An astronaut of mass 82.0 kg is taking a space walk to work on the International

Question

An astronaut of mass 82.0 kg is taking a space walk to work on the International Space Station. Because of a malfunction with the booster rockets on his spacesuit, he finds himself drifting away from the station with a constant speed of 0.510 m/s. With the booster rockets no longer working, the only way for him to return to the station is to throw the 7.25 kg wrench he is holding.

He throws the wrench with speed 15.70 m/s WITH RESPECT TO HIMSELF. (Hint: That is NOT the speed an observer on the space station would see. Think of a traveler walking on a moving walkway at the airport. The velocity of the traveler with respect to the ground equals the velocity of the traveler with respect to the moving walkway plus the velocity of the moving walkway with respect to the ground. For this problem, the velocity of the wrench with respect to the space station equals the velocity of the wrench with respect to the astronaut plus the FINAL velocity of the astronaut with respect to the space station.)

After he throws the wrench, how fast is the astronaut drifting toward the space station?


What is the speed of the wrench with respect to the space station?

I got the answer wrong several times and I looked for help but they didn't get it right either?? can somebody help clarify? thank you

Explanation / Answer

The momentum of the system will be conserved.

so, (m+M)u = mv - MV

v = 15.7 + 0.51 = 16.21 m/s

this is the speed of the wrench with respect to the space station.

so, (7.25 + 82)(0.51) = 7.25(16.21) - 82V

=> V = 0.878 m/s

this is the speed of the astronaut with respect to the space station.

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