Chapter 9 - Impulse and Momentum - Exercises and Problems - Page 244: 75

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We will treat this problem as a momentum conserved momentum at which we need to find if the firing speed of the bullet from Mr.Smith's gun is enough to move the chair the 3 centimeters or not. If it is enough, so the client, Mr.Smith, is guilty, but if not, he is innocent. We need to use momentum conservation during the collision between the bullet and the chair. We need to find the velocity of the chair just after the bullet hits it. Noting that the bullet embedded inside the chair which means that they will move as one unit after the collision. $$p_{ix}=p_{fx}$$ $$m_{bullet}v_{ix,bullet}+m_{chair} \overbrace{v_{ix,chair}}^{0}=(m_{bullet}+m_{chair} )v_{fx}$$ $$m_{bullet}v_{ix,bullet} =(m_{bullet}+m_{chair} )v_{fx}$$ Solving for $v_{fx}$ which is the velocity of the chair after the collision. $$v_{fx}=\dfrac{m_{bullet}v_{ix,bullet} }{(m_{bullet}+m_{chair} )}$$ Plugging the known; $$v_{fx}=\dfrac{0.01\times 450}{(0.01+20 )}=\bf0.225\;\rm m/s$$ Now we need to find the distance traveled by the chair when its initial speed was 0.225 m/s. So we need to find its acceleration. We know that the only force exerted on it while sliding forward is the friction force that is directed backward. Applying Newton's second law on the chair. Thus, $$\sum F_x=-f_k=(m_{bullet}+m_{chair} )a_x$$ and we know that the friction force is given by $\mu_k F_n$. Thus, $$-\mu_k F_n=(m_{bullet}+m_{chair} )a_x\tag 1$$ $$\sum F_y=F_n-(m_{bullet}+m_{chair} )g=(m_{bullet}+m_{chair} )a_x=(m_{bullet}+m_{chair} )(0)=0$$ Thus, $$F_n=(m_{bullet}+m_{chair} )g$$ Plugging into (1); $$-\mu_k(m_{bullet}+m_{chair} )g=(m_{bullet}+m_{chair} )a_x $$ So, $$a_x= \mu_kg\tag 2$$ Therefore the distance traveled by the chair is given by $$v_{fx,chair}^2=v_{ix,chair}^2+2a_x\Delta x$$ we know that finally, the chair stops. So that $$0^2=v_{ix,chair}^2+2a_x\Delta x$$ Hence, $$\Delta x=\dfrac{-v_{ix,chair}^2}{2a_x}$$ Plugging from (2) and then plug the known; $$\Delta x=\dfrac{-v_{ix,chair}^2}{2 \times-\mu_kg}=\dfrac{0.225^2}{2\times 0.2\times 9.8}=\bf 0.013\;\rm m$$ $$\Delta x=\color{red}{\bf 1.29}\;\rm cm$$ Therefore, Mr.Smith is innocent.