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Question Number 132102 by mr W last updated on 14/Feb/21
Commented by mr W last updated on 16/Feb/21
a ball is thrown from point A with speed 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 e, find the minimum speed u the ball must have and at which angle 𝛉 the ball should be thrown.
$${a}\:{ball}\:{is}\:{thrown}\:{from}\:{point}\:{A}\:{with} \\ $$$${speed}\:\boldsymbol{{u}}\:{and}\:{strikes}\:{at}\:{a}\:{point}\:{B} \\ $$$${on}\:{the}\:{semispherical}\:{surface}\:{with} \\ $$$${radius}\:{R}\:{and}\:{returns}\:{back}\:{to}\:{point}\:{A}. \\ $$$${if}\:{the}\:{restitution}\:{coefficient}\:{of} \\ $$$${the}\:{collision}\:{is}\:\boldsymbol{{e}},\:{find}\:{the}\:{minimum} \\ $$$${speed}\:\boldsymbol{{u}}\:{the}\:{ball}\:{must}\:{have}\:{and}\:{at} \\ $$$${which}\:{angle}\:\boldsymbol{\theta}\:{the}\:{ball}\:{should}\:{be} \\ $$$${thrown}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 14/Feb/21
You are the world best no one can challenge you!
$$\mathrm{You}\:\mathrm{are}\:\mathrm{the}\:\mathrm{world}\:\mathrm{best}\:\mathrm{no}\:\mathrm{one}\:\mathrm{can}\: \\ $$$$\mathrm{challenge}\:\mathrm{you}! \\ $$
Commented by ajfour last updated on 15/Feb/21
For u_(min) I got φ=30° t_1 =(√((2(√3)R)/g)){1±(√((1/e^2 )−1))} u_(min) ^2 =(((R(√3))/(2t))+((gt)/4))^2 +((3(√3))/2)gR((1/e^2 )−1)
$${For}\:{u}_{{min}} \:\:\:{I}\:\:{got}\:\:\:\phi=\mathrm{30}° \\ $$$${t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}{R}}{{g}}}\left\{\mathrm{1}\pm\sqrt{\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{1}}\right\} \\ $$$${u}_{{min}} ^{\mathrm{2}} =\left(\frac{{R}\sqrt{\mathrm{3}}}{\mathrm{2}{t}}+\frac{{gt}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}{gR}\left(\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$
Commented by mr W last updated on 15/Feb/21
do you mean that u_(min) is always at φ=30° indepentently from e? i can′t confirm this sir.
$${do}\:{you}\:{mean}\:{that}\:{u}_{{min}} \:{is}\:{always}\:{at} \\ $$$$\phi=\mathrm{30}°\:{indepentently}\:{from}\:{e}? \\ $$$${i}\:{can}'{t}\:{confirm}\:{this}\:{sir}. \\ $$
Answered by mr W last updated on 25/Feb/21
Commented by mr W last updated on 12/Feb/21
Commented by mr W last updated on 16/Feb/21
