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?

          

Answers (1)
D Devendra Khairwa

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 

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