Question

A $0.098\text{kg}$ block slides down a frictionless track as shown (Assume no energy loss on the entire track)
The vertical component of the velocity of block at $A$ is

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

Answer 1

The correct option is A $\sqrt{g}$


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 $V_{Ay}=V_{A}sin{30}^{\circ }=\sqrt{g}$