Physics Center of Mass problem

A 55 kg woman and an 85 kg man stand 7.0 m apart on 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 1.1 m, how far from the woman will he be now?

(c) How far will the man have moved when he collides with the woman?

Center of mass will be where the mass of the man times his distance to the CM will equal the mass of the woman times her distance from the CM

if x is the distance from man to CM
then

85x = 55( 7 - x )
85x = 385 - 55x
140x = 385
x = 2.75 m
so woman is 7-2.75 = 4.25 m from their CM ANSWER a)

in a frictionless system the center of mass will not move when the masses approach each other.

85 (2.75 -1.1) = 55 (4.25 - y)
85 (1.65) = 233.75 -55 y
85 (1.65) = 233.75 -55 y
140.25 - 233.75 = -55y
-93.5 =-55y
y = 1.7 m ANSWER b)

ANSWER c)
As their center of mass of their individual bodies occurs within their bodies of unknown dimensions, we do not know exactly when collision will occur. The man will move slightly less than 2.75 m as their center of mass of the combined system will not have moved.

You're right. I figured how far the woman moved, not how far they were apart when done. Which was what the question called for.
Answer b should be 7 - 1.1 - 1.7 = 4.2 m

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