Question

A 52-kg woman and a 72-kg man stand 10.0 m apart on nearly frictionless ice.

(a) How far from the woman is their CM?

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 2.5 m, how far from the woman will he be now?

(c) How far will the man have moved when he collides with the woman?

Short answer

(a)The center of mass from the woman is\(5.8\;{\rm{m}}\).

(b) The distance between the man and the woman is\({\rm{4}}{\rm{.0}}\;{\rm{m}}\).

(c) The distance the man has moved from his initial position is \(4.2\;{\rm{m}}\).

Step-by-step solution

Define the center of mass (CM)

The CM of any group of bodies or any single large body obeys the following principles:

(i)Newton's laws of motion

(ii)Constant acceleration equation

(iii)Work-energy theorem

Given information

The mass of the woman is\({m_{\rm{w}}} = 52\;{\rm{kg}}\).

The mass of the man is\({m_{\rm{m}}} = 72\;{\rm{kg}}\).

The distance of the man from the woman is \({x_{\rm{m}}} = 10.0\;{\rm{m}}\).

Calculate the distance of the center of mass from the woman

(a)

Consider the position of the woman as the origin. Then, the initial position of the woman will be\({x_{\rm{w}}} = 0\).

The distance of the center of mass from the woman can be calculated as shown below:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{{m_{\rm{w}}}{x_{\rm{w}}} + {m_{\rm{m}}}{x_{\rm{m}}}}}{{{m_{\rm{w}}} + {m_{\rm{m}}}}}\\{x_{{\rm{CM}}}} = \frac{{\left( {52\;{\rm{kg}}} \right)\left( 0 \right) + \left( {72\;{\rm{kg}}} \right)\left( {10\;{\rm{m}}} \right)}}{{\left( {52\;{\rm{kg}}} \right) + \left( {72\;{\rm{kg}}} \right)}}\\{x_{{\rm{CM}}}} = 5.8\;{\rm{m}}\end{array}\)

Thus, the distance of the center of mass from the woman is \(5.8\;{\rm{m}}\).

Calculate the distance between the man and the woman

(b)

The man moves a distance of\(2.5\;{\rm{m}}\). Hence, his distance from the origin is

\(\begin{array}{c}{x_{\rm{m}}} = \left( {10.0\;{\rm{m}}} \right) - \left( {2.5\;{\rm{m}}} \right)\\{x_{\rm{m}}} = 7.5\;{\rm{m}}{\rm{.}}\end{array}\)

The net external force on the system is zero. Therefore, the center of mass does not change.

The position of the woman can be calculated as shown below:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{{m_{\rm{w}}}{x_{\rm{w}}} + {m_{\rm{m}}}{x_{\rm{m}}}}}{{{m_{\rm{w}}} + {m_{\rm{m}}}}}\\{x_{\rm{w}}} = \frac{{{x_{{\rm{CM}}}}\left( {{m_{\rm{w}}} + {m_{\rm{m}}}} \right) - {m_{\rm{m}}}{x_{\rm{m}}}}}{{{m_{\rm{w}}}}}\\{x_{\rm{w}}} = \frac{{\left( {5.8\;{\rm{m}}} \right)\left[ {\left( {52\;{\rm{kg}}} \right) + \left( {72\;{\rm{kg}}} \right)} \right] - \left( {72\;{\rm{kg}}} \right)\left( {7.5\;{\rm{m}}} \right)}}{{\left( {52\;{\rm{kg}}} \right)}}\\{x_{\rm{w}}} = 3.46\;{\rm{m}}\end{array}\)

The distance between the man and the woman can be calculated as shown below:

\(\begin{array}{c}d = {x_{\rm{m}}} - {x_{\rm{w}}}\\d = \left( {7.5\;{\rm{m}}} \right) - \left( {3.46\;{\rm{m}}} \right)\\d = 4.038\;{\rm{m}} \approx {\rm{4}}{\rm{.0}}\;{\rm{m}}\end{array}\)

Thus, the distance between the man and the woman is \({\rm{4}}{\rm{.0}}\;{\rm{m}}\).

Calculate the distance the man has moved from his initial position

(c)

The man collides with the woman. The net external force on the system is zero. Hence, the location of the man is at the center of the mass of the system.

The distance the man moves from his initial position can be calculated as shown below:

\(\begin{array}{c}d = \left| {{x_{{\rm{m,final}}}} - {x_{{\rm{m,initial}}}}} \right|\\d = \left| {\left( {4.038\;{\rm{m}}} \right) - \left( {10.0\;{\rm{m}}} \right)} \right|\\d = \left| { - 4.19\;{\rm{m}}} \right|\\d \approx 4.2\;{\rm{m}}\end{array}\)

Thus, the distance the man has moved from his initial position is \(4.2\;{\rm{m}}\).