10판/4. 2차원운동과 3차원운동

4-2 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 13. 21:39

{px=5.0[m]py=8.0[m]pz=0.0[m]\begin{cases} p_x = -5.0\ut{m}\\ p_y = 8.0\ut{m}\\ p_z = 0.0\ut{m}\\ \end{cases}
(a)p=?\vec p=? p=(5.0[m])i^+8.0[m]j^ \vec p = (-5.0\ut{m})\i+8.0\ut{m}\j
(b)p=?p=? p=(5)2+82=899.4[m] \begin{aligned} p =& \sqrt{(-5)^2+8^2}\\ =& \sqrt{89}\\ \approx&9.4\ut{m} \end{aligned}
(c)ϕpi^=?\phi_{p|\i}=? ϕpi^=cos1(pi^pi^)=cos1(589)2.1[rad] \begin{aligned} \phi_{\vec p|\i}=&\cos^{-1}\(\frac{\vec p \cdot \i}{\abs{p}\abs{\i}}\)\\ =&\cos^{-1}\(\frac{-5}{\sqrt{89}}\)\\ \approx&2.1\ut{rad} \end{aligned}
(d)


(e)Δp=?\Delta\vec p=? p=(3.00[m])i^ \vec p = (3.00\ut{m})\i
(f)Δp=?\Delta p=? p=3[m]p=3\ut{m}
(g)ϕΔpi^=?\phi_{\Delta\vec p|\i}=? ϕΔpi^=0\phi_{\Delta\vec p|\i}=0