A55.8kgspacewalkingastronautpushesoffa666.0kgsatellite,exertinga106.0Nforceforthe0.659sittakeshimtostraightenhisarms.Howfarapartaretheastronautandthesatelliteafter1.11min?

Question

A 55.8 kg spacewalking astronaut pushes off a 666.0 kg satellite,exerting a 106.0 N force for the 0.659 s it takes him to straighten his arms.How far apart are the astronaut and the satellite after 1.11 min?

Answer 1

由动量定理,宇航员推出卫星的过程中,有
m1*v1=F*t
其中m1,v1为卫星的质量与速度,得
v1=F*t/m1
而整个过程动量守恒,故有
m1*v1+m2*v2=0
其中m2,v2为宇航员的质量与速度,故
v2=-F*t/m2
因此1.11min后,宇航员与卫星的距离
s=x1-x2=(v1-v2)*1.11min
=F*t*(1/m1+1/m2)*1.11min
=106.0N*0.659s*1.11min*60s/min*(1/55.8kg+1/666.0kg)
=90.4 (m)