had enough linear velocity for boy to reach and catch on time.. But if it was thrown exactly whe disc started rotating, you migh get what you want if the ball is thrown with rw in the opposite direction of rotation, making a required angle in that direction to cancel out velocity component due to rotating disc
@@YouNight69 Same bro! And I got v = g pi/ root(2) w 😅
@@snehapeter4686 hmm i get what you are trying to say. If you throw the ball to catch it diametrically while the disc is rotating. the trajectory of the ball might me similar to how the sun rotates the earth wrt us. not a perfectly vertical parabola, a slighty titlted one.
Sir, my doubt is that in rotating disk the velocity on ball due to rotation will act in opposite direction due to inertia......wright or wrong??
Thanks Sir I thought same way Full confident 🙏🏻
Sir maine angles ka use Kiya sirf or mera xota soln aaya
Velocity vector ka angle v/√2 se @ mana or tan @ nikala
Phir circle ke centre pr angle π -2@ substend hua jab ball catch hui or same time me boy ne π-2@ angle ko cover kia
So
π-2@=wt =>
t=√2v/g
tan@=√2Rw/v
2(π/2 -2@)=√2vw/g
2arccot√2Rw/v=√2vw/g
Finally
v= √2 Rw tan(w v/√2g)
Sir equating angles are more convenient then equating distances
Inequality se kese solve kare. Like range cannot exceed 2R. Distance travelled by rw velocity cannot exceed R etc.
This video tells that , still work is needed for advance...........
I have never seen that pyq but still got the right answer😊
Sir when you will start jee advance 2023 series? Please tell
Yes I really think this 😊
Sir have you assumed that the resultant velocity of ball is sufficient to make it land on the circumfrence and not inside or outside the circle?
Sir I was able to solve the question correctly, just a silly mistake, even I had taken it as cos instead of sign. But I was able to solve it completely thank you sir!
Can we do this in this way
Along line OP
R-RCos(wRt)=Vt/root2
And then Rearrange for V???
Anyone Please Tell🙏
Sir my approach is different i am equating displacement in x and z direction and qeta equal to wt and finding t from z direction dispalacement
AM I CORRECT
sir need your help in chemistry and maths pls sir your experience required a lot.....I think you would guide us ......thanku
Sir I got the answer , it took 8 minutes of thinking 😫
Sir, can we solve the second question in the following way?
As triangle PMO is isosceles, angle OPM = 90 - theta/2
So, tan (90 - theta/2) = Rw√2 / v
Where theta = w × time of flight of ball.
Solving this, I got : tan (wv/g√2) = v/Rw√2
Sir correct me if i am wrong but if we take that man rotated by an angle theta and then write his x and y direction displacement in terms of Rcostheta and Rsintheta then correlating these displacements with the displacement of our projectile as final coordinate will be same , we can then use sin2theta + cos2theta = 1
And find the value of v in proper terms of w , R and g .
Exactly my method
We equate x and y displacements of ball and man and then get t from time of flight
But here problem is we have more equation(3) than variable(2-time of flight and v)
Hence it is never possible to catch the ball
@@gourik1929 no but i am getting the answer, time of flight isnt a variable
Sir plzz give info about pys advance batch 😅😅😅....we are waiting eagerly and need it
What was Done by Pogi ? 6:22
Pogi ko uski salary time p nhi mili isliye usne apna Kam nhi kiya
Jiska naam hi pogi hai usse kya expect Kiya tha 😂
@@ankita5078 pogi k pass job to h karne ko. Tere pass to kuch karne ko bhi nhi h 😂😂 Don't hate pogi😠
@@ma.hinder_ 😔 sorry pogi. Kitna cute naam hai apka I luv you pogi ❤️ from my soul....
