A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 m/s so as to have same K.E. as that of boy. The original speed of the man will be

The correct option is C.

Let m = mass of the boy, M = mass of the man, v = velocity of the boy and V = velocity of the man

Initial kinetic energy of man = $\frac{1}{2}M{V}^2=\frac{1}{2}\left[\frac{1}{2}m{v}^2\right]=\frac{1}{2}\left[\frac{1}{2}\left(\frac{M}{2}\right)v^2\right]$

[As $m=\frac{M}{2}$]

$\Rightarrow V^2=\frac{v^2}{4}\Rightarrow V=\frac{v}{2}....\left(i\right)$

When the man speeds up by 1 m/s,

$\frac{1}{2}M(V+1{)}^2=\frac{1}{2}m{v}^2=\frac{1}{2}\left(\frac{M}{2}\right)v^2$

$\Rightarrow (V+1{)}^2=\frac{v^2}{2}$
$\Rightarrow V+1=\frac{v}{\sqrt{2}}\left(ii\right)$

From (i) and (ii) we get speed of the man $V=\frac{1}{\sqrt{2}-1}m/s$.