Question

A wedge of height \( h \) is released from rest with a light particle \( P \) placed on it as shown. The wedge slides down an incline which makes an angle \( \theta \) with the horizontal. All the surface are smooth. \( P \) will reach the surface of the incline in time

(1) \( \sqrt{\frac{2 h}{g \sin ^{2} \theta}} \)

(2) \( \sqrt{\frac{2 h}{g \sin ^{2} \theta \cos \theta}} \)

(3) \( \sqrt{\frac{2 h}{g \tan \theta}} \)

(4) \( \sqrt{\frac{2 h}{g \cos ^{2} \theta}} \)

Answer 1

Correct option is (1) \(\sqrt{\frac{2h }{g \sin^2 \theta }}\)

As no force other than gravity is acting on the particle P so it will not move in horizontal direction but due to movement of wedge it will reach the inclined plane.

Horizontal acceleration = gsinθcosθ

Distance to move = hcotθ

\(s = ut + \frac 12 at^2\)

\(h \cot\theta = \frac 12 g \sin \theta \cos\theta t^2\)

\(t = \sqrt{\frac{2h \cos \theta}{g \sin \theta \times \sin \theta .\cos\theta}}\)

\(t = \sqrt{\frac{2h }{g \sin^2 \theta }}\)