Question Number 132102 by mr W last updated on 14/Feb/21
Commented bymr W last updated on 16/Feb/21
$${a}\:{ball}\:{is}\:{thrown}\:{from}\:{point}\:{A}\:{with} \\ $$
$${speed}\:\boldsymbol{{u}}\:{and}\:{strikes}\:{at}\:{a}\:{point}\:{B} \\ $$
$${on}\:{the}\:{semispherical}\:{surface}\:{with} \\ $$
$${radius}\:{R}\:{and}\:{returns}\:{back}\:{to}\:{point}\:{A}. \\ $$
$${if}\:{the}\:{restitution}\:{coefficient}\:{of} \\ $$
$${the}\:{collision}\:{is}\:\boldsymbol{{e}},\:{find}\:{the}\:{minimum} \\ $$
$${speed}\:\boldsymbol{{u}}\:{the}\:{ball}\:{must}\:{have}\:{and}\:{at} \\ $$
$${which}\:{angle}\:\boldsymbol{\theta}\:{the}\:{ball}\:{should}\:{be} \\ $$
$${thrown}. \\ $$
$${speed}\:\boldsymbol{{u}}\:{and}\:{strikes}\:{at}\:{a}\:{point}\:{B} \\ $$
$${on}\:{the}\:{semispherical}\:{surface}\:{with} \\ $$
$${radius}\:{R}\:{and}\:{returns}\:{back}\:{to}\:{point}\:{A}. \\ $$
$${if}\:{the}\:{restitution}\:{coefficient}\:{of} \\ $$
$${the}\:{collision}\:{is}\:\boldsymbol{{e}},\:{find}\:{the}\:{minimum} \\ $$
$${speed}\:\boldsymbol{{u}}\:{the}\:{ball}\:{must}\:{have}\:{and}\:{at} \\ $$
$${which}\:{angle}\:\boldsymbol{\theta}\:{the}\:{ball}\:{should}\:{be} \\ $$
$${thrown}. \\ $$
Commented byotchereabdullai@gmail.com last updated on 14/Feb/21
$$\mathrm{You}\:\mathrm{are}\:\mathrm{the}\:\mathrm{world}\:\mathrm{best}\:\mathrm{no}\:\mathrm{one}\:\mathrm{can}\: \\ $$
$$\mathrm{challenge}\:\mathrm{you}! \\ $$
$$\mathrm{challenge}\:\mathrm{you}! \\ $$
Commented byajfour last updated on 15/Feb/21
$${For}\:{u}_{{min}} \:\:\:{I}\:\:{got}\:\:\:\phi=\mathrm{30}° \\ $$
$${t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}{R}}{{g}}}\left\{\mathrm{1}\pm\sqrt{\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{1}}\right\} \\ $$
$${u}_{{min}} ^{\mathrm{2}} =\left(\frac{{R}\sqrt{\mathrm{3}}}{\mathrm{2}{t}}+\frac{{gt}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}{gR}\left(\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$
$${t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}{R}}{{g}}}\left\{\mathrm{1}\pm\sqrt{\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{1}}\right\} \\ $$
$${u}_{{min}} ^{\mathrm{2}} =\left(\frac{{R}\sqrt{\mathrm{3}}}{\mathrm{2}{t}}+\frac{{gt}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}{gR}\left(\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$
Commented bymr W last updated on 15/Feb/21
$${do}\:{you}\:{mean}\:{that}\:{u}_{{min}} \:{is}\:{always}\:{at} \\ $$
$$\phi=\mathrm{30}°\:{indepentently}\:{from}\:{e}? \\ $$
$${i}\:{can}'{t}\:{confirm}\:{this}\:{sir}. \\ $$
$$\phi=\mathrm{30}°\:{indepentently}\:{from}\:{e}? \\ $$
$${i}\:{can}'{t}\:{confirm}\:{this}\:{sir}. \\ $$
Answered by mr W last updated on 25/Feb/21
Commented bymr W last updated on 12/Feb/21
Commented bymr W last updated on 16/Feb/21
$$\underline{\boldsymbol{{motion}}\:\boldsymbol{{from}}\:\boldsymbol{{A}}\:\boldsymbol{{to}}\:\boldsymbol{{B}}:} \\ $$
$${t}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta} \\ $$
$${R}\:\mathrm{sin}\:\phi=−{u}\:\mathrm{sin}\:\theta×\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}+\frac{{g}}{\mathrm{2}}×\frac{{R}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$
$$\frac{\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} }×\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)−\mathrm{tan}\:\theta \\ $$
$${let}\:\lambda=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$
$$\frac{\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)−\mathrm{tan}\:\theta \\ $$
