A neutron travelling with a velocity v and K.E. E collides perfectly elastically head on with nucleu

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5.Work, Energy, Power and Collision

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

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

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

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

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

Solution

(a) ${\left( {\frac{{\Delta k}}{k}} \right)_{{\rm{retained}}}} = {\left( {\frac{{{m_1} – {m_2}}}{{{m_1} + {m_2}}}} \right)^2} = {\left( {\frac{{1 – A}}{{1 + A}}} \right)^2}$

Standard 11

Physics

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

Solution

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$U = \frac{M}{{{r^6}}} – \frac{N}{{{r^{12}}}}$,

$M$ and $N$ being positive constants, then the potential energy at equilibrium must be