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 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

C

Explanation

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=1/2mv_1^2`...(i)
and `mgh=1/2mv_2^2`...(ii)
From (i) and (ii) , we get `v_1=v_2`
As is clear from figure, acceleration of the two stones are `a_1=gsintheta_1` and `a_2=gsintheta_2` respectively.
As `theta_2 gt theta_1 " " therefore " " a_2 gt a_1`
From v=u+at=0 + at or t=`v/a`
As `tprop 1/a`, and `a_2 gt a_1 therefore t_2 lt t_1`
Hence, stone II will take lesser time and reach the bottom earlier than stone I.