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: 30

Answer

The final speed of the proton is $1.7\times 10^7~m/s$ to the left. The final speed of the carbon atom is $3.1\times 10^6~m/s$ to the right.

Work Step by Step

Let $m_A$ be the mass of the proton and let $m_B$ be the mass of the carbon atom. Let $v_A$ and $v_B$ be the initial velocity of each particle. Let $v_A'$ and $v_B'$ be the final velocity of each particle. 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. $m_A~v_A + 0 = m_A~v_B' - m_Av_A + m_B~v_B'$ $v_B' = \frac{2m_A~v_A}{m_A+m_B}$ $v_B' = \frac{2m_A~v_A}{m_A+12m_A}$ $v_B' = \frac{2~v_A}{13}$ $v_B' = \frac{(2)(2.0\times 10^7~m/s)}{13}$ $v_B' = 3.1\times 10^6~m/s$ We can use $v_B'$ to find $v_A'$. $v_A' = v_B' - v_A$ $v_A' = 3.1\times 10^6~m/s - 2.0\times 10^7~m/s$ $v_A' = -1.7\times 10^7~m/s$ The final speed of the proton is $1.7\times 10^7~m/s$ to the left. The final speed of the carbon atom is $3.1\times 10^6~m/s$ to the right.
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