# Ricardo of mass 80 kg and Carmelita 2

Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe’s center. If the canoe moves 40 cm horizontally relative to a pier post, what is Carmelita’s mass? Designed by Freepik

### Solution:

Let $L$ represent the length of the canoe. Let the mass of the canoe be represented by $M_{canoe}$ and the distance from the center of the canoe to the center of mass of the two people be represented by $x$. The mass of Ricardo is $M_R$ and mass of Carmelita is $M_C$.

Using these variables, we can write the following:

$M_R(\dfrac{{L}}{2}-x)=M_{canoe}x+M_C(\dfrac{{L}}{2}+x)$

When they switch positions, we know that the canoe moves 2 times the distance from it’s initial position. Using this fact, we can figure out the value of $x$.

$x=40/2=20\,\text{cm}=0.2$ m

We can now substitute all the values given to us in the question and solve for $M_C$.

$M_C=\dfrac{{M_R(\dfrac{{L}}{2}-x})-(M_{canoe})(x)}{\dfrac{{L}}{2}+x}$

$M_C=\dfrac{{(80)(3/2-0.2)}-(30)(0.2)}{(3/2)+0.2}$

$M_C=57.64$ kg

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