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 135 kg and the bag of tools has a mass of 22.0 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

Use conservation of momentum to solve this problem:

m1*u1 + m2*u2 = m1*V1 + m2*V2

m1 = mass of astronaut = 135 Kg
m2 = mass of tool bag = 22 Kg
u1 = u2 = 1.5 m/s before the astronaut throw the tool bag.

In order to keep the astronaut from drifting away forever, his velocity V1 must be 0 after he throw the tool bag away. So,
135 * 1.5 + 22 * 1.5 = 0 + 22 * V2

V2 = 235.5/22 = 10.7045 m/s.

10.7045 m/s 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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