Question

Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. What is their separation after the heavier astronaut has moved 12 m?24 m16 m9.0 m28 m21 m

Answer 1

We will have the following:

First, we recall that momentum is given by:

$m_1v_0+m_2v_0=m_1v_1+m_2v_2$

So:

$\begin{gathered} (60kg)(0m/s)+(80kg)(0m/s)=(60kg)v_1+(80kg)v_2 \\ \\ \Rightarrow v_1=\frac{-(80kg)v_2}{(60kg)}\Rightarrow v_1=-\frac{4}{3}v_2 \end{gathered}$

Then:

[Let's assume that when astronaut 2 moved 12 meters = t]

$\begin{gathered} x_1=v_1t\Rightarrow x_1=-\frac{4}{3}v_2t \\ \\ x_2=v_2t\Rightarrow t=\frac{12}{v_2} \end{gathered}$

Then:

$x_1=-\frac{4}{3}v_2\ast\frac{12}{v_2}\Rightarrow x_1=-16m$

Finally, the total distance will be:

$\begin{gathered} d=|x_1|+|x_2|\Rightarrow d=16m+12m \\ \\ \Rightarrow d=28m \end{gathered}$