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

Answer

A, B

Explanation

If speed of the block is V, then using conservation of linear momentum
`V=(mv)/(M)`
We can also use conservation of energy to write the following equation: `MgR=(1)/(2) mv^(2)+(1)/(2) M((mv)/(M))^(2)`
`rArr mgR=(1)/(2) mv^(2)(1+(m)/(M))`
`rArr v= sqrt((2gR)/(1+(m)/(M)))`
Hence option (a) is correct, When point mass slides down then block moves to the left. Displacement of point mass with respect to block is R towards right. Let block moves by an amount x towards left then net displacement of point mass becomes (R - x) towards right. Since displacement of centre of mass is zero hence we can write the following:
`Mx=x(R-x)`
`rArr x=(mR)/(M+m)`
We can see that option() is also correct. Negative sign is applied because it has moved towards negative side of the origin.