Two spherical bodies of masses $m$ and $5 M$ and radii $R$ and $2 R$ , respectively, are released in free space with initial separation between their centres equal to $12 R$ . If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

2.5 R

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4.5 R

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7.5 R

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1.5 R

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Solution

The correct option is C $7.5 R$
$Let at 'o' there will be a collision.if smaller sphere moves x distance to reach at O.$
$Then bigger sphere will move a distance of (9 R - x) .$
$F = \frac{G M \times 5 M}{12 R - x)^{2}}$
$a_{small} = \frac{F}{M} = \frac{G M}{(12 R - x)^{2}}$
$a_{big}$ $x = \frac{1}{2} a_{small} t^{2} = \frac{1}{2} \frac{G \times 5 M}{(12 - x)^{2}} t^{2} . . . . . . . . . . . . . . . . . . . . . (1)$
$(9 R - x) = \frac{1}{2} a_{big} t^{2} = \frac{1}{2} \frac{G M}{(12 - x)^{2}} t^{2} . . . . . . . . . . . . . . . . . . . . . . . . . (2)$
Thus,dividing eq. $(1)$ by eq $(2),$ we get ,
$∴ \frac{x}{9 R - x} = 5$
or $x = 45 R - 5 x$
or $6 x = 45 R$
or $x = 7.5 R$

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Two spherical bodies of masses m and 5 M and radii R and 2R, respectively, are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is