Physics - W.E.P. Question with Solution | TestHub

Let$v_M$be velocity of man,$v_B$of boy, then kinetic energy according to question ,

$ie K=\frac{1}{2}M{v}_M^2=\frac{1}{2}.\frac{M}{2}.v_B^2$

Or$v_M^2=\frac{V_B^2}{2}$

Or$\sqrt{ 2}{v}_M=v_B$

When man speeds up$2 \mathrm{m} {\mathrm{s}}^{-1}$and boy changes his speed by$x \mathrm{m} {\mathrm{s}}^{-1}$.Then ,

$\frac{1}{2}M{\left(v_M+2\right)}^2=\frac{1}{2}\cdot \frac{M}{2}\cdot {\left(v_B+x\right)}^2$

Or${\left(v_M+2\right)}^2=\frac{{\left(v_B+x\right)}^2}{2}$

$2{\left(v_M+2\right)}^2={\left(\sqrt{2}{v}_M+x\right)}^2$$\left(\because v_B=\sqrt{2}{v}_m\right)$

Or$\sqrt{2}\left(v_M+2\right)=\sqrt{2}{v}_M+x$

Or +$2\sqrt{2}=x$