Question

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by $1\,m/s$ , so as the have same kinetic energy as that of the boy. The original speed of the man is:

• A. $\left( \sqrt{2}-1 \right)m/s$
• B. $\sqrt{2}\,m/s$
• C. $\frac{1}{\sqrt{2}-1}\,m/s$
• D. $\frac{1}{\sqrt{2}}\,m/s$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{2}-1}\,m/s$

Answer 2

The kinetic energy of a moving body is equal to half the product of the mass $(m)$ of the body and the square of its speed $\left(v^{2}\right)$.
Kinetic energy $=\frac{1}{2} \times \text { mass } \times(\text { speed })^{2}$
i.e., $=\frac{1}{2} m v^{2}$
Let mass of man is $M\, kg$ and speed is $v$ and speed of boy is $v_{1}$, then
$\frac{1}{2} M v^{2}=\frac{1}{2}\left(\frac{1}{2} \frac{M}{2} v_{1}^{2}\right)\,\,\,...(1)$
When man speeds up by $1 m / s$, then
$v =v+1$
$\therefore \frac{1}{2} M(v+1)^{2} =\frac{1}{2}\left(\frac{M}{2}\right) \cdot v_{1}^{2}\,\,\,...(2)$
Dividing E (1) by (2), we get
$\frac{v^{2}}{(v+1)^{2}} =\frac{1}{2}$
$\Rightarrow \frac{v}{v+1} =\frac{1}{\sqrt{2}}$
$\Rightarrow \sqrt{2} v =v+1$
$\Rightarrow v =\frac{1}{\sqrt{2}-1} \,m / s$
Note : In the formula for kinetic energy speed occurs in the second power and so has a larger effect compared to mass.