motion from A to B: t=((R(1+cos φ))/(u cos θ)) R sin φ=−u sin θ×((R(1+cos φ))/(u cos θ))+(g/2)×((R^2 (1+cos φ)^2 )/(u^2 cos^2 θ)) ((sin φ)/(1+cos φ))=((gR)/(2u^2 ))×(1+cos φ)(1+tan^2 θ)−tan θ let λ=((gR)/(2u^2 )) ((sin φ)/(1+cos φ))=λ(1+cos φ)(1+tan^2 θ)−tan θ λ(1+cos φ)tan^2 θ−tan θ+((λ(1+cos φ)^2 −sin φ)/(1+cos φ))=0 tan θ=((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ))) ⇒θ=tan^(−1) [((1±(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] at point B: U_x =u cos θ U_y =gt−u sin θ=((gR(1+cos φ))/(u cos θ))−u sin θ U_⊥ =U_x cos φ+U_y sin φ U_⊥ =u cos θcos φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]sin φ ⇒U_⊥ =u cos (φ+θ)+((gR(1+cos φ)sin φ)/(u cos θ)) U_∥ =−U_x sin φ+U_y cos φ U_∥ =−u cos θ sin φ+[((gR(1+cos φ))/(u cos θ))−u sin θ]cos φ ⇒U_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_⊥ =eU_⊥ ⇒V_⊥ =eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ)) V_∥ =U_∥ ⇒V_∥ =−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ)) V_x =V_⊥ cos φ+V_∥ sin φ V_x =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]cos φ+[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]sin φ V_x =eu cos (φ+θ)cos φ+((egR(1+cos φ)sin φ cos φ)/(u cos θ))−u sin (φ+θ)sin φ+((gR(1+cos φ)sin φcos φ)/(u cos θ)) ⇒V_x =u[(1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ))] V_y =V_⊥ sin φ−V_∥ cos φ V_y =[eu cos (φ+θ)+((egR(1+cos φ)sin φ)/(u cos θ))]sin φ−[−u sin (φ+θ)+((gR(1+cos φ)cos φ)/(u cos θ))]cos φ ⇒V_y =u[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e) sin^2 φ−1))/(cos θ))] return from B to A: t=((R(1+cos φ))/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+cos φ)(1+e) sin 2φ)/(cos θ))))×(1/u) ⇒((tu)/R)=ξ=((1+cos φ)/((1+e)cos (φ+θ)cos φ−cos θ+((λ(1+e)(1+cos φ) sin 2φ)/(cos θ)))) R sin φ=t{u[(1+e)cos (φ+θ)sin φ+sin θ]+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((gt)/2)} sin φ=((tu)/R)[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)(e sin^2 φ−cos^2 φ))/(cos θ))−((λtu)/R)] sin φ=ξ[(1+e)cos (φ+θ)sin φ+sin θ+((2λ(1+cos φ)((1+e)sin^2 φ−1))/(cos θ))−λξ] or sin φ=ξ[e′cos (φ+θ) sin φ+sin θ+((2λ(1+cos φ)(e′sin^2 φ−1))/(cos θ))−λξ] ...(I) with e′=1+e λ=((gR)/(2u^2 )) θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] ∗) ξ=((1+cos φ)/(e′cos (φ+θ) cos φ−cos θ+((λe′(1+cos φ) sin 2φ)/(cos θ)))) for a given e ∈(0,1] we can find solution(s) for φ in (I) for some values of parameter λ. the maximum value of λ such that a solution for φ exists can be determined numerically. this corresponds to the minimum speed the ball must have: u_(min) =(√((gR)/(2λ_(max) ))) examples: e=0.5 ⇒u_(min) ≈2.7682(√(gR)) e=0.75 ⇒u_(min) ≈1.6082(√(gR)) e=1 ⇒u_(min) ≈0.9098(√(gR)) (= Q65589) ∗) with the other solution θ=tan^(−1) [((1−(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] 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 θ=tan^(−1) [((1+(√(1+4λ sin φ−4λ^2 (1+cos φ)^2 )))/(2λ(1+cos φ)))] see examples with e=0.5 and 0.75.
