Block A is released on the circular track of bigger block B from rest from the position as shown in figure. Friction in absent everywhere. Choose the correct statement

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When A reaches Q acceleration of centre of mass of A and B is zero

Acceleration of B is maximum when A reaches Q

Acceleration of center of mass of A and B is zero initially

Normal force of the ground on the wedge is greater than combined weight of A and B. when A reaches at point Q

answer is C.

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Detailed Solution

There is external force acting on the system of “A” and ‘B” in vertical direction, also ‘A’ has acceleration it means the centre of mass of system will also accelerate.

2) when A reaches Q the acceleration of A is $\frac{v_{0}^{2}}{R}$ in frame of ‘B’ so acceleration of centre of mass will not be zero
3) From F.B.D of the block at Q
$N - m g = \frac{m v_{0}^{2}}{R} \Rightarrow N = m g + \frac{m v_{0}^{2}}{R}$

So net normal force will be greater than combined weight
4) When ‘A’ is at Q acceleration of B is zero as no external force acts on the system in horizontal direction

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