A man running has half the kinetic energy of a boy of half his mass. The man speeds up by 1 m/s and then has kinetic energy as that of the boy. What were the original speeds of man and the boy ? 

Options

A. `sqrt(2)m//s : 2sqrt(2) m//s`
B. `(sqrt(2) - 1) m//s : 2(sqrt(2)-1) m//s`
C. `(sqrt(2) + 1) m//s : 2(sqrt(2) +1) m//s`
D. `1/(sqrt2) m//s , sqrt(2)m//s`

Answer

C

Explanation

Let the mass of boy be m.
Therefore, mass of man is 2m.
As KE of man `= 1/2 KE` of boy
`:. 1/2 (2m) u^2 = 1/2 xx 1/2 m u.^2`
or `u^2 = (u.^(2))/(4) or u = (u.)/(2)`
When man speeds up by 1 m/s.
KE of man = KE of boy
`:. 1/2 (2m) (u + 1)^2 = 1/2 m u.^(2) = 1/2 m (2u)^2 " "` (using (i))
or `(u +1)^2 = 2u^2 or u + 1 = sqrt(2) u`
or `u = 1/(sqrt(2) - 1) = (sqrt(2) + 1)/((sqrt(2) - 1)(sqrt(2) + 1)) = (sqrt(2) + 1) m//s`
`:. u. = 2u = 2(sqrt(2) + 1) m//s` .