Calculating Space Shuttle Velocity: Astronaut Melroy's Historic Flight

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Homework Help Overview

The problem involves calculating the velocities of the space shuttle and the International Space Station (ISS) as they separate after undocking. It is situated within the context of momentum conservation principles in physics.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of momentum conservation to determine the velocities of the shuttle and the ISS. There are questions regarding the interpretation of relative speeds and whether the initial velocities should be considered in the calculations.

Discussion Status

Some participants have provided insights into the need to account for the motion of both spacecraft during separation. There is a recognition of different interpretations regarding the calculation of relative speed, with some guidance offered on combining the velocities of both the shuttle and the ISS.

Contextual Notes

There appears to be confusion regarding the initial conditions and the expected outcomes, as indicated by differing answers and references to a textbook solution.

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Homework Statement



Problem Statement: Astronaut Pam Melroy, history's 3rd woman space shuttle pilot, flew the space shuttle discovery to the International Space Station to complete construction in October of 2000. To undock from the space station, Pilot Melroy released goooks holding the 2 spacecraft together and the 68 000-kg shuttle pushed away from the space station with the aid of 4 large springs.
A) if the 73 000-kg space station moved back at a speed of 0.50 m/s, how fast and in what direction did the space shuttle move?
B) What was the relative speed of the 2 spacecraft as they separated?M_{1}= 68 000-kg
V_{1F}= ?
M_{2}= 73 000-kg
V_{2F}= -0.50 m/s
V_{o}= 0 m/s

Homework Equations



(M_{1} + M_{2}) V_{o} = M_{1}V_{1F} + M_{2}V_{2F}

The Attempt at a Solution



(68 000-kg + 73 000-kg)0 m/s = ((68 000-kg)( V_{1F} ) + (73 000-kg)(-0.50 m/s)

0 = (68 000-kg)V_{1F} - 36 500-kg \timesm/s
add -36 500-kg \timesm/s to 0

36 500-kg \timesm/s = (68 000-kg)V_{1F}

= 0.53676 m/s

BUT the back of the book for this odd says
1.04 m/s
 
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It is asking you the speed they separate, remember that the ISS is also moving so it is the total speed between them.
You have worked out the speed of the shuttle from the starting point.
 
mgb_phys said:
It is asking you the speed they separate, remember that the ISS is also moving so it is the total speed between them.
You have worked out the speed of the shuttle from the starting point.

so my answer is the initial velocity??
 
No the answer is 0.5m/s for the iss plus 0.53676 m/s for the shuttle
So the velocity they are moving apart is 0.5 + 0.53676 = 1.04m/s
 

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