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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 124 kg and the bag of tools has a mass of 16.0 kg. If the astronaut is moving away from the space station at ????i=1.80 m/s initially, what is the minimum final speed of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?

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Answer:

15.75 m/s

Explanation:

v = Velocity of the combined mass of astronaut and tools = 1.8 m/s

[tex]m_1[/tex] = Mass of astronaut = 124 kg

[tex]m_2[/tex] = Mass of tools = 16 kg

[tex]v_1[/tex] = Velocity of astronaut = 0

[tex]v_2[/tex] = Velocity of tools

As linear momentum is conserved

[tex]m_1v_1 + m_2v_2 =(m_1 + m_2)v\\\Rightarrow v_2=\frac{(m_1 + m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(124+16)\times 1.8-124\times 0}{16}\\\Rightarrow v_2=15.75\ m/s[/tex]

The velocity of the tools is 15.75 m/s

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