Question 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 θ1
= 30°,
θ2
= 60°,
and h =
10 m, what are the speeds and times taken by the two stones?
Solution:
No; the stone moving down the steep plane will reach the bottom first
Yes; the
stones will reach the bottom with the same speedNo; the stone moving down the steep plane will reach the bottom first
vB = vC = 14 m/s
t1 = 2.86 s; t2 = 1.65 s
The given situation can be shown as in the following figure:

As per the law of conservation of energy, the kinetic energy of the stones at points B and C will also be the same, i. e.,
(1/2)mv12 = (1/2)mv22
v1 = v2 = v, say
Where,
m = Mass of each stone
v = Speed of each stone at points B and C
Hence, both stones will reach the bottom with the same speed, v.
For stone I:
Net force acting on this stone is given by:
Fnet = ma1 = mgSinθ1
a1 = g Sinθ1
For stone II:
a2 = gSin θ2
∵ θ2 > θ1
∴ Sin θ2 > Sin θ1
∴ a2 > a1
Using the first equation of motion, the time of slide can be obtained as:
v = u +at
∴ t = v / a (∵ u = 0)
For stone I:
t1 = v / a1
For stone II:
t2 = v / a2
∵ a2 > a1
∴ t2 < t1
Hence, the stone moving down the steep plane will reach the bottom first.
The speed (v) of each stone at points B and C is given by the relation obtained from the law of conservation of energy.
mgh = (1/2) mv2
∴ v = (2gh)1/2
= (2 X 9.8 X 10)1/2
= (196)1/2 = 14 m/s
The times are given as:
t1 = v / a1 = v / (g Sinθ1) = 14 / (9.8 X Sin 30) = 14 / (9.8 X (1/2)) = 2.86 s
t2 = v / a2 = v / (g Sinθ2)= 14 / (9.8 X Sin 60) = 14 / (9.8 X (√3/2)) = 1.65 s
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