question_answer

Two spherical bodies of mass 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

A) \[2.5\] R                              

B) \[4.5\] R                              

C) \[7.5\] R                              

D) \[1.5\] R

Correct Answer: C

Solution :

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).             By universal law of gravitation,                     \[F=\frac{GM\times 5M}{{{(12R-x)}^{2}}}\] Acceleration of smaller sphere,                     \[{{a}_{small}}=\frac{F}{M}=\frac{G\times 5M}{{{(12R-x)}^{2}}}\] Acceleration of bigger sphere,                     \[{{a}_{big}}=\frac{F}{5M}=\frac{GM}{{{(12R-x)}^{2}}}\] \[\Rightarrow \]   \[x=\frac{1}{2}{{a}_{small}}{{t}^{2}}=\frac{1}{2}\frac{G\times 5M}{{{(12R-x)}^{2}}}{{t}^{2}}\]      ?. (i) \[(9R-x)=\frac{1}{2}{{a}_{big}}{{t}^{2}}=\frac{1}{2}\frac{GM}{{{(12R-x)}^{2}}}{{t}^{2}}\]     ?. (ii) Thus, dividing Eq. (i) by Eq. (ii), we get     \[\frac{x}{9R-x}=5\]        \[\Rightarrow \]               \[x=45R-5x\] \[\Rightarrow \]               \[6x=45R\,\,\,\,\Rightarrow \,\,\,\,x=7.5R\]