Question

Suresh and Shyam of masses $60\text{kg}$ and $70\text{kg}$ respectively are standing at the two extremes of a long boat of mass $50\text{kg}$ and length $6\text{m}$ standing still in water. After some time, they come to the middle of the boat. Neglecting water resistance, how far does the boat move on the water during the process?

• A. $\frac{2}{3}\text{m}$
• B. $\frac{1}{6}\text{m}$
• C. $\frac{5}{6}\text{m}$
• D. $\frac{4}{3}\text{m}$

Answer 1

The correct option is B $\frac{1}{6}\text{m}$

Mass of boat $=m_1=50\text{kg}$
Mass of Suresh $=m_2=60\text{kg}$
Mass of Shyam $=m_3=70\text{kg}$

When they come to middle,
Let displacement $\left(\Delta x_1\right)$ of boat w.r.t ground $=d$
Displacement $\left(\Delta x_2\right)$ of Suresh w.r.t ground = displacement of Suresh w.r.t boat + displacement of boat w.r.t ground $=(3+d)$
Displacement$\left(\Delta x_3\right)$ of Shyam w.r.t ground = displacement of Shyam w.r.t boat + displacement of boat w.r.t ground $=(-3+d)$


Since, initially all were at rest and no external force is acting in horizontal direction,
$\Rightarrow \Delta x_{\text{COM}}=0$

$\Rightarrow \frac{m_1\Delta x_1+m_2\Delta x_2+m_3\Delta x_3}{m_1+m_2+m_3}=0$

$\Rightarrow \frac{50d+60(d+3)+70(d-3)}{50+60+70}$
$\Rightarrow 50d+60d+70d+180-210=0$
$180d=30$
$d=\frac{1}{6}\text{m}$