A man of mass 70kg is standing at one end of a stationary,floating barge of mass 210kg.He then walks to the other end of the barge , a distance of 90 metres.Ignore any frictional effects between the barge and the water.
1) how far will the barge move?
2) if the man walks at an average velocity of 8m/s,what is the average velocity of the barge?

$$\begin{gathered} 1. Since there are no external forces acting on the man + barge system, the centre of mass ( cm) of the system cannot accelerate. \\ In particular, \sin ce the system is originally at rest, the centre of mass cannot move. \\ Let x = 0 denote the midpoint of the barge (which is its own centre of mass , assuming it is uniform), we can figure out the centre of mass \\ of the man + barge system: \\ {X}_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=\frac{\left[70\times \left(-45\right)\right]+\left[210\times 0\right]}{70+210}=\frac{-3150}{280}=-11.25 m \\ So the centre of mass is a dis\tan ce of 11.25 m from the midpoint of the barge, and \sin ce the man’s mass is originally at the left end, \\ the centre of mass is a dis\tan ce of 11.25 m to the left of the barge’s midpoint. \\ When the man reaches the other end of the barge, the centre of mass will, by symmetry, be 11.25 m to the right of the midpoint of the barge. \\ But, \sin ce the position of the centre of mass cannot move, this means the barge itself must have moved a dis\tan ce of 11.25 + 11.25 = 22.5 m to the left. \\ 2. Let the time it takes the man to walk across the barge be denoted by t; then , \\ t=\frac{s}{v}=\frac{90m}{8m/s} . \\ In this amount of time, the barge moves a dis\tan ce of 22.5 meters in the opposite direction, so the velocity of the barge is , \\ {v}_{barge}=\frac{-22.5}{t}=\frac{-22.5 m}{90 m}=\frac{-180}{90}=-2 m/s \end{gathered}$$