Question

A 92-kg astronaut and a 2000-kg satellite are at rest relative to a space station. the astronaut pushes on the satellite, giving it a speed of 0.14 m/s directly away from the station. seven and a half seconds later the astronaut comes into contact with the station. part a what was the initial distance from the station to the astronaut?

Answer 1

Momentum is conserved. 

Momentum of the satellite is 1600*0.14=224 kg-m/sec 
Resulting momentum of the astronaut will be equal -- hence 
86*v=224 or the velocity of the astronaut is 2.60 m/sec 

In 7.5 seconds the astronaut would travel 19.53 meters

Answer 2

Answer:

22.8 s

Explanation:

First we calculate the change in momentum of the satellite.

ΔP = 2000 kg (0.14 m/s 0 m/s) = 280 kg·m/s

The change in momentum of the satellite is equal to the change in momentum of the astronaut as a result of Newton's Third law. Therefore:

280 kg·m/s = 92kg * v

                 v = 3.04 m/s

d = speed * time = 3.04 m/s * 7.5 s = 22.8 m