JEE Advanced 2017 · Paper 1 · Physics Q2

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PhysicsMultiple correct+4 / -2Question 2 of 54

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 block M on a frictionless table with a quarter-circle cut of radius R at its top-right corner. The point mass m starts at the top of the cut (a distance R to the left of, and R above, the bottom end of the arc). The cut ends at the block's right edge, which is initially at x = 0; axes x (right) and y (up).

The position of the point mass

m

is:

x = -\sqrt{2}\,\dfrac{mR}{M+m}

The velocity of the point mass

m

is:

v = \sqrt{\dfrac{2gR}{1 + \frac{m}{M}}}

The

x

component of displacement of the center of mass of the block

M

is:

-\dfrac{mR}{M+m}

The velocity of the block

M

is:

V = -\dfrac{m}{M}\sqrt{2gR}

Answer and worked solution

Official answer: B, C

Conservation laws. There is no external horizontal force, and the surfaces are frictionless. When $m$ leaves the block at the bottom of the arc, both move horizontally:
$mv=MV,\qquad mgR=\tfrac12mv^2+\tfrac12MV^2.$

Speed of $m$ . $mgR=\tfrac12mv^2\left(1+\tfrac mM\right)$ , so
$v=\sqrt{\frac{2gR}{1+\frac mM}}.$
(B) is true. Then $V=\dfrac mMv$ , not $\dfrac mM\sqrt{2gR}$ , so (D) is false.

Displacements. Relative to the block, $m$ moves $R$ to the right. With the centre of mass fixed horizontally, $mx_m+MX_M=0$ and $x_m-X_M=R$ , so
$X_M=-\frac{mR}{M+m}.$
(C) is true.

Position of $m$ . It starts at $x=-R$ and ends at the block's edge, which is now at $-\dfrac{mR}{M+m}$ . That is not $-\sqrt2\dfrac{mR}{M+m}$ , so (A) is false.

(B), (C)

Topics:Conservation of linear momentum Centre of mass Conservation of mechanical energy

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JEE Advanced 2017 · Paper 1, Physics question 2