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Two uniform spherical stars made of same material have radii `R` and `2R`. Mass of the smaller planet is `m`. They start moving from rest towards each from a large distance under mutual force of gravity. The collision between the starts is inelastic with coefficient of restitution `1//2`.
Kinetic energy of the system just after the collision is
A. `(8Gm^(2))/(3R)`
B. `(2Gm^(2))/(3R)`
C. `(4Gm^(2))/(3R)`
D. cannot be determined

1 Answer

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Best answer
Correct Answer - B
From conservation of linear momentum
Velocity before collision
`vec(P)_(1)+vec(P)_(2)=0 rArr P_(1)+P_(2)`
`m u_(1)+8m u_(2) =0 rArr u_(1) =-8u_(2)` ….(i)
From energy conservation
`U_(i)+K_(i)=U_(i)+K_(f) rArr 0+U=-(G(m)(8m))/(3R)+K_(f) rArr K_(f)=(8)/(3)(Gm^(2))/(R )`
This is the kinetic energy before collision.
For kinetic energy after collision from conservatgion of linear of linear momentum.
`0 = m_(1)v_(1)+m_(2)v_(2) rArr v_(1)+8 v_(2) = 0` ....(ii)
`e = ("velocity of sepration")/("velocity of approach") rArr (1)/(2) = (v_(2)-v_(1))/((u_(1)+(u_(1))/(8)))`
`rArr(9u_(1))/(16)=v_(2)-v_(1)` ....(iii)
From (ii) and (iii) `9 v_(2)=(9u_(1))/(16)`
`v_(2) = (9u_(1))/(16)` and `v_(1) = (-u_(1))/(2)` or `v_(2) = (u_(2))/(2)` and `v_(1) =-(u_(1))/(2)`
It means total KE after collision `= (1)/(4)xx` total KE before collision
`=(1)/(4)xx(8)/(3)(Gm^(2))/(R )=(2)/(3)(Gm^(2))/(R )`

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