Two skaters of mass m1=50 kg and m2=70 kg are standing motionless on a horizontal ice surface. They are initially a distance L= 8.0 meters apart. They hold a massless rope between them. After pulling the rope, the skater of mass m1 has moved a distance l= 2.0 meters away from his initial position. We can completely neglect friction in this problem.

Question

Two skaters of mass m1=50 kg and m2=70 kg are standing motionless on a horizontal ice surface. They are initially a distance L= 8.0  meters apart. They hold a massless rope between them. After pulling the rope, the skater of mass m1 has moved a distance l= 2.0  meters away from his initial position. We can completely neglect friction in this problem.

What is the distance L′ between the two skaters when the skater of mass m1 has moved a distance l? (in meters)

bobpursley · Answer

The cneter of gravity has to remain at the same point.

measuring from the center of gravity point, then if d is the distance the m1 is frm the cg, then
50d=70(8-d) and after the jerk, then we know
50(d-2)=70(L'-(d-2)

you have two equations, two unknowns

solve the first equation for d
50d=560-70d
120d=560
d= 4 2/3
now the second equation..
50(d-2)=70(L'-d+2)
50(2 2/3)=70L' -70(4 2/3-2)

L'= 120(2 2/3)/70=4.7 m check my math, done in my head.