Question

A block of mass $m$ lies on wedge of mass $M$ as shown in figure.
With what minimum acceleration must the wedge be moved towards right horizontally so that block $m$ falls freely.

Answer 1

To have a free-fall motion, the block should lose contact with the wedge.

$\therefore a\sin \theta \ge g\cos \theta$

$\therefore a=g\cot \theta$

motion along incline

$a^{\prime }=g\sin \theta +a\cos \theta$

$=g\sin \theta +g\cot \theta \cos \theta$

$=g[\sin \theta +\frac{{\cos }^2\theta }{\sin \theta }]$

$=g\left[\frac{1}{\sin \theta }\right]$

$\therefore a^{\prime }\sin \theta =g$

Acceleration in vertically downward direction is g.