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A man whose mass is 80kg and a woman whose mass is 50kgsit at opposite ends of a canoe 5m long, whose mass is 30kg. (a) Relative to the man, where is the mass of the system consisting of man-woman, and canoe? (Hint: Choose a specific coordinate system with a specific origin.) (b) Suppose that the man moves quickly to the center of the canoe and sits down there. How far does the canoe move in the water? Explain your work and your assumptions.

Short Answer

Expert verified
  1. The center of mass of the system consisting of man-woman, and canoe, Relative to the man is 2.03m
  2. The canoe moves 1.72m in the downward direction.

Step by step solution

01

Identification of given data

  • The mass of a man is m1=80kg
  • The mass of a woman is m2=50kg
  • The mass of a canoe is m3=30kg
  • The length of a canoe is 5m long
02

Concept of the mass of the system

The mass of the system is defined by considering the average positions of all the objects acting in the system.

03

(a) Determination of the mass of the system consisting of man-woman, and canoe, Relative to the man

The center of mass of the system can be,

MxCM=m1x1+m2x2+m3x3

Where,

x1=0m,x2=2.5m,x3=5mM=m1+m2+m3=80+50+30=160kg

Substitute these values in above expression,

MxCM=m1x1+m2x2+m3x3=(80×0)+(30×2.5)+(50×5)(kg·m)=325M·1kg·m1kg

Substituting M values in above expression

xCM=325160·1kg·m1kg=2.03·1m=2.03m

Hence, the center of mass of the system is2.03m

04

(b) Determination of the new center of mass

We know that,

MxCM=m1x1+m2x2+m3x3=(80×2.5)MxCM=200M·1kg·m1kg

Substituting M values in above expression

xCM=2001601kg·m1kgxCM=1.25m

The distanceto find how the canoe moved will be foundby;

=2.5m+1.25m-2.03m=1.72m

Here the negative sign indicates the downward movement of the canoe. The canoe moves 1.72min the downward direction in the water when man moves quickly to the center of the canoe and sits down there.

Hence, the canoe moves 1.72min the downward direction.

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