Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - Problems: 32

Answer

(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.

Work Step by Step

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.