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Three solid spheres each of mass \mathrm{m}  and radius \mathrm{R} are released from the position shown in figure. The speed of any one sphere at the time of collision would be


 

Option: 1

\mathrm{\sqrt{\operatorname{Gm}\left(\frac{1}{\mathrm{~d}}-\frac{3}{\mathrm{R}}\right)}}

 


Option: 2

\mathrm{\sqrt{\mathrm{Gm}\left(\frac{3}{\mathrm{~d}}-\frac{1}{\mathrm{R}}\right)}}
 


Option: 3

\mathrm{\sqrt{\operatorname{Gm}\left(\frac{2}{\mathrm{R}}-\frac{1}{\mathrm{~d}}\right)}}
 


Option: 4

\mathrm{\sqrt{\mathrm{Gm}\left(\frac{1}{\mathrm{R}}-\frac{2}{\mathrm{~d}}\right)}}


Answers (1)

best_answer

From conservation of mechanical energy

\mathrm{3\left \{ {\frac{1}{2}mv}^{2} \right \}=\left\{^{2}\frac{\mathrm{Gm}^2}{2 \mathrm{R}}-\frac{\mathrm{Gm}^2}{\mathrm{~d}}\right\} }

\mathrm{ \mathrm{v}^2=\mathrm{Gm}\left\{\frac{1}{\mathrm{R}}-\frac{2}{\mathrm{~d}}\right\} }

\mathrm{ \therefore \mathrm{v}=\sqrt{\mathrm{Gm}\left\{\frac{1}{\mathrm{R}}-\frac{2}{\mathrm{~d}}\right\}}}

Hence option 4 is correct.



 

Posted by

rishi.raj

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