Two boats float on a lake: the first has mass m_1 = 200 kg…
Two boats float on a lake: the first has mass $m_1 = 200\ \mathrm{kg}$, the second $m_2 = 450\ \mathrm{kg}$. A rope is stretched between them, and a person in the first boat pulls it with a force $F = 200\ \mathrm{N}$. Find the speed of the first boat relative to the shore and relative to the second boat $t = 2\ \mathrm{s}$ after the pulling starts. The boats start at rest; neglect water resistance.
Explanation of Method
Two boats pulled together by a rope.
Step 1: Accelerations
The rope pulls both boats toward each other with the same force $F$ (the mass $m_1$ includes the person): $a_1 = \frac{F}{m_1} = 1\ \mathrm{m/s^2}$.
\[ a_2 = \frac{F}{m_2} \approx 0.444\ \mathrm{m/s^2}. \]
Step 2: Speeds after 2 s (relative to the shore)
\[ v_1 = a_1 t = 2\ \mathrm{m/s}, \]
\[ v_2 = a_2 t \approx 0.89\ \mathrm{m/s} \ \text{(toward each other)}. \]
\[v_1 = 2\ \mathrm{m/s} \ \text{relative to the shore}\]
Step 3: Relative to the second boat
The boats move in opposite directions, so their speeds add: $v_{\text{rel}} = v_1 + v_2 \approx 2.89\ \mathrm{m/s}$.
Check
Total momentum $ 200 \cdot 2 - 450 \cdot 0.889 = 400 - 400 = 0\ \checkmark $ (the rope's forces are internal)
Result
\[\boxed{\,v_1 = 2\ \mathrm{m/s} \ \text{relative to the shore}\,}\]
\[\boxed{\,v_{\text{rel}} = v_1 + v_2 \approx 2.89\ \mathrm{m/s}\,}\]
Remark
Check. The original results are correct (its text calls the relative speed "the difference"; it is the difference of the velocities, which is the sum of the speeds because they point in opposite directions).