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?

Options

A. The velocity of the point mass m is: `v = sqrt((2gR)/(1 + m/M))`
B. The velocity of the block M is: `v = -m/M sqrt(2 gR)`
C. The position of the point mass is: `x = -sqrt(2) (mR)/(M + m)`
D. The x component of displacement of the center of mass of the block M is: `- (mR)/(M + m)`

Answer

A, D

Explanation

Net external force acting on the system along the x-axis is zero.
`:. `Along the x-axis
Momentum is conserved `implies mv = MV`
From conservation of energy
Loss in GPE of particle of mass m = Gain in kinetic energy of both the masses.
`implies mgh = 1/2 mv^2 + 1/2 MV^2 implies mgh = 1/2 mv^2 + 1/2 M ((mv)/(M))^(2)`
`implies 2 gh = (1 + m/M)v^2 implies v = sqrt((2gh)/(1 + m/M))` hence option (A) is correct.
`V = m/M v = (m sqrt(2gh))/(sqrt(M(m+M)))` along - ve x -axis.
Hence option [B] is incorrect.
Since the location of center of mass does not change along the x-axis.
`m Delta x_(m//G) + M Deltax_(M//G) = 0 " "(x + 1)^(2) + (y + 2)^(2) = 1 " "m(R - x) - Mx = 0f`
`x = (mR)/(m + M)" " Deltax_(M//G) = (-mR)/(m + M)`
Hence option (D) is correct.
`Delta x_(m//G) = R - x = (MR)/(M + m)`
Final position of `m = -x = - (m R)/(m +M)`
Hence option (C) is incorrect.