An astronaut of mass 77.0 kg is taking a space walk to work on the International

Question

An astronaut of mass 77.0 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 0.300 m/s. With the booster rockets no longer working, the only way for him to return to the station is to throw the 8.35 kg wrench he is holding.

He throws the wrench with speed 7.67 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.)

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

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

Explanation / Answer

The person moves one way as the wrench is thrown the other.

The only thing that remains constant is the centre of mass.

So as the wrench and the person are together we know that the centre of mass is moving at 0.300 m/s relative to the space station.

And that the space station has no forces on it so it remains aninertial frame.

OK so the person moves backwards at - V and the wrench moves forwards at u

We know that u + V = 7.67 ie the relative speed is 7.67

But we ALSO know that u* 8.35 = V * 77

momentum is conserved ( here I am assuming that the astronaut,space suit, rocket boosters but NOT the wrench have a mass of 77kg)

so from these two equations u *8.35 u / 80 = 7.67m/s

u* (1+8.35/77) = 7.67m/s

u = 7.67/ (1+8.35/77) = 63.16 m/s from the centre of mass.

and he throws it away from the space station

so its final velocity relative to that station is 0.300 + 63.16 =63.46 m/s

Now you probably made the same error in part b but "got away with it" due to the larger mass of the astronaut.

V *(1+77/8.35) = 7.67m/s

V = 7.67 /(1+77/8.35) = 0.821

So his speed relative to the ship becomes

0.821 - 0.300 = 0.521 m/s towards the ship

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