Correct Answer - A::B::C
By applying low of conservation of momentum
`mv_(1)-4mv_(2)=0impliesv_(1)=4v_(2)`
by applying conservation of energy `(1)/(2)mv_(1)^(2)+(1)/(2)4mv_(2)^(2)=(Gm4m)/(r)implies10mv_(2)^(2)=(G4m^(2))/(r)impliesv_(2)=2sqrt((Gm)/(10r))`
`therefore` total kinetic energy `=(4Gm^(2))/(r)`
Relative velocity for the particle `impliesv_(rel)=|vecv_(1)-vecv_(2)|=5v_(2)=sqrt((10Gm)/(r))`
Mechanical energy of system `=0=` constant by using reduced mass concept
`(1)/(2)muv_(rel)^(2)=(Gm(4m))/(r)` where `mu=((m)(4m))/(m+4m)=(4)/(5)mimpliesv_(rel)=sqrt((10Gm)/(r))`
Also total KE system `=(G(m)(4m))/(r)=(4Gm^(2))/(r)`