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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 ????a=124 kgma=124 kg and the bag of tools has a mass of ????b=10.0 kg.mb=10.0 kg. If the astronaut is moving away from the space station at ????i=1.20 m/svi=1.20 m/s initially, what is the minimum final speed ????b,fvb,f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?

asked
User Logcat
by
8.6k points

1 Answer

1 vote

Answer:

The velocity of the bag will be 16.08 m/s

Step-by-step explanation:

Given:

Mass of the astronaut, M = 124 kg

Mass of the bag, m = 10 lg

Initial speed of the astronaut with bag,
v_i = 1.20 m/s

Applying the concept of conservation of momentum

we have

(M + m)
v_i = Mv₁ + mv₂

here,

v₁ is the final velocity of the astronaut = 0

v₂ is final velocity of the bag

thus, on substituting the values, we have

(124 + 10)1.20 = 124 × 0 + 10v₂

or

v₂ = 16.08 m/s

Hence, the velocity of the bag will be 16.08 m/s

answered
User Dheer
by
8.0k points
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