Conservation of Momentum of astronaut leaving ship

[TyroneTheDino]

Homework Statement

An astronaut is pushed from her ship at a velocity of 2m/s. Her weight including her tool belt is 120kg. Remembering Newtons 3rd law, she takes 5 seconds to decide to detach her belt and throw it away from her. The 20kg tool belt leaves her suit as she throws it in front of her 10m/s. If she has 20 minutes of air left in her tank, does she make it?

Homework Equations

Conservation of Momentum
mv_i=mv_f

Distance
d=vt

The Attempt at a Solution

I say that the astronaut survives.

Because of distance equation= (2)(5)=10m
Her initial momentum is 240kg m/s
And by throwing the belt in front of her she is propelled at -.4m/s
Since she only traveled 10m she would make it back on time.

 

[Wily Willy]

I would start by revisiting conservation of momentum and assigning each of the velocity vectors a sign depending on whether they are directed towards the ship or away from the ship. I think that will really clear things up.

 

[gneill]

The question as stated seems to be a bit ambiguous to me. Is the speed that the tool belt is thrown meant to be taken relative to the astronaut or the initial reference frame? If she happened to be pushed out at a speed of 10 m/s instead of 2 m/s, would that mean she couldn't throw the tool belt away from herself at all?

 

[haruspex]

The question as stated seems to be a bit ambiguous to me. Is the speed that the tool belt is thrown meant to be taken relative to the astronaut or the initial reference frame? If she happened to be pushed out at a speed of 10 m/s instead of 2 m/s, would that mean she couldn't throw the tool belt away from herself at all?

There are three reference frames to choose from for the 10m/s: the ship, the astronaut before throwing the belt, the astronaut after throwing the belt. For the question to be at least a bit interesting, I think it has to be the second.
Tyrone, how do you get the -.4m/s?