$$\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{tan}^{\mathrm{2}} \:\theta−\mathrm{tan}\:\theta+\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} −\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\mathrm{0} \\ $$
$$\mathrm{tan}\:\theta=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)} \\ $$
$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$
$$ \\ $$
$$\underline{\boldsymbol{{at}}\:\boldsymbol{{point}}\:\boldsymbol{{B}}:} \\ $$
$${U}_{{x}} ={u}\:\mathrm{cos}\:\theta \\ $$
$${U}_{{y}} ={gt}−{u}\:\mathrm{sin}\:\theta=\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta \\ $$
$${U}_{\bot} ={U}_{{x}} \mathrm{cos}\:\phi+{U}_{{y}} \mathrm{sin}\:\phi \\ $$
$${U}_{\bot} ={u}\:\mathrm{cos}\:\theta\mathrm{cos}\:\phi+\left[\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta\right]\mathrm{sin}\:\phi \\ $$
$$\Rightarrow{U}_{\bot} ={u}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$${U}_{\parallel} =−{U}_{{x}} \mathrm{sin}\:\phi+{U}_{{y}} \mathrm{cos}\:\phi \\ $$
$${U}_{\parallel} =−{u}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi+\left[\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta\right]\mathrm{cos}\:\phi \\ $$
$$\Rightarrow{U}_{\parallel} =−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$$ \\ $$
$${V}_{\bot} ={eU}_{\bot} \\ $$
$$\Rightarrow{V}_{\bot} ={eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$${V}_{\parallel} ={U}_{\parallel} \\ $$
$$\Rightarrow{V}_{\parallel} =−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$$ \\ $$
$${V}_{{x}} ={V}_{\bot} \mathrm{cos}\:\phi+{V}_{\parallel} \mathrm{sin}\:\phi \\ $$
$${V}_{{x}} =\left[{eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{cos}\:\phi+\left[−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{sin}\:\phi \\ $$
$${V}_{{x}} ={eu}\:\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi\:\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$$\Rightarrow{V}_{{x}} ={u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+{e}\right)\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}\right] \\ $$
$${V}_{{y}} ={V}_{\bot} \mathrm{sin}\:\phi−{V}_{\parallel} \mathrm{cos}\:\phi \\ $$
$${V}_{{y}} =\left[{eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{sin}\:\phi−\left[−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{cos}\:\phi \\ $$
$$\Rightarrow{V}_{{y}} ={u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\left(\mathrm{1}+{e}\right)\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}\right] \\ $$
$$ \\ $$
$$\underline{\boldsymbol{{return}}\:\boldsymbol{{from}}\:\boldsymbol{{B}}\:\boldsymbol{{to}}\:\boldsymbol{{A}}:} \\ $$
$${t}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+{e}\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}}×\frac{\mathrm{1}}{{u}} \\ $$
$$\Rightarrow\frac{{tu}}{{R}}=\xi=\frac{\mathrm{1}+\mathrm{cos}\:\phi}{\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+{e}\right)\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}} \\ $$
$${R}\:\mathrm{sin}\:\phi={t}\left\{{u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta\right]+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{\mathrm{cos}\:\theta}−\frac{{gt}}{\mathrm{2}}\right\} \\ $$
$$\mathrm{sin}\:\phi=\frac{{tu}}{{R}}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{\mathrm{cos}\:\theta}−\frac{\lambda{tu}}{{R}}\right] \\ $$
$$\mathrm{sin}\:\phi=\xi\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\left(\mathrm{1}+{e}\right)\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}−\lambda\xi\right] \\ $$
