Question

Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the process?

Answer 1

It is given that:
Mass of Mr. Verma, m₁ = 50 kg
Mass of Mr. Mathur, m₂ = 60 kg
Mass of boat, m₃ = 40 kg

Let A be the origin of the system (boat plus two men).

Initially, Mr. Verma and Mr. Mathur were at two extremes of the boat.



∴ Distance of the centre of mass:

$$\begin{gathered} X_{\mathrm{cm}}=\frac{m_1\times x_1+m_2\times x_2+m_3\times x_3}{m_1+m_2+m_3} \\ =\frac{60\times 0+50\times 4+40\times 2}{60+50+40} \\ =\frac{280}{150}=1.87\mathrm{m}\mathrm{from}\mathrm{A} \end{gathered}$$

As no external force is experienced in longitudinal direction, the centre of mass would not shift.

Initially, the centre of mass lies at a distance of 2 m from A.
When the men move towards the middle of the boat, the centre of mass shifts and lies at 1.87 m from A.

Therefore, the shift in centre of mass = 2 − 1.87 = 0.13 m, towards right
Hence, the boat moves 13 cm or 0.13 m towards right.