A neutron moving with velocity u collides ela | vedclass

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(B) Let the mass of the neutron be $m$. Then the mass of the atom is $M = Am$.From the conservation of linear momentum:$mu + Am(0) = mv_1 + Amv_2$$u =

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

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(B) Let the mass of the neutron be $m$. Then the mass of the atom is $M = Am$.From the conservation of linear momentum:$mu + Am(0) = mv_1 + Amv_2$$u = v_1 + Av_2$  .....$(1)$For an elastic head-on collision,the coefficient of restitution $e = 1$:$v_2 - v_1 = u - 0 = u$  .....$(2)$From $(2)$,$v_2 = u + v_1$. Substituting this into $(1)$:$u = v_1 + A(u + v_1)$$u = v_1 + Au + Av_1$$u(1 - A) = v_1(1 + A)$$v_1 = u \left( \frac{1 - A}{1 + A} \right) = -u \left( \frac{A - 1}{A + 1} \right)$The final kinetic energy of the neutron is:$E' = \frac{1}{2} mv_1^2 = \frac{1}{2} m \left[ -u \left( \frac{A - 1}{A + 1} \right) \right]^2$$E' = \left( \frac{1}{2} mu^2 \right) \left( \frac{A - 1}{A + 1} \right)^2$Since $E = \frac{1}{2} mu^2$,we get:$E' = E \left( \frac{A - 1}{A + 1} \right)^2$

$A$ neutron moving with velocity $u$ collides elastically with an atom of mass number $A$. If the collision is head-on and the initial kinetic energy of the neutron is $E$,then the final kinetic energy of the neutron after the collision is

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