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Question

Two inclined planes OA and OB having inclination (with horizontal) 30 and 60, respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity u=10(3) ms1 along a direction perpendicular to the plane OA. If the particle strikes the plane OB perpendicular at Q, calculate the displacement PQ.

A
20 m
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B
10 m
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C
5 m
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D
2.5 m
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Solution

The correct option is A 20 m
Two given planes are mutually perpendicular and the particle is projected perpendicularly from plane OA. It means u is parallel to plane OB.
At the instant of collision of the particle with OB, its velocity is perpendicular to OB or the velocity component parallel to OB is zero.

First considering motion of particle parallel to plane OB,
Initial velocity along OB, u=10(3) ms1
Component of acceleration due to gravity along OB, gOB=gsin60
gOB =53 ms2
The final velocity of the particle along OB is
vOB=0
The time of flight (t) of particle is
vOB=ugOBt
0=1035(3)t
t=2 s

The distance between OQ is
s=ut12gOBt2
OQ=10(3)×2125(3)(2)2
OQ=10(3)m

Now considering motion of the particle normal to the plane OB.
Initial velocity along OA=0
Acceleration along OA, gOA=gcos60=5 ms2
The time of flight will remain same,
t=2 s
The distance between PO is
s=12gOAt2
s=12×5×22
s=PO=10 m
Distance between PQ is
PQ=(PO)2+(OQ)2
PQ=(102)+(103)2
PQ=20 m

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