A neutron, with kinetic energy KE and moving with velocity $v$ , collides head on with a nucleus of mass number $A$ perfectly elastically. The fraction of total energy possessed by the neutron transferred to nucleus is

\frac{4 A}{(A + 1)^{2}}

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

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

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

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Solution

The correct option is A $\frac{4 A}{(A + 1)^{2}}$
$C o n s e r v a t i o n o f l i n e a r m o m e n t u m m v = A m v_{1} + m v_{2} a n d d e f i n t i o n o f c o e f f i c i e n t o f R e s t i t u t i o n \frac{v_{2} - v_{1}}{v - 0} = - 1 \Rightarrow v_{2} - v_{1} = - v s o l v i n g t h e s e e q u a t i o n s w e g e t v_{1} = \frac{2}{A + 1} v K E o f t h e n e u t r o n = \frac{1}{2} A m {(\frac{2}{A + 1} v)}^{2} = \frac{4 A}{{(A + 1)}^{2}} K E$

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The correct option is A 4A(A+1)2Conservationoflinearmomentummv=Amv1+mv2anddefintionofcoefficientofRestitutionv2−v1v−0=−1⇒v2−v1=−vsolvingtheseequationswegetv1=2A+1vKEoftheneutron=12Am(2A+1v)2=4A(A+1)2KE

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