In the previous problem , the maximum loss in `K.E.` will be

Options

A. `(1)/(2) mv_(0)^(2)`
B. `(1)/(3) mv_(0)^(2)`
C. `(2)/(3) mv_(0)^(2)`
D. `(3)/(4) mv_(0)^(2)`

Answer

B

Explanation

For loss of `K.E`. to be maximum , the collision should be completely inelastic , i.e. the block should stick together and move with same velocity after the collision.
`(Delta K)_(max) = (1)/(2) (m_(1) m_(2))/(m_(1) + m_(2)) (u_(1) - u_(2))^(2)`
`= (1)/(2) ( m xx 2m)/( m + 2m) (v_(0) - 0)^(2)`
`= (1)/(3) mv_(0)^(2)`