Question

Two inclined friction less 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 ${\theta }_1={30}^0,{\theta }_2={60}^0$, and $h=10m$, 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 1 and 2 respectively. As height of both the planes is the same, therefore, both the stones will reach the bottom with same speed.

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

$mgh=1/2m{v}_1^2=1/2m{v}_2^2$

$v_1=v_2$

As it is clear from fig. above, acceleration of the two blocks are

$a_1=gsin{\theta }_1$

$a_2=gsin{\theta }_2$

As ${\theta }_2>{\theta }_1$

$a_2>a_1$

From $v=u+at=0+at$

or,$t=v/a$

As

$t\propto \frac{1}{a}and{a}_2>a_1$

$t_2<t_1$

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