Body `A` of mass `4m` moving with speed `u` collides with another body `B` of mass `2m` at rest, the collision is head on and elastic in nature. After the collision the fraction of energy lost by colliding body `A` is :

Options

A. `1/9`
B. `8/9`
C. `4/9`
D. `5/9`

Answer

A

Explanation

Let `v_(1) and v_(2) ` be the velocities of bodies after collision as shown in figure

Conservation of linear momentum .
`4mv_(1)+2mv_(2)=4"mu"`
Coefficient of elasticity can be used as follows :
` - 1= (v_(2)-v_(1))/(0-u) rArr v_(1)-v_(1)=u`
On solving the above two equations we get
`v_(1) = u//3 `
Fraction of energy lost ` = (1/2 (4m)u^(2)-1/2(4m)(u/3)^(2))/(1/2(4m)u^(2))`
` = 1-1/9 = 8/9 `