$${or} \\ $$
$$\mathrm{sin}\:\phi=\xi\left[{e}'\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}'\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}−\lambda\xi\right]\:\:\:…\left({I}\right) \\ $$
$${with} \\ $$
$${e}'=\mathrm{1}+{e} \\ $$
$$\lambda=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$
$$\left.\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right]\:\:\:\:\:\:\:\ast\right) \\ $$
$$\xi=\frac{\mathrm{1}+\mathrm{cos}\:\phi}{{e}'\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda{e}'\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}} \\ $$
$${for}\:{a}\:{given}\:{e}\:\in\left(\mathrm{0},\mathrm{1}\right]\:{we}\:{can}\:{find} \\ $$
$${solution}\left({s}\right)\:{for}\:\phi\:{in}\:\left({I}\right)\:{for}\:{some} \\ $$
$${values}\:{of}\:{parameter}\:\lambda.\:{the}\:{maximum} \\ $$
$${value}\:{of}\:\lambda\:{such}\:{that}\:{a}\:{solution}\:{for}\:\phi \\ $$
$${exists}\:{can}\:{be}\:{determined}\:{numerically}. \\ $$
$${this}\:{corresponds}\:{to}\:{the}\:{minimum} \\ $$
$${speed}\:{the}\:{ball}\:{must}\:{have}: \\ $$
$${u}_{{min}} =\sqrt{\frac{{gR}}{\mathrm{2}\lambda_{{max}} }} \\ $$
$$ \\ $$
$${examples}: \\ $$
$${e}=\mathrm{0}.\mathrm{5}\:\Rightarrow{u}_{{min}} \approx\mathrm{2}.\mathrm{7682}\sqrt{{gR}} \\ $$
$${e}=\mathrm{0}.\mathrm{75}\:\Rightarrow{u}_{{min}} \approx\mathrm{1}.\mathrm{6082}\sqrt{{gR}} \\ $$
$${e}=\mathrm{1}\:\Rightarrow{u}_{{min}} \approx\mathrm{0}.\mathrm{9098}\sqrt{{gR}}\:\:\left(=\:{Q}\mathrm{65589}\right) \\ $$
$$ \\ $$
$$\left.\ast\right) \\ $$
$${with}\:{the}\:{other}\:{solution} \\ $$
$$\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$
$${we}\:{can}\:{get}\:{a}\:{solution}\:{for}\:{the}\:{case} \\ $$
$${that}\:{the}\:{ball}\:{strikes}\:{the}\:{surface} \\ $$
$${upwards},\:{i}.{e}.\:{counterclockwise}. \\ $$
$${but}\:{in}\:{this}\:{case}\:{more}\:{energy}\:{is} \\ $$
$${needed},\:{i}.{e}.\:{u}_{{min}} \:{is}\:{larger}\:{than}\:{with} \\ $$
$$\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$
$${see}\:{examples}\:{with}\:{e}=\mathrm{0}.\mathrm{5}\:{and}\:\mathrm{0}.\mathrm{75}. \\ $$
$${t}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta} \\ $$
$${R}\:\mathrm{sin}\:\phi=−{u}\:\mathrm{sin}\:\theta×\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}+\frac{{g}}{\mathrm{2}}×\frac{{R}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$
$$\frac{\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} }×\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)−\mathrm{tan}\:\theta \\ $$
$${let}\:\lambda=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$
$$\frac{\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)−\mathrm{tan}\:\theta \\ $$
$$\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{tan}^{\mathrm{2}} \:\theta−\mathrm{tan}\:\theta+\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} −\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\mathrm{0} \\ $$
$$\mathrm{tan}\:\theta=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)} \\ $$
$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$
$$ \\ $$
$$\underline{\boldsymbol{{at}}\:\boldsymbol{{point}}\:\boldsymbol{{B}}:} \\ $$
$${U}_{{x}} ={u}\:\mathrm{cos}\:\theta \\ $$
$${U}_{{y}} ={gt}−{u}\:\mathrm{sin}\:\theta=\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta \\ $$
$${U}_{\bot} ={U}_{{x}} \mathrm{cos}\:\phi+{U}_{{y}} \mathrm{sin}\:\phi \\ $$
