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

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 $V_{astronaut}$:

$V_{astronaut}=-\frac{m_{object}V_{object}}{m_{astronaut}}$ (3)

$V_{astronaut}=-\frac{(0.5 kg)(8 m/s)}{80 kg}$ (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:

$V_{astronaut}=\frac{d}{t}$ (6)

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

$t=\frac{d}{V_{astronaut}}$ (7)

$t=\frac{15 m}{0.05 m/s}$

Finally:

$t=300 s$