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.
2 answers
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.
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
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
5 answers