Q. 253.6( 22 Votes )

# 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 θ_{1} = 30^{0}, θ_{2} = 60^{0}, and h = 10 m, what are the speeds and times taken by the two stones?

Answer :

It is given that AB and AC are two frictionless planes inclined to the horizontal at angles θ_{1} and θ_{2} respectively.

The kinetic energy at B will be equal to the kinetic energy at C which will be the potential energy at A due to law of conservation of energy since there are no losses.

Let v_{1} and v_{2} be the final velocities of the stones on the planes AB and AC respectively and m be the mass of each stone.

Then, mgh = (1/2)mv_{1}^{2} = (1/2)mv_{2}^{2}

⇒ v_{1} = v_{2} = v (say)

If a_{1} and a_{2} are the accelerations of the two stones along the planes AB and AC respectively, then

a_{1} = gsinθ_{1}

a_{2} = gsinθ_{2}

As θ_{2}>θ_{1}, a_{2}>a_{1}.

Initial velocities of the two stones are zero.

From Newton’s first equation of motion,

v = u + at

⇒ v = 0 + at

⇒ v = at

⇒ t = v/a

Since v is constant for both the stones,

t ∝ 1/a

Since a_{2}>a_{1}, t_{2}<t_{1}.

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

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Give example of a situation in which an applied force does not result in a change in kinetic energy.

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