Question about center of mass, physics?

Taking moment about the center of mass

M*2 = 0.1*[ 8-2]

M = 0.3 kg

the mass of the particle at the origin is 0.3 kg

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Sum of the momentum of the particles = the momentum of the center of mass.

[0.3 + 0.1]*5 = 2 kg m/s

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0.3 u + 0.1*0 = 2

u = 2/0.3 = 6.67m/s.

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Momentum of the center of system = addition of momentum of all the particles of the system

Let the mass of the particle at origin is m kg and velocity u m/s, than

(m+0.10) x 5 = m x u + 0.10 x 0

=>5m - 0.50 = mu ----------------------------(i)

For center of mass, equating the masses at x = 2m

=>m x 2 = 0.10 x (8-2)

=>m = 0.30 kg(ans for 1)

by putting this value in (i):-

=>5 x 0.30 - 0.50 = 0.30u

=>1 = 0.30u

=>u = 1/0.30 = 3.33 m/s(ans for 3)

Total momentum = (0.30+0.10) x 5 = 2 N-s

Hmmm, I'm not an expert but perhaps the location of the center of mass doesn't change with respect to the pier but does change with respect to the canoe's center. The boat moves 46.5 cm in order to keep the location of the center of mass unchanged with respect to the pier. If this is so, calculating Carmelita's mass should be straightforward.