Two identical boats, each of mass m = 180 kg (including one…

Two identical boats, each of mass $m = 180\ \mathrm{kg}$ (including one package), move on parallel courses toward each other, each at $v_0 = 3\ \mathrm{m/s}$ relative to the shore. (1) At one meeting, a package of mass $m_1 = 20\ \mathrm{kg}$ is passed from the first boat to the second, and then the same package is passed back. (2) At another meeting, the boats exchange identical packages simultaneously. Water resistance is neglected, and each package is handed across sideways (it keeps the velocity of the boat it leaves). After which exchange are the boats' speeds greater?

Explanation of Method

Two boats exchanging packages.

Step 1: (1) One after the other

Take boat 1's direction as positive. The package leaves boat 1 at $+3\ \mathrm{m/s}$ (boat 1's speed does not change) and lands in boat 2: $v_2' = \frac{180 \cdot (-3) + 20 \cdot 3}{200} = -2.4\ \mathrm{m/s}$.

It is then passed back at $-2.4\ \mathrm{m/s}$ into boat 1, which has 160 kg without it: \[v_1' = \frac{160 \cdot 3 + 20 \cdot (-2.4)}{180} = \frac{432}{180} = 2.4\ \mathrm{m/s}.\]

\[\text{Both boats: } 2.40\ \mathrm{m/s}\]

Step 2: (2) Simultaneous exchange

Each boat (160 kg without its package) receives a package moving the opposite way at 3 m/s: $v_1'' = \frac{160 \cdot 3 + 20 \cdot (-3)}{180} = \frac{420}{180}$.

\[\text{Both boats: } 2.33\ \mathrm{m/s}\]

$0$ $1$ $2$ $3$ $v\ (\mathrm{m/s})$ $\text{before}$ $3.00$ $\text{1st: one after the other}$ $2.40$ $\text{2nd: simultaneous}$ $2.33$

Speeds of the boats after each kind of exchange, both starting from 3 m/s. A package that arrives from the other boat carries momentum in the opposite direction, so every exchange slows the boats; the simultaneous exchange slows them more.

Step 3: Answer

\[\begin{gathered}\text{Greater after exchange (1):} \\ 2.40 \gt 2.33\ \mathrm{m/s} \end{gathered}\]

In (1) the package that returns to boat 1 has already been slowed to 2.4 m/s, so it brings less opposite momentum than a package arriving directly at 3 m/s in (2).

Result

\[\boxed{\,\text{Both boats: } 2.40\ \mathrm{m/s}\,}\]

\[\boxed{\,\text{Both boats: } 2.33\ \mathrm{m/s}\,}\]

\[\boxed{\,\begin{gathered}\text{Greater after exchange (1):} \\ 2.40 > 2.33\ \mathrm{m/s} \end{gathered}\,}\]

Remark

About the original solution. Its equations are not momentum balances (they assume a boat keeps its full momentum $mv_0$ while changing mass), and its conclusion, that the speeds increase with every exchange and are greater after the second, is wrong. A package from the other boat always carries momentum in the opposite direction, so each exchange slows both boats. If the 180 kg is taken without the package, the numbers change slightly (2.46 and 2.40 vs 2.40 m/s), but exchange (1) still leaves the boats faster.