JEE Advanced 2017 Paper 1

Section: Physics

Question : 1 of 54

Marks: +1, -0

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?

[JEE Adv 2017 P1]

The velocity of the point mass m is: v=√
2gR
1+
m
M
The velocity of the block M is : V=−
m
2
m√2gR
The position of the point mass is :x=−√2
mR
M+m
The x component of displacement of the center of mass of the block M is : −
mR
M+m

Validate

Go to Question:

Next Question