$${U}_{\bot} ={u}\:\mathrm{cos}\:\theta\mathrm{cos}\:\phi+\left[\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta\right]\mathrm{sin}\:\phi \\ $$
$$\Rightarrow{U}_{\bot} ={u}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$${U}_{\parallel} =−{U}_{{x}} \mathrm{sin}\:\phi+{U}_{{y}} \mathrm{cos}\:\phi \\ $$
$${U}_{\parallel} =−{u}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi+\left[\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta\right]\mathrm{cos}\:\phi \\ $$
$$\Rightarrow{U}_{\parallel} =−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$$ \\ $$
$${V}_{\bot} ={eU}_{\bot} \\ $$
$$\Rightarrow{V}_{\bot} ={eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$${V}_{\parallel} ={U}_{\parallel} \\ $$
$$\Rightarrow{V}_{\parallel} =−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$$ \\ $$
$${V}_{{x}} ={V}_{\bot} \mathrm{cos}\:\phi+{V}_{\parallel} \mathrm{sin}\:\phi \\ $$
$${V}_{{x}} =\left[{eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{cos}\:\phi+\left[−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{sin}\:\phi \\ $$
$${V}_{{x}} ={eu}\:\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi\:\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$
$$\Rightarrow{V}_{{x}} ={u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+{e}\right)\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}\right] \\ $$
$${V}_{{y}} ={V}_{\bot} \mathrm{sin}\:\phi−{V}_{\parallel} \mathrm{cos}\:\phi \\ $$
$${V}_{{y}} =\left[{eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{sin}\:\phi−\left[−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{cos}\:\phi \\ $$
$$\Rightarrow{V}_{{y}} ={u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\left(\mathrm{1}+{e}\right)\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}\right] \\ $$
$$ \\ $$
$$\underline{\boldsymbol{{return}}\:\boldsymbol{{from}}\:\boldsymbol{{B}}\:\boldsymbol{{to}}\:\boldsymbol{{A}}:} \\ $$
$${t}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+{e}\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}}×\frac{\mathrm{1}}{{u}} \\ $$
$$\Rightarrow\frac{{tu}}{{R}}=\xi=\frac{\mathrm{1}+\mathrm{cos}\:\phi}{\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+{e}\right)\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}} \\ $$
$${R}\:\mathrm{sin}\:\phi={t}\left\{{u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta\right]+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{\mathrm{cos}\:\theta}−\frac{{gt}}{\mathrm{2}}\right\} \\ $$
$$\mathrm{sin}\:\phi=\frac{{tu}}{{R}}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{\mathrm{cos}\:\theta}−\frac{\lambda{tu}}{{R}}\right] \\ $$
$$\mathrm{sin}\:\phi=\xi\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\left(\mathrm{1}+{e}\right)\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}−\lambda\xi\right] \\ $$
$${or} \\ $$
$$\mathrm{sin}\:\phi=\xi\left[{e}'\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}'\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}−\lambda\xi\right]\:\:\:…\left({I}\right) \\ $$
$${with} \\ $$
$${e}'=\mathrm{1}+{e} \\ $$
$$\lambda=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$
$$\left.\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right]\:\:\:\:\:\:\:\ast\right) \\ $$
$$\xi=\frac{\mathrm{1}+\mathrm{cos}\:\phi}{{e}'\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda{e}'\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}} \\ $$
$${for}\:{a}\:{given}\:{e}\:\in\left(\mathrm{0},\mathrm{1}\right]\:{we}\:{can}\:{find} \\ $$
