A $750 \mathrm{~kg}$ boat is $10 \mathrm{~m}$ long and is floating without motion on still water. A man of mass $80 \mathrm{~kg}$ is at one end and if he runs to another end of the boat and stops, the displacement of boat is

• A

  1.8 m in the direction of displacement of man
• Correct answer
  B

  0.96 m in the direction of opposite to the displacement of man
• C

  0.96 m in the direction of displacement of the man
• D

  1.8 m in the direction opposite to displacement of man

Solution

Mass of boat, $M=750 \mathrm{~kg}$ Mass of man, $m=80 \mathrm{~kg}$ Length, $L=10 \mathrm{~m}$
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 $=x^{\prime} \mathrm{cm}$ By conservation of linear momentum along the boat, distance of centre of mass from reference line remains same. $$ \begin{aligned} X_{\mathrm{CM}} & =X^{\prime}{ }_{\mathrm{CM}} \\ \frac{m \times 0+M \frac{L}{2}}{m+M} & =\frac{m(L-x)+M\left(\frac{L}{2}-x\right)}{m+M} \\ \frac{M L}{2} & =m L-m x+\frac{M L}{2}-M x \\ (M+m) x & =m L \\ x & =\frac{m L}{M+m}=\frac{80 \times 10}{750+80}=0.96 \mathrm{~m} \end{aligned} $$ Hence, the displacement of boat is $0.96 \mathrm{~m}$ in opposite to the displacement of man.

Hence, the correct answer is option: 'B'.