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?

The minimum final speed of the bag of tools (with respect to the space station) that will keep the astronaut from drifting away forever is 15.81 m/s.

Explanation:

To solve the question, we list out the variables as follows,

Mass of astronaut, m₁ = 124 kg

Mass of the bag of tools, m₂ = 19.0 kg

Initial velocity of astronaut, v₁ = 2.10 m/s = Initial velocity of bag of tools v₂

Final velocity of astronaut, v₃ = 0 m/s, assuming the astronaut is brought to                     a stop

Velocity of the bag of tools = v₄

We can observe that the question is about conservation of linear momentum. Therefore we have, from the principle of conservation of linear momentum.

Initial total momentum = Final Total momentum

We then have

m₁v₁ + m₂v₂ = m₁v₃ - m₂v₄

Since v₁ = v₂ we have

(m₁ + m₂) × v₁ = m₁v₃ - m₂v₄

Plugging the values and solving for the required unknown variable we have

(124 kg + 19.0 kg) × 2.10 m/s = 124 kg × v₃ - 19.0 kg × v₄

Since v₃ = 0 m/s, we have

300.3 kg·m/s = 0 kg·m/s - 19.0 kg × v₄

∴ v₄ = = 15.81 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 = 15.81 m/s.