A person walking on a boat

Problem

A person of $60\;\text{kg}$ walks $2{,}0\;\text{m}$ from one end to the other of a boat of $180\;\text{kg}$, at rest on a lake and $4{,}0\;\text{m}$ long. Neglecting the friction between water and boat, by how much does the boat move?

Solution

The person + boat system is isolated horizontally: the centre of mass stays fixed relative to the water. The person moves $s = 2{,}0\;\text{m}$ relative to the boat; let $\Delta x_p$ and $\Delta x_b$ be the displacements relative to the water, with $\Delta x_p - \Delta x_b = s$. From $m_p \Delta x_p + m_b \Delta x_b = 0$ we get $\Delta x_p = -3\,\Delta x_b$ (since $m_p/m_b = 60/180 = 1/3$), hence $-3\,\Delta x_b - \Delta x_b = 2{,}0 \;\Rightarrow\; \Delta x_b = -\frac{2{,}0}{4} = -0{,}50\;\text{m}$ $\ev{\Delta x_b = -0{,}50\;\text{m}}$ The boat moves $0{,}50\;\text{m}$ in the direction opposite to the person’s walk.

Links

Topics: Centre of mass Concepts: Centre of mass · Conservation of momentum Skills: Calculating the centre of mass