The kinetic energy of a man is half that of a | vedclass

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(C) Let the mass of the man be $M$ and his initial speed be $V$. Let the mass of the boy be $m = M/2$ and his speed be $v$.Given,kinetic energy of the

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

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(C) Let the mass of the man be $M$ and his initial speed be $V$. Let the mass of the boy be $m = M/2$ and his speed be $v$.Given,kinetic energy of the man is half that of the boy:$\frac{1}{2} M V^2 = \frac{1}{2} \left( \frac{1}{2} m v^2 \right)$Substituting $m = M/2$:$\frac{1}{2} M V^2 = \frac{1}{2} \left( \frac{1}{2} \cdot \frac{M}{2} \cdot v^2 \right) \implies V^2 = \frac{v^2}{4} \implies v = 2V$ ... $(i)$When the man increases his speed by $1 \ m/s$,his kinetic energy becomes equal to the boy's kinetic energy:$\frac{1}{2} M (V + 1)^2 = \frac{1}{2} m v^2$$\frac{1}{2} M (V + 1)^2 = \frac{1}{2} \left( \frac{M}{2} \right) v^2$$(V + 1)^2 = \frac{v^2}{2}$Taking the square root: $V + 1 = \frac{v}{\sqrt{2}}$ ... (ii)Substitute $v = 2V$ from $(i)$ into (ii):$V + 1 = \frac{2V}{\sqrt{2}} = \sqrt{2} V$$1 = V(\sqrt{2} - 1)$$V = \frac{1}{\sqrt{2} - 1} \ m/s$.

The kinetic energy of a man is half that of a boy whose mass is half that of the man. If the man increases his speed by $1 \ m/s$,his kinetic energy becomes equal to that of the boy. What is the initial speed of the man?

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