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 tracks 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

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 is 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
$\therefore mgh=\frac{1}{2}m{v}_1^2....\left(i\right)$
and $mgh=\frac{1}{2}m{v}_2^2....\left(ii\right)$
From (i) and (ii), we get $v_1=v_2$
As is clear from figure, acceleration of the two stones are $a_1=g\sin {\theta }_1$ and $a_2=g\sin {\theta }_2$ respectively.
As ${\theta }_2>{\theta }_1.\therefore a_2>a_1$
From $v=u+at=0+at$ or $t=\frac{v}{a}$
As $t\propto \frac{1}{a}$, and $a_2>a_2\therefore t_2<t_1$
Hence, stone II will take lesser time and reach the bottom earlier than stone I.