What is the Final Velocity of the Astronaut After Throwing Away the Power Pack?

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Homework Help Overview

The discussion revolves around a physics problem involving an astronaut who throws away a power pack while drifting in space. The scenario includes initial velocities and mass considerations, focusing on the conservation of momentum to determine the astronaut's new velocity relative to the Space Shuttle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore different frames of reference to analyze the problem, questioning how to incorporate initial velocities into the momentum conservation equation. There are discussions about the correct setup of the momentum equation, including the need for three terms to account for all objects involved.

Discussion Status

Some participants have offered alternative approaches and corrections to the original poster's method. There is acknowledgment of the complexity of the problem, with some expressing confusion and others providing insights that lead to a better understanding of the conservation of momentum in this context.

Contextual Notes

Participants note the importance of correctly defining the initial conditions and the reference frame used in the calculations. There is mention of the original poster's feelings of discouragement regarding their physics studies, highlighting the supportive nature of the forum.

vertex78
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Homework Statement


An astronaut runs out of fuel in his power pack while drifting at 1.0 m/s away from the Space Shuttle. The astronaut removes his 25.0 kg power pack and throws it away at 2.9 m/s relative to the shuttle and in the direction away from the shuttle. The mass of the astronaut plus the space suit, minus the power pack is 103.0 kg. Calculate the new relative speed of the astronaut towards the shuttle. Take the velocity away from the shuttle to be minus


Homework Equations


m1*v1(final) + m2*v2(final) = 0


The Attempt at a Solution


m1 = 103.0kg (astronaut)
m2 = 25.0kg (power pack)
v1(final) = ? (astronaut)
v1(initial) = -1.0 m/s (astronaut)
v2(final) = -2.9 m/s (power pack)

So I am solving for v1(final):
v1(final) = (-m2*v2)/m1
= (-25.0kg*-2.9m/s)/103.0 kg
= .703 m/s

So I thought this would be the astronauts new velocity but it is not. I'm really not sure how to tie in the astronauts initial velocity of -1.0 m/s into the momentum equation.
 
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Hmmm. There's an error in there that I explain this way (then you can go back and correct that error):

Let's switch to a new frame of reference. The astronaut is stationary to start with, and the shuttle is moving away from him at 1.0 m/s. I'll use to the left as positive and to the right as a negative direction (to help visualize it.) Thus, the shuttle is moving toward the left at 1.0 m/s. The astronaut throws his pack away from the shuttle at a speed of 2.9m/s *relative to the shuttle*, thus it's only moving away from him at -1.9m/s (to the right). Should be relatively simple to calculate the astronaut's new velocity.

m1v1+m2v2=0
103kg v1 = - 25kg * -1.9m/s
v1 = .46 m/s (since it's positive, the astronaut is moving toward the shuttle) Remember, the shuttle is moving at 1.0 m/s in the same direction. Thus, the astronaut is in trouble.
 
Without switching to a new frame of reference, there should be three terms in your conservation of momentum equation. Initial of astronaut plus power pack, final of power pack and final of astronaut. And, yes, he's still in trouble.
 
i tried what you said dr pizza and I could not get the correct answer. So i tried without switching the frame of reference and did this:

v1i = initial velocity of astronaut
v1f = what i am looking for
v2i = initial velocity of the power pack
v2f = final velocity of the power pack

using the conservation of momentum formula and solving for v1f i got this equation:

v1f = (m1*v1i + m2*v2i - m2*v2f)/m1

This got me the correct answer! thanks for the help guys, I am glad I came across this forum. I am attempting to get a minor in physics and I am getting very discouraged, but now at least I have one more resource to get help from!
 
Last edited:
You might want to try it from drpizza's frame as well. It should get you the same answer. It would be good practice.
 

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