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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

Physics

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