Question

Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centers equal to 12R. If they attract each other due to gravitational force only. then the distance covered by the smaller body just before collision is
(a). 1.5R
(b). 2.5R
(c). 4.5R
(d). 7.5R

Answer 1

Hint: This question talks about attraction and that due to gravitational force. So, we have to apply here universal law of gravitation to arrive at the answer.

Complete step by step solution:
The radius of first body is R and its mass is M
The radius of second body is 2R and its mass is 5m
Since the bodies are spherical so in order to find the distance between the two bodies we need to measure it from the center of one body to the center of another. But these are not point objects and hence in order to apply our universal law of gravitation we need to have the separation between the two surfaces.
Distance between their surfaces =12R−R−2R=9R
If \[{{x}_{1}}\] and \[{{x}_{2}}\] are the distance covered by the two bodies, then \[{{x}_{1}}\] + \[{{x}_{2}}\]=9R
Suppose they both meet at point O, If smaller sphere moves x distance to reach at O, then bigger sphere will move a distance of (9R−x)
Using universal law of gravitation,
\[F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]
And F=ma
So \[a{}_{small}=\dfrac{F}{m}=\dfrac{G\times 5m}{{{(12R-x)}^{2}}}\]-----------------(1)
\[{{a}_{big}}=\dfrac{F}{5m}=\dfrac{G\times m}{{{(12R-x)}^{2}}}\]--------------------(2)
Since initially both the bodies were at rest, using second equation of motion
\[x=\dfrac{a{{t}^{2}}}{2}\]
\[x=\dfrac{{{a}_{small}}{{t}^{2}}}{2}=\dfrac{5\times Gm\times {{t}^{2}}}{2{{(12R-x)}^{2}}}\]----------------(3)
\[9R-x=\dfrac{{{a}_{big}}{{t}^{2}}}{2}=\dfrac{Gm\times {{t}^{2}}}{2{{(12R-x)}^{2}}}\]-------------(4)
Dividing eq (3) by eq (4) we get
x=45R−5x
6x=45R
x=7.5R

Thus the correct answer is option D.

Note: The distance covered by the smaller body just before collision is 7.5R