Question

Two astronauts, each of mass $75kg$, are floating next to each other in space, outside the space shuttle. They push each other through a distance of an arm's length $=1m$ each with a force of $300N$.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 $1m$$)$

Answer 1

$Work=Force\times .Displacement$

$=300\times (1+1)=600N-m$, which is converted to kinetic energy.

Thus $2\times \frac{1}{2}\times 75\times v^2=600or,v=\sqrt{8}m/s$

Relative velocity will thus be $2v$ (because both move with $v$ in opposite direction).Thus,$\frac{{V_o}^2}{4}=\frac{{(2\times \sqrt{8})}^2}{4}=8$