A man who is running has half the kinetic energy of a boy of half his mass the man speeds up by 1 `ms^(-1)` and then has the same kinetic energy as the boy the initial speed of the boy is

Options

A. `sqrt(2) + 1 ms^(-1)`
B. `2(sqrt(2)+ 1 ) ms^(-1)`
C. `sqrt(2) ms^(-1)`
D. `2 ms^(-1)`

Answer

B

Explanation

According to the question let the mass of man be m
So mass fo boy is `(m)/(2)`
Let speed of man be `v_(1)` and that of boy be `v_(2)`
`:. ` kinetic energy of man `(1)/(2)` kinetic energy of by
Or `(1)/(2) mv_(1)^(2) = (1)/(2) xx (1) /(2) xx (m)/(2) v_(2)^(2) ` or `(v_(1)^(2))/(v_(2)^(2)) = (1)/(4)`
`rArr (V_(1))/(v_(2)) = (1)/(2)`
`:. , v_(1) = (1)/(2) v_(2)`
If the velocity of man increased by 1 `m//s` then new velocity of man `v_(1) = ((v_(2))/(2)+ )`
Now kinetic energy of man (KE) - kinetic energy of boy `(KE_(B))`
`:. , (1)/(2) xx m xx ((v_(2))/(2) + 1) ^(2) = (1)/(2 ) xx (m)/(2) xx v_(2)^(2)`
`(v_(2))/(2) + 1 = (v_(2))/(sqrt(2))`
`2sqrt(2) = 2v_(2) - sqrt(2)v_(2)`
`v_(2) = (2)/((sqrt(2)-1))`
`rArr v_(2) = (2)/((sqrt(2)-1)) xx ((sqrt(2)+1))/((sqrt(2) + 1)) = (2 (sqrt(2)+1))/((2-1))`
So the intial speed of the boy `v_(2) = 2(sqrt(2)+1)`