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Q25 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 $\theta _1 = 30 \degree , \theta = 60 \degree ,$ and h = 10 m, what are the speeds and times taken by the two stones?

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The FBD of the track is shown in the figure below :

Work, power energy, 20324

Using the law of conservation of energy we have :

$\frac{1}{2}mv_1^2\ =\ \frac{1}{2}mv_2^2$

or $v_1\ =\ v_2$

Hence both stones will reach the bottom with the same speed.

For stone 1 we can write :

$F\ =\ mg\sin \Theta_1$

or $a_1\ =\ g\sin \Theta_1$

For stone 2 we have :

$a_2\ =\ g\sin \Theta_2$

Also, using the equation of motion,

$v\ =\ u\ +\ at$

or $t\ =\ \frac{v}{a}$

It is given that $\Theta _2 > \Theta_1$

or $a _2 > a_1$

Thus $t _1 > t_2$

Hence, the stone travelling on the steep plane will reach before.

For finding speed and time we can use conservation of energy.

$mgh\ =\ \frac{1}{2}mv^2$

or $v\ =\ \sqrt{2gh}$

or $=\ \sqrt{2\times 9.8\times 10}$

or $=\ 14\ m/s$

And the time is given by :

$t_1\ =\ \frac{v}{a_1}\ =\ \frac{14}{9.8\times \sin 30^{\circ}}\ =\ 2.86\ s$

and $t_2\ =\ \frac{v}{a_2}\ =\ \frac{14}{9.8\times \sin 60^{\circ}}\ =\ 1.65\ s$

Posted by

Devendra Khairwa

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