Two particles of masses m and 2m are kept at a distance a. Find their relative velocity of approach when separation becomes a/2.
A
2√3a2GM
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B
√a2GM
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C
2√2GM3a
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D
√6GMa
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Solution
The correct option is D√6GMa In arrangements (1) and (2), Momentum conservation, 0=mv1−2mv2 ...........(i) Energy conseravtion −Gm⋅2ma=12mv21+122mv22−Gm⋅2ma/2 ..........(ii) From Eq. (i), we get v1=2v2 Putting this value (v1=2v2) in Eq. (ii), we get −2Gm2a=12m(2v2)2+12⋅2mv22−4Gm2a 3mv22=2Gm2a v2=√2Gm3a v1/2=v1+v2 =3√2Gm3a =√6Gma.