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 20m/s towards left, then the velocity v of the sphere at this instant will be:
A
20m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
15m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
16m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B15m/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(90−θ)=20cosθ ⇒vcos37∘=20cos53∘ ∴v=(20×35)45=15m/s