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(C) Let $M$ be the mass of the man and $m$ be the mass of the boy. Given $m = M/2$.Let $V$ be the initial velocity of the man and $v$ be the velocity

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $M$ be the mass of the man and $m$ be the mass of the boy. Given $m = M/2$.Let $V$ be the initial velocity of the man and $v$ be the velocity of the boy.The kinetic energy of the man is $K_M = \frac{1}{2} M V^2$ and the kinetic energy of the boy is $K_b = \frac{1}{2} m v^2$.According to the problem,$K_M = \frac{1}{2} K_b$,so $\frac{1}{2} M V^2 = \frac{1}{2} (\frac{1}{2} m v^2) = \frac{1}{4} m v^2$. Substituting $m = M/2$,we get $\frac{1}{2} M V^2 = \frac{1}{4} (M/2) v^2$,which simplifies to $V^2 = v^2/4$,or $v = 2V$.When the man increases his speed by $1 \, m/s$,his new kinetic energy is $\frac{1}{2} M (V+1)^2$. This is equal to the boy's kinetic energy $K_b = \frac{1}{2} m v^2 = \frac{1}{2} (M/2) (2V)^2 = M V^2$.Thus,$\frac{1}{2} M (V+1)^2 = M V^2$.Dividing by $M/2$,we get $(V+1)^2 = 2V^2$.Taking the square root,$V+1 = \sqrt{2} V$.Rearranging gives $1 = V(\sqrt{2} - 1)$,so $V = \frac{1}{\sqrt{2} - 1} \, m/s$.

$A$ running man has half the kinetic energy of a boy who has half the man's mass. The man speeds up by $1 \, m/s$ so that his kinetic energy becomes equal to that of the boy. The original speed of the man is:

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