Question

In the figure shown, a particle is released from position $\text{A}$ on a smooth track, when the particle reaches $\text{B}$, then the velocity of the particle is

• A. $2\sqrt{gh}$
• B. $\sqrt{gh}$
• C. $\sqrt{3gh}$
• D. $\sqrt{2gh}$

Answer 1

The correct option is D $\sqrt{2gh}$

Since, smooth track is given in the question, so the mechanical energy of the particle remains conserved throughout the motion.

Applying law of conservation of energy between $\text{A}$ and $\text{B}$

$(PE{)}_A+(KE{)}_A=(PE{)}_B+(KE{)}_B$

Ground is taken as reference and velocity of particle at position $\text{A}$ is zero, so

$(mg\times 3h)+0=(mg\times 2h)+\frac{1}{2}m{v}^2$

$v^2=2gh$

$\therefore v=\sqrt{2gh}$

Therefore, option $\left(D\right)$ is the correct answer.