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A neutron moving with speed v and kinetic energy E undergoes a perfectly elastic, head-on collision with a stationary atomic nucleus of mass number A. The approximate fraction of the neutron’s total energy that remains after the collision is

1. [(A − 1)/(A + 1)]²
2. [(A + 1)/(A − 1)]²
3. [(A − 1)/A]²
4. [(A + 1)/A]²

Correct answer: [(A − 1)/(A + 1)]²

Solution

For a neutron (mass ~1) in a head-on elastic collision with a nucleus of mass A, the fraction of energy retained is [(m-M)/(m+M)]^2 = [(1-A)/(1+A)]^2 = [(A-1)/(A+1)]^2.

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