Posted by **Christy** on Saturday, March 1, 2008 at 6:34pm.

I am definitely lost on this problem. Can someone help me out?

A(n) 51.3 kg astronaut becomes separated

from the shuttle, while on a spacewalk. She

finds herself 50.3 m away from the shuttle

and moving with zero speed relative to the

shuttle. She has a(n) 0.537 kg camera in her

hand and decides to get back to the shuttle

by throwing the camera at a speed of 12 m/s

in the direction away from the shuttle.

How long will it take for her to reach the

shuttle? Answer in units of min.

- Physical science question 2 -
**Quidditch**, Saturday, March 1, 2008 at 6:57pm
Think conservation of momentum here. Relative to the shuttle, the momentum of the astronaut and camera is 0 (zero velocity realtive to the shuttle). After she throws the camera, her momentum plus the momentum of the camera is still 0.

- Physical science question 2 -
**drwls**, Saturday, March 1, 2008 at 7:00pm
Consider a coordinate sytaem moving with the shuttle. Apply the law of conservation of momentumin that coordinate system.

Since total momentum will be conserved when she throws away the camera, Mastronaut*Vrecoil= = Mcamera*Vcamera

Vrecoil = (.537)/51.3)*12 m/s

= 0.1256 m/s

Time to reach shuttle = 50.3 m/0.1256 m/s = 400 seconds = 6 2/3 minutes

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