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 statement is correct ?
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Options

A. Both the stones reach the botton 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 ealier than stone I
D. Both the stones reach the bottom at different times and with different speeds

Answer

C

Explanation

(c ) As the given trackes are frictionless, hence, mechanical energy will be conserved. As both the tracks having common height, h.
From conservation of mechanical energy,
`(1)/(2)mv^(2)=mgh " " ("for both tracks I and II")`
`v=sqrt(2gh)`
Hence, speed is same for both stones. For stone I,
`a_(1)="acceleration along inclined plane"=g sin theta_(1)`
Similarly, for stone II `a_(2)=g sin theta_(2) " as " theta_(2) gt theta_(1)` hence, `a_(2) gt a_(1)`.
And length for track II is also less hence, stone II reaches earliet than stone I.