Answer
(a) After the collision, the speed of the neutron is $\frac{v_A}{3}$, where $v_A$ was the original speed.
(b) $K_2 = \frac{K_1}{9}$
(c) Ten collisions should reduce the speed of a neutron to 1/59000 of its original value.
Work Step by Step
Let $v_A'$ be the final velocity of the neutron.
Let $v_B'$ be the final velocity of the deuteron.
Let $m_A = 1u$ and let $m_B = 2u$
We can use conservation of momentum to set up an equation.
$m_A~v_A + 0 = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can use equation (8.27) to set up another equation.
$v_A - 0 = v_B' - v_A'$
$v_A' = v_B' - v_A$
We can use this expression for $v_A'$ in the first equation.
$m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$
$v_B' = \frac{2m_A~v_A}{m_A+m_B}$
$v_B' = \frac{(2)(1u)(v_A)}{(1u)+(2u)}$
$v_B' = \frac{2v_A}{3}$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A$
$v_A' = \frac{2v_A}{3} - v_A = \frac{-v_A}{3}$
After the collision, the speed of the neutron is $\frac{v_A}{3}$, where $v_A$ was the original speed.
(b) $K_1 =\frac{1}{2}uv_A^2$
$K_2 = \frac{1}{2}u(\frac{v_A}{3})^2$
$K_2 = \frac{1}{9}\times \frac{1}{2}uv_A^2$
$K_2 = \frac{K_1}{9}$
(c) $(\frac{1}{3})^n = \frac{1}{59000}$
$3^n = 59000$
$n~ln(3)= ln(59000)$
$n = \frac{ln(59000)}{ln(3)}$
$n = 10.0$
Ten collisions should reduce the speed of a neutron to 1/59000 of its original value.
