Question

On smooth curved track shown in the diagram, an small particle is projected from the point O , that it just manages to reach the point P. Assume that particle always remain contact with track, The velocity of the particle on reaching the point P will be equal to :

• A. Zero
• B. $\sqrt{2gH}$
• C. $\sqrt{6gH}$
• D. $\sqrt{4gH}$

Answer 1

The correct option is D $\sqrt{4gH}$

The particle should have initial velocity (kinetic energy) such that it just reaches highest height of trajectory(here i.e., $3H$)

So,

At $h=3H$,

$v=0$

Height of point $P=H$

Let velocity of particle at P be vp

Then,

Work done by gravity from highest point till point $p=mg(3H-H)$

$w_{gravity}=2mgH$

As here only gravity is acting,

Applying work energy theorem,

$\Delta KE=w_{gravity}$

$\Rightarrow \frac{m}{2}[v{p}^2-0^2]=2mgH$

$\Rightarrow vp=\sqrt{4gH}$

Option D is correct.