An astronaut of mass 84.0 kg is taking a space walk to work on the International

Question

An astronaut of mass 84.0 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 0.550 m/s. With the booster rockets no longer working, the only way for him to return to the station is to throw the 7.65 kg wrench he is holding.
In which direction should he throw the wrench?

away from the stationtoward the station    

He throws the wrench with speed 16.47 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?

Explanation / Answer

a) away from the station

b) let v is the speed of astronaut towards the space station.

let M = 84 kg
m = 7.65 kg

Apply conservation of momentum

initial momentum = final momentum

(m + M)*0.55 = m*(0.55 + 16.47) + M*v

(7.65 + 84)*0.55 = 7.65*(0.55 + 16.47) + 84*v

==> 84*v = (7.65 + 84)*0.55 - 7.65*(0.55 + 16.47)

= -79.8

v = -79.8/84

= -0.95 m/s

speed, |v| = 0.95 m/s <<<<<<-----Answer

speed of wrench with respect to space station = 0.55 + 16.47

= 17.02 m/s <<<<<<-----Answer

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