**Problem Description:**An astronaut comes loose while outside the spacecraft and finds himself floating away from the spacecraft with only a big bag of tools. In desperation, the astronaut throws his bag of tools in the direction of his motion (away from the space station). The astronaut has a mass of 102 kg and the bag of tools has a mass of 10.0 kg. If the astronaut is moving away from the space station at 1.80 m/s initially, what is the minimum final speed of the bag of tools (with respect to the space station) that will keep the astronaut from drifting away forever?**Solution:**To find the minimum final speed of the bag of tools, apply the principle of conservation of momentum. The total momentum of the system (astronaut + tools) before and after the bag is thrown must be equal.**Given:**- Mass of astronaut, \( m_a = 102 \, \text{kg} \)- Mass of tools, \( m_t = 10.0 \, \text{kg} \)- Initial speed of astronaut + tools, \( v_i = 1.80 \, \text{m/s} \)**Conservation of Momentum:**\[ m_a \cdot v_{a_i} + m_t \cdot v_{t_i} = m_a \cdot v_{a_f} + m_t \cdot v_{t_f} \]The initial velocity of both astronaut and tools is the same, \( v_i = 1.80 \, \text{m/s} \). Therefore:\[ (m_a + m_t) \cdot v_i = m_a \cdot v_{a_f} + m_t \cdot v_{t_f} \]To stop drifting, the astronaut’s final velocity with respect to the space station should be \( v_{a_f} = 0 \).Substituting, we get:\[ (102 \, \text{kg} + 10.0 \, \text{kg}) \cdot 1.80 \, \text{m/s} = 102 \, \text{kg} \cdot 0 + 10.0 \, \text{kg} \cdot v_{t_f} \]\[ 112 \cdot 1.80 = 10.0 \cdot v_{t_f} \]\[

Given: The mass of the astronaut is M = 102 kg The mass of the tool bag is m = 10 Kg The initial…

Answered: An astronaut comes loose while outside…

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