Each person holds the end of a rope and the man pulls on the rope so that he moves toward the woman. How far will the man have moved when he collides with the woman?

Full Question

A 52-kg woman and a 85-kg man stand 11 m apart on frictionless ice.

Each person holds the end of a rope and the man pulls on the rope so that he moves toward the woman. How far will the man have moved when he collides with the woman (in meters)

4.14m

Explanation:

Given

m = mass of the woman = 52kg

M = Mass of the man = 86kg

Taking the position of the woman as the origin

r = position of the woman = 0m

R = position of the man = 11m

d = Distance between them = 11m

First we need to calculate their centre of mass (COM)

this is calculated by

Rm = (mr + MR)/(m + M)

By substituton

Rm = (52(0) + 86(11))/(52 + 86)

Rm = 6.86m

Collision will happen between the woman and the man at there centre of mass.

So, the man will move a distance of (11m - 6.86m)

Distance moved by the man = 4.14m

Question:

A 52-kg woman and a 85-kg man stand 11 m apart on frictionless ice.

A) How far from the woman is their CM (in meters)?

B) Each person holds the end of a rope and the man pulls on the rope so that he moves toward the woman. How far will the man have moved when he collides with the woman (in meters)?

The man has to move 4.175 m before he collides with the woman

Explanation:

The we have

m₁v₁ = m₂v₂

Where:

m₁ = Mass of the man

m₂ = Mass of the woman

v₁ = Speed of the man

v₂ = Speed of the woman

v₁/v₂ = 52/85 Therefore in 11 m, the man has to move 52/(85+52) of 11 m while the woman moves 85/(52+85) of 11 m in the same time

The center of mass is given by

Where:

= Center of mass from the man and

d = Distance between the man and the woman

Therefore,

Therefore, the man has to move 4.175 m before he collides with the woman.

1/0.14nm = 10^10/1.4 a lot