6.25 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 = 30 , = 60 , and h = 10 m, what are the speeds and times taken by the two stones ?

6.25 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 = 30 , = 60 , and h = 10 m, what are the speeds and times taken by the two stones ?
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1 Answer
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6.25 From the law of conservation of energy,
the potential energy at the top = Kinetic energy at the bottom
mgh = (1/2)m ….(1)
and
mgh = (1/2)m ….(2)
= , Both the stones will reach with the same speed
For stone 1, the force acting on the stone 1 is given by , = m = mg
= g
For stone 2, = g
As , >
From v = u + at, we get t = v/a
Therefore < Stone 2 will reach faster than stone 1
From the law of conservation of energy
mgh = (1/2) mv2
v = = 14 m/s ( Given h = 10 m)
The time taken by two stones given as
&
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