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 9 - Impulse and Momentum - Exercises and Problems - Page 243: 59

Answer

$14\;\rm u$

Work Step by Step

We can consider the proton plus the golden target as an isolated system. So, the momentum is conserved. $$p_{ix}=p_{fx}$$ $$m_pv_{ix,p}+m_t\overbrace{v_{ix,t}}^{=0}=m_pv_{fx,p}+m_tv_{fx,t}$$ where $p$ refers to the proton and $t$ refers to the target. The initial speed of the target is zero since it was at rest. Now we need to solve for $m_{t}$; $$m_pv_{ix,p} -m_pv_{fx,p}=m_tv_{fx,t}$$ $$m_t =\dfrac{m_pv_{ix,p} -m_pv_{fx,p}}{v_{fx,t}} $$ We know that the recoil speed of the proton is reduced by 25%, so its final speed is now about 0.75 its initial speed but in the opposite direction. so $v_{fx,p}=-0.75v_{ix,p}$ $$ m_t =\dfrac{m_pv_{ix,p} -(-0.75m_pv_{ix,p})}{v_{fx,t}} $$ $$ m_t =\dfrac{1.75m_pv_{ix,p}}{v_{fx,t}} $$ Plugging the known; $$ m_t =\dfrac{1.75(1)( 2.5\times 10^6)}{(3.12\times 10^5)} =\color{red}{\bf 14.0}\;\rm u$$
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