Question

A $0.098$-kg block slides down a frictionless track as shown in Fig.
The vertical component of the velocity of block at A is:

• A. $\sqrt{g}$
• B. $2\sqrt{g}$
• C. $3\sqrt{g}$
• D. $4\sqrt{g}$

Answer 1

The correct option is A $\sqrt{g}$

$\frac{1}{2}m{v}^2=mg(3-1)=2mg$ [from conservation of energy betwn A and B]
or $v=\sqrt{4g}=2\sqrt{g}$
Vertical component at $A$ is $2\sqrt{g}\sin {30}^{\circ }=\sqrt{g}$