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Question

Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

A
[2G(m1m2)r]1/2
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B
[2Gr(m1+m2)]1/2
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C
[r2G(m1m2)]1/2
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D
[2Grm1m2]1/2
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Solution

The correct option is D [2Gr(m1+m2)]1/2
By applying law of conservation of momentum,
m1v1m2v2=0m1v1=m2v2........ (i)
Where v1 and v2 are the velocities of masses m1 and m2 at a distance r from each other.
By conservation of energy,
Change in P.E= change in K.E.
Gm1m2r=12m1v21+12m2v22........ (ii)
Solving eqn. (i) and (ii) we get
v1= 2Gm22r(m1+m2) and v2= 2Gm21r(m1+m2)
Relative velocity of approach, vR
=|v1|+|v2|=2Gr(m1+m2)

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