The boy walking on the boat

Problem

A boat of mass $M = 80\;\text{kg}$ and length $L = 4{,}0\;\text{m}$ floats at rest on a calm lake (friction with the water negligible). A boy of mass $m = 20\;\text{kg}$ walks from one end of the boat to the other. By how much does the boat move relative to the water while the boy walks?

Solution

Key idea. The boy + boat system is horizontally isolated (zero friction): the total momentum stays zero, so the centre of mass does not move relative to the water.

Let $\Delta x_b$ be the displacement of the boat (and of the boy along with it) relative to the water. The boy moves by $L$ relative to the boat, so relative to the water by $\Delta x_r = L + \Delta x_b$. Imposing that the CM stays still: $m\,\Delta x_r + M\,\Delta x_b = 0 \;\Rightarrow\; m(L + \Delta x_b) + M\,\Delta x_b = 0$ $\Delta x_b = -\frac{m\,L}{m + M} = -\frac{20\cdot 4{,}0}{100} = -0{,}80\;\text{m}$

$\ev{|\Delta x_b| = 0{,}80\;\text{m}}$ The boat moves back by $0{,}80\;\text{m}$, in the direction opposite to the boy’s walk.

Links

Topics: Centro di massa Concepts: Centro di massa · Conservazione della quantità di moto Skills: Calcolo del centro di massa