A block of mass `m` moving with speed `v_(0)` strikes another particle of mass `2 m` at rest. If collision is head - on and the coefficient of restitution `e = 1//2` , then the loss in kinetic energy 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

Use direct formula ,
Loss in `K.E. ,Delta K = (1)/(2) (m_(1) m_(2))/(m_(1) + m_(2)) (u_(1) - u_(2))^(2) (1 - e^(2))`
`Delta K = (1)/(2) (m xx 2m)/(m + 2m) (v_(0) - 0)^(2) [ 1 - (1//2)^(2)]`
`= (1)/(4) mv_(0)^(2)`