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Question

The point B lies on a smooth plane inclined at 30 to the horizontal. A particle of mass 17 kg is dropped from a point A which lies 10 m vertically above B. The particle rebounds from the plane in the direction BC with speed v m/s at an angle of 45 to the plane. Find the impulse exerted by the plane on the particle (in Ns)


A
1+3
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B
13
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C
2+3
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D
223
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Solution

The correct option is A 1+3
As the plane is smooth, the impulse must be perpendicular to the plane.
Velocity of particle just before the collision is u=2gh=2×9.8×10=14 m/s



Since A is vertically above B, θ=30
Let v be the velocity of particle after collision.
No momentum exchange parallel to the plane.
Hence, 14sin30=vcos45
or v=72 m/s

Impulse |J|=[mvsin45m(ucos30)]
|J|=17[72×12+14×32]
|J|=(1+3) Ns

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