Floating boat

Classical Mechanics Level 2

Two boats $A$ and $B,$ connected with a string, are floating on a calm lake. The mass of $A$ and $B$ are $100 \text{ kg}$ and $150 \text{ kg},$ respectively, and the distance between them is $25 \text{ m}.$ If the string is pulled from boat $A$ with a constant force and the two boats meet in $20 \text{ seconds},$ what is the distance that boat A moves ?

Suppose that there is no friction whatsoever.

15 m 10 m 20 m 5 m

1 solution

Tushar Marda
Jun 6, 2014

The time in which they meet is inconsequential. Since there is no external force on the (two boat and string) system, the center of mass will not move. Center of mass is located by the product of mass and distance and the sum of masses. Sum of masses remain the same, hence the product of mass of each boat with the distance moved by that boat must remain the same. Hence,

$100 \times d_{a} = 150 \times d_{b}$

$\frac{d_a}{d_b} = \frac{150}{100} = \frac{3}{2}$

Also,

$d_{a} + d_{b} = 25$

Solving these gives us

$\boxed{d_a = 15}$

Floating boat

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