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 (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 300, θ₂ = 600, and h = 10 m, what are the speeds and times taken by the two stones ?

Answer 1

Given, the angle of gradual inclined track is θ 1 =30°, the angle of steep inclined track is θ 2 =60° and height of the inclined track is 10 m.

The following diagram shows two stones at the inclined tracks.

Here, 1 and 2 indicate stones at the two tracks.

The initial potential energy of the stones is same for both and is given by,

U=mgh

Here, U is the potential energy, m is the mass of the stone and h is the height of the track.

This potential energy of each stone is converted into the final kinetic energy of the stones as there is no frictional resistance on the tracks, therefore,

K 1 = K 2 1 2 m v 1 2 = 1 2 m v 2 2 v 1 = v 2

Here, K 1 is the final kinetic energy of stone 1, K 2 is the final kinetic energy of stone 2, v 1 is the final speed of stone 1 and v 2 is the final speed of stone 2.

The above expression shows that the final speed of both the stones will be same.

Let F 1 be the net downward force acting on stone 1 along the track, then

F 1 =mgsin θ 1 m a 1 =mgsin θ 1 a 1 =gsin θ 1 …… (1)

Here, a 1 is the acceleration with which stone 1 will come down from the track.

Let F 2 be the net downward force acting on stone 2 along the track, then

F 2 =mgsin θ 2 m a 2 =mgsin θ 2 a 2 =gsin θ 2 …… (2)

Here, a 2 is the acceleration with which stone 2 will come down from the track.

It is given that θ 2 > θ 1 , hence sin θ 2 >sin θ 1 , now from equation (1) and (2),

a 2 > a 1 …… (3)

Let u be the initial velocity, a be the acceleration, v be the final velocity and t be the time taken.

Then from equation of motion,

v=u+at

When initial velocity is 0, then

v=0+at v=at t= v a …… (4)

From the above expression, time is inversely dependent on acceleration. Hence, from equation (4) and equation (3),

t 2 < t 1

Here, t 1 is the time taken by the stone 1 and t 2 is the time taken by the stone 2 to reach downwards.

Thus, the above expression shows that time taken by stone 2 is less than that by stone 1.

Hence, as the time taken by stone at steeper track is less than that of the other, it will reach the bottom earlier and the speed of both the stones at the bottom will be same ( v).