Question

Two astronauts, each of mass 75 kg, are floating next to each other in space, outside the space shuttle. They push each other through a distance of an arm's length = 1 m each with a force of 300 N. If the final relative velocity of the two, w.r.t. each other is $V_0$ m/s, find the value of $\frac{(V_0{)}^2}{4}$. (Note that both astronauts are displaced by 1 m.)

Answer 1

Mass of each astronaut $M=75$ kg

$\therefore$ Acceleration of each astronaut $a=\frac{F}{M}=\frac{300}{75}=4$ $m/s^2$ (In opposite directions)

For astronaut A : Distance by which he is displaced $S=1m$

Initial Velocity of the astronaut A $u=0$

Using $v^2-u^2=2aS$

$\therefore$ $v^2-0=2\left(4\right)\times 1$ $⟹v=2\sqrt{2}$ $m/s$

Relative velocity of A w.r.t B, $V_o=v+v=2v=4\sqrt{2}$ $m/s$

$\therefore$ $\frac{(V_o{)}^2}{4}=\frac{(4\sqrt{2}{)}^2}{4}=8$