A block of mass m slides down on an inclined plane of a wedge of same mass m (see fig). Friction is absent everywhere. Which of the following statement(s) is/are correct?

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Vertical component of acceleration of block is $4 g / 5$ downwards

Acceleration of centre of mass of (block + wedge) is $g / 5$ downwards

Vertical component of acceleration of block is $2 g / 5$ downwards

Acceleration of centre of mass (block + wedge) is $3 g / 5$ downwards

answer is B, D.

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Detailed Solution

Let a be the acceleration of wedge leftwards and a_r be the relative acceleration of block down the plane. Then, absolute acceleration of block in horizontal direction will be $(a_{r} \cos θ - a)$ rightwards.

$\begin{array}{l} {\bar{F}}_{h o r i z o n t a l} = 0 \Rightarrow {\bar{a}}_{h o r i z o n t a l} = 0 \\ \Rightarrow a_{r} \cos θ - a = a \\ (o r) 2 a = a_{r} \cos θ .... (i) \\ FBD of wedge FBD of block w.r.t wedge \end{array}$

N $\sin θ = m a$

Net force on block perpendicular to plane is zero

So , $N + m a \sin θ = m g \cos θ ..... (i i)$

On solving eqs. (i), (ii), and (iii), we get

$a_{r} = \frac{2 g \sin θ}{1 + \sin^{2} θ} = \frac{2 g \sin 30}{1 + \sin^{2} 30} = \frac{4 g}{5}$

For block, $a_{y} = a_{r} \sin θ = \frac{2 g \sin^{2} θ}{1 + \sin^{2} θ} = \frac{2 g (\frac{1}{4})}{1 + \frac{1}{4}} = \frac{2 g}{5}$

$a_{C O M} = \frac{m (0) + m [\frac{2 g \sin^{2} θ}{1 + \sin^{2} θ}]}{m + m} = \frac{g \sin^{2} θ}{1 + \sin^{2} θ} = \frac{g \sin^{2} 30}{1 + \sin^{2} 30} = \frac{g}{5}$