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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?
Solution
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:
\[X_{cm} = \frac{m_1 \times x_1 + m_2 \times x_2 + m_3 \times x_3}{m_1 + m_2 + m_3}\]
\[ = \frac{60 \times 0 + 50 \times 4 + 40 \times 2}{60 + 50 + 40}\]
\[ = \frac{280}{150} = 1 . 87 \text{ m from A}\]
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.
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