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Asked by Vikash Last Modified
Answer 
Tasneem
No; the stone moving down the steep plane will reach the bottom first
Yes; the stones will reach the bottom with the same speed
vB = vC = 14 m/s
t1 = 2.86 s; t2 = 1.65 s
The given situation can be shown as in the following figure:
Here, the initial height (AD) for both the stones is the same (h). Hence, both will have the same potential energy at point A.
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.,
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:
For stone II:
∴
Using the first equation of motion, the time of slide can be obtained as:
For stone I:
For stone II:
∴
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.
The times are given as:
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