A running man has the same kinetic energy as that of a boy of half his mass. The man speeds up by 2 `ms^(-1)` and the boy 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

Let `v_(M)` is velocity of man, `v_(B)` of boy, then kinetic energy
`" "K=1/2Mv^(2)""_(M)=1/2.M/2*v_(B)""^(2)`
`rArr" "v_(M)^(2)=(v_(B)""^(2))/2" ...(1)"`
`rArr" "sqrt(2)v_(M)=v_(B)`
When man speeds up 2 `ms^(-1)` and boy changes his speed by x `ms^(-1)`. Then,
`" "1/2M(v_(M)+2)^(2)=1/2*M/2*(v_(B)+x)^(2)`
`rArr" "(v_(M)+2)^(2)=(v_(B)+x)^(2)/2" ...(2)"`
From Eq. (1), we have
`" "2(v_(M)+2)^(2)=(sqrt(2)v_(M)+x)^(2)`
`rArr" "sqrt(2)(v_(M)+2)=sqrt(2)v_(M)+x`
`rArr" "+2sqrt(2)=x`