An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion, away from the space station. The astronaut has a mass of ma=102 kgma=102 kg and the bag of tools has a mass of mb=10.0 kg.mb=10.0 kg. If the astronaut is moving away from the space station at vi=2.10 m/svi=2.10 m/s initially, what is the minimum final speed vb,fvb,f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?

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codiepienagoya codiepienagoya · Answer 1 · 2021-02-24T12:27:33+01:00

Answer:

The answer is "[tex]2.352 \ \frac{m}{s}[/tex]"

Explanation:

[tex]\to mass(m_1)=102 \ kg\\\\\to mass(m_2)=10 \ kg \\\\\to v=2.10\ \frac{m}{s}\\\\[/tex]

momentum before:

[tex]\to p=(m_1+m_2)v[/tex]

[tex]=(102+10)2.10\\\\=(102\times 2.10 +10 \times 2.10)\\\\=214.2+21\\\\=235.2[/tex]

momentum After:

[tex]\to p=(m_1+m_2)v[/tex]

[tex]=(102\times 0 +10 \times v)\\\\ =(0 +10v)\\\\=10v\\[/tex]

Calculating the conservation of momentum:

[tex]\to \text{momentum before = momentum After}[/tex]

[tex]\to 235.2=10v\\\\\to v= \frac{235.2}{10}\\\\ \to v=2.352 \ \frac{m}{s}[/tex]