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Imagine an astronaut floating 15 meters away from their spacecraft in the vacuum of space. To return to their ship, they can't just swim back like in water; they must rely on the laws of physics that govern motion, specifically the Conservation of Momentum. This situation creates a fascinating spectacle of physics. This problem mirrors what happens when two ice skaters push apart from each other. Although the settings are vastly different, the principles hold. Here, the astronaut uses a wrench, throwing it away from the spacecraft to propel themselves towards it. This act conserves momentum since there are no external forces acting on the astronaut-wrench system. In essence, the act of throwing the wrench gives it momentum in one direction which—due to the conservation—imparts an equal and opposite momentum to the astronaut. This principle of action and reaction helps our astronaut find their way back to safety.

To understand how our astronaut moves towards their spacecraft, we need to figure out their velocity after throwing the wrench. This is where the Conservation of Momentum becomes crucial.The momentum before and after the throw is maintained because no outside force is acting on the isolated system. Mathematically, this is represented as:- Initial total momentum: 0 (since both are at rest)- Resultant momentum: \(m_1v_1 + m_2v_2 = 0\)Here, \(m_1\) and \(v_1\) represent the mass and velocity of the astronaut, while \(m_2\) and \(v_2\) are for the wrench. Solving for \(v_1\) once the wrench is thrown allows us to determine how fast the astronaut moves towards their target.By substituting the known values:- Mass of astronaut: 128 kg- Mass of wrench: 1.10 kg- Velocity of the wrench: 5.40 m/sWe can calculate the astronaut's velocity: \[v_1 = \frac{-1.10 \times 5.40}{128} \, \text{m/s} = -0.04640625 \, \text{m/s}\]This negative sign indicates the direction is opposite to the wrench's throw.

Having determined the astronaut's velocity, the next concern is figuring out how long it will take to reach the spacecraft. This involves using the formula for time, which is distance divided by velocity:\[t = \frac{d}{v_1}\]The distance, \(d\), is the 15 meters initially separating the astronaut from the spacecraft, and \(v_1\) is their calculated velocity. Plugging in the values gives:- Distance: 15.0 m- Velocity: 0.04640625 m/sTherefore, the time \(t\) can be calculated as:\[t = \frac{15.0}{0.04640625} \approx 323.32 \, \text{seconds}\]This solution shows how long it would take to cover the distance without any initial extra movement. If the astronaut initially had any velocity away from the spacecraft, calculating the effective velocity will lead to a longer time to return.This helpful time computation allows astronauts to make informed decisions and plans when maneuvering in space, emphasizing how physics is essential even outside Earth's atmosphere.