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(B) Let the mass of the man be $m$. Then the mass of the boy is $m_b = \frac{m}{2}$.Let the initial speed of the man be $v_m$ and the speed of the boy

Vedclass · Accepted Answer

$2(\sqrt{2}+1) \,ms^{-1}$

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(B) Let the mass of the man be $m$. Then the mass of the boy is $m_b = \frac{m}{2}$.Let the initial speed of the man be $v_m$ and the speed of the boy be $v_b$.According to the problem,the kinetic energy of the man is half the kinetic energy of the boy:$\frac{1}{2} m v_m^2 = \frac{1}{2} (\frac{1}{2} m_b v_b^2) = \frac{1}{2} (\frac{1}{2} \cdot \frac{m}{2} v_b^2) = \frac{1}{8} m v_b^2$.Simplifying,$v_m^2 = \frac{1}{4} v_b^2$,which gives $v_m = \frac{v_b}{2}$.When the man speeds up by $1 \,ms^{-1}$,his new speed is $v_m' = v_m + 1 = \frac{v_b}{2} + 1$.Now,his kinetic energy equals the boy's kinetic energy:$\frac{1}{2} m (\frac{v_b}{2} + 1)^2 = \frac{1}{2} (\frac{m}{2}) v_b^2$.Dividing both sides by $\frac{m}{2}$,we get $(\frac{v_b}{2} + 1)^2 = \frac{v_b^2}{2}$.Taking the square root of both sides: $\frac{v_b}{2} + 1 = \frac{v_b}{\sqrt{2}}$.$1 = v_b (\frac{1}{\sqrt{2}} - \frac{1}{2}) = v_b (\frac{2 - \sqrt{2}}{2\sqrt{2}})$.$v_b = \frac{2\sqrt{2}}{2 - \sqrt{2}} = \frac{2\sqrt{2}(2 + \sqrt{2})}{4 - 2} = \frac{4\sqrt{2} + 4}{2} = 2\sqrt{2} + 2 = 2(\sqrt{2} + 1) \,ms^{-1}$.

$A$ man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by $1 \,ms^{-1}$ and then has the same kinetic energy as the boy. The initial speed of the boy is

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