An unfortunate astronaut loses his grip during a spacewalk and finds himself flo

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 102 kg and the bag of tools has a mass of 13.0 kg. If the astronaut is moving away from the space station at 1.50 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?

Explanation / Answer

let's just treat it as a simple linear momentum problem.
initial p = final p
(102 + 13)kg x 1.50 m/s = 102kg * u + 13kg * v

Now in order to "keep the astronaut from drifting away forever," we need to have his velocity be reduced to zero. (Ideally more; but at 0 m/s he will no longer be "drifting away.") So set u = 0 m/s in the preceding equation and solve:
115kg x 1.50m/s = 0 + 13kg * v
v = 13.27 m/s

this is the minimum final speed of the bag of tools.

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