Question Number 213221 by mr W last updated on 01/Nov/24
Commented by mr W last updated on 01/Nov/24
$${Q}\mathrm{212936} \\ $$
Commented by ajfour last updated on 01/Nov/24
https://youtu.be/tB8F0OnMlAU?si=fmSM55gGzwSidaU-
Commented by aleks041103 last updated on 01/Nov/24
$${I}\:{saw}\:{the}\:{solution}\:{and}\:{it}\:{seems}\:{you}\: \\ $$$${neglect}\:{the}\:{action}\:{of}\:{gravity}\:{on}\:{the} \\ $$$${right}\:{halfplane}. \\ $$$${If}\:{you}\:{acount}\:{for}\:{g}\:{there}\:{too},\:{the}\:{problem} \\ $$$${becomes}\:{really}\:{difficult},\:{since}\:{you}\:{would} \\ $$$${have}\:{to}\:{solve}\:{a}\:{system}\:{of}\:{equations} \\ $$$${which}\:{are}\:{transcendental}. \\ $$
Answered by mr W last updated on 01/Nov/24
![v_1 =u cos α h=((u^2 sin^2 α)/(2g)) a=((2h)/(tan α))=((u^2 sin^2 α)/(g tan α)) v_(2x) =v_1 =u cos α r=((mv_1 )/(qB))=((mu cos α)/(gB)) h−b=((2mu cos α)/(qB))+(g/2)(((πm)/(qB)))^2 v_(2y) =(√(2g(h−b)))=(√(2g[((2mu cos α)/(qB))+(g/2)(((πm)/(qB)))^2 ])) b=v_(2y) ×(a/(u cos α))+((ga^2 )/(2u^2 cos^2 α)) ((u^2 sin^2 α)/(2g))−((2mu cos α)/(qB))−(g/2)(((πm)/(qB)))^2 =(√(2g[((2mu cos α)/(qB))+(g/2)(((πm)/(qB)))^2 ]))×(((u^2 sin^2 α)/(2g))/(u cos α))+((g(((u^2 sin^2 α)/(2g)))^2 )/(2u^2 cos^2 α)) ((sin^2 α)/2)−((2mg cos α)/(uqB))−(1/2)(((πmg)/(uqB)))^2 =(√(((4mg cos α)/(uqB))+(((πmg)/(uqB)))^2 ))×((sin^2 α)/(2 cos α))+((sin^4 α)/(8 cos^2 α)) let λ=((mg)/(uqB)) (4λ cos α+π^2 λ^2 )+sin α tan α (√(4λ cos α+π^2 λ^2 ))+(((tan^2 α)/4)−1)sin^2 α=0 (√(4λ cos α+π^2 λ^2 ))=((−sin α tan α+cos α)/2)=((cos 2α)/(2 cos α)) π^2 λ^2 +4λ cos α−((cos^2 2α)/(4 cos^2 α))=0 ⇒λ=((mg)/(uqB))=((−4cos^2 α+(√(16 cos^4 α+π^2 cos^2 2α)))/(2π^2 cos α))](https://www.tinkutara.com/question/Q213224.png)
$${v}_{\mathrm{1}} ={u}\:\mathrm{cos}\:\alpha \\ $$$${h}=\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{2}{g}} \\ $$$${a}=\frac{\mathrm{2}{h}}{\mathrm{tan}\:\alpha}=\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha}{{g}\:\mathrm{tan}\:\alpha} \\ $$$${v}_{\mathrm{2}{x}} ={v}_{\mathrm{1}} ={u}\:\mathrm{cos}\:\alpha \\ $$$${r}=\frac{{mv}_{\mathrm{1}} }{{qB}}=\frac{{mu}\:\mathrm{cos}\:\alpha}{{gB}} \\ $$$${h}−{b}=\frac{\mathrm{2}{mu}\:\mathrm{cos}\:\alpha}{{qB}}+\frac{{g}}{\mathrm{2}}\left(\frac{\pi{m}}{{qB}}\right)^{\mathrm{2}} \\ $$$${v}_{\mathrm{2}{y}} =\sqrt{\mathrm{2}{g}\left({h}−{b}\right)}=\sqrt{\mathrm{2}{g}\left[\frac{\mathrm{2}{mu}\:\mathrm{cos}\:\alpha}{{qB}}+\frac{{g}}{\mathrm{2}}\left(\frac{\pi{m}}{{qB}}\right)^{\mathrm{2}} \right]} \\ $$$${b}={v}_{\mathrm{2}{y}} ×\frac{{a}}{{u}\:\mathrm{cos}\:\alpha}+\frac{{ga}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha} \\ $$$$\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{2}{g}}−\frac{\mathrm{2}{mu}\:\mathrm{cos}\:\alpha}{{qB}}−\frac{{g}}{\mathrm{2}}\left(\frac{\pi{m}}{{qB}}\right)^{\mathrm{2}} =\sqrt{\mathrm{2}{g}\left[\frac{\mathrm{2}{mu}\:\mathrm{cos}\:\alpha}{{qB}}+\frac{{g}}{\mathrm{2}}\left(\frac{\pi{m}}{{qB}}\right)^{\mathrm{2}} \right]}×\frac{\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{2}{g}}}{{u}\:\mathrm{cos}\:\alpha}+\frac{{g}\left(\frac{{u}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{2}{g}}\right)^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha} \\ $$$$\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{2}}−\frac{\mathrm{2}{mg}\:\mathrm{cos}\:\alpha}{{uqB}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi{mg}}{{uqB}}\right)^{\mathrm{2}} =\sqrt{\frac{\mathrm{4}{mg}\:\mathrm{cos}\:\alpha}{{uqB}}+\left(\frac{\pi{mg}}{{uqB}}\right)^{\mathrm{2}} }×\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{2}\:\mathrm{cos}\:\alpha}+\frac{\mathrm{sin}^{\mathrm{4}} \:\alpha}{\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\alpha} \\ $$$${let}\:\lambda=\frac{{mg}}{{uqB}} \\ $$$$\left(\mathrm{4}\lambda\:\mathrm{cos}\:\alpha+\pi^{\mathrm{2}} \lambda^{\mathrm{2}} \right)+\mathrm{sin}\:\alpha\:\mathrm{tan}\:\alpha\:\sqrt{\mathrm{4}\lambda\:\mathrm{cos}\:\alpha+\pi^{\mathrm{2}} \lambda^{\mathrm{2}} }+\left(\frac{\mathrm{tan}^{\mathrm{2}} \:\alpha}{\mathrm{4}}−\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \:\alpha=\mathrm{0} \\ $$$$\sqrt{\mathrm{4}\lambda\:\mathrm{cos}\:\alpha+\pi^{\mathrm{2}} \lambda^{\mathrm{2}} }=\frac{−\mathrm{sin}\:\alpha\:\mathrm{tan}\:\alpha+\mathrm{cos}\:\alpha}{\mathrm{2}}=\frac{\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{2}\:\mathrm{cos}\:\alpha} \\ $$$$\pi^{\mathrm{2}} \lambda^{\mathrm{2}} +\mathrm{4}\lambda\:\mathrm{cos}\:\alpha−\frac{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\alpha}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{{mg}}{{uqB}}=\frac{−\mathrm{4cos}\:^{\mathrm{2}} \alpha+\sqrt{\mathrm{16}\:\mathrm{cos}^{\mathrm{4}} \:\alpha+\pi^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha}}{\mathrm{2}\pi^{\mathrm{2}} \:\mathrm{cos}\:\alpha} \\ $$
Commented by ajfour last updated on 01/Nov/24
$${yes},\:{v}\:{nice},\:{overall}\:{i}\:{followed}\:{and}\:{agree} \\ $$$${sir}.\:{Lets}\:{say}\:\mathrm{cos}\:\alpha={s} \\ $$$${B}=\frac{\mathrm{2}{mg}}{{qu}}\left\{\frac{\mathrm{4}{s}^{\mathrm{2}} +\sqrt{\mathrm{16}{s}^{\mathrm{4}} +\pi^{\mathrm{2}} \left(\mathrm{2}{s}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}{\frac{\mathrm{1}}{{s}}\left(\mathrm{2}{s}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$
Commented by mr W last updated on 01/Nov/24
$${thanks}\:{for}\:{checking}\:{sir}! \\ $$