Chapter 7 - Linear Momentum - Problems - Page 193: 32

(a) The neutron lost all of its original kinetic energy. (b) The neutron lost 0.89 of its original kinetic energy. (c) The neutron lost 0.29 of its original kinetic energy. (d) The neutron lost 0.02 of its original kinetic energy.

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.