A running man has the same kinetic energy as that of a boy of half his mass. The man speed up by $2\ {ms}^{- 1}$ and the boy changes his speed by $x = {ms}^{- 1}$, so that the kinetic energies of the boy and the man are again equal. Then $x$ in ${ms}^{- 1}$ is

Question

Dr. Abubacker · Accepted Answer

(b) Let $v_{M}$is velocity of man,$v_{B}$of boy, then kinetic energy according to question ,
$ie\ K = \frac{1}{2}Mv_{M}^{2} = \frac{1}{2}.\frac{M}{2}.v_{B}^{2}$
Or $v_{M}^{2} = \frac{V_{B}^{2}}{2}$
Or $\ \ \sqrt{\ 2}v_{M} = v_{B}$
When man speeds up 2 $ms^{- 1}$and boy changes his speed by $x\ ms^{- 1}$.Then ,
$\frac{1}{2}M\left( v_{M} + 2 \right)^{2} = \frac{1}{2} \bullet \frac{M}{2} \bullet \left( v_{B} + x \right)^{2}$
Or $\left( v_{M} + 2 \right)^{2} = \frac{\left( v_{B} + x \right)^{2}}{2}$
$2\left( v_{M} + 2 \right)^{2} = \left( \sqrt{2}v_{M} + x \right)^{2}$ $\left( \because v_{B} = \sqrt{2}v_{m} \right)$
Or $\sqrt{2}\left( v_{M} + 2 \right) = \sqrt{2}v_{M} + x$
Or +$2\sqrt{2} = x$

Q. A running man has the same kinetic energy as that of a boy of half his mass. The man speed up by $2\ {ms}^{- 1}$ and the boy changes his speed by $x = {ms}^{- 1}$, so that the kinetic energies of the boy and the man are again equal. Then $x$ in ${ms}^{- 1}$ is

Explanation

Reviewed by Dr. Abubacker

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