An astronaut of mass 75.0 kg is taking a space walk to work on the International

Question

An astronaut of mass 75.0 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 0.450m/s. With the booster rockets no longer working, the only way for him to return to the station is to throw the 7.95 kg wrench he is holding.

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

For every force there is an equal and opposite reaction, so the astronaut must throw the wrench away from the

station, so that he can move towards it.

v_wrench - v_astronaut = 11.74 m/s

v_wrench = v_astronaut + 11.74 m/s

Let s find the momentums.

p_astronaut = (75 kg)(v_astronaut)

p_wrench = (7.95 kg)(v_wrench)

Now, momentum must be conserved, and the law of conservation of momentum states that

p_initial = p_final, for any closed system

We assumed that the astronaut and the wrench start at rest, so p_initial = 0. So

0 = p_astronaut + p_wrench

0 = (75 kg)(v_astronaut) + (7.95 kg)(v_wrench)

0 = (75 kg)(v_astronaut) + (7.95 kg)(v_astronaut + 11.74 m/s)

0 = 82.95*v_astronaut + 93.333 kg*m/s

v_astronaut = -1.125 m/s, with respect to the frame in which he was originally at rest

Now let's find his velocity with respect to the space station. He was originally drifting away from the station at 0.450

m/s. So that is the velocity of the frame in which he was originally at rest with respect to the space station. To find his

velocity with respect to the station, let s add the two velocities.

0.450 m/s + (-1.125 m/s) = -0.675 m/s, with respect to the station

Now for the wrench,

v_wrench = v_astronaut + 11.74 m/s

v_wrench = -0.675 m/s + 11.74 m/s

v_wrench = 11.065 m/s, with respect to the station

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