# Two inclined fric

It is given that AB and AC are two frictionless planes inclined to the horizontal at angles θ1 and θ2 respectively.

The kinetic energy at B will be equal to the kinetic energy at C which will be the potential energy at A due to law of conservation of energy since there are no losses.

Let v1 and v2 be the final velocities of the stones on the planes AB and AC respectively and m be the mass of each stone.

Then, mgh = (1/2)mv12 = (1/2)mv22

v1 = v2 = v (say)

If a1 and a2 are the accelerations of the two stones along the planes AB and AC respectively, then

a1 = gsinθ1

a2 = gsinθ2

As θ21, a2>a1.

Initial velocities of the two stones are zero.

From Newton’s first equation of motion,

v = u + at

v = 0 + at

v = at

t = v/a

Since v is constant for both the stones,

t 1/a

Since a2>a1, t2<t1.

So, the stone on plane AC will take less time than the stone on plane AB to reach the bottom.

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