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

The Correct Option is C

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.