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 12m

Answer 1

Answer:

The astronauts are separated by 28 m.

Explanation:

The separation of the astronauts can be found by conservation of linear momentum:

$p_{i} = p_{f}$

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}$

$m_{1}*0 + m_{2}*0 = m_{1}v_{1f} + m_{2}v_{2f}$

$m_{1}v_{1f} = -m_{2}v_{2f}$

$v_{1f} = -\frac{m_{2}v_{2f}}{m_{1}} = -\frac{80v_{2f}}{60}$

Now, the distance (x) is:      

$x = \frac{v}{t}$  

The distance traveled by the astronaut 1 is:

$x_{1} = v_{1f}*t = -\frac{80v_{2f}}{60}*t$    (1)

And, the distance traveled by the astronaut 2 is:

$x_{2} = v_{2f}*t$  (2)

From the above equation we have:  

$t = \frac{x_{2}}{v_{2f}}$    (3)                                    

By entering equation (3) into (1) we have:    

$x_{1} = -\frac{80v_{2f}}{60}*(\frac{x_{2}}{v_{2f}})$

$x_{1} = -\frac{4*12}{3} = -16 m$    

The minus sign is because astronaut 1 is moving in the opposite direction of the astronaut 2.      

Finally, the separation of the astronauts is:

$x_{T} = |x_{1}| + x_{2} = (16 + 12)m = 28 m$

Therefore, the astronauts are separated by 28 m.

   

I hope it helps you!