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 Fig.

Which of the following statement 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

The correct option is C Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I

In the given situation as stones slides they convert potential energy into kinetic energy. And velocity at bottom depends only on the height not the path followed, because gravitational force is conservative.

So both of them will have same velocity.

Now about time taken, acceleration on the inclined plane is goven by $gsin\theta$

So time taken to cover $hcosec{\theta }_1$ with acceleration $gsin{\theta }_1$ and initial velocity is 0. $t_1=\sqrt{\frac{2hcosec{\theta }_1}{gsin{\theta }_1}}=\sqrt{\frac{2h}{gsi{n}^2{\theta }_1}}$

Similarly $t_2=\sqrt{\frac{2hcosec{\theta }_2}{gsin{\theta }_2}}=\sqrt{\frac{2h}{gsi{n}^2{\theta }_2}}$

So it is clear as $theta$ increases $sin\theta$ increases as a result t decreases.

Hence option C is best possible.