Question

A running man has half the kinetic energy that a running boy of half his mass has. The man speeds up by $1.0$ m/s and now has the same kinetic energy as the boy. The original speed of the man expressed in m/s must be

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

Answer 1

Answer: (A)

$\sqrt{2}+1$

Let the mass, energy and the speed of the man be $m,\frac{k}{2},V_{man}$

and the mass, energy and speed of the boy be $\frac{m}{2},k,V_{boy}$

$\frac{k}{2}=\frac{1}{2}m(V_{man}{)}^2$ and $k=\frac{1}{2}\times \frac{m}{2}\times \left(V_{boy}\right)$

$\Rightarrow V_{man}.V_{boy}=1:2$

Now $k=\frac{1}{2}m(V_{man}+1{)}^2$

$\Rightarrow \frac{1}{4}m{V}_{boy}^2=\frac{1}{2}m(V_{man}+1{)}^2$

$m{V}_{man}^2=\frac{1}{2}m(V_{man}+1{)}^2$

$2V_{man}^2=(V_{man}+1{)}^2$

$\sqrt{2}{V}_{man}={\sqrt{V}}_{man}+1$

$\Rightarrow V_{man}=\sqrt{2}+1$