Chal be pogi asli I'd se aa😂😂
@@ankita5078 pogi guine h , agar naraz ho gya to jee main m fail kar dega 😂
Can this be done by keeping the disc in rest and applying the velocity and relative acceleration. As in this case we will get velocity only along y and x axis and acceleration in -x axis ( the centripetal acceleration in reverse direction applied to the ball). In this way it will simply become standard projectile question but with horizontal component of acceleration too.
Pogi rocking every lecture 🤣😂
Yes sir l am able to grasp the concept of this question.
why resultant velocity is not included in time of flight it seems that tangential velocity only affect range and direction of projectile not time of flight🤔🤔🤔🤔🤔🤔🤔
Can anyone tell that advanced illustration uploaded on channel and 700+ jee advanced illustration book, are both having the same questions amd illustrations?
Hy sir,
Sir mane is question mai ball ki xy plane mai velocity ke do components leiya or point p ka net displacement liya in time T or fir ball ke displacement se equate karke mera expression for v is (2w2R2 +2Rg)’0.5 aagya plz tell is it right
sirrrr but previously you had told that we should follow d PYQ analysis book by finger reading method but in this video you are telling that just doing that is not usefull please tell properly which method to follow thank you
Question is based on a flawed logic that i realised while reading, I will attach another person's comment to convey:
Actually, for the boy to catch the ball, both of their x and y coordinates at time t must be identical. However, the boy's velocity in the negative y direction is always decreasing while the ball's vertical component of velocity is constant. This implies that the displacement in the negative y direction of the ball will forever be greater than that of the boy and hence, they can never be at the same position at time t which is not 0. I have mathematically analysed the result ashish sir derived in this video and can confidently say that the derived equation has no solution(v) for all real values of w and r which are constants in our equation.
Sir i am prapring with my fill energy sir par advance ka form fee hi 2900 hai so price dekh k sir but mere 87.5 bane they pahle attempt me 2 month ke gap me major familiar issue me padh nhi paye drop ka v soch rhe bcz sir form feee v dal de and result v na bane ( my syllabus in physic 85 % and chem 30 and in math 40 cokplete hai and phy chem wale chapter ke accha grip hai unme sir kya kru Mai
sir for the first question, if both the bodies are undergoing projectile motion, it wont be deflected from its path right when in air ?
Sir I did half of the question then I thought that the question is wrong... actually I misunderstood that boy will remain on P
I have a doubt and any sort of assistance and constructive criticism will be appreciated. How is it even possible that the boy catches the thrown ball? Considering the plane of the board as a plane of a compass, as the ball is thrown, due to the translational velocity which has towards the south, it moves south a certain distance during this flight. The southward velocity of the ball remains constant. in the same time interval, P has to move a greater distance than the ball(in the south direction) whilst moving with the same speed as the ball to reach the point where the ball falls.. in such a situation, how is it even possible that P and the ball reach that point at the same time??
It is extraordinary respected sir
ashish arora sir op 🥰🥰
Sir i also thought of the same approach and even the equations written by you were same written by me
Thank you so much sir😊
Sir, Please make a video for guiding us 2025 JEE Aspirants as well...
🙂🙂🙂 confidence level badhta ja rha
concept matched !!
Sir please listen!!
I'm going to take drop this year and I'm confused to purchase which book of physics galaxy
So please help me to figure out!!
Take all 5 volume books and follow it chapterwise, I'm doing same
PLS tell my mistake i tried to resolve in components and made inequality on the basis of that, and got 2 cases for 1st half and 2nd half .
Love you sir you stand for us every time
Truly great sir👏👍👌
I was able to solve thnx sir
Sir, can we add another constraint,
Since the ball falls on the disc the final coordinates(assuming P as the origin) must lie on the circle - ( x - R)²+ y² = R²?
Since this doesn't take into consideration that the boy catches the ball back, is this another condition the velocity should satisfy?
Sir, glad to solve both the questions
Sir ,agar opposite end PE boy ball ko receive kar rha ha wo nh assume kar sakte ha
Sir my first questions answer is wrong but when you tell about these facts my second questions answer is absolutely right ❤