Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released. Assuming that only mutual gravitational force are acting, find the speeds of the particles when the separation decreases to 0.5 m.

Answer

A, B, D

Explanation

The linear momentum of 2 bodies is 0 initially. Since gravitastioN/Al force is interN/Al fiN/Al momentum is also zero.
So, `(10kg)v_1=(20kg)v_2`
or `v_1=2v_2` ……….i
`Since P.E. is conserverd
intial P.E.=(-6.67xx10611xx10xx20)/1`
-13.34xx10^-9J`
When separation is 0.5 m`
`rarr13.34xx10^-9+0`
`=(-13.34xx10^-9)/(1/2)+(1/2)xx10v_1^2+(1/2)xx20v_2^2` ........2
`rarr -13.34xx10^-9=26.68xx10^-9+5v_1^2+10v_2^2`
`rarr -13.34xx10^-9=26.68xx10^-9+30v_2^2`
`rarr v_2^2=(-13.34xx10^-9)/30`
=4.44xx10^-10`
`rarr v^2=2.1xx10^-5m/s`
So, `v_1=4.2x10^-5`