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 to have same kinetic energy as that of the boy. The original speed of the man will be

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

Answer 1

The correct option is C $\frac{1}{(\sqrt{2}-1)}m/s$

Let M$=$mass of the man;

$m=$mass of the boy.

$V=$initial velocity of the man;

$v=$intial velocity of boy

According to Qs $\frac{1}{2}M{V}^2=\frac{1}{2}\left(\frac{1}{2}m{v}^2\right)=\frac{1}{4}m{v}^2$

$m=\frac{M}{2}$

$\frac{1}{2}M{V}^2=\frac{1}{4}\frac{M}{2}{v}^2$

$V=\frac{v}{2}$

In the second case,$\frac{1}{2}M(V+1{)}^2=\frac{1}{2}m{v}^2=\frac{1}{2}\frac{M}{2}.4V^2=M{V}^2$

$V^2-2V-1=0$

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

Hence, option C is correct.