A neutron travelling with a velocity v and kinetic energy E collides elastically head on with the nucleus of an atom  of mass number A  at rest. The fraction of total  energy retained by the neutron is

  1. (A-1 / A+1)2
  2. (A+1 / A-1)2
  3. (A-1 / A)2
  4. (A+1 / A)2


Let say that a neutron with mass 'm1' and initial speed 'u1' strikes the nucleus (which is at rest, so u2 = 0) and undergoes a perfectly elastic collision. The final velocity (v1) of the neutron would be given as (from combining the Laws of Conservation of Kinetic energy and Momentum)



In this case, u2 = 0, thus  


taking m1 common and canceling it , we get



here A is the mass number of the nucleus.

Now, the fraction of kinetic energy retained would be  




So, the answer would be option (1)

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