TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Mass of boat, $M=750\,\text{kg}$ Mass of man, $m=80\,\text{kg}$ Length, $L=10\,\text{m}$Let $C$ be the centre of mass of boat-man system when mass was at point $A$. When man reaches at point $B$. Let the boat displaces by $x$ distance in opposite to displacement of man. Then, new position of centre of mass from the reference line $=x'\,\text{cm}$ By conservation of linear momentum along the boat, distance of centre of mass from reference line remains same. $\therefore \;\; X_{\text{CM}} \;\; =X_{\text{CM}}'$ $\;\frac{m \times 0 + M \frac{L}{2}}{m+M} \;\; =\;\frac{m(L-x)+M\left(\frac{L}{2}-x\right)}{m+M}$ $\;\frac{M L}{2} \;\; =m L-m x+\;\frac{M L}{2}-M x$ $(M+m) x \;\; =m L$ $x \;\; =\;\frac{m L}{M+m}=\;\frac{80 \times 10}{750+80}=0.96\,\text{m}$ Hence, the displacement of boat is $0.96\,\text{m}$ in opposite to the displacement of man.