Question

An 80-kg astronaut becomes separated from his spaceship. He is 15.0 m away from it and at rest relative to it. In an effort to get back, he throws a 500 gram object with a speed of 8.0 m/s in a direction away from the ship. How long does it take him to get back to the ship?

Answer 1

Answer: 300 s

Step-by-step explanation:

The momentum
`p` is given by the following equation:

`p=m.V`

Where
`m` is the mass of the object and
`V` is the velocity.

In addition, according to the conservation of linear momentum, we have:

`p_(1)=p_(2)` (1)

Where:

`p_(1)=0` is the initial momentum of the astronaut, which is initially at rest

`p_(2)=m_(object)V_(object)+m_(astronaut)V_(astronaut)` is the final momentum, being
`m_(object)=500g=0.5 kg`,
`V_(object)=8 m/s` and
`m_(astronaut)=80 kg`

Then (1) is rewritten as:

`0=m_(object)V_(object)+m_(astronaut)V_(astronaut)` (2)

Finding te velocity of the astronaut
`V_(astronaut)`:

`V_(astronaut)=-(m_(object)V_(object))/(m_(astronaut))` (3)

`V_(astronaut)=-((0.5 kg)(8 m/s))/(80 kg)` (4)

`V_(astronaut)=-0.05 m/s` (5) The negative sign of the velocity indicates it is directed towards the spaceship, however its speed (the magnitude of the velocity vector) is positive
`0.05 m/s`

On the other hand we have the following:

`V_(astronaut)=(d)/(t)` (6)

Where
`d=15 m` the distance between the astronaut and the spacheship and
`t` the time. So, we have to find
`t`:

`t=(d)/(V_(astronaut))` (7)

`t=(15 m)/(0.05 m/s)`

Finally:

`t=300 s`