A body of mass `m_(1)` moving at a constant speed undergoes an elastic head on collision with a body of mass `m_(2)` initially at rest. The ratio of the kinetic energy of mass `m_(1)` after the collision to that before the collision is -

Options

A. `((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)`
B. `((m_(1)+m_(2))/(m_(1)-m_(2)))^(2)`
C. `((2m_(1))/(m_(1)+m_(2)))^(2)`
D. `((2m_(2))/(m_(1)+m_(2)))^(2)`

Answer

A

Explanation

`m_(1)v_(0)=m_(2)v_(2)-m_(1)v_(1)" ".........(1)`
`e=(v_(1)+v_(2))/v_(0)=1" ".........(2)`
Solving (1) and (2)
`(v_(1)/v_(0))^(2)=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)`