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Q.

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

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a

$\frac{1}{\sqrt{2}} m / s$

b

$\frac{1}{(\sqrt{2} - 1)} m / s$

c

$(\sqrt{2} - 1) m / s$

d

$\sqrt{2} m / s$

answer is C.

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Detailed Solution

Let m = mass of boy, M = mass of man
v = velocity of boy, V = velocity of man
$\frac{1}{2} M V^{2} = \frac{1}{2} [\frac{1}{2} m v^{2}] . . . . . . (i)$
$\frac{1}{2} M {(V + 1)}^{2} = 1 [\frac{1}{2} m v^{2}] . . . . . . (i i)$
Putting $m = \frac{M}{2}$ and solving $V = \frac{1}{\sqrt{2} - 1}$

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