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Center of Mass, Man, Woman, and Frictionless Ice

by PeachBanana
Tags: frictionless, mass, woman
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PeachBanana
#1
Mar7-12, 08:42 AM
P: 191
1. The problem statement, all variables and given/known data
A 58 kg woman and an 78 kg man stand 8.00 m apart on frictionless ice. How far from the woman is their CM? 4.6 m

If each holds one end of a rope, and the man pulls on the rope so that he moves 2.6 m, how far from the woman will he be now? Use two significant figures in answer. 1.9 m

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

2. Relevant equations

mass man * Δx man = mass woman * -Δx woman

3. The attempt at a solution


This can't be right because it's too close to letter B's answer.

I'm assuming because the ice is frictionless that they'll just keep going until the hit each other.

(78 kg)(x) = -(58 kg)(4.6 m)

x = -3.42 m

8.00 m - 2.6 m - 3.42 m = 1.98 m
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Doc Al
#2
Mar7-12, 09:11 AM
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Think this way: When they finally hit each other, where will they both end up? How far from their starting points is that?
PeachBanana
#3
Mar7-12, 03:30 PM
P: 191
DocAl:

The man starts 8.00 m apart from the woman and I knew he was 4.6 m from her CM. 8.00m - 4.6m = 3.4m

Thank you

Doc Al
#4
Mar8-12, 07:14 AM
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Center of Mass, Man, Woman, and Frictionless Ice

I'll rephrase: The man was initially 3.4 m from the center of mass, so when they collide he will have traveled 3.4 m.


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