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Question

A sphere of radius R is in contact with a wedge. The point of contact is at a distance R5 from the ground as shown in the figure. The wedge is moving with velocity 20 m/s towards left, then the velocity v of the sphere at this instant will be:


A
20 m/s
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B
15 m/s
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C
16 m/s
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D
12 m/s
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Solution

The correct option is B 15 m/s
Velocity of sphere and wedge at the given instant are shown in the figure given below. θ is the angle made by normal reaction N on the sphere from the wedge, with the horizontal.


Normal reaction will act in a direction perpendicular to the inclined surface and passes through the centre of the sphere. Hence, from the geometry of the figure:
sinθ=(4R5)R=45
θ=53

To maintain contact at the surface, the component of velocity of the wedge and sphere should be equal along the direction perpendicular to the inclined surface (or normal direction).
v2=v1 (from the diagram)
vcos(90θ)=20cosθ
vcos37=20cos53
v=(20×35)45=15 m/s

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