Ncert Exercises & Solutions

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 as shown in the figure.

Which of the following statement is correct?

1.	Both stones reach the bottom at the same time but not at the same speed.
2.	Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
3.	Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
4.	Both stones reach the bottom at different times and at different speeds.

Step 1: Find the final velocities of the stones at the bottom.

As the given tracks are frictionless. hence. energy will be conserved. As
both the tracks having common height, h.

From the conservation of mechanical energy.

$\begin{array}{l} \frac{1}{2} {mv}^{2} = mgh (for both tracks I and II) \\ v = \sqrt{2 gh} \end{array}$

Step 2: Find the time after which the stones reaches the bottom.

Hence, speed is the same for both stones. For stone l, a, = acceleration along with the inclined plane $= g \sin θ_{1}$

Similarly, for stone $∥ a_{2} = gsin θ_{2} as θ_{2} > θ_{1} hence, a_{2} > a_{1}$ .

And both length tor track II is also less hence, stone II reaches earlier than stone l.

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