Question

A running man has half the kinetic energy than a boy of half his mass has. The man speed up by $1.0\, ms^{-1}$ and then he has the same energy as the boy. The original speeds of the man and boy respectively are

• A. $2.4\, ms^{-1}$, $1.2\, ms^{-1}$
• B. $1.2\, ms^{-1}$, $4.4\, ms^{-1}$
• C. $2.4\, ms^{-1}$, $4.8\, ms^{-1}$
• D. $4.8\, ms^{-1}$, $2.4\, ms^{-1}$

Answer 1

Answer: (C)

$2.4\, ms^{-1}$, $4.8\, ms^{-1}$

Answer 2

As the kinetic energy of the man is half the kinetic energy of the boy $\therefore \frac{1}{2}MV^{2} = \frac{1}{2}\left(\frac{1}{2}mv^{2}\right)$ or $\frac{1}{2}MV^{2} = \frac{1}{2} \frac{1}{2} \left(\frac{M}{2}\right)v^{2}\,\,\left(\because m = \frac{M}{2}\right)$ or $v^{2} = 4V^{2}$ or $v = 2V$. Here, $M$ and $V$ for man and $m$ and $v$ for boy. Now when the velocity of the man is increased to $\left(V + 1\right)$, then kinetic energies become equal $\frac{1}{2}M\left(V+1\right)^{2}=\frac{1}{2}\left(\frac{M}{2}\right)v^{2}$= $\frac{1}{4}M\left(2V\right)^{2}$ $V^{2} + 2V+ 1 = 2V^{2}$ or $V^{2} - 2V -1 = 0$ $V = \sqrt{2}+1= 2.4\,ms^{-1}$ and $v = 2\left(\sqrt{2}+1\right) = 4.8\,ms^{-1}$.