AP EAMCET202021 Sep 2020Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual
When a moving body collides with a stationary body of (n ) times its mass, then the amount of kinetic energy transferred to the stationary body is
Options
- A( 4 n (1+n)^2 )
- B( n (1+n)^2 )
- C( n^2 (1+n)^2 )
- D( 4 n^2 (1+n)^2 )
Correct answer
A. ( 4 n (1+n)^2 )
Step-by-step solution
Suppose, mass of moving body is (m₁ ). Mass of stationary body, (M₂=n m₁ ) For elastic collision, velocity of separation = velocity of approach ( aligned v₂-v₁ & =u-0 v₂-u & =v₁ (i) aligned ) By the law of conservation of momentum, ( aligned m₁ u & =m₁ v₁+M₂ v₂ m₁ u & =m₁ v₁+n m₁ v₂ u & =v₁+n v₂ (ii) u & =v₂-u+n v₂ [From Eq. (i)] 2 u & =(n+1) v₂ v₂ & = 2 u n+1 aligned ) Putting this value in Eq. (i) (v₁=v₂-u= 2 u n+1 -u= 2 u-n u-u n+1 = (1-n) u n+1 ) ( ) Kinetic energy of moving body of mass (m₁ ), before collision