$${solution}\left({s}\right)\:{for}\:\phi\:{in}\:\left({I}\right)\:{for}\:{some} \\ $$
$${values}\:{of}\:{parameter}\:\lambda.\:{the}\:{maximum} \\ $$
$${value}\:{of}\:\lambda\:{such}\:{that}\:{a}\:{solution}\:{for}\:\phi \\ $$
$${exists}\:{can}\:{be}\:{determined}\:{numerically}. \\ $$
$${this}\:{corresponds}\:{to}\:{the}\:{minimum} \\ $$
$${speed}\:{the}\:{ball}\:{must}\:{have}: \\ $$
$${u}_{{min}} =\sqrt{\frac{{gR}}{\mathrm{2}\lambda_{{max}} }} \\ $$
$$ \\ $$
$${examples}: \\ $$
$${e}=\mathrm{0}.\mathrm{5}\:\Rightarrow{u}_{{min}} \approx\mathrm{2}.\mathrm{7682}\sqrt{{gR}} \\ $$
$${e}=\mathrm{0}.\mathrm{75}\:\Rightarrow{u}_{{min}} \approx\mathrm{1}.\mathrm{6082}\sqrt{{gR}} \\ $$
$${e}=\mathrm{1}\:\Rightarrow{u}_{{min}} \approx\mathrm{0}.\mathrm{9098}\sqrt{{gR}}\:\:\left(=\:{Q}\mathrm{65589}\right) \\ $$
$$ \\ $$
$$\left.\ast\right) \\ $$
$${with}\:{the}\:{other}\:{solution} \\ $$
$$\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$
$${we}\:{can}\:{get}\:{a}\:{solution}\:{for}\:{the}\:{case} \\ $$
$${that}\:{the}\:{ball}\:{strikes}\:{the}\:{surface} \\ $$
$${upwards},\:{i}.{e}.\:{counterclockwise}. \\ $$
$${but}\:{in}\:{this}\:{case}\:{more}\:{energy}\:{is} \\ $$
$${needed},\:{i}.{e}.\:{u}_{{min}} \:{is}\:{larger}\:{than}\:{with} \\ $$
$$\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$
$${see}\:{examples}\:{with}\:{e}=\mathrm{0}.\mathrm{5}\:{and}\:\mathrm{0}.\mathrm{75}. \\ $$
Commented bymr W last updated on 15/Feb/21
Commented bymr W last updated on 13/Feb/21
Commented bymr W last updated on 13/Feb/21
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Commented bymr W last updated on 12/Feb/21
Commented bymr W last updated on 14/Feb/21
$${following}\:{diagrams}\:{show}\:{different} \\ $$
$${situations}\:{for}\:{e}=\mathrm{0}.\mathrm{75}. \\ $$
$$ \\ $$
$${generally}: \\ $$
$${for}\:{a}\:{given}\:{e}\in\left(\mathrm{0},\mathrm{1}\right]\:{there}\:{exists}\:{an} \\ $$
$${u}_{{min}} .\:{for}\:{any}\:{u}>{u}_{{min}} \:{there}\:{are}\:{four}\: \\ $$
$${possible}\:{ways}\:{how}\:{the}\:{ball}\:{returns} \\ $$
$${back}\:{to}\:{its}\:{starting}\:{position}:\:{two} \\ $$
$${ways}\:{clockwise}\:{and}\:{two}\:{counter}− \\ $$
$${clockwise}. \\ $$
$${situations}\:{for}\:{e}=\mathrm{0}.\mathrm{75}. \\ $$
$$ \\ $$
$${generally}: \\ $$
$${for}\:{a}\:{given}\:{e}\in\left(\mathrm{0},\mathrm{1}\right]\:{there}\:{exists}\:{an} \\ $$
$${u}_{{min}} .\:{for}\:{any}\:{u}>{u}_{{min}} \:{there}\:{are}\:{four}\: \\ $$
$${possible}\:{ways}\:{how}\:{the}\:{ball}\:{returns} \\ $$
$${back}\:{to}\:{its}\:{starting}\:{position}:\:{two} \\ $$
$${ways}\:{clockwise}\:{and}\:{two}\:{counter}− \\ $$
$${clockwise}. \\ $$
Commented bymr W last updated on 15/Feb/21
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Commented bymr W last updated on 14/Feb/21
Commented bymr W last updated on 13/Feb/21
Commented bymr W last updated on 14/Feb/21
Answered by mr W last updated on 14/Feb/21
Commented bymr W last updated on 14/Feb/21
$$\boldsymbol{{background}}\:\boldsymbol{{knowledge}} \\ $$
$$ \\ $$
$${what}\:{happens}\:{when}\:{a}\:{ball}\:{strikes}\:{on} \\ $$
$${a}\:{wall}\:{and}\:{the}\:{restitution}\:{coefficient} \\ $$
$${of}\:{the}\:{collision}\:{is}\:\boldsymbol{{e}}? \\ $$
$$ \\ $$
$${U}={coming}\:{speed} \\ $$
$${V}={leaving}\:{speed} \\ $$
$$ \\ $$
$${U}_{\bot} ={U}\:\mathrm{cos}\:\delta_{\mathrm{1}} \\ $$
$${U}_{\parallel} ={U}\:\mathrm{sin}\:\delta_{\mathrm{1}} \\ $$
