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Two astronauts, one of mass 55 kg and the other 85 kg, are initially at rest together in outer space. They then push each other apart. How far apart are they when the lighter astronaut has moved 12 m?

Short Answer

Expert verified

The distance between the astronauts is \(D = 19.76\;{\rm{m}}\).

Step by step solution

01

Centre of mass

Hence, there is only internal force acting on the two astronauts; the centre of mass of the system does not change.

02

Given data

Given data:

The mass of one of the astronauts is\(m = 55\;{\rm{kg}}\).

The mass of the other astronaut is\(m' = 85\;{\rm{kg}}\).

The distance moved by the lighter astronaut is \(d = 12\;{\rm{m}}\).

03

Determine the distance parted

The formula of centre of mass is as follows:

\({x_{{\rm{CM}}}} = \frac{{md + m'd'}}{{m + m'}}\)

Here,\(d'\)is the distance moved by the other astronaut.

On plugging the values in the above equation,

\(\begin{array}{l}0 = \left[ {\frac{{\left( {55\;{\rm{kg}}} \right)\left( {12\;{\rm{m}}} \right) + \left( {85\;{\rm{kg}}} \right)d'}}{{\left( {55\;{\rm{kg}}} \right) + \left( {85\;{\rm{kg}}} \right)}}} \right]\\d' = - 7.76\;{\rm{m}}\end{array}\).

The distance part is calculated as follows:

\(\begin{array}{l}D = d - d'\\D = \left( {12\;{\rm{m}}} \right) - \left( { - 7.76\;{\rm{m}}} \right)\\D = 19.76\;{\rm{m}}\end{array}\)

Thus, \(D = 19.76\;{\rm{m}}\) is the distance between the two astronauts.

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