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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from 60 ° to 30 ° at point B . The block is initially at rest at A . Assume that collisions between the block and the incline are totally inelastic ( g = 10 m s - 2 ). Figure: ​ The speed of the block at point C , immediately before it leaves the second incline is -

Options

  1. A1 2 0  m s - 1
  2. B1 0 5  m s - 1
  3. C9 0  m s - 1
  4. D7 5  m s - 1

Correct answer

B. 1 0 5  m s - 1

Step-by-step solution

The height of the point B is 3 3 m tan 3 0 ∘ = 3 3 m 1 3 = 3 m The energy of the block at B is KE + PE = 1 2 M 4 5   m   s - 1 2 + Mg 3 m The energy at the block at C is KE = 1 2 M V 2 The principle of conservation of energy gives 1 2 M V 2 = 1 2 M 4 5   m   s - 1 2 + M 1 0   m   s - 2 3 m This gives V = 1 0 5   m   s - 1 Therefore, the choice ( 1 0 5   m s - 1 ) is correct.

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