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.,
(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:
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:
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: