Two inclined frictionless tracks, one gradual and the other steep meet at A from where to stones are allowed to slide down from rest, one on each track (fig.) Will hte stones reach the bottom at the same time? Will they reach there with the same speed? Explain, given `theta_(1)=30^(@)`, `theta_92)=60^(@)` and h=10m. What are the speeds and time taken by the two stones?

Explanation

Let AB and AC be two smooth planes inclined at angles `theta_(1) and theta_(2)` to the hrozontal . As the height of both the stones is same , therefore both will reach the bottom with same speed .
As the stones are intially at same height therefore potential energy at A = kinetic energy at B or C
`mgh = 1/2 mv^(2)`
`rArr " " v = sqrt(2gh)`
` = sqrt(2xx10xx10) = 14.14 " m "s^(-1)`
The acceleration of two stones are :
`a_(1) = g sin theta_(1) and a_(2)= g sin theta_(2)`
As `teta_(2) gt theta_(1) " " :. a_(2) gt a_(1)`
As `a prop 1/t`
` :. t_(2) lt t_(1)`
Thus second stone will reach earlier and will take less time than the Ist stone ( i.e `t_(2) ltt_(1)` ) .
Let the time taken by the stone moving along AB to reach the ground be `t_(1)` and that moving along AC be `t_(2)`
`AB = h/(sin 30^(@)) = (10 m)/(sin 30^(@)) = 20 m `
AC ` = h /(sin 60^(@)) = (10 m ) / (cos 30^(@)) = 20/(sqrt(3)) m `
The times `t_(1) and t_(2) ` can be calculated as :
`t_(1) = sqrt((AB)/( g sin 30^(@))) = sqrt((40)/(10 xx1 //2)) = 2sqrt(2)s `
`t_(2)= sqrt((2AC)/(g sin 60^(@)) ) = sqrt((40sqrt(3))/(10 xx (sqrt(3)//2)) ) = 2 sqrt(2/3) s `