InterviewSolution
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InterviewSolution
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Two stars each of mass `M` and radius `R` are approaching each other for a head-on collision. They start approaching each other when their separation is `rgt gtR`. If their speed at this separation are negligible, the speed `v` with which they collide would beA. `v=sqrt(GM(1/R-1/r))`B. `v=sqrt(GM(1/(2R)-1/r))`C. `v=sqrt(GM(1/R+1/r))`D. `v=sqrt(GM(1/(2R)+1/r))` |
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Answer» Correct Answer - B Since the speeds of the stars are negligible when they are at distance `r`, hence the initial kinetic energy of the system is zero. Therefore, the initial total energy of the system is `E_(i)=KE+PE=0+(-(GMM)/r)=-(GM^(2))/r` Where `M` represents the mass of each star and `r` is initial separation between them, twice the radius of a star i.e. `2R`. Let `v` be the speed with which two stars collides. Then total energy of the system at the instant of their collision is given by `E_(f)=2xx(1/2 Mv^(2))+(-(GM_(E)m)/(2R))=Mv^(2)-(GM^(2))/(2R)` According to law of conservation of mechanical energy `E_(f)-E_(i)` `Mv^(2)-(GM^(2))/(2R)=-(GM^(2))/r` or `v^(2)=GM(1/(2R)-1/r)` or `v=sqrt(GM(1/(2R)-1/r))` |
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