Question

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

• A. $4\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\sqrt{2}$
• D. $2$

Answer 1

Answer: (B)

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