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Two point masses of mass 4m and m respec...

Two point masses of mass 4m and m respectively separated by d distance are revolving under mutual force of attraction. Ratio of their kinetic energies will be

A

`1 : 2`

B

`1 : 5`

C

`1 : 1`

D

`1 : 2`

Text Solution

Verified by Experts

The correct Answer is:
A

They will revolue about this centre of mass
`0 = 4m(-x)+m(d-x)`
`x=(d)/(5)`
They will same `omega`
`(K_(4m))/(K_(m))=((1)/(2)I_(4m)omega^(2))/((1)/(2)I_(m)omega^(2)) rArr (K_(4m))/(K_(m))=(I_(4m))/(I_(m))`
`(K_(4m))/(K_(m)) = ((1)/(2)(4m)(d//5)^(2))/((1)/(2)(m)(4d//5)^(2)) rArr (K_(4m))/(K_(m))=(1)/(4)`
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