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Two particles of the same mass m are moving in circular orbits because of force, given by F(r)= \frac{-16}{r}-r^{3} The first particle is at a distance r=1, and the second, at r=4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :

  • Option 1)

    6\times 10^{-2}

  • Option 2)

    3\times 10^{-3}

  • Option 3)

    10^{-1}

  • Option 4)

    6\times 10^{2}

 

Answers (1)

As we have learned

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

m_{1},m_{2}:masses

u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}

u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}

 

 

F(r)= \frac{-16}{r}+r^{3}

\Rightarrow \frac{mv^{2}}{r}=\frac{16}{r}+r^{3}K.E= 1/2 mv^{2}=8+1/2r^{4}

\frac{k_{1}}{k_{2}}= \frac{8+1/2}{8+(1/2)^{256}}=17/272\approx 6\times 10^{-2}

 

 

 

 


Option 1)

6\times 10^{-2}

This is correct

Option 2)

3\times 10^{-3}

This is incorrect

Option 3)

10^{-1}

This is incorrect

Option 4)

6\times 10^{2}

This is incorrect

Posted by

Vakul

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