Question

You are with 17 other people on a boat at rest in frictionless water. The group's total mass is 1326kg , and the boat's mass is 1.2×104kg . The entire party walks the 6.3m distance from bow to stern.

How far does the boat move?

Answer 1

Looking at the center of mass of the system beforehand, we can see
\[X_{1 \ CoM}=\frac{m_b x_1 + m_g x_0}{m_b+m_g}=\frac{(1.2\times 10^4 kg)(3.15m)+(1326 kg)(0)}{(1326+1.2\times 10^4 kg)}\]
\[X_{1 \ CoM}=2.84m\]
So the center of mass in our coordinate (attached to the boat) system is originally 2.84 meters away from the left hand side of the boat.

After they move, we have
\[X_{2 \ CoM} = \frac{m_b x_1 + m_g x_2}{m_b + m_g}=\frac{(1.2\times 10^4 kg)(3.15m)+(1326 kg)(6.3m)}{(1326+1.2\times 10^4 kg)}\]
\[X_{2 \ CoM}=3.46m\]
So after they move, in the coordinate system defined the center of mass is 3.26m away from the left hand side of the boat.

But there's been no net force on the boat! That means the Center of Mass could not have moved! To a reference frame that's not on the boat the center of mass remains stationary, so the boat itself moves the distance
\[X_{2 \ CoM}-X_{1 \ CoM}=.62m\]
in the negative x direction (in the direction of the bow).