NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
Three identical blocks A , B and C are placed on a horizontal frictionless surface. The blocks B and C are at rest but A is approaching towards B with a speed 10 m s - 1 . The coefficient of restitution for all collisions is 0.5 . The speed of the block C just after the collision is
Options
- A5.6   m   s - 1
- B6   m   s - 1
- C8   m   s - 1
- D10   m   s - 1
Correct answer
A. 5.6   m   s - 1
Step-by-step solution
For collision between blocks A and B , e = v B - v A u A - u B = v B - v A 10 - 0 = v B - v A 10 ∴ v B - v A = 10 e = 10 × 0 .5 = 5          … . ( i ) From principle of momentum conservation, m A u A + m B u B = m A v A + m B v B m × 10 + 0 = mv A + mv B ∴    v A + v B = 10                                 … . ( ii ) Adding eqs. (i) and (ii