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A cube of mass m and side length 4R has a groove made inside it in a vertical plane. The groove is a circle having radius R. A particle of mass m is moving down with a velocity $v = \sqrt{2 g R} .$ Assume all surfaces are smooth. The correct statement(s) at this instant.

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Normal reaction on particle is mg

Normal reaction on cube by the ground is mg

Acceleration of block w.r.t ground is g

Acceleration of particle with respect to ground is $g \sqrt{2}$

answer is A, B, C, D.

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Detailed Solution

$N_{2} - m g = 0 \Rightarrow N_{2} = m g N_{1} = m a N_{1} + m a = m (\frac{V^{2}}{R}) N_{1} + m a = m 2 g \Rightarrow N_{1} = m g a n d a = g$

${\vec{a}}_{p, g} = {\vec{a}}_{p, c} + {\vec{a}}_{c, g} \to g \begin{array}{l} ↓ = g \sqrt{2} \\ g \end{array}$

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