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 - Search and Learn - Page 197: 5

Answer

(a) $v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A$ $v_B' = (\frac{2m_A}{m_a+m_B})~v_A$ (b) $v_A' \approx -v_A$ $v_B' \approx 0$ (c) $v_A' \approx v_A$ $v_B' \approx 2v_A$ (d) $v_A' = 0$ $v_B' = v_A$

Work Step by Step

(a) 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. $m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$ $v_B' = \frac{2m_A~v_A}{m_a+m_B} = (\frac{2m_A}{m_a+m_B})~v_A$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A = (\frac{2m_A~v_A}{m_a+m_B}) - v_A$ $v_A' = (\frac{2m_A~v_A}{m_a+m_B}) - (\frac{m_A~v_A + m_B~v_A}{m_a+m_B})$ $v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A$ (b) When $m_A$ is much smaller than $m_B$ $v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A \approx -v_A$ $v_B' = (\frac{2m_A}{m_a+m_B})~v_A \approx 0$ A common example of this would be a tennis ball bouncing back off a wall. (c) When $m_A$ is much larger than $m_B$ $v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A \approx v_A$ $v_B' = (\frac{2m_A}{m_a+m_B})~v_A \approx 2v_A$ A common example of this would be a person's foot kicking a tiny pebble on the sidewalk. (d) When $m_A$ is equal to $m_B$ $v_A' = (\frac{m_A - m_B}{m_a+m_B})~v_A = 0$ $v_B' = (\frac{2m_A}{m_a+m_B})~v_A = v_A$ A common example of this would be a moving billiard ball hitting head on another billiard ball which is at rest.
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