Q. A man running has half the kinetic energy of a boy of half his mass. The man speeds up by $1\ ms^{- 1}$ and then has KE as that of the boy. What were the original speeds of man and the boy?
A) $\sqrt{2}ms^{- 1};2\sqrt{2} - 1ms^{- 1}$
B) $(\sqrt{2} - 1)ms^{- 1},2(\sqrt{2} - 1)ms^{- 1}$
C) $\left( \sqrt{2} + 1 \right)ms^{- 1};2(\sqrt{2} + 1)ms^{- 1}$
D) None of the above
Explanation
(c) Let mass of boy be $m$. Therefore, mass of man $= 2\ m$, as
KE of man $= \ \frac{1}{2}$ KE of boy
$\therefore\ \frac{1}{2}(2m)u^{2} = \frac{1}{2} \times \frac{1}{2}mu^{'2}$
$u^{2} = \frac{u^{'2}}{4},u = \frac{u'}{2}$
When man speeds up to 1 $ms^{- 1},$
KE of man = KE of boy
$\frac{1}{2}(2m)(u + 1)^{2} = \frac{1}{2}m{u'}^{2} = \frac{1}{2}m(2u)^{2}\ $
$(u + 1)^{2} = 2u^{2}$
$u + 1 = \sqrt{2}u$
$u = \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)}$
$u = \left( \sqrt{2} + 1 \right)ms^{- 1}$
$u^{'} = 2u = 2\left( \sqrt{2} + 1 \right)ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.