Conservation of Momentum of astronaut leaving ship

AI Thread Summary
An astronaut is pushed from her ship at 2 m/s and weighs 120 kg, including a 20 kg tool belt. After 5 seconds, she throws the belt forward at 10 m/s, which affects her momentum. The initial momentum is calculated as 240 kg m/s, and after throwing the belt, her new velocity is determined to be -0.4 m/s. The discussion raises questions about reference frames for the belt's velocity and whether the astronaut can still throw it effectively if her initial speed were higher. The conclusion suggests that she can return to the ship in time, but clarity on the reference frame is needed for accurate calculations.
TyroneTheDino
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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
mvi=mvf

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.
 
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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.
 
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?
 
gneill said:
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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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