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Home » 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. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given , , and , what are the speeds and times taken by the two stones?
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. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given \theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}, and \mathrm{h}=10 \mathrm{~m}, what are the speeds and times taken by the two stones?
CBSE | Class 11 | NCERT | Physics | Work, Energy, and Power
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. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given \theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}, and \mathrm{h}=10 \mathrm{~m}, what are the speeds and times taken by the two stones?

Solution:

The sides A B and A C of the figure are both inclined to the horizontal at \theta_{1} and \theta_{2}, respectively. According to the law of mechanical energy conservation,

Potential energy at the top = Kinetic Energy at the bottom

\therefore \mathrm{mgh}=(1 / 2) \mathrm{mv}_{1}^{2}-(1)

and \mathrm{mgh}=(1 / 2) \mathrm{mv}_{2}^{2}-(2)

Due to the fact that both sides are the same height, both stones will reach the bottom at the same time.

From (1) and (2), we get \mathrm{v}_{1}=\mathrm{v}_{2}

So, both the stones will reach the bottom with the same velocity.

For stone 1:

Net force acting on the stone can be calculated as,

\begin{array}{l} \mathrm{F}=\operatorname{ma}_{1}=\mathrm{mg} \sin \theta_{1} \\ \mathrm{a}_{1}=\mathrm{g} \sin \theta_{1} \end{array}

For stone 2:

\begin{array}{l} \mathrm{a}_{2}=\mathrm{g} \sin \theta_{2} \\ \text { As } \theta_{2}>\theta_{1} \end{array}

So, a_{2}>a_{1}

From v=u+a t=0+a t

\Rightarrow \mathrm{t}=\mathrm{v} / \mathrm{a}

For stone 1: t_{1}=v / a_{1}

For stone 2: t_{2}=v / a_{2}

As t \propto 1 / \mathrm{a}, and \mathrm{a}_{2}>\mathrm{a}_{1}

So, t_{2}<t_{1}

Hence, stone 1 will reach slower than stone 2.

By applying the law of conservation of energy we get

\mathrm{ngh}=(1 / 2) \mathrm{mv}^{2}

When the height, \mathrm{h}=10 \mathrm{~m}, the speed of the stones is given as

v=\sqrt{2 g h}=\sqrt{2} \times 9.8 \times 10=14 \mathrm{~m} / \mathrm{s}

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