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  • Help me solve this An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is :

An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is : 

  • Option 1)

     4 m 

  • Option 2)

    1.5 m 

  • Option 3)

    2 m

  • Option 4)

     3.5 m 

Answers (1)

best_answer

 

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}

 

 


Option 1)

 4 m 

Option 2)

1.5 m 

Option 3)

2 m

Option 4)

 3.5 m 

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