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(D) Let the mass of the neutron be $m$ and its initial velocity be $v$. The mass of the deuterium atom (deuteron) is $M = 2m$. Since the collision is

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$\frac{8}{9}$

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(D) Let the mass of the neutron be $m$ and its initial velocity be $v$. The mass of the deuterium atom (deuteron) is $M = 2m$. Since the collision is elastic,both linear momentum and kinetic energy are conserved.Let $v_1$ be the final velocity of the neutron and $v_2$ be the final velocity of the deuteron.By conservation of momentum: $mv = mv_1 + (2m)v_2$  => $v = v_1 + 2v_2$  ... $(i)$By conservation of kinetic energy: $\frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2$ => $v^2 = v_1^2 + 2v_2^2$ ... $(ii)$From $(i)$,$2v_2 = v - v_1$. Substituting into $(ii)$:$v^2 = v_1^2 + \frac{1}{2}(v - v_1)^2$$2v^2 = 2v_1^2 + v^2 - 2vv_1 + v_1^2$$3v_1^2 - 2vv_1 - v^2 = 0$$(3v_1 + v)(v_1 - v) = 0$Since $v_1 
eq v$,we have $v_1 = -\frac{v}{3}$.The initial kinetic energy is $K_i = \frac{1}{2}mv^2$. The final kinetic energy is $K_f = \frac{1}{2}mv_1^2 = \frac{1}{2}m(-\frac{v}{3})^2 = \frac{1}{9}(\frac{1}{2}mv^2) = \frac{1}{9}K_i$.The change in kinetic energy is $\Delta K = K_f - K_i = \frac{1}{9}K_i - K_i = -\frac{8}{9}K_i$. The magnitude of the change is $\frac{8}{9}$ of the initial kinetic energy.

$A$ neutron with velocity $v$ strikes a stationary deuterium atom. By what factor does its kinetic energy change?

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