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Question

A particle is projected at an angle 60o with speed 103 m/s, from the point A, as shown in the figure. At the same time the wedge is made to move with speed 103 m/s towards right as shown in the figure. Then the time after which particle will strike with wedge is
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A
2 s
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B
23 s
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C
43 s
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D
none of these
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Solution

The correct option is A 2 s
velocity of particle is given as VP=53^i+15^j
velocity of wedge is given as Vwedge=103^i
The relative velocity of particle w.r.t. wedge is given as

VP/wedge=VPVwedge
VP/wedge=53^i+15^j103^i
VP/wedge=53^i+15^j

Now it can be analysed as projectile motion on a stationary inclined plane, where the time period of its flight is given as :

T=2usin(αβ)gcosβ

where α=60,β=30,u=103

T=2(103)sin(6030)10cos30=2s

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