Two persons of masses 55 kg and 65 kg are at the opposite ends of a boat. The length of the boat is 3.0 m and its mass is 100 kg. The 55 kg man walks upto the 65 kg man and sits with him. If the boat is in still water, the centre of mass of the system shifts by
As the boat was initially at rest and no external force acts on it centre of mass will remain at rest.
A boat of length \( 12\text{ m} \) and mass \( 840\text{ kg} \) is floating without motion in still water. A man of mass \( 60\text{ kg} \) standing at one end of it walks to the other end of it and stops. The magnitude of displacement of the boat relative to the ground is:
Since no external horizontal force acts on the boat-man system, the center of mass does not move. The displacement of the boat is \( x = \frac{m L}{m + M} = \frac{60 \times 12}{60 + 840} = 0.8\text{ m} = 80\text{ cm} \).
Two bodies with masses \( m_1 \) and \( m_2 \) (\( m_1 > m_2 \)) are joined by a string passing over a fixed pulley. Assuming masses of the pulley and thread are negligible. Then the acceleration of the centre of mass of the system is:
The acceleration of each block is \( a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \). The acceleration of the center of mass is \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \).