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 ????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?

Answer 1

Answer:

The velocity of the bag will be 16.08 m/s

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