Question

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 figure. Which of the following statements is correct?

• A. Both the stones reach the bottom at the same time but not with the same speed.
• B. Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
• C. Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
• D. Both the stones reach the bottom at different times and with different speeds.

Answer 1

Answer: (C)

Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.

Answer 2

In figure, AB and AC are two smooth planes inclined to the horizontal at $\angle\theta_{1}and\angle\theta_{2}$respectively.

As height of both the planes in the same, therefore, both the stones will reach the bottom with same speed.

According to law of conservation of mechanical energy,

PE at the top = KE at the bottom

$\therefore mgh = \frac{1}{2}mv_{1}^{2}$ ……(i)

And mgh$= \frac{1}{2}mv_{2}^{2}$ ……(ii)

From (i) and (ii), we get $v_{1} = v_{2}$

As is clear from figure, acceleration of the two stones are

respectively.

As $\theta_{2} > \theta_{1}\therefore a_{2} > a_{1}$

From v = u +at = 0$+ atort = \frac{v}{a}$

As $t \propto \frac{1}{a},$ and $a_{2} > a_{1}\therefore t_{2} < t_{1}$

Hence, stone II will take lesser time and reach the bottom earlier than stone I.