• # question_answer A man is running along a road with speed u. On his chest there is a paper of mass m and area S. There is a wind blowing against the man at speed V. Density of air is$\rho$. Assume that the air molecules after striking the paper come to rest relative to the man. The minimum coefficient of friction between the paper and the chest so that the paper does not fall is A) $\frac{mg}{\rho S{{(V-u)}^{2}}}$        B)        $\frac{mg}{\rho S{{(2V-u)}^{2}}}$ C) $\frac{mg}{\rho S{{(V+u)}^{2}}}$      D)        $\frac{mg}{\rho S{{(2V+u)}^{2}}}$

[c] Volume of air striking the paper in unit time $=S(V+u)$ Mass of air striking the paper in unit time $=\rho S(V+u)$ In reference frame of man, the air molecules strike at a speed $(V+u)$ and comes to rest. $\therefore$ Rate of change of momentum for air particles $=\rho S(V+u)\,\,(V+u)$ $\therefore$ Force on paper due to air $=\rho S{{(V+u)}^{2}}$ For the paper to not fall, friction on it must balance its weight. ${{f}_{\max }}\ge mg$ $\mu \rho S\,\,{{(V+u)}^{2}}\ge mg$ $\mu \ge \frac{mg}{\rho S{{(V+u)}^{2}}}$