Question

A running man has half the kinetic energy that a boy of half his mass has.The man speeds up by $1ms$${}^{-1}$ and then has the same kinetic energy as that of the boy. The original speeds of man and boy in $ms$${}^{-1}$ are:

• A. $(\sqrt{2}+1),(\sqrt{2}-1)$
• B. $(\sqrt{2}+1),2(\sqrt{2}+1)$
• C. $\sqrt{2},\sqrt{2}$
• D. $(\sqrt{2}+1),2(\sqrt{2}-1)$

Answer 1

The correct option is B $(\sqrt{2}+1),2(\sqrt{2}+1)$

Let,

$m$ be the mass of the man,

$m^{\prime }$ be the mass of the boy,

$v$ be the velocity of the man,

$v^{\prime }$ be the velocity of the boy.

As per the question,

$\frac{1}{2}m{v}^2=\frac{1}{2}\left(\frac{1}{2}{m}^{\prime }{v}^{\prime 2}\right)$

$\frac{1}{2}m{v}^2=\frac{1}{4}{m}^{\prime }{v}^{\prime 2}$

$\frac{1}{2}m{v}^2=\frac{1}{4}\frac{m}{2}{v}^{\prime 2}$

$v=\frac{v^{\prime }}{2}$

$v^{\prime }=2v$

In later case

$\frac{1}{2}m(v+1{)}^2=\frac{1}{2}{m}^{\prime }{v}^{\prime 2}$

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

$\frac{1}{2}m(v+1{)}^2=m{v}^2$

$v^2+2v+1=2v^2$

$v^2-2v-1$

$v^2-2v=1$

$v^2-2v+1=2$

$(v-1{)}^2=2$

$v-1=\sqrt{2}$

$v=\sqrt{2}+1$

and $v^{\prime }=2v$

$v^{\prime }=2(\sqrt{2}+1)$

Therefore, the speed of man is $\sqrt{2}+1$ and the speed of boy is $2(\sqrt{2}+1)$.