Question

a person falls flat on their back on some slippery ice and is unconscious for a second. the horizontal position of their legs torso and head from their feet is given by 0.45 m, 1.11 m and 1.66 m respectively and their weights are 30 kg 50 kg and 10 kg. they try to sit up on the ice which is so slippery they are not able to move their horizontal centre of mass. the new positions of their legs, torso and head from their feet are 0.22 m, 0.57 m and 0.76 m. what is the magnitude of the change in position of the persons feet?

Explanation:

Step1: Calculate the initial center of mass

The formula for the center of mass \(x_{cm}=\frac{m_1x_1 + m_2x_2+m_3x_3}{m_1 + m_2+m_3}\).
Let \(m_1 = 30\space kg\), \(x_1=0.45\space m\); \(m_2 = 50\space kg\), \(x_2 = 1.11\space m\); \(m_3=10\space kg\), \(x_3 = 1.66\space m\).

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Step2: Calculate the final center of mass

Let the new positions of legs, torso and head be \(x_1'=0.22\space m\), \(x_2'=0.57\space m\), \(x_3'=0.76\space m\) and masses \(m_1 = 30\space kg\), \(m_2 = 50\space kg\), \(m_3=10\space kg\)

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Step3: Find the change in position of the feet

Since the center of mass does not move (\(x_{cm1}=x_{cm2}+ \Delta x\)), \(\Delta x=x_{cm1}-x_{cm2}\)
\(\Delta x = 0.951-0.474 = 0.477\approx0.48\space m\)

Answer:

\(0.48\)