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 m a = 124 kg and the bag of tools has a mass of m b = 13.0 kg. If the astronaut is moving away from the space station at v i = 2.10 m / s initially, what is the minimum final speed v b , f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?

Answer 1

Answer:

vb = 22.13 m/s

Explanation:

ma = 124 kg

mb = 13 kg

vi = 2.10 m/s

According to the property of conservation of momentum, and considering that, initially, both the astronaut and the bag moved together at 2.10 m/s:

$(m_a+m_b)v_i=m_av_a+m_bv_b$

The minimum final velocity of the bag, vb, the will keep the astronaut from drifting away forever occurs when va = 0:

$(124+13)2.10=124*0+13v_b\\v_b=\frac{287.7}{13}\\v_b= 22.13\ m/s$

The minimum final velocity of the bag is 22.13 m/s.