Source: High school physics (Chinese)
Problem Sets:
Problem
A person with mass $m_p=80$ kg stands at the stern of an iceboat of length $L=24$ m and mass $m_b=400$ kg. The boat is on smooth ice.
- If the iceboat is initially at rest, how far does it slide when the person walks from the stern to the bow?
- If the boat is initially moving at $v_0 = 4.0$ m/s, and the person walks toward the bow at a speed $v_{rel} = 2$ m/s relative to the boat, how far does the boat slide on the ice during this time?
[Q1] 4.0 m [Q2] 44 m
The system is the person and the boat. Since there are no external horizontal forces, the center of mass (CM) of the system does not accelerate. Let $\Delta x_p$ and $\Delta x_b$ be the displacements of the person and boat relative to the ice. The person's displacement relative to the boat is $L$.
$$ m_p \Delta x_p + m_b \Delta x_b = (m_p+m_b)\Delta x_{CM} $$ $$ \Delta x_p - \Delta x_b = L $$Solving for $\Delta x_b$ gives $\Delta x_b = \Delta x_{CM} - \frac{m_p L}{m_p+m_b}$.
[Q1] The system is initially at rest, so the CM does not move, $\Delta x_{CM} = 0$. The boat's displacement is $\Delta x_b = - \frac{m_p L}{m_p+m_b}$. The magnitude is the distance it slides. [Q2] The system moves with initial velocity $v_0$. The time for the person to walk is $t=L/v_{rel}$. The CM's displacement is $\Delta x_{CM} = v_0 t$. The boat's displacement is $\Delta x_b = v_0 \frac{L}{v_{rel}} - \frac{m_p L}{m_p+m_b}$.