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 - Page 276: 74

Answer

$-0.2\;\rm m/s,\;1.4\;m/s$

Work Step by Step

Let's assume that the two carts are moving without friction and that the spring is massless and obeying Hooke's law; which means that the system [the two carts+the spring] is an isolated system. So the momentum and energy of the system are conserved. $$p_{ix}=p_{fx}$$ where $p_{ix}$ is before the string being cut, end $p_{fx}$ is after cutting the string. $$(m_1+m_2)v_i=m_1v_1+m_2v_2$$ Plugging the known; $$(0.1+0.3)\times 1=0.1v_1+0.3v_2$$ $$0.4=0.1v_1+0.3v_2$$ Multiplying both sides by 10; $$4= v_1+3v_2$$ Thus, $$v_1=4-3v_2\tag 1$$ $$E_i=E_f$$ $$K_i+U_{is}=K_f+U_{fs}$$ The massless spring finally is set free with no compression and no stretching. $$ \frac{1}{2}(m_1+m_2)v_i^2+ \frac{1}{2}kx_i^2= \frac{1}{2} m_1v_1^2+ \frac{1}{2} m_2 v_2^2+\frac{1}{2}k(0)^2$$ $$\color{red}{\bf\not}\frac{1}{2}(m_1+m_2)v_i^2+\color{red}{\bf\not}\frac{1}{2}kx_i^2=\color{red}{\bf\not}\frac{1}{2} m_1v_1^2+\color{red}{\bf\not}\frac{1}{2} m_2 v_2^2 $$ Plugging the known; $$ (0.1+0.3)(1)^2+120(0.04)^2= 0.1v_1^2+ 0.3 v_2^2 $$ $$0.592= 0.1v_1^2+ 0.3 v_2^2 $$ Multiplying both sides by 10; $$5.92= v_1^2+ 3 v_2^2 $$ Plugging from (1); $$5.92= (4-3v_2)^2+ 3 v_2^2 $$ $$5.92= 16-24v_2+9v_2^2 + 3 v_2^2 $$ Thus, $$ 12 v_2^2 -24v_2+10.08=0$$ Therefore, $v_{2}=0.6\;\rm m/s$ or, $v_{2}=1.4\;\rm m/s$ Hence, plugging these into (1); $v_{1}=2.2\;\rm m/s$ or, $v_{2}=-0.2\;\rm m/s$ Now we have two results and we do not know which one is the right one. 1- If $v_1=2.2\;\rm m/s$, then $v_{2}=0.6\;\rm m/s$ 2- If $v_1=-0.2\;\rm m/s$, then $v_{2}=1.4\;\rm m/s$ Let's discuss this situation, we know that when the string is cut, the spring will push $m_2$ forward while pushing $m_1$ backward at the same time. This means that the velocity of cart 2 must be increased to be greater than the initial speed of the system of 1 m/s and hence the velocity of cart 1 must be decreased to be less than the initial speed. This means that the second result is the right answer since cart 2's velocity increases and cart 1 velocity decreases to a limit that makes it moves in the opposite direction. From all the above; $$v_1=\color{red}{\bf -0.2}\;\rm m/s$$ $$v_2=\color{red}{\bf 1.4}\;\rm m/s$$
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