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 as shown in the figure. 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

both stone starts sliding downward from same height ‘ h ’ so, the initial energy of both stones will be same. e.g., mgh , where m is mass of each stone.

after reaching at the bottom of inclined track, final energy of each stone = kinetic energy of stone.

in case of the stone which moves in AB,
mgh = kinetic energy = 1/2m$v_1^2$.......(1)
in case of the stone which moves in AC,
mgh = kinetic energy = 1/2m$v_2^2$......(2)

from equations (1) and (2),

$v_1=v_2$

now, If $a_1$ and $a_2$ are the accelerations of the two stones along the planes AB and AC respectively, then
$a_1=gsin\theta_1$
$a_2=gsin\theta_2$

As $\theta_2>\theta_1$
so, $a_2>a_1$

Initial velocities of the two stones are zero.
From kinematics equation of motion,
v = u + at
 or, v = 0 + at
or,  v = at
or,  t = v/a
Since v is constant for both the stones,
t ∝ 1/a
Since $a_2 > a_1$
so, $t_1 < t_2$

hence, the stone on plane AC will take less time than the stone on plane AB to reach the bottom.