Mr. Sharma ( 50 kg ) and Mr. Gupta ( 60 kg ) are seated at the two...

Since there is no external force acting on the system in the horizontal direction, the center of mass of the system remains stationary.

Let the initial position of the center of the boat be at the origin, $x_B = 0$. The length of the boat is $L = 4 \text{ m}$.

Let Mr. Sharma ($m_S = 50 \text{ kg}$) be at one end, $x_S = -2 \text{ m}$, and Mr. Gupta ($m_G = 60 \text{ kg}$) be at the other end, $x_G = 2 \text{ m}$. The mass of the boat is $m_B = 40 \text{ kg}$.

The initial position of the center of mass of the system is:

$x_{cm} = \dfrac{m_S x_S + m_G x_G + m_B x_B}{m_S + m_G + m_B}$

$x_{cm} = \dfrac{50(-2) + 60(2) + 40(0)}{50 + 60 + 40}$

$x_{cm} = \dfrac{-100 + 120}{150} = \dfrac{20}{150} = \dfrac{2}{15} \text{ m}$

When both men move to the middle of the boat, their new positions coincide with the new center of the boat, $x'_B$. The final position of the center of mass is:

$x'_{cm} = \dfrac{(m_S + m_G + m_B) x'_B}{m_S + m_G + m_B} = x'_B$

Since the center of mass does not move, $x'_{cm} = x_{cm}$.

$x'_B = \dfrac{2}{15} \text{ m}$

The magnitude of displacement of the boat is:

$\Delta x_B = \left| x'_B - x_B \right| = \dfrac{2}{15} \text{ m} = \dfrac{2}{15} \times 100 \text{ cm} = \dfrac{40}{3} \text{ cm}$