## Physics: Principles with Applications (7th Edition)

Let $m_A$ be the mass of the neutron. Let $v_A'$ be the final velocity of the neutron. 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 7-7 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. $m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$ $v_B' = (\frac{2m_A}{m_A+m_B})~v_A$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A = \frac{2m_A~v_A}{m_A+m_B} - \frac{m_A~v_A+m_B~v_A}{m_A+m_B}$ $v_A' = (\frac{m_A - m_B}{m_A+m_B})~v_A$ We can find the original KE of the neutron. $KE_0 = \frac{1}{2}(1.01~u)v_A^2$ $KE_0 = (0.505~u)~v_A^2$ (a) Let $m_A = 1.01 ~u$ and let $m_B = 1.01~u$ $v_A' = (\frac{m_A - m_B}{m_A+m_B})~v_A$ $v_A' = 0$ $KE_f = 0$ The neutron lost all of its original kinetic energy. (b) Let $m_A = 1.01 ~u$ and let $m_B = 2.01~u$ $v_A' = (\frac{1.01~u - 2.01~u}{1.01~u+2.01~u})~v_A = -0.33~v_A$ We can find the neutron's KE after the collision. $KE_f = \frac{1}{2}(1.01~u)(-0.33~v_A)^2$ $KE_f = (0.055~u)~v_A^2$ $\frac{KE_f}{KE_0} = \frac{0.055}{0.505} = 0.11$ The neutron lost 0.89 of its original kinetic energy. (c) Let $m_A = 1.01 ~u$ and let $m_B = 12.00~u$ $v_A' = (\frac{1.01~u - 12.00~u}{1.01~u+12.00~u})~v_A = -0.845~v_A$ We can find the neutron's KE after the collision. $KE_f = \frac{1}{2}(1.01~u)(-0.845~v_A)^2$ $KE_f = (0.36~u)~v_A^2 = KE_0$ $\frac{KE_f}{KE_0} = \frac{0.36}{0.505} = 0.71$ The neutron lost 0.29 of its original kinetic energy. (d) Let $m_A = 1.01 ~u$ and let $m_B = 208~u$ $v_A' = (\frac{1.01~u - 208~u}{1.01~u+208~u})~v_A = -0.99~v_A$ We can find the neutron's KE after the collision. $KE_f = \frac{1}{2}(1.01~u)(-0.99~v_A)^2$ $KE_f = (0.495~u)~v_A^2$ $\frac{KE_f}{KE_0} = \frac{0.495}{0.505} = 0.98$ The neutron lost 0.02 of its original kinetic energy.