Respuesta :
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)
Answer:
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)?
Answer:
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
[tex]x_{cm} = \frac{m_1 x_1 + m_2 x_2 }{x_1 + x_2}[/tex]
[tex]x_{cm1} =\frac{m_2}{m_1 + m_2} d[/tex]
Where:
[tex]x_{cm1}[/tex] = Center of mass from the man and
d = Distance between the man and the woman
Therefore,
[tex]x_{cm1} =\frac{52}{85 + 52} \times 11 = \frac{52}{137} \times 11 = 4.175 m[/tex]
Therefore, the man has to move 4.175 m before he collides with the woman.
