Question

A $0.098\text{kg}$ block slides down a frictionless track as shown (Assume no energy loss on the entire track)
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

The correct option is D $\frac{1+\sqrt{3}}{\sqrt{g}}$


Applying conservation of mechanical energy between point D and A,
Initial Potential energy (P.E) + initial kinetic enegy (K.E) = final potential energy (P.E) + kinetic energy (K.E)
$mg{h}_D+\frac{1}{2}m{V}_D^2=mg{h}_A+\frac{1}{2}m{V}_A^2$
$mg\left(3\right)+\frac{1}{2}m(0{)}^2=mg(1)+\frac{1}{2}m{V}_A^2$
Solving the above equation we get,
$V_A=2\sqrt{g}$
Now vertcal component of $V_A$ i.e $V_{Ay}=V_{A}sin{30}^{\circ }=\sqrt{g}$
Applying 2nd equation of motion between poitns A and C,
$S=ut+\frac{1}{2}a{t}^2$
Here $S=-1,u=\sqrt{g},a=-g$
$\Rightarrow -1=\sqrt{g}t+\frac{1}{2}(-g)t^2$
Solving the above equation we get,
$t=\frac{1+\sqrt{3}}{\sqrt{g}}$