A fisherman in a small fishing boat at rest in a lake hooks a giant log floating in the lake 30 meters away. The fisherman reels the log in. During this process, the boat moves 12 meters in the direction of the log. If the mass of the boat and fisherman is 400 kg, what is the mass of the log? Assume frictionless.

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Solution

Because there are no external forces on the system, and the center of mass is initially at rest,, the center of mass must remain at rest. Set $x = 0$ as the center of mass of the fisherman/log system
$x_{C M} = \frac{x_{f i s h e r m a n} m_{f i s h e r m a n} + x_{l o g} m_{l o g}}{m_{f i s h e r m a n} + m l o g} = 0$

$x_{f i s h e r m a n} m_{f i s h e r m a n} + x_{l o g} m_{l o g} = 0$

$m_{l o g} = \frac{- x_{f i s h e r m a n} m_{f i s h e r m a n}}{x_{l o g}}$

$m_{l o g} = \frac{(12 m) (400 k g)}{(18 m)} = 267 k g$

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