JEE Advanced 2017 | Physics Paper 1 | Question 2 Solution

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 position of the point mass m is: x = −√2 (mR)/(M+m) [B] The velocity of the point mass m is: v = √(2gR / (1 + m/M)) [C] The x component of displacement of the center of mass of the block M is: −mR/(M+m) [D] The velocity of the block M is: V = −(m/M)√(2gR) Kleppner and Kolenkow | Problem 4.4 :    • Kleppner and Kolenkow Problem 4.4 | An Int...