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 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

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

Let m = mass of boy, M = mass of man
v = velocity of boy, V = velocity of man
$\frac{1}{2}M{V}^2=\frac{1}{2}[\frac{1}{2}m{v}^2]....\left(i\right)$
$\frac{1}{2}M(V+1{)}^2=1[\frac{1}{2}m{v}^2].....(ii)$
Putting $m=\frac{M}{2}$ and solving $V=\frac{1}{\sqrt{2}-1}$