Ricardo, of mass 80 kg , and Carmelita, who is lighter, is enjoying Lake Merced at dusk in a30 kgcanoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 mapart and symmetrically located with respect to the canoe’s center. If the canoe moves 40 cmhorizontally relative to a pier post, what is Carmelita’s mass?

Short Answer

Expert verified

The mass of the Carmelita is m2=58kg.

Step by step solution

01

Listing the given quantities:

The mass of Ricardo ism1=80kg.

The mass of the canoe is m3=30kg.

The separation distance between Ricardo and Carmelita isL=3.0m.

The canoe shifted by the distance is 2x=0.40m.

02

Understanding the concept of center of mass:

The center of gravity is a position defined relative to an object or system of objects. It is the average position of all parts of the system, weighted by their weight. For simple rigid objects of uniform density, the center of mass is located at the center of gravity.

You can use the concept of the center of mass of the system.

Formula:

Rcm=m1r1+m2r2m1+m2

Here, Rcm is the distance to the center of mass, r1 is the distance of mass m1, and r2 is the distance of mass m2.

03

Calculations for the mass of Carmelita:

The acceleration of the center of mass of the cart-block system:

You can apply the center of mass of the system concept.

Consider the mass of the Carmelita as .

Rcm=m1r1+m2r2+m3r3m1+m2+m3

The center of the mass is not moving, hence

role="math" localid="1661238681972" 0=m1L2-x-m2L2-x-m3xm1+m2+m3m3x=m1L2-x-m2L2-x........(1)

When they change their positions, the center of the canoe is shifted by 2x from its initial position.

2x=0.40mx=0.20m

Equation (1) becomes,

m2L2-x=m1L2-x-m3xm2=m1L2-x-m3xm2L2-x

Substitute known values in the above equation.

m2=80kg3.0m2-0.20m-30kg×0.20m3.0m2+0.20m=104kg.m-6kg.m1.7m=98kg.m1.7m=58kg

Hence, the mass of the Carmelita is 58 kg .

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