Solved Problem on Linear Momentum
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A man of mass m is sitting on the stern of a resting boat in a lake. The mass of the boat is M = 3m, and its length is L. The man rises and walks toward the bow. Neglecting the water resistance determine the distance D that the boat travels when the man walks from the stern to the bow.


Problem Data:
  • Mass of man:    m;
  • Mass of boat:    M = 3m;
  • Length of boat:    L.
Problem diagram:

Since the man-boat system is isolated from external forces, the system interaction force is internal to the system, and it is valid the Law of Conservation of Linear Momentum.

Figure 1

We choose a frame of reference in the boat (R') the man walks the length L of the boat. When the frame of reference (R) is fixed in the water, when the man walks forward, considering the conservation of linear momentum, the boat goes backward. The boat moves from a distance of D to determining, then, relative to the reference in the water the man walks at a distance of L−D.

Solution

The linear momentum is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mv} \end{gather} \]
The linear momentum of man Qm must be equal to the linear momentum of the boat Qb
\[ \begin{gather} Q_{m}=Q_{b}\\[5pt] m v=M V \end{gather} \]
the speeds of man and boat will be respectively   \( v=\frac{\Delta s}{\Delta t} \)   and   \( V=\frac{\Delta S}{\Delta t} \),   substituting in the expression above
\[ \begin{gather} m\frac{\Delta s}{\Delta t}=M\frac{\Delta S}{\Delta t}\\[5pt] m\Delta s=M\Delta S \end{gather} \]
regarding the frame of reference in the water, the man's displacement will be Δs = L−D (Figure 1), and the boat displacement will be ΔS = D, substituting these values, ​​and the mass of the boat given in the problem, into the expression above
\[ \begin{gather} m(L-D)=3mD\\[5pt] L-D=3D\\[5pt] 3D-D=L\\[5pt] 4D=L \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {D=\frac{L}{4}} \end{gather} \]
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