8. When two identical masses are placed at a separation of 1 cm, thegravitational attractive force between them is 0.00667 N. Mass of each body is
(i)1 kg
(ii)10 kg
(iii)100 kg
(iv)1000 kg
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8. When two identical masses are placed at a separation of 1 cm, thegravitational attractive force between them is 0.00667 N. Mass of each body is
(i)1 kg
(ii)10 kg
(iii)100 kg
(iv)1000 kg
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Answer:
Solution :
The linear momentum of 2 bodies is 0 initially. Since gravitastioN/Al force is interN/Al fiN/Al momentum is also zero.
So, (10kg)v1=(20kg)v2
or v1=2v2 ……….i
sinceP.E.isconserverd∫ialP.E.=−6.67×10611×10×201
-13.34xx10^-9JWhenseparationis0.5m
→13.34×10−9+0
=−13.34×10−912+(12)×10v21+(12)×20v22 ........2
→−13.34×10−9=26.68×10−9+5v21+10v22
→−13.34×10−9=26.68×10−9+30v22
→v22=−13.34×10−930
=4.44xx10^-10rarr v^2=2.1xx10^-5m/sSo,v_1=4.2x10^-5`