Question Below...?

Question

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 cm apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moves 40 cm horizontally relative to a pier post during the exchange and calculates Carmelita's mass. What is it?

Answer 1

Let coordinates of Carmelita, center of the canoe and Ricardo are 0 m, 1.5 m and 3 m, respectively.

Their coordinates after the exchange will be 0.4 m (Ricardo), 1.9 m (center of the canoe) and 3.4 (Carmelita).

Let “x” is the mass of Carmelita.

The total mass of the system is 80+30+x = 110 + x kilograms.

The coordinate of the center of mass in meters before the exchange is
(0*x + 1.5*30 + 3*80) / (110 + x)

The coordinate of the center of mass in meters after the exchange is
(0.4*80 + 1.9*30 + 3.4*x) / (110 + x)

The center of mass did not move, therefore both the above values must be equal:
(0*x + 1.5*30 + 3*80) / (110 + x) = (0.4*80 + 1.9*30 + 3.4*x) / (110 + x)
0*x + 1.5*30 + 3*80 = 0.4*80 + 1.9*30 + 3.4*x

Solution of this equation is

x = 57.65 kg.

For better understanding see
http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html
or
http://en.wikipedia.org/wiki/Center_of_mass

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Answer 2

Are you sure it is 3.0 cm? I'm thinking you might mean 3m

To solve the problem, use the fact that the center of mass of the canoe should not have moved during this exchange.

if D is the distance each moved, then the change in the center of mass due to their movement is 2*D*(80-C)

The change of center of mass due to the canoe's moving is, of course (0.4)*(30+80+C)

Set these two equal to each other to solve for C.

Answer 3

thanks to each and everyone of you for the replies!