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 that he has same kinetic energy as that of the boy. The original speed of the man is?

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

$\text{Step 1: Initial situation}$

Let mass of man in $Mkg$ and speed is $V$ , mass of boy is $M/2$ and speed of boy is $V_1$

$K.E_{man}=\frac{1}{2}(K.E_{boy})$

$\frac{1}{2}M{V}^2=\frac{1}{2}\left(\frac{1}{2}\frac{M}{2}{V}_1^2\right)$ $....\left(1\right)$

$\text{Step 2: Final situation}$

Man speeds up by $1m/s$ so new speed of man $=V+1$

$K.E_{man}=K.E_{boy}$

$\frac{1}{2}M(V+1{)}^2=\frac{1}{2}\left(\frac{M}{2}\right)V_1^2$ $....\left(2\right)$

$\text{Step 3: Divide equation (1) by equation (2)}$

$\frac{V^2}{(V+1{)}^2}=\frac{1}{2}$

$\Rightarrow V=\frac{1}{\sqrt{2}-1}$

Hence option (C) is correct.