Two spherical bodies of masses M and SM 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,

1.5 R

2.5 R

4.5 R

7.5 R

Solution

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 ( 9R - x )

$F = \frac{GM \times 5 M}{{(12 R - x)}^{2}}$

a_small = $\frac{F}{M}$

a_small = $\frac{G \times 5 M}{{(12 R - x)}^{2}}$

a_big_{= $\frac{F}{5 M}$}

a_big = $\frac{GM}{{(12 R - x)}^{2}}$

$x = \frac{1}{2}$ a_small t²

$x = \frac{1}{2} \frac{G \times 5 M}{{(12 R - x)}^{2}} t^{2}$ .....(i)

( 9R - x ) = $\frac{1}{2}$ a_big t²

(9R - x) = $\frac{1}{2} \frac{GM}{{(12 R - x)}^{2}} t^{2}$ ....(ii)

Thus dividing equation (i) by equation (ii), we get

$\frac{x}{9 R - x} = 5$

⇒ x = 45 R - 5x

⇒ 6x = 45 R

∴ x = 7.5 R

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