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}=124mathrm{kg} ) and the bag of tools has a mass of ( m_{b}=10.0mathrm{kg} ). the astronaut is moving away from the space station at ( v_{i}=1.20mathrm{m}/mathrm{s} ). 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?

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the astronaut - tool system is \(P_i=(m_a + m_b)v_i\). The final momentum is \(P_f=m_av_{a,f}+m_bv_{b,f}\). According to the law of conservation of momentum \(P_i = P_f\), so \((m_a + m_b)v_i=m_av_{a,f}+m_bv_{b,f}\). To prevent the astronaut from drifting away forever, the minimum condition is \(v_{a,f} = 0\) (astronaut stops moving relative to the space - station).

Step2: Solve for \(v_{b,f}\)

Substitute \(v_{a,f} = 0\) into \((m_a + m_b)v_i=m_av_{a,f}+m_bv_{b,f}\). We get \((m_a + m_b)v_i=m_bv_{b,f}\). Then \(v_{b,f}=\frac{(m_a + m_b)v_i}{m_b}\).

Substitute \(m_a = 124\space kg\), \(m_b = 10.0\space kg\) and \(v_i = 1.20\space m/s\) into the formula:

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Answer:

\(16.1\space m/s\)