A wire of `rho_(L)=10^(-6)Omega//m` is turned in the form of a circle of diameter 2m. A piece of same material is connected. In diameter AB. Then find resistance in between A and B.

Explanation

System of these two particles is in free, therefore no extenal forces act on them. There total linear momentum remains conserved and their mass center moves with constant velocity relative to an inertial frame.
Velocity of the mass center
`v_c=(Sigmam_ivecv_i)/(Sigmam_i)=(2(3hati-2hatj+hatk)+(3i+j-2hatk))/(2+3) = (3hat(i) - hat(j) - 4 hat(k))/(5) m//s`
Location `vec(r)_(co)`of the mass center at the instant `t = 0 s`

`barr_c=(Sigmam_ibarr_i)/(Sigmami)` rarr " " `barr_infty=(2(-2hati+hatj+4hatk)+3(2hati-3hatj+6hatk))/(2+3)=(2hati-7hatj+26hatk)/(5)`
New location `vec(r )_c` of the mass center at the instant t=10s
`vec(r )_c= vec(r )_infty+vecv_(c) t rarr " " vec(r )_(c) = (2hati-7hatj+26hatk)/(5)=(3i-hatj-4hatk)/5xx10=(32hati-17hatj-14hatk)/5`
New location (x, y, z) of the second particle.
`vec(r)_(c)= (Sigmam_1vecr_1)/(Sigmam_i)rarr` " " `(32hati-17hatj-14hatk)/5=(2(6hati+8hatj-6hatk)+3(xbari+yhatj+zbark))/(2+3)`
Solving the above equation, we obtain the corrdinates of the second particle `(20//3, - 11, -2//3)`