Momentum of an astronaut floating in space

[nathancurtis11]

Homework Statement

You have been hired to check the technical correctness of an
upcoming sci-fi thriller film. The movie takes place in the space
shuttle. In one scene, an astronaut's safety line is damaged
while she is on a space walk, so she is no longer connected to
the space shuttle. She checks and finds that her thruster pack
has also been damaged and no longer works.
She is 200 m from the shuttle and moving with it (i.e., she is not
moving with respect to the shuttle). She is drifting in space with
only 4 minutes of air remaining. To get back to the shuttle, she
decides to unstrap her 10-kg tool kit and throw it away with all
her strength, so that it has a speed of 8 m/s relative to her. In
the script, she survives, but is this correct? Her mass, including
her space suit, is 80 kg.

Homework Equations

m1 * Δv1 = m2 * Δv2 (I think (law of conservation of momentum?))

Time = Distance / Velocity

The Attempt at a Solution

So I did:
80 * x = 10 * 8 (Law of conservation of momentum?)
x= 1 m/s

Time = 200 / 1 = 200 seconds

So she would make it with 40 seconds remaining in air, but this seems way to easy and very very wrong.

 

[Doc Al]

I'd say you got it just right. Glad you thought it was easy!

Edit: I misread the problem. See gneill's comments.

 

[gneill]

Hmm. The given speed is the separation speed between her and the tool kit, not the speed of the tool kit in the center of mass ("stationary") frame of reference.

If she's moving towards the shuttle at speed v1, and the tool kit is moving away at speed v2, then v1 + v2 = 8 m/s.

I think her margin of safety is going to be somewhat smaller than 40 seconds.

 

[Doc Al]

Hmm. The given speed is the separation speed between her and the tool kit, not the speed of the tool kit in the center of mass ("stationary") frame of reference.

Oops! You're right. I misread the problem. :uhh:

 

[nathancurtis11]

Hmm. The given speed is the separation speed between her and the tool kit, not the speed of the tool kit in the center of mass ("stationary") frame of reference.

If she's moving towards the shuttle at speed v1, and the tool kit is moving away at speed v2, then v1 + v2 = 8 m/s.

I think her margin of safety is going to be somewhat smaller than 40 seconds.

So given this information how do I go about finding v1 and v2? I know that now 80*v1 = 10*v2 given the law of conservation of motion correct? Do I need to find a v1 and a v2 that will satisfy both the v1+v2 = 8 m/s and the 80*v1 = 10 * v2 ??

 

[gneill]

So given this information how do I go about finding v1 and v2? I know that now 80*v1 = 10*v2 given the law of conservation of motion correct?

Correct.

Do I need to find a v1 and a v2 that will satisfy both the v1+v2 = 8 m/s and the 80*v1 = 10 * v2 ??

Yes, two equations in two unknowns. And you only need one of them

 

[nathancurtis11]

I did this is this right? ;

80v1 - 10v2 = 0 (multiply both by .1)

8v1 - v2 = 0
v1 + v2 = 8

-7v1 = -8

v1 = 7/8 m/s

Time = 200 / (7/8) ≈ 228.57 seconds

240 - 228.57 ≈ 11.43 seconds

So she will make it with 11.43 seconds of air left, is that better?

 

[gneill]

I did this is this right? ;

80v1 - 10v2 = 0 (multiply both by .1)

8v1 - v2 = 0
v1 + v2 = 8

-7v1 = -8

How does that follow? Why not just add the two equation to eliminate v2?

 

[nathancurtis11]

Yeah I noticed a stupid algebra error there -_- .

I multiplied the .1 to make it so the v2 's would cancel when adding them together

so;

8v1 - v2 = 0 (this step is okay right?)
v1 + v2 = 8

9v1 = 8

v1 = 8/9

Time = 200 / (8/9) = 225 seconds

240 - 225 = 15

So she gets there with 15 seconds to spare. I think that's better hopefully!

 

[gneill]

Yeah I noticed a stupid algebra error there -_- .

I multiplied the .1 to make it so the v2 's would cancel when adding them together

so;

8v1 - v2 = 0 (this step is okay right?)
v1 + v2 = 8

9v1 = 8

v1 = 8/9

Time = 200 / (8/9) = 225 seconds

240 - 225 = 15

So she gets there with 15 seconds to spare. I think that's better hopefully!

It's certainly better mathematically, if not cinematographically!