Question

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 19.0 kg. If the astronaut is moving away from the space station at 2.10 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?

Answer 1

Mass of astronaut is \(m_{\mathrm{A}}=124 \mathrm{~kg}\) Mass of tool is \(m_{\mathrm{T}}=19.0 \mathrm{~kg}\)

The combined speed of astronaut and tool is \(v_{0}=2.10 \mathrm{~m} / \mathrm{s}\)

From the law of conservation of momentum,

$$ \begin{array}{l} \left(m_{\mathrm{A}}+m_{\mathrm{T}}\right) v_{0}=m_{\mathrm{A}} v_{\mathrm{A}}+m_{\mathrm{T}} v_{\mathrm{T}} \\ (124 \mathrm{~kg}+19.0 \mathrm{~kg})(2.10 \mathrm{~m} / \mathrm{s})=(124 \mathrm{~kg})(0)+(19.0 \mathrm{~kg}) v_{\mathrm{T}} \\ v_{\mathrm{T}}=15.80 \mathrm{~m} / \mathrm{s} \end{array} $$

Thus, the speed of bag tool is \(v_{\mathrm{T}}=15.80 \mathrm{~m} / \mathrm{s}\).