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 . Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30°, θ₂ = 60°, and h = 10 m, what are the speeds and times taken by the two stones?

Answer 1

The given situation can be shown as in the following figure:

AB and AC are two smooth planes inclined to the horizontal at ∠θ₁ and ∠θ₂ respectively. As height of both the planes is the same, therefore, noth the stones will reach the bottom with same speed.

As P.E. at O = K.E. at A = K.E. at B

∴ mgh = 1/2 mv₁² = 1/2 mv₂²

∴ v₁ = v₂

As it is clear from fig. above, accleration of the two blocks are a₁ = g sin θ₁ and a₂ = g sin θ₂

As θ₂ > θ₁

∴ a₂ > a₁

From v = u + at = 0 + at

or, t = v/a

As t ∝ 1/a, and a₂ > a₁

∴ t₂ < t₁

i.e., Second stone will take lesser time and reach the bottom earlier than the first stone.

Answer 2

1/2mv^2 = mgh, v = `sqrt(2gh)`

`=sqrt(2xx10xx10)  ms^(-1) = 14.14 ms^(-1)`

`v_B = v_C = 14.14 ms^(-1), l = 1/2(g sin theta) t^2`

`sin theta = h/l , l = h/(sin theta)`

`h/sin theta = 1/2 g sin theta t^2`  or `t =sqrt(2h/g).1/(sin theta)`

`t_B = sqrt((2xx10)/10). "" 1/sin 30^@ = 2sqrt2s`

`t_C = sqrt(2xx10)/10 . 1/sin  60^@ = (2sqrt2)/sqrt(3) s.`