Question

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​

Answer 1

HOPE ITS HELP

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`