A52-kg woman and a 72-kg man stand 10.0 m apart on nearly frictionless ice. a) how far from the woman is their cm? b) if each holds one end of a rope, and the man pulls on the rope so that he moves 2.5 m, how far from the woman will he be now? c) how far will the man have moved when he collides with the woman?

The distance of the woman from their center mass is 5.8 m.

The distance of the man from the woman is 5.15 m.

When the man collides with the woman, the distance moved by the man is 5.3 m.

The given parameters;

Let the distance of the woman from their center mass = x

take moment about their center;

52x = 72(10 - x)

52x = 720 - 72x

52x + 72x = 720

124x = 720

The new distance of the man from the woman when he moves 2.5 m is calculated as;

when the man moves 2.5 m, his position = 7.5. m

the distance of the man from the center mass = (7.5 - 5.8) = 1.7 m

the distance moved by the woman = x

52x = 72(1.7)

52x = 122.4

The distance of the man from the woman = 7.5 - 2.35 = 5.15 m

When the man collides with the woman, the distance moved by the man is calculated as follows;

distance = 10 - (maximum distance moved by the woman)

distance = 10 - (2.35 + 2.35)

distance = 5.3 m

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Part a)

let the position of woman is taken as origin

so by the formula of COM

so COM will be at distance 5.81 m from woman

Part b)

since there is no external force on the system of man and woman

so the displacement of COM will be zero here

so we will write

so woman will move to distance 3.46 m

Part c)

they both will collide at their COM position as COM will not shift

So the man will reach to his COM

so distance moved will be d = (10 - 5.81) m

d = 4.19 m

the rearrangement of particles in a physical change. hope this