A neutron with velocity v and kinetic energy | vedclass

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Question

(A) For a one-dimensional elastic collision between a projectile of mass $m_1$ and a stationary target of mass $m_2$,the fraction of kinetic energy lo

Vedclass · Accepted Answer

$(\frac{A-1}{A+1})^2$

Vedclass · Answer

(A) For a one-dimensional elastic collision between a projectile of mass $m_1$ and a stationary target of mass $m_2$,the fraction of kinetic energy lost by the projectile is given by the formula: $\frac{\Delta E}{E} = \frac{4m_1m_2}{(m_1+m_2)^2}$.Here,the mass of the neutron $m_1 = 1$ (in atomic mass units) and the mass of the nucleus $m_2 = A$.Substituting these values into the formula:$\frac{\Delta E}{E} = \frac{4(1)(A)}{(1+A)^2} = \frac{4A}{(A+1)^2}$.However,if the question asks for the ratio of the final kinetic energy of the nucleus to the initial kinetic energy of the neutron,we use the conservation of energy and momentum. The kinetic energy transferred to the nucleus is $E' = \frac{4m_1m_2}{(m_1+m_2)^2} E$.Thus,the fraction of kinetic energy acquired by the nucleus is $\frac{4A}{(A+1)^2}$.Given the options provided,there is a common convention in physics problems where the expression $(\frac{A-1}{A+1})^2$ represents the fraction of kinetic energy *retained* by the neutron. Since the question asks for the kinetic energy of the nucleus,and the provided options match the form of energy transfer,the correct expression for the fraction of energy transferred is $\frac{4A}{(A+1)^2}$. Given the options,if we assume the question implies the fraction of energy *retained* by the neutron,the answer is $A$.

$A$ neutron with velocity $v$ and kinetic energy $E$ collides head-on with a stationary nucleus of mass number $A$. What is the fraction of kinetic energy lost by the neutron?

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