Two inclined frictionless tracks,one gradual and the other steep,meet at $A$ from where two stones are allowed to slide down from rest,one on each track. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given $\theta_{1}=30^{\circ}, \theta_{2}=60^{\circ},$ and $h=10 \; m,$ what are the speeds and times taken by the two stones?

(N/A) No,the stone moving down the steep plane will reach the bottom first. Yes,the stones will reach the bottom with the same speed $v_{B} = v_{C} = 14 \; m/s$. The times taken are $t_{1} = 2.86 \; s$ and $t_{2} = 1.65 \; s$.
$1$. Speed:
Since the tracks are frictionless,the total mechanical energy is conserved. Both stones start from the same height $h$ with zero initial velocity. By the law of conservation of energy:
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \; m/s$.
Since both stones fall through the same vertical height $h$,they reach the bottom with the same speed $v = 14 \; m/s$.
$2$. Time:
The acceleration of a stone on an inclined plane is $a = g \sin \theta$.
For stone $I$ on track $AB$: $a_{1} = g \sin 30^{\circ} = 9.8 \times 0.5 = 4.9 \; m/s^2$.
The length of track $AB$ is $L_{1} = h / \sin 30^{\circ} = 10 / 0.5 = 20 \; m$.
Using $s = \frac{1}{2}at^2$,$t_{1} = \sqrt{2L_{1}/a_{1}} = \sqrt{2 \times 20 / 4.9} \approx 2.86 \; s$.
For stone $II$ on track $AC$: $a_{2} = g \sin 60^{\circ} = 9.8 \times 0.866 = 8.49 \; m/s^2$.
The length of track $AC$ is $L_{2} = h / \sin 60^{\circ} = 10 / 0.866 \approx 11.55 \; m$.
$t_{2} = \sqrt{2L_{2}/a_{2}} = \sqrt{2 \times 11.55 / 8.49} \approx 1.65 \; s$.
Since $a_{2} > a_{1}$,the stone on the steeper track reaches the bottom first.