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 K.E. 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

Answer: (C)

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

Answer 2

Suppose mass and speed of man is $M$ and $V$ respectively. Suppose the speed of the boy is $v$ Then. $\frac{1}{2} M V^{2}=\frac{1}{2}\left[\frac{1}{2} \cdot\left(\frac{M}{2} \cdot\right) v^{2}\right]$ ...(1) $\frac{1}{2} M(V+1)^{2}=\frac{1}{2}\left(\frac{M}{2}\right) v^{2}$ ...(2) Dividing equation (1) by equation (2) we obtain $\frac{V^{2}}{(V+1)^{2}} =\frac{1}{2}$ or $\frac{V}{(V+1)}=\frac{1}{\sqrt{2}}$ or $\sqrt{2} V =V+1$ $V(\sqrt{2}-1) =1$ $V =\frac{1}{\sqrt{2}-1} m / s$