__Question 6.25__:*θ*

_{1}= 30°,

*θ*

_{2}= 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

*v*

_{B}=

*v*

_{C}= 14 m/s

*t*

_{1}= 2.86 s;

*t*

_{2}= 1.65 s

The given situation can be shown as in the following figure:

*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)mv

_{1}

^{2}= (1/2)mv

_{2}

^{2}

*v*

_{1}=

*v*

_{2}=

*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:

F

_{net}= ma

_{1}= mgSinθ

_{1}

a

_{1}= g Sinθ

_{1}

_{}

**For stone II:**a

_{2}= gSin θ

_{2}

∵ θ

_{2}> θ

_{1}

∴ Sin θ

_{2}> Sin θ

_{1}

∴ a

_{2 }> a

_{1}

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:**t

_{1}= v / a

_{1}

_{}

**For stone II:**t

_{2}= v / a

_{2}

∵ a

_{2}> a

_{1}

∴ t

_{2}< t

_{1}

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) mv

^{2}

∴ v = (2gh)

^{1/2}

= (2 X 9.8 X 10)

^{1/2}

= (196)

^{1/2}= 14 m/s

The times are given as:

t

_{1 }= v / a

_{1}= v / (g Sinθ

_{1}) = 14 / (9.8 X Sin 30) = 14 / (9.8 X (1/2)) = 2.86 s

t

_{2 }= v / a

_{2}= v / (g Sinθ

_{2})= 14 / (9.8 X Sin 60) = 14 / (9.8 X (√3/2)) = 1.65 s

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