Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 10 - Energy - Exercises and Problems: 29

Answer

The final speed of the 50-g marble is 0.9 m/s The final speed of the 20-g marble is 2.9 m/s

Work Step by Step

Let $m_A = 0.050~kg$ and let $m_B = 0.020~kg$. Let $v_A$ and $v_B$ be the initial velocity of each marble. Let $v_A'$ and $v_B'$ be the final velocity of each marble. 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 set up another equation. $v_A - v_B = v_B' - v_A'$ $v_A - 0 = v_B' - v_A'$ $v_A' = v_B' - v_A$ We can use this expression for $v_A'$ in the first equation. $m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$ $v_B' = \frac{2m_A~v_A}{m_A+m_B}$ $v_B' = \frac{(2)(0.050~kg)(2.0~m/s)}{(0.050~kg)+(0.020~kg)}$ $v_B' = 2.9~m/s$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A$ $v_A' = 2.9~m/s - 2.0~m/s$ $v_A' = 0.9~m/s$ The final speed of the 50-g marble is 0.9 m/s The final speed of the 20-g marble is 2.9 m/s
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