NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
Two blocks A and B of equal mass are released on two sides of a fixed wedge C as shown in the figure. Find the acceleration of the centre of mass of blocks A and B . Neglect friction.
Options
- Ag 2 , downwards
- Bg , downwards
- Cg 2 , upwards
- D3 4 g , downwards
Correct answer
A. g 2 , downwards
Step-by-step solution
Acceleration of both the blocks will be g   sin 45 ° or g 2 at right angles to each other. Now, a → com = m A a → A + m B a → B m A + m B Here, m A   =   m B ∴     a com = 1 2 a A + a B = 1 2 g (downwards)