Question

An 80-kg person stands at one end of a 130-kg boat. He then walks to the other end of the boat so that the boat moves 80 cm with respect to the bottom of the lake. (Hint: Note that the total momentum of the person-boat system remains constant.) How much did the center of mass of the person-boat system move when the person walked from one end to the other?

Answer 1

Explanation:

We assume that length of boat is L cm. And, we suppose to have a coordinate system relative to the bottom of the lake with:

Boat's position initially from x = 0 cm to x = L cm

Let person (S) is initially at x = L cm (right hand end of boat)

and, let boat's center of mass at x = C cm

Therefore, initially center of mass of overall (system's) is at $X_{1}$ where (taking moments about x = 0).

             $(80 + 130)X_{1}$

             = 130C + 80L ........... (i)  

After S walks left to the other end of the boat, the boat moves right 80 cm.

Hence,  

Boat's position is now from x = 80 cm to x = (L+ 80) cm

S's position is x = 80 (left end of boat)

Boat's center of mass is at (C + 80).

Therefore, overall (system's) final center of mass is at $X_{2}$ where (taking moments about x = 0).

                $(80 + 130)X_{2}$

             = $(130)(C + 80) + 80 \times 80$ ............. (ii)

As no external force is acted on the system, the system's center of mass has not moved, so $X_{1} = X_{2}$.

This means the left hand sides of both equations (i) and (ii) are equal. So, we can equate the right hand sides giving: the equation as follows.

                130C + 80L = $(130)(C+80) + 80 \times 80$

                           80L = 16800 cm

                           L = 210 cm

                             = 2.1 m

Length of boat = 2.1 m

When the person is walking from one end to the other as there was no external force then center of mass of the person-boat system did not move.