<p>An astronaut is working in space on the space station when he notices that hi

Question

<p>An astronaut is working in space on the space station when he notices that his safety rope tying him to the station has come undone. He is 5.00 metres away and drifting further away at 0.100m/s. Fortunately he is holding a 1.00kg hammer as well as a 15.00kg antenna (that costs $100,000). The mass of the astronaut by himself is 95.00kg. He can throw the hammer at 10.00m/s and the antenna at 2.00m/s. (Significant digits are important for this question!!)</p>
<p>(1) Clearly explain how (why) he can get back to safety by throwing the object(s). Which direction should he throw them? (detailed explanation please.)</p>
<p>(2) How long (if ever) will it take him to get back to the space station by throwing the hammer?</p>
<p>(3) How long will it take him if he throws away the hammer and the expensive antenna? (neglect any time difference between throwing the two objects)</p>
<p>Detailed solution please :(</p>

Explanation / Answer

1) He can get back to safety by throwing the antenna or both the hammer and the antenna in the direction he is drifting away from the space station, but not just the hammer (explained below), due to the law of conservation of momentum. 2) He cannot get back by simply throwing the hammer. His current momentum is his total mass times his velocity: (mass of himself + mass of hammer + mass of antenna)*(velocity) = (95.00+1.00+15.00)*(.100) = 11.1 kg*m/s (3 sig figs) If he throws the hammer in the positive direction, the direction he's traveling, then the equation becomes (mass total)*(.100 m/s) = (mass of hammer)*(10.00m/s) + (mass of himself and antenna)*(final velocity). The final velocity becomes .0100 m/s, still drifting away from the station. 3) Throwing both at the same time will make the equation (mass total)*(.100 m/s) = (mass of hammer)*(10.00m/s) + (mass of antenna)*(2.00 m/s) + (mass of himself)*(final velocity), where final velocity becomes about -.3 m/s, which is .3 m/s going toward the station. It will then take about 5 m / (.3 m/s) = 16.4 seconds to get back.

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