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 B is

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

Answer 1

The correct option is B $\frac{2}{\sqrt{g}}$


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 $V_{Ay}=V_{A}sin{30}^{\circ }=\sqrt{g}$
When the block leaves the track at A it will undergo projectile motion.
Time of flight for projectile from A to B is given by,
$T=\frac{2V_{A}sin\theta }{g}$
Since $V_{A}sin\theta =\sqrt{g}$ $(\because \theta ={30}^{\circ })$
$T=\frac{2\sqrt{g}}{g}=\frac{2}{\sqrt{g}}$