Question

A particle is released from top of a smooth inclined plane of inclination $\theta .$ Let $v$ be the speed of the particle after moving a distance s. Then the quantity which is constant is

• A. $v^{2}-2gs \sin \,\theta$
• B. $v^{2}-\sqrt{2 g s} \sin \theta$
• C. $v^{2}+2 g s \sin \theta$
• D. $v^{2}+\sqrt{2 g s} \sin \theta$

Answer 1

Answer: (A)

$v =\sqrt{2gH }=\sqrt{2gs \sin \theta}$
so, $v^{2}-2gs \sin \theta=0$