University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 8 - Momentum, Impulse, and Collision - Problems - Exercises - Page 266: 8.48

Answer

(a) The magnitude of the 10.0-g marble is 0.500 m/s, and it is moving to the right. The magnitude of the 30.0-g marble is 0.100 m/s and, it is moving to the left. (b) The change in momentum is equal in magnitude but in the opposite directions for the two marbles. The change in momentum for the 10.0-g marble is $0.00900~kg~m/s$, while the change in momentum for the 30.0-g marble is $-0.00900~kg~m/s$, (c) The change in kinetic energy is equal in magnitude for the two marbles. However the 10.0-g marble gained 0.000450 J of kinetic energy, while the 30.0-g marble lost 0.000450 J of kinetic energy.

Work Step by Step

Let $v_A'$ be the final velocity of the 10.0-g marble. Let $v_B'$ be the final velocity of the 30.0-g marble. Let $m_A = 0.0100~kg$ and let $m_B = 0.0300~kg$ We can use conservation of momentum to set up an equation. $m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$ Since the collision is elastic, we can use equation (8.27) to set up another equation. $v_A - v_B = v_B' - v_A'$ $v_A' = v_B' - v_A + v_B$ We can use this expression for $v_A'$ in the first equation. $m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$ $v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$ $v_B’ = \frac{(2)(0.0100~kg)(-0.400~m/s)+(0.0300~kg)(0.200~m/s)- (0.0100~kg)(0.200~m/s)}{(0.0100~kg)+(0.0300~kg)}$ $v_B' = -0.100~m/s$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A + v_B$ $v_A' = -0.100~m/s - (-0.400)~m/s + 0.200~m/s$ $v_A' = 0.500~m/s$ The magnitude of the 10.0-g marble is 0.500 m/s and it is moving to the right. The magnitude of the 30.0-g marble is 0.100 m/s and it is moving to the left. (b) $\Delta p_A = (0.0100~kg)(0.500~m/s)-(0.0100~kg)(-0.400~m/s)$ $\Delta p_A = 0.00900~kg~m/s$ $\Delta p_B = (0.0300~kg)(-0.100~m/s)-(0.0300~kg)(0.200~m/s)$ $\Delta p_B = -0.00900~kg~m/s$ The change in momentum is equal in magnitude but in the opposite directions for the two marbles. The change in momentum for the 10.0-g marble is $0.00900~kg~m/s$ while the change in momentum for the 30.0-g marble is $-0.00900~kg~m/s$ (c) $\Delta K_A = K_2-K_1$ $\Delta K_A = \frac{1}{2}(0.0100~kg)(0.500~m/s)^2-\frac{1}{2}(0.0100~kg)(0.400~m/s)^2$ $\Delta K_A = 0.000450~J$ $\Delta K_B = K_2-K_1$ $\Delta K_B = \frac{1}{2}(0.0300~kg)(0.100~m/s)^2-\frac{1}{2}(0.0300~kg)(0.200~m/s)^2$ $\Delta K_B = -0.000450~J$ The change in kinetic energy is equal in magnitude for the two marbles. However the 10.0-g marble gained 0.000450 J of kinetic energy while the 30.0-g marble lost 0.000450 J of 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.