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Question

# Particle A is released from a point P on a smooth inclined plane inclined at an angle α with the horizontal. At the same instant another particle B is projected with initial velocity u making an angle β with the horizontal. Both the particles meet again on the inclined plane. Find the relation between α and β

A
α+β=45
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B
α+β=30
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C
α+β=90
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D
αβ=30
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Solution

## The correct option is C α+β=90∘ Consider motion of B along the plane, initial velocity =ucos(α+β) and acceleration =gsinα Using second equation of motion along the inclined plane, ∴OP=ucos(α+βt)+12gsinαt2(1) For motion of particle A along the plane, Initial velocity = 0 Acceleration =gsinα Using second equation of motion along the inclined plane, ∴OP=12gsinαt2(2) From Equation (1) and (2), we get ucos(α+β)t=0 so, either t=0 or α+β=π2 Thus, the condition for the particles to collide is α+β=π2

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