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. Fir

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 𝑚a=113 kg and the bag of tools has a mass of 𝑚b=16.0 kg. If the astronaut is moving away from the space station at 𝑣i=2.10 m/s initially, what is the minimum final speed 𝑣bf of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?

Answer 1

Use the conservation of momentum.

P₁ = P₂

P₁ is when the astronaut is drifting away at 2.10 m/s and is carrying his bag of tools. Since momentum equal mass time velocity then:

P₁ = mv

P₁ = (113 + 16)(2.10) = 270.9 kg•m/s

If we assume that "not drifting away forever" means the astronaut's final velocity equals zero, then:

P₂ = m_a(0) + m_t(v_bf)

P₂ = 0 + (16)(v_bf)

But, since P₁ = P₂:

270.9 = 0 + (16)(v_bf)

270.9/16 = v_bf

v_bf = 16.93125 m/s so, since we want the astronaut to not drift away, I would answer v_bf = 17.0 m/s