$$\underline{\boldsymbol{{motion}}\:\boldsymbol{{from}}\:\boldsymbol{{A}}\:\boldsymbol{{to}}\:\boldsymbol{{B}}:} \\ $$$${t}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta} \\ $$$${R}\:\mathrm{sin}\:\phi=−{u}\:\mathrm{sin}\:\theta×\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}+\frac{{g}}{\mathrm{2}}×\frac{{R}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\frac{\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} }×\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)−\mathrm{tan}\:\theta \\ $$$${let}\:\lambda=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)−\mathrm{tan}\:\theta \\ $$$$\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{tan}^{\mathrm{2}} \:\theta−\mathrm{tan}\:\theta+\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} −\mathrm{sin}\:\phi}{\mathrm{1}+\mathrm{cos}\:\phi}=\mathrm{0} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)} \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$$$ \\ $$$$\underline{\boldsymbol{{at}}\:\boldsymbol{{point}}\:\boldsymbol{{B}}:} \\ $$$${U}_{{x}} ={u}\:\mathrm{cos}\:\theta \\ $$$${U}_{{y}} ={gt}−{u}\:\mathrm{sin}\:\theta=\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta \\ $$$${U}_{\bot} ={U}_{{x}} \mathrm{cos}\:\phi+{U}_{{y}} \mathrm{sin}\:\phi \\ $$$${U}_{\bot} ={u}\:\mathrm{cos}\:\theta\mathrm{cos}\:\phi+\left[\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta\right]\mathrm{sin}\:\phi \\ $$$$\Rightarrow{U}_{\bot} ={u}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$$${U}_{\parallel} =−{U}_{{x}} \mathrm{sin}\:\phi+{U}_{{y}} \mathrm{cos}\:\phi \\ $$$${U}_{\parallel} =−{u}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi+\left[\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\theta\right]\mathrm{cos}\:\phi \\ $$$$\Rightarrow{U}_{\parallel} =−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$$$ \\ $$$${V}_{\bot} ={eU}_{\bot} \\ $$$$\Rightarrow{V}_{\bot} ={eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$$${V}_{\parallel} ={U}_{\parallel} \\ $$$$\Rightarrow{V}_{\parallel} =−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$$$ \\ $$$${V}_{{x}} ={V}_{\bot} \mathrm{cos}\:\phi+{V}_{\parallel} \mathrm{sin}\:\phi \\ $$$${V}_{{x}} =\left[{eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{cos}\:\phi+\left[−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{sin}\:\phi \\ $$$${V}_{{x}} ={eu}\:\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi\:\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{V}_{{x}} ={u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+{e}\right)\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}\right] \\ $$$${V}_{{y}} ={V}_{\bot} \mathrm{sin}\:\phi−{V}_{\parallel} \mathrm{cos}\:\phi \\ $$$${V}_{{y}} =\left[{eu}\:\mathrm{cos}\:\left(\phi+\theta\right)+\frac{{egR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{sin}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{sin}\:\phi−\left[−{u}\:\mathrm{sin}\:\left(\phi+\theta\right)+\frac{{gR}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\mathrm{cos}\:\phi}{{u}\:\mathrm{cos}\:\theta}\right]\mathrm{cos}\:\phi \\ $$$$\Rightarrow{V}_{{y}} ={u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\left(\mathrm{1}+{e}\right)\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}\right] \\ $$$$ \\ $$$$\underline{\boldsymbol{{return}}\:\boldsymbol{{from}}\:\boldsymbol{{B}}\:\boldsymbol{{to}}\:\boldsymbol{{A}}:} \\ $$$${t}=\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\mathrm{1}+{e}\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}}×\frac{\mathrm{1}}{{u}} \\ $$$$\Rightarrow\frac{{tu}}{{R}}=\xi=\frac{\mathrm{1}+\mathrm{cos}\:\phi}{\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda\left(\mathrm{1}+{e}\right)\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}} \\ $$$${R}\:\mathrm{sin}\:\phi={t}\left\{{u}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta\right]+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{\mathrm{cos}\:\theta}−\frac{{gt}}{\mathrm{2}}\right\} \\ $$$$\mathrm{sin}\:\phi=\frac{{tu}}{{R}}\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}\:\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{cos}^{\mathrm{2}} \:\phi\right)}{\mathrm{cos}\:\theta}−\frac{\lambda{tu}}{{R}}\right] \\ $$$$\mathrm{sin}\:\phi=\xi\left[\left(\mathrm{1}+{e}\right)\mathrm{cos}\:\left(\phi+\theta\right)\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left(\left(\mathrm{1}+{e}\right)\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}−\lambda\xi\right] \\ $$$${or} \\ $$$$\mathrm{sin}\:\phi=\xi\left[{e}'\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{sin}\:\phi+\mathrm{sin}\:\theta+\frac{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\left({e}'\mathrm{sin}^{\mathrm{2}} \:\phi−\mathrm{1}\right)}{\mathrm{cos}\:\theta}−\lambda\xi\right]\:\:\:…\left({I}\right) \\ $$$${with} \\ $$$${e}'=\mathrm{1}+{e} \\ $$$$\lambda=\frac{{gR}}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$$$\left.\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right]\:\:\:\:\:\:\:\ast\right) \\ $$$$\xi=\frac{\mathrm{1}+\mathrm{cos}\:\phi}{{e}'\mathrm{cos}\:\left(\phi+\theta\right)\:\mathrm{cos}\:\phi−\mathrm{cos}\:\theta+\frac{\lambda{e}'\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\:\mathrm{sin}\:\mathrm{2}\phi}{\mathrm{cos}\:\theta}} \\ $$$${for}\:{a}\:{given}\:{e}\:\in\left(\mathrm{0},\mathrm{1}\right]\:{we}\:{can}\:{find} \\ $$$${solution}\left({s}\right)\:{for}\:\phi\:{in}\:\left({I}\right)\:{for}\:{some} \\ $$$${values}\:{of}\:{parameter}\:\lambda.\:{the}\:{maximum} \\ $$$${value}\:{of}\:\lambda\:{such}\:{that}\:{a}\:{solution}\:{for}\:\phi \\ $$$${exists}\:{can}\:{be}\:{determined}\:{numerically}. \\ $$$${this}\:{corresponds}\:{to}\:{the}\:{minimum} \\ $$$${speed}\:{the}\:{ball}\:{must}\:{have}: \\ $$$${u}_{{min}} =\sqrt{\frac{{gR}}{\mathrm{2}\lambda_{{max}} }} \\ $$$$ \\ $$$${examples}: \\ $$$${e}=\mathrm{0}.\mathrm{5}\:\Rightarrow{u}_{{min}} \approx\mathrm{2}.\mathrm{7682}\sqrt{{gR}} \\ $$$${e}=\mathrm{0}.\mathrm{75}\:\Rightarrow{u}_{{min}} \approx\mathrm{1}.\mathrm{6082}\sqrt{{gR}} \\ $$$${e}=\mathrm{1}\:\Rightarrow{u}_{{min}} \approx\mathrm{0}.\mathrm{9098}\sqrt{{gR}}\:\:\left(=\:{Q}\mathrm{65589}\right) \\ $$$$ \\ $$$$\left.\ast\right) \\ $$$${with}\:{the}\:{other}\:{solution} \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$$${we}\:{can}\:{get}\:{a}\:{solution}\:{for}\:{the}\:{case} \\ $$$${that}\:{the}\:{ball}\:{strikes}\:{the}\:{surface} \\ $$$${upwards},\:{i}.{e}.\:{counterclockwise}. \\ $$$${but}\:{in}\:{this}\:{case}\:{more}\:{energy}\:{is} \\ $$$${needed},\:{i}.{e}.\:{u}_{{min}} \:{is}\:{larger}\:{than}\:{with} \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \left[\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}\lambda\:\mathrm{sin}\:\phi−\mathrm{4}\lambda^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\phi\right)^{\mathrm{2}} }}{\mathrm{2}\lambda\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}\right] \\ $$$${see}\:{examples}\:{with}\:{e}=\mathrm{0}.\mathrm{5}\:{and}\:\mathrm{0}.\mathrm{75}. \\ $$
Commented by mr W last updated on 15/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 12/Feb/21
Commented by mr W last updated on 12/Feb/21
Commented by mr W last updated on 14/Feb/21
following diagrams show different situations for e=0.75. generally: for a given e∈(0,1] 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.
