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 242: 50

Answer

$27.8\;\rm m/s$

Work Step by Step

First of all, we need to find the speed of Brian just before he carries Ashely and then apply the principle of momentum to find their speed as one unit after carrying her. So we need to find his acceleration before and after he hits his friend. As we see below, and according to Newtin's second law, his acceleartion is given by $$\sum F_x=m_Bg\sin\theta-f_{k1}=m_Ba_{x1}$$ Hence, $$a_{x1}=\dfrac{m_Bg\sin\theta-f_{k1}}{m_B}$$ where $f_{k1}=\mu_k F_{n1}$; and here $\mu_k$ is for wood on ice. $$a_{x1}=\dfrac{m_Bg\sin\theta-\mu_k F_{n1}}{m_B}\tag 1$$ and the normal force is given by $$\sum F_y=F_{n1}-m_Bg\cos\theta=m_Ba_y=m_B(0)=0$$ thus, $$F_{n1}= m_Bg\cos\theta $$ Plugging into (1); $$a_{x1}=\dfrac{ \color{red}{\bf\not} m_Bg\sin\theta-\mu_k \color{red}{\bf\not} m_Bg\cos\theta}{ \color{red}{\bf\not} m_B} $$ $$a_{x1}= g\left[\sin\theta- \mu_k \cos\theta \right]$$ Plugging the known; $$a_{x1}= 9.8\left[\sin20^\circ- 0.06 \cos20^\circ \right]=\bf 2.8\;\rm m/s^2\tag 2$$ Now we can use the kinematic formula of velocity squared to find his velocity just before he meets her. $$v_{x1}^2=v_{x0}^2+2a_{x1}d_1$$ Brian starts from rest, so $$v_{x1}^2=0^2+2a_{x1}d_1$$ where $d_1=\frac{1}{2}L$ and $L$ is the length of the ramp. Thus, $$v_{x1} =\sqrt{ a_{x1}L}=\sqrt{2.8\times 200}=\bf 23.66\;\rm m/s$$ Applying the conservation of momentum principle, $$p_{ix}=p_{fx}$$ $$m_Bv_{x1,B}+m_A\overbrace{v_{x1,A}}^{=0}=(m_A+m_B)v_{x2}$$ Solving for $v_{x2}$ which is the velocity of the new system just after brian carried Ashley. And we assume that the distance traveled during this process is negligible. Thus, $$ v_{x2}=\dfrac{m_Bv_{x1,B}}{m_A+m_B}=\dfrac{80\times 23.66}{80+50}=\bf 14.56\;\rm m/s$$ Now we have the two of them are sliding down the incline from an itial speed of 14.56 m/s a distance of 100 m; and we need to find their final speed at the end of the incline. So that we need to find their acceleration and to do so we need to draw their force diagram as we did for Brian alone. See the figures below. $$\sum F_x=(m_B+m_A)g\sin\theta-f_{k2}=(m_B+m_A)a_{x2}$$ Thus, $$(m_B+m_A)g\sin\theta-\mu_k F_{n2}=(m_B+m_A)a_{x2}$$ and the normal force is given by $$ F_{n2}-(m_B+m_A)g\cos\theta =(m_B+m_A)a_y=0$$ Thus, $ F_{n2}=(m_B+m_A)g\cos\theta$ Plugging into the formula above this to find $a_x$ $$(m_B+m_A)g\sin\theta-\mu_k (m_B+m_A)g\cos\theta=(m_B+m_A)a_{x2}$$ $$a_{x2}= g\sin\theta-\mu_k g\cos\theta=g\left[ \sin\theta-\mu_k \cos\theta \right]$$ $$a_{x2}=9.8\left[ \sin20^\circ-0.06 \cos20^\circ \right]=a_{x1}=\bf 2.8\;\rm m/s^2$$ Therfore, the finalvelocity is given by $$v_{fx}^2=v_{x2}^2+2a_{x2}d_2$$ $$v_{fx} =\sqrt{v_{x2}^2+2a_{x2}d_2}$$ $$v_{fx} =\sqrt{ 14.56^2+(2\times 2.8\times 100 )}=\color{red}{\bf 27.8}\;\rm m/s$$
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