$${V}_{\bot} ={V}\:\mathrm{cos}\:\delta_{\mathrm{2}} \\ $$
$${V}_{\parallel} ={V}\:\mathrm{sin}\:\delta_{\mathrm{2}} \\ $$
$$ \\ $$
$${V}_{\bot} ={e}×{U}_{\bot} \:\Rightarrow{V}\:\mathrm{cos}\:\delta_{\mathrm{2}} ={e}×{U}\:\mathrm{cos}\:\delta_{\mathrm{1}} \\ $$
$${V}_{\parallel} ={U}_{\parallel} \:\:\:\Rightarrow\:{V}\:\mathrm{sin}\:\delta_{\mathrm{2}} ={U}\:\mathrm{sin}\:\delta_{\mathrm{1}} \\ $$
$$\mathrm{tan}\:\delta_{\mathrm{2}} =\frac{\mathrm{tan}\:\delta_{\mathrm{1}} }{{e}} \\ $$
$$\delta_{\mathrm{2}} \geqslant\delta_{\mathrm{1}} \\ $$
$$ \\ $$
$${KE}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{mU}^{\mathrm{2}} \\ $$
$${KE}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mU}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\delta_{\mathrm{1}} +{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\delta_{\mathrm{1}} \right) \\ $$
$$\Rightarrow{KE}_{\mathrm{2}} =\left(\mathrm{sin}^{\mathrm{2}} \:\delta_{\mathrm{1}} +{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\delta_{\mathrm{1}} \right){KE}_{\mathrm{1}} \\ $$
$${we}\:{see}\:{KE}_{\mathrm{2}} ={KE}_{\mathrm{1}} \:{only}\:{when}\:{e}=\mathrm{1}, \\ $$
$${otherwise}\:{KE}_{\mathrm{2}} <{KE}_{\mathrm{1}} . \\ $$
$$ \\ $$
$${the}\:{collision}\:{is}\:{called}\:{elastic}\:{if}\:{e}=\mathrm{1}. \\ $$
$$ \\ $$
$${what}\:{happens}\:{when}\:{a}\:{ball}\:{strikes}\:{on} \\ $$
$${a}\:{wall}\:{and}\:{the}\:{restitution}\:{coefficient} \\ $$
$${of}\:{the}\:{collision}\:{is}\:\boldsymbol{{e}}? \\ $$
$$ \\ $$
$${U}={coming}\:{speed} \\ $$
$${V}={leaving}\:{speed} \\ $$
$$ \\ $$
$${U}_{\bot} ={U}\:\mathrm{cos}\:\delta_{\mathrm{1}} \\ $$
$${U}_{\parallel} ={U}\:\mathrm{sin}\:\delta_{\mathrm{1}} \\ $$
$${V}_{\bot} ={V}\:\mathrm{cos}\:\delta_{\mathrm{2}} \\ $$
$${V}_{\parallel} ={V}\:\mathrm{sin}\:\delta_{\mathrm{2}} \\ $$
$$ \\ $$
$${V}_{\bot} ={e}×{U}_{\bot} \:\Rightarrow{V}\:\mathrm{cos}\:\delta_{\mathrm{2}} ={e}×{U}\:\mathrm{cos}\:\delta_{\mathrm{1}} \\ $$
$${V}_{\parallel} ={U}_{\parallel} \:\:\:\Rightarrow\:{V}\:\mathrm{sin}\:\delta_{\mathrm{2}} ={U}\:\mathrm{sin}\:\delta_{\mathrm{1}} \\ $$
$$\mathrm{tan}\:\delta_{\mathrm{2}} =\frac{\mathrm{tan}\:\delta_{\mathrm{1}} }{{e}} \\ $$
$$\delta_{\mathrm{2}} \geqslant\delta_{\mathrm{1}} \\ $$
$$ \\ $$
$${KE}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{mU}^{\mathrm{2}} \\ $$
$${KE}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mU}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\delta_{\mathrm{1}} +{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\delta_{\mathrm{1}} \right) \\ $$
$$\Rightarrow{KE}_{\mathrm{2}} =\left(\mathrm{sin}^{\mathrm{2}} \:\delta_{\mathrm{1}} +{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\delta_{\mathrm{1}} \right){KE}_{\mathrm{1}} \\ $$
$${we}\:{see}\:{KE}_{\mathrm{2}} ={KE}_{\mathrm{1}} \:{only}\:{when}\:{e}=\mathrm{1}, \\ $$
$${otherwise}\:{KE}_{\mathrm{2}} <{KE}_{\mathrm{1}} . \\ $$
$$ \\ $$
$${the}\:{collision}\:{is}\:{called}\:{elastic}\:{if}\:{e}=\mathrm{1}. \\ $$
Commented byAr Brandon last updated on 13/Feb/21
$$ \\ $$
Wow ! 😃 Where did you learn your geometry Sir ? I feel you have this unique skill here. \\n
Wow ! 😃 Where did you learn your geometry Sir ? I feel you have this unique skill here. \\n
Commented byAr Brandon last updated on 14/Feb/21
😃\\n
Commented bymr W last updated on 14/Feb/21
$${the}\:{best}\:{teacher}\:{is}\:{the}\:{own}\:{interest}… \\ $$
Answered by ajfour last updated on 15/Feb/21
Commented byajfour last updated on 15/Feb/21