NTA Abhyas JEE Main2020PhysicsMotion in Two DimensionsPractice
A block is held stationary at the position shown in the figure over the surface of a solid paraboloid. What should be the magnitude of the velocity, needed to be given to the block at this point such that it moves along the surface of the paraboloid without having any normal reaction anywhere? g = 10 m s - 2 r = 2 h , h = 10 3 m . Consider motion only along the plane of the paper.
Options
- A20   m   s - 1
- B10 2 3   m   s - 1
- C10   m   s - 1
- DCannot be calculated
Correct answer
C. 10   m   s - 1
Step-by-step solution
Considering that the block is thrown horizontally with velocity u from the top point O. - h = r tan ⁡ θ - 1 2 g r 2 u 2 cos 2 ⁡ θ - h = g r 2 2 u 2   ⇒     u 2 = g r 2 2 h Velocity at point P v 2 = u 2 + 2 g h = g r 2 2 h + 2 g h = g r 2 + 4 h 2 2 h = g . 3 h = 10 × 3 × 10 3 = 100 v = 10   m   s - 1