A man is running on horizontal road has half the kinetic energy of a boy of half of his mass.When the man speeds up by $1 m / s$ then his kinetic energy becomes equal to kinetic energy of the boy, the original speed of the man is

\sqrt{2} m / s

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

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2 m / s

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(\sqrt{2} + 1) m / s

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Solution

The correct option is D $(\sqrt{2} + 1) m / s$
$\frac{1}{2} M V^{2} = \frac{1}{4} \times \frac{1}{2} M V_{1}^{2}$
$4 M =$ Mass of man.
$V^{2} = \frac{V_{1}^{2}}{4}$
$V = \frac{V_{1}}{2}$
where $V =$ Velocity of mass
$V_{1} =$ Velocity of boy.
$\frac{1}{2} M (V + 1)^{2} = \frac{1}{2} (\frac{M}{2}) V_{1}^{2}$
$(V + 1)^{2} = \frac{(V_{1})^{2}}{2}$
$(V + 1)^{2} = {(\sqrt{2} V_{o})}_{o}^{2}$
$V + 1 =_{-}^{+} \sqrt{2} V$
$V = \sqrt{2} + 1$

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A man is running on horizontal road has half the kinetic energy of a boy of half of his mass.When the man speeds up by 1m/s then his kinetic energy becomes equal to kinetic energy of the boy, the original speed of the man is

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A man is running on horizontal road has half the kinetic energy of a boy of half of his mass.When the man speeds up by $1 m / s$ then his kinetic energy becomes equal to kinetic energy of the boy, the original speed of the man is