Two spherical bodies of mass m and 5m, radius R and 2R respectively are released in free space with an initial separation of 12 R between their centres. If they attract each other due to gravitational force alone, then the distance covered by the smaller body just before the 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

$F = \frac{G m (5 m)}{(12 R - x)^{2}}$
$∴ a_{s m a l l} = \frac{F}{m} = \frac{G (5 m)}{(12 R - x)^{2}}$
$a_{b i g} = \frac{F}{5 m} = \frac{G m}{(12 R - x)^{2}}$
$∴ x = \frac{1}{2} a_{s m a l l} t^{2} = \frac{1}{2} \frac{G 5 m}{(12 R - x)^{2}} t_{2}^{2} \dots (1)$
$9 R - x = \frac{1}{2} a_{b i g} t^{2} = \frac{1}{2} \frac{G (m)}{(12 R - x)^{2}} t^{2} \dots (2)$
$(1) \div (2) \Rightarrow x = 7.5 R$

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Two spherical bodies of mass M and 5 M and radii R and 2 R 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 before collision is :

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