A neutron moving with velocity u collides ela | vedclass

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Question

(B) Let the mass of the neutron be $m$.Then the mass of the stationary atom is $M = Am$.By the law of conservation of linear momentum: $mu + Am(0) = m

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$A$ neutron moving with velocity $u$ collides elastically with a stationary 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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