A 50 kg woman and an 80 kg man stand 12.0 m apart on frictionless ice.

(a) How far from the woman is their CM?

m

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?

m

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

m

Given that

m₁ = 50 kg

m₂=80 kg

d = 12 m

a)

We know that center of mass given as

X = (x₁m₁+x₂m₂) / (m₁+m₂)

Lets take distance of CM from woman is X

So now by putting the value

X = (0 x 50+12 x 80) / (50+80)

x=7.38 m

b)

There is no any external force so the CM will not move.

So we can say that

x₁m₁+x₂m₂ = 0

50 (x) - 80 (1.3) = 0

x=2.08

So the distance move by woman d=12-2.08-1.3=8.62 m

d=8.62 m

c) lets take distance move by man is x

50 (x) - 80 (12-x) = 0

x=7.38

So the distance move by woman d=12-7.38

d=4.62 m