Question

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

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

Answer 1

Answer: (D)

$\frac{1+\sqrt{3}}{\sqrt{g}}$

$\frac{1}{2}m{v}^2=mg(3-1)=2mg$ [From conservation of energy]

or

$v=\sqrt{4g}=2\sqrt{g}$

Vertical component at $A$ is $2\sqrt{g}\sin {30}^o=\sqrt{g}$

Time of flight,

$T=\frac{2v\sin \theta }{g}=\frac{2\sqrt{g}}{g}=\frac{2}{\sqrt{g}}$

Using $S=ut+\frac{1}{2}a{t}^2,$ we get

$-1=\sqrt{gt}-\frac{1}{2}g{t}^2$ or $\frac{1}{2}g{t}^2-\sqrt{gt}-1=0$

$t=\frac{\sqrt{g}\pm \sqrt{g+4\times \frac{1}{2}g}}{2\times \frac{g}{2}}t=\frac{1+\pm \sqrt{3}}{\sqrt{g}}$