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 244: 74

Answer

$8\;\rm bullets$

Work Step by Step

First, we need to unify the units of speeds of the rocket and the 5-kg bullets. It is easy to convert one speed rather than 2 of them. Thus, the speed of one bullet in km/h is $$v_b=\rm \left(\dfrac{139\;000\;\color{red}{\bf\not} m}{1\;\color{red}{\bf\not} s}\right) \left(\dfrac{1.0\;km}{1000\;\color{red}{\bf\not} m}\right)\left(\dfrac{60^2\;\color{red}{\bf\not} s}{1.0\;h}\right)$$ $$v_b=\bf 500\;400\;\rm km/h$$ This speed isnrelative to the rocket, not the planet. This means that its speed relative to the planet is $$v_{fb}=v_{f,rocket}+500\;400\tag 1$$ Now we need to find how many bullets we should fire to reduce the rocket's speed from 25,000 km/h to 15,000 km/h. The momentum is conserved here, and the bullets are shoted forward to reduce the speed of the rocket because if we shot them backward it will increase its speed. $$p_i=p_f$$ $$m_{rocket}v_i=(m_{rocket}-Nm_b)v_{f,rocket}+Nm_bv_{fb}$$ where $N$ is the number of fired bullets. Plug from (1); $$m_{rocket}v_i=(m_{rocket}-Nm_b)v_{f,rocket}+Nm_b(v_{f,rocket}+500\;400)$$ Solving for $N$; $$m_{rocket}v_i= m_{rocket}v_{f,rocket}-Nm_b v_{f,rocket}+Nm_b v_{f,rocket}+500\;400Nm_b $$ $$m_{rocket}v_i= m_{rocket}v_{f,rocket} +500\;400Nm_b $$ $$N=\dfrac{m_{rocket}v_i- m_{rocket}v_{f,rocket}}{500\;400 m_b} =\dfrac{m_{rocket}(v_i- v_{f,rocket})}{500\;400 m_b}$$ Plugging the known; $$N=\dfrac{ 2000 (25000 - 15000)}{500 400 \times 5}\approx \color{red}{\bf 8}\;\rm bullets$$ Therefore, 8 bullets are enough to reduce the rocket's speed as needed.
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