The Displacement of the Platform When the Man and Woman Meet
The Scenario
The Problem
Find the displacement s of the platform when the two meet if the displacement x1 of the man relative to the platform is 9 m. The length l of the platform is 15.5 m.The Solution
Answer: Δs = 0.0679 mExplanation:
- m₀ = 103 kg
- m₁ = 82 kg
- m₂ = 80 kg
- x₀ = (L/2) = (15.5/2) m = 7.75 m
Initial positions
- x₁ = 0 m
- x₂ = L = 15.5 m
Final positions
- x₁ = 9 m
- x₂ = L – x₁ = 15.5 m – 9 m = 6.5 m
Xcm = (m₀*X₀+m₁*X₁+m₂*X₂) / (m₀+m₁+m₂)
We get the center of mass in every case
Xcm initial = (103 Kg *7.75 m +82 Kg*0 m+ 80 Kg*15.5 m)/(103+82+80) Kg = 7.6915 m
Xcm final = (103 Kg *7.75 m +82 Kg*9 m+ 80 Kg*6.5 m)/(103+82+80) Kg = 7.7594 m
Variation: Δs = Xcm final - Xcm initial = 7.7594 m - 7.6915 m = 0.0679 m
What are the initial positions of the man and woman on the platform?
The man is at position 0 m and the woman is at position 15.5 m on the platform.