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Two particles at a distance 5 m apart, are thrown towards each other on an inclined smooth plane with equal speeds 'v'.One particle is projected up the plane and the other is projected down the plane . Inclined plane is inclined at an angle of $30^{0}$ with the horizontal. It is known that both particle move along the same straight line. The particles collide at the point from
where the lower particle is thrown. Find the value of v. [take g = 10 $m / s^{2}$ ]

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2 m/s

1.5 m/s

2.5 m/s

1 m/s

answer is D.

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Detailed Solution

Down the plane,

$5 = v . t + \frac{1}{2} (gsin θ) t^{2} - - - - - - (i)$

At the plane, 0 = v-g $\sin θ t_{1} \Rightarrow t_{1} = \frac{v}{g \sin θ}$

t = $2 t_{1} = \frac{2 v}{g \sin θ}$

[time taken by power particle coming back to initial position]

5 = $\frac{2 . v^{2}}{g sinθ} + \frac{1}{2} \frac{g \sin θ . 4 v^{2}}{g^{2} \sin^{2} θ}$

10g $\sin θ = 8 v^{2}$

v = $\sqrt{\frac{5 g \sin θ}{4}} = \frac{5}{2} m / s$

Alternate sol.

$2 vt = 5$

$\frac{2 v}{g \sin θ} = t \Rightarrow 2 v \times \frac{2 v}{g \sin θ} = 5$

$\Rightarrow v = \sqrt{\frac{5 g \sin θ}{4}} = \frac{5}{2} m / s$

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