$${following}\:{diagrams}\:{show}\:{different} \\ $$$${situations}\:{for}\:{e}=\mathrm{0}.\mathrm{75}. \\ $$$$ \\ $$$${generally}: \\ $$$${for}\:{a}\:{given}\:{e}\in\left(\mathrm{0},\mathrm{1}\right]\:{there}\:{exists}\:{an} \\ $$$${u}_{{min}} .\:{for}\:{any}\:{u}>{u}_{{min}} \:{there}\:{are}\:{four}\: \\ $$$${possible}\:{ways}\:{how}\:{the}\:{ball}\:{returns} \\ $$$${back}\:{to}\:{its}\:{starting}\:{position}:\:{two} \\ $$$${ways}\:{clockwise}\:{and}\:{two}\:{counter}− \\ $$$${clockwise}. \\ $$
Commented by mr W last updated on 15/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 14/Feb/21
Commented by mr W last updated on 13/Feb/21
Commented by mr W last updated on 14/Feb/21
Answered by mr W last updated on 14/Feb/21
Commented by mr W last updated on 14/Feb/21
background knowledge what happens when a ball strikes on a wall and the restitution coefficient of the collision is e? U=coming speed V=leaving speed U_⊥ =U cos δ_1 U_∥ =U sin δ_1 V_⊥ =V cos δ_2 V_∥ =V sin δ_2 V_⊥ =e×U_⊥ ⇒V cos δ_2 =e×U cos δ_1 V_∥ =U_∥ ⇒ V sin δ_2 =U sin δ_1 tan δ_2 =((tan δ_1 )/e) δ_2 ≥δ_1 KE_1 =(1/2)mU^2 KE_2 =(1/2)mV^2 =(1/2)mU^2 (sin^2 δ_1 +e^2 cos^2 δ_1 ) ⇒KE_2 =(sin^2 δ_1 +e^2 cos^2 δ_1 )KE_1 we see KE_2 =KE_1 only when e=1, otherwise KE_2 <KE_1 . the collision is called elastic if e=1.
$$\boldsymbol{{background}}\:\boldsymbol{{knowledge}} \\ $$$$ \\ $$$${what}\:{happens}\:{when}\:{a}\:{ball}\:{strikes}\:{on} \\ $$$${a}\:{wall}\:{and}\:{the}\:{restitution}\:{coefficient} \\ $$$${of}\:{the}\:{collision}\:{is}\:\boldsymbol{{e}}? \\ $$$$ \\ $$$${U}={coming}\:{speed} \\ $$$${V}={leaving}\:{speed} \\ $$$$ \\ $$$${U}_{\bot} ={U}\:\mathrm{cos}\:\delta_{\mathrm{1}} \\ $$$${U}_{\parallel} ={U}\:\mathrm{sin}\:\delta_{\mathrm{1}} \\ $$$${V}_{\bot} ={V}\:\mathrm{cos}\:\delta_{\mathrm{2}} \\ $$$${V}_{\parallel} ={V}\:\mathrm{sin}\:\delta_{\mathrm{2}} \\ $$$$ \\ $$$${V}_{\bot} ={e}×{U}_{\bot} \:\Rightarrow{V}\:\mathrm{cos}\:\delta_{\mathrm{2}} ={e}×{U}\:\mathrm{cos}\:\delta_{\mathrm{1}} \\ $$$${V}_{\parallel} ={U}_{\parallel} \:\:\:\Rightarrow\:{V}\:\mathrm{sin}\:\delta_{\mathrm{2}} ={U}\:\mathrm{sin}\:\delta_{\mathrm{1}} \\ $$$$\mathrm{tan}\:\delta_{\mathrm{2}} =\frac{\mathrm{tan}\:\delta_{\mathrm{1}} }{{e}} \\ $$$$\delta_{\mathrm{2}} \geqslant\delta_{\mathrm{1}} \\ $$$$ \\ $$$${KE}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{mU}^{\mathrm{2}} \\ $$$${KE}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mU}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\delta_{\mathrm{1}} +{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\delta_{\mathrm{1}} \right) \\ $$$$\Rightarrow{KE}_{\mathrm{2}} =\left(\mathrm{sin}^{\mathrm{2}} \:\delta_{\mathrm{1}} +{e}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\delta_{\mathrm{1}} \right){KE}_{\mathrm{1}} \\ $$$${we}\:{see}\:{KE}_{\mathrm{2}} ={KE}_{\mathrm{1}} \:{only}\:{when}\:{e}=\mathrm{1}, \\ $$$${otherwise}\:{KE}_{\mathrm{2}} <{KE}_{\mathrm{1}} . \\ $$$$ \\ $$$${the}\:{collision}\:{is}\:{called}\:{elastic}\:{if}\:{e}=\mathrm{1}. \\ $$
Commented by Ar Brandon last updated on 13/Feb/21
Wow ! 😃 Where did you learn your geometry Sir ? I feel you have this unique skill here.
