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 - Page 193: 27

Answer

The final speed of the tennis ball is 2.50 m/s in the original direction. The final speed of the other ball is 5.00 m/s in the original direction.

Work Step by Step

Let $m_A = 0.060~kg$ and let $m_B = 0.090~kg$. 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 7-7 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: $m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$ $v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_a+m_B}$ $v_B' = \frac{(2)(0.060~kg)(5.50~m/s)+(0.090~kg)(3.00~m/s)- (0.060~kg)(3.00~m/s)}{(0.060~kg)+(0.090~kg)}$ $v_B' = 5.00~m/s$ We can use $v_B'$ to find $v_A'$: $v_A' = v_B' - v_A + v_B$ $v_A' = 5.00~m/s - 5.50~m/s + 3.00~m/s = 2.50~m/s$
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