jee advanced 2017 solution | A block of mass M has a circular cut with a frictionless surface as sho
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- Опубліковано 8 лют 2025
- A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x = 0 , in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct?
(A) The x component of displacement of the center of mass of the block M is: mR M+m
(B) The position of the point mass i_{s} / x = - sqrt(2) * (mR)/(M + m)
(C) The velocity of the point mass m s / v = sqrt((2gR)/(1 + m/M))
(D) The velocity of the block M is: V = - m/M * sqrt(2gR)
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Who came here as jee 2024 advance aspirant😊
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Sir u are amazing
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Sir why not kinetic energy initial =0 ?
It's is zero check again
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Why there is no work done by normal force ?(internal forces can also do work) and in this case magnitude of normal force is equal but due to it displacement of both of them is different
Nice question... We can say that internal forces can do work done on the cases where they are responsible for the motion like spring n two blocks or motion of two particle under there gravity but in this case gravity is responsible for motion
@AbsolutePhysicsbyAshwiniJha sir can we prove mathematically that work done by normal on system Is zero in this case?
I was not able to do it mathematically as normal vector was variable
Yes we can solve it but it will be very lengthy
Sir yaha ques m given h oi fixed table.. toh vo move kaise kr skta h
Table ismai M k niche vala surface ko bola gaya h (read 1 line carefully
Ok thank you sir