Question

A $0.098kg$ block slides down a frictionless track as shown in Fig.
The horizontal distance $x$ traveled by the block in moving from $A$ to $C$ is:

• A. $(1+\sqrt{3})m$
• B. $(1-\sqrt{3})m$
• C. $(\sqrt{3}+3)m$
• D. $9meter$

Answer 1

Answer: (C)

$(\sqrt{3}+3)m$

$\frac{1}{2}m{v}^2=mg(3-1)=2mg$ [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
$T=\frac{2v\sin \theta }{g}=\frac{2\sqrt{g}}{g}=\frac{2}{\sqrt{g}}$
Using $S=ut+\frac{1}{2}a{t}^2$, we get
$-1=\sqrt{g}t-\frac{1}{2}g{t}^2$ or $\frac{1}{2}g{t}^2-\sqrt{g}t-1=0$
$t=\frac{\sqrt{g}\pm \sqrt{g+4\times \frac{1}{2}g}}{2\times \frac{g}{2}}=\frac{1\pm \sqrt{3}}{\sqrt{g}}$
Neglecting negative time, $t=\frac{1+\sqrt{3}}{\sqrt{g}}$
$x=2\sqrt{g}\cos {30}^{\circ }\left[\frac{1\pm \sqrt{3}}{\sqrt{g}}\right]$ or $x=(\sqrt{3}+3)m$.