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 = 102 kg and the bag of tools has a mass of mo = 13.0 kg. If the astronaut is moving away from the space station at v; = 1.80 m/s initially, what is the minimum final speed Ub,f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever? Ub,f = m/s

Answer 1

Using law of conservation of momentum,

m1 v1 + m2 v2 = (m1 + m2) v

(102 x 0) + (13 x v2) = (102 + 13) x 1.80

Velocity, v2 = 15.92 m/s

Comment in case any doubt please rate my answer....