Question

A $0.098-kg$ block slides down a frictionless track as shown in Fig.
The time taken by the block to move from $A$ to $B$ is:

• A. $\frac{1}{\sqrt{g}}$
• B. $\frac{2}{\sqrt{g}}$
• C. $\frac{3}{\sqrt{g}}$
• D. $\frac{4}{\sqrt{g}}$

Answer 1

Answer: (B)

$\frac{2}{\sqrt{g}}$

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