Respuesta :
Part a)
let the position of woman is taken as origin
so by the formula of COM
[tex]r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}[/tex]
[tex]r_{cm} = \frac{52(0) + 72(10)}{52 + 72}[/tex]
[tex]r_{cm} = 5.81 m[/tex]
so COM will be at distance 5.81 m from woman
Part b)
since there is no external force on the system of man and woman
so the displacement of COM will be zero here
so we will write
[tex]\Delta r_{cm} = 0 = m_1\Delta r_1 + m_2\Delta r_2[/tex]
[tex]0 = 52\Delta r_1 - 72(2.5)[/tex]
[tex]\Delta r_1 = 3.46 m[/tex]
so woman will move to distance 3.46 m
Part c)
they both will collide at their COM position as COM will not shift
So the man will reach to his COM
so distance moved will be d = (10 - 5.81) m
d = 4.19 m
The distance of the woman from their center mass is 5.8 m.
The distance of the man from the woman is 5.15 m.
When the man collides with the woman, the distance moved by the man is 5.3 m.
The given parameters;
- mass of the woman, = 52 kg
- mass of the man = 72 kg
- distance between the man and the woman = 10 cm = 0.1 m
Let the distance of the woman from their center mass = x
take moment about their center;
52x = 72(10 - x)
52x = 720 - 72x
52x + 72x = 720
124x = 720
[tex]x = \frac{720}{124} \\\\x = 5.8 \ m[/tex]
The new distance of the man from the woman when he moves 2.5 m is calculated as;
when the man moves 2.5 m, his position = 7.5. m
the distance of the man from the center mass = (7.5 - 5.8) = 1.7 m
the distance moved by the woman = x
52x = 72(1.7)
52x = 122.4
[tex]x = \frac{122.4}{52} \\\\x = 2.35 \ m[/tex]
The distance of the man from the woman = 7.5 - 2.35 = 5.15 m
When the man collides with the woman, the distance moved by the man is calculated as follows;
distance = 10 - (maximum distance moved by the woman)
distance = 10 - (2.35 + 2.35)
distance = 5.3 m
Learn more here:https://brainly.com/question/13759143
