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Question Number 71311 by ajfour last updated on 13/Oct/19

Commented by ajfour last updated on 13/Oct/19

Assuming elastic collision, find d_(max) and corresponding value for x.

$${Assuming}\:{elastic}\:{collision},\: \\ $$$${find}\:{d}_{{max}} \:\:{and}\:{corresponding} \\ $$$${value}\:{for}\:{x}. \\ $$

Answered by mr W last updated on 13/Oct/19

Commented by mr W last updated on 13/Oct/19

α=(π/2)−ϕ θ=ϕ−α=2ϕ−(π/2) ((mu^2 )/2)=mg(h−R sin ϕ) u=(√(2g(h−R sin ϕ))) −R sin ϕ=u sin θ t−(1/2)gt^2 0=−R sin ϕ+u cos 2ϕ t+(1/2)gt^2 t=((−u cos 2ϕ+(√(u^2 cos^2 2ϕ+2gR sin ϕ)))/g) t=((−u cos 2ϕ+(√(2g(h−R sin ϕ)cos^2 2ϕ+2gR sin ϕ)))/g) d+R−R cos ϕ=u cos θ t d+R−R cos ϕ=u sin 2ϕ t d=u sin 2ϕ t−(1−cos ϕ)R d=2R sin 2ϕ (√((h/R)−sin ϕ)){−cos 2ϕ(√((h/R)−sin ϕ))+(√(((h/R)−sin ϕ)cos^2 2ϕ+sin ϕ))}−(1−cos ϕ)R with η=(h/R) (d/R)=2 sin 2ϕ (√(η−sin ϕ)){−cos 2ϕ(√(η−sin ϕ))+(√((η−sin ϕ)cos^2 2ϕ+sin ϕ))}+cos ϕ−1 S=(d/R)=2 sin 2ϕ{−cos 2ϕ(η−sin ϕ)+(√([(η−sin ϕ)cos^2 2ϕ+sin ϕ](η−sin ϕ)))}+cos ϕ−1 (dS/dϕ)=0 ⇒ eqn. for ϕ example: η=(h/R)=4 ⇒(d_(max) /R)=6.4529 at ϕ=64.7775°

$$\alpha=\frac{\pi}{\mathrm{2}}−\varphi \\ $$$$\theta=\varphi−\alpha=\mathrm{2}\varphi−\frac{\pi}{\mathrm{2}} \\ $$$$\frac{{mu}^{\mathrm{2}} }{\mathrm{2}}={mg}\left({h}−{R}\:\mathrm{sin}\:\varphi\right) \\ $$$${u}=\sqrt{\mathrm{2}{g}\left({h}−{R}\:\mathrm{sin}\:\varphi\right)} \\ $$$$−{R}\:\mathrm{sin}\:\varphi={u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$$\mathrm{0}=−{R}\:\mathrm{sin}\:\varphi+{u}\:\mathrm{cos}\:\mathrm{2}\varphi\:{t}+\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${t}=\frac{−{u}\:\mathrm{cos}\:\mathrm{2}\varphi+\sqrt{{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\varphi+\mathrm{2}{gR}\:\mathrm{sin}\:\varphi}}{{g}} \\ $$$${t}=\frac{−{u}\:\mathrm{cos}\:\mathrm{2}\varphi+\sqrt{\mathrm{2}{g}\left({h}−{R}\:\mathrm{sin}\:\varphi\right)\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\varphi+\mathrm{2}{gR}\:\mathrm{sin}\:\varphi}}{{g}} \\ $$$${d}+{R}−{R}\:\mathrm{cos}\:\varphi={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$${d}+{R}−{R}\:\mathrm{cos}\:\varphi={u}\:\mathrm{sin}\:\mathrm{2}\varphi\:{t} \\ $$$${d}={u}\:\mathrm{sin}\:\mathrm{2}\varphi\:{t}−\left(\mathrm{1}−\mathrm{cos}\:\varphi\right){R} \\ $$$${d}=\mathrm{2}{R}\:\mathrm{sin}\:\mathrm{2}\varphi\:\sqrt{\frac{{h}}{{R}}−\mathrm{sin}\:\varphi}\left\{−\mathrm{cos}\:\mathrm{2}\varphi\sqrt{\frac{{h}}{{R}}−\mathrm{sin}\:\varphi}+\sqrt{\left(\frac{{h}}{{R}}−\mathrm{sin}\:\varphi\right)\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\varphi+\mathrm{sin}\:\varphi}\right\}−\left(\mathrm{1}−\mathrm{cos}\:\varphi\right){R} \\ $$$${with}\:\eta=\frac{{h}}{{R}} \\ $$$$\frac{{d}}{{R}}=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\varphi\:\sqrt{\eta−\mathrm{sin}\:\varphi}\left\{−\mathrm{cos}\:\mathrm{2}\varphi\sqrt{\eta−\mathrm{sin}\:\varphi}+\sqrt{\left(\eta−\mathrm{sin}\:\varphi\right)\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\varphi+\mathrm{sin}\:\varphi}\right\}+\mathrm{cos}\:\varphi−\mathrm{1} \\ $$$${S}=\frac{{d}}{{R}}=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\varphi\left\{−\mathrm{cos}\:\mathrm{2}\varphi\left(\eta−\mathrm{sin}\:\varphi\right)+\sqrt{\left[\left(\eta−\mathrm{sin}\:\varphi\right)\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\varphi+\mathrm{sin}\:\varphi\right]\left(\eta−\mathrm{sin}\:\varphi\right)}\right\}+\mathrm{cos}\:\varphi−\mathrm{1} \\ $$$$\frac{{dS}}{{d}\varphi}=\mathrm{0}\:\Rightarrow\:{eqn}.\:{for}\:\varphi \\ $$$${example}: \\ $$$$\eta=\frac{{h}}{{R}}=\mathrm{4} \\ $$$$\Rightarrow\frac{{d}_{{max}} }{{R}}=\mathrm{6}.\mathrm{4529}\:{at}\:\varphi=\mathrm{64}.\mathrm{7775}° \\ $$

Commented by mr W last updated on 13/Oct/19

Commented by ajfour last updated on 13/Oct/19

Awfully complicated it gets. Thanks a lot Sir.Nevertheless, you are amazing sir!

$${Awfully}\:{complicated}\:{it}\:{gets}. \\ $$$${Thanks}\:{a}\:{lot}\:{Sir}.{Nevertheless}, \\ $$$${you}\:{are}\:{amazing}\:{sir}! \\ $$

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