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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

Respuesta :

Answer:

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

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