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In the figure shown in text, m(1) = m, m...

In the figure shown in text, `m_(1) = m`, `m_(2) = 2m` and initial distance between them is `r_(0)`. Find velocities of the masses when separation between them becoms `r_(0)/(2)`.

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Let their velocites are `upsilon_(1)` and`upsilon_(2)`. From conservation of linear momentum.
`p_(i) = p_(f)`
`:. 0 = m upsilon_(1)-2m upsilon_(2)` ..(i)
From conservation of mechanical energy,
`E_(i) = E_(f)`
or `K_(i)+U_(i) = K_(F)+U_(F)`
or `0 - (G(m)(2m))/(r_(0))= (1)/(2) m upsilon_(1)^(2) + (1)/(2) xx 2m xx upsilon_(2)^(2) - (G(m) (2m))/((r_(2)//2))` ..(ii)
Solving Eqr.(i) and (ii), we get
`upsilon_(1)=2 sqrt((2Gm)/(3r_(0))),upsilon_(2) = sqrt((2Gm)/(3r_(0)))`
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