Question

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by $1m/s$ , 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)

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 $Mkg$ 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 $1m/s$, 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 Eq. (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.