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

Options

A. `sqrt(2) ms^(-1), 2sqrt(2) - 1 ms^(-1)`
B. `(sqrt(2) - 1) ms^(-1), 2 (sqrt(2) - 1) ms^(-1)`
C. `(sqrt(2) + 1) ms^(-1) , 2(sqrt(2) + 1) ms^(-1)`
D. None of these

Answer

C

Explanation

Let mass of boy be m, therefore, mass of man = 2m .
As , KE of man = `1/2 ` KE of boy `therefore 1/2 (2m) u^2 = 1/2 xx 1/2 m u^2`
`u^2 = (u.^2)/4, u = (u.)/2`
When man speeds upto 1 m/s.
KE of man = KE = boy `1/2 (2m) (u + 1)^2 = 1/2 m u.^2 = 1/2 m (2u)^2 , (u + 1)^2 = 2u^2`
`u + 1 = sqrt(2) u , u = 1/(sqrt(2) - 1) = (sqrt(2) + 1)/((sqrt(2) - 1)(sqrt(2) + 1)), u = (sqrt(2) + 1) ms^(-1)`
`u. = 2u = 2 (sqrt(2) + 1) ms^(-1)`