A running man has the same kinetic energy as that of a boy of half his mass. The man speed up by `2 ms^(-1)` and the body changes his speed by `x ms^(-1)` so that the kinetic energies of the boy and the man are again equal. Then `x` in `ms^(-1)` is

Options

A. `-2sqrt(2)`
B. `+2sqrt(2)`
C. `sqrt(2)`
D. `2`

Answer

B

Explanation

If `m` is mass of boy, the mass of man `=2m`
As K.E. of man `=` K.E. of boy
`= :. (1)/(2)(2m)upsilon^(2)=(1)/(2)m upsilon^(2)`
`upsilon^(')=sqrt(2)upsilon`
In the second case,
`(1)/(2)(2m) (upsilon+2)^(2)=(1)/(2)m(upsilon^(')+x)^(2)`
`2 (upsilon+2)^(2)=(sqrt(2)upsilon+x)^(2)`
`sqrt(2)upsilon+x=sqrt(2)(upsilon+2)=sqrt(2)upsilon+2sqrt(2)`
`x=2sqrt(2)m//s`