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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