In: Physics
Robot No. 5 (mass 273kg) and Prisoner No. 6 (mass 66kg) stand on a frictionless ice rink, both holding a rope. The robot is at x = −3m and the prisoner is at x = 17m. The robot starts winding up the rope. Considering the two of them plus the rope as a system, there are no external horizontal forces on either of them. When the rope is fully wound, where do No. 5 and No. 6 meet?
here robot no-5 is standing at , x= -3 m ,
prisoner-6 is standing at , x= 17 m ,
mass of robot no-5 = 273 kg
mass of prisoner-6 = 66 kg
so here we can draw a simple diagram ,
for two mass system , we have to find center of the mass of the system where they both meet ,
we have formula for finding center of mass of two system = m1 x1 + m2 x2 / m1 + m2
here for robots and prisoner system,
Xcm = mrobot x1 + mprisoner x2 / mrobot + mprisoner
here mrobot = 273 kg
mprisoner = 66 kg
here X1 = here robot no-5 is standing at , x= -3 m , X1 = -3 m
here X2 = prisoner-6 is standing at , x= 17 m , X2 = 17 m
so here center of the mass of system where they both meet ,
Xcm = mrobot x1 + mprisoner x2 / mrobot + mprisoner
= 273* (-3) + 66 (17) / (273+66)
Xcm =303 / 339
Xcm = 0.89 m
so they both are meet at Xcm = 0.89 m
here when robot-5 and prisoner-6 meet's at Xcm = 0.89 m , then rope is fully wound.