$$ \\ $$Wow ! 😃 Where did you learn your geometry Sir ? I feel you have this unique skill here.
Commented by Ar Brandon last updated on 14/Feb/21
😃
😃
Commented by mr W last updated on 14/Feb/21
the best teacher is the own interest...
$${the}\:{best}\:{teacher}\:{is}\:{the}\:{own}\:{interest}… \\ $$
Answered by ajfour last updated on 15/Feb/21
Commented by ajfour last updated on 15/Feb/21
First just assuming (without proof) that u is u_(min) when least energy is lost upon collision. Let just before collision velocity v_1 ^� =−qi^� −pj^� and just after v_2 ^� =−qi^� +epj^� −△K =(p^2 /2)(1−e^2 ) This is minimum if p is minimum. But ball has to reach back ⇒ (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ) say 2φ=θ , and f(θ)=sin θ(1+cos θ) f ′(θ)=cos θ+cos^2 θ−sin^2 θ =2cos^2 θ+cos θ−1 =0 ⇒ cos 2φ=cos θ=−(1/4)±(√((1/(16))+(1/2))) cos 2φ=cos θ=(1/2) ⇒ φ=30° Now from (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ) ⇒ p_(min) ^2 =((2/e^2 ))(((g(√3))/2))(((3R)/2)) p_(min) =(√((3(√3)gR)/(2e^2 ))) u_(y,min) ^2 =p_(min) ^2 −2(gsin 2φ)R(1+cos 2φ) =((3(√3)gR)/(2e^2 ))−((3(√3)gR)/2) u_(y,min) =(√((3(√3)gR(1−e^2 ))/(2e^2 ))) Rsin 2φ=u_x t_1 −(((gcos 2φ)/2))t_1 ^2 u_x =((R(√3))/(2t_1 ))+((gt_1 )/4) t_1 =((p_(min) −u_(y,min) )/(gsin 2φ))=2(((p_(min) −u_(y,min) )/(g(√3)))) t_1 =(2/(g(√3)))((√((3(√3)gR)/(2e^2 ))) )(1±(√(1−e^2 )) ) u_(min) ^2 =(u_x ^2 +u_y ^2 ) with φ=30° =(((R(√3))/(2t_1 ))+((gt_1 )/4))^2 +((3(√3)gR(1−e^2 ))/(2e^2 ))
$${First}\:{just}\:{assuming}\:\left({without}\right. \\ $$$$\left.{proof}\right)\:{that}\:{u}\:{is}\:{u}_{{min}} \:{when} \\ $$$${least}\:{energy}\:{is}\:{lost}\:{upon}\:{collision}. \\ $$$${Let}\:{just}\:{before}\:{collision}\: \\ $$$${velocity}\:\:\:\bar {{v}}_{\mathrm{1}} =−{q}\hat {{i}}−{p}\hat {{j}} \\ $$$${and}\:{just}\:{after}\:\:\bar {{v}}_{\mathrm{2}} =−{q}\hat {{i}}+{ep}\hat {{j}} \\ $$$$−\bigtriangleup{K}\:=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}−{e}^{\mathrm{2}} \right) \\ $$$${This}\:{is}\:{minimum}\:{if}\:{p}\:{is} \\ $$$${minimum}. \\ $$$${But}\:{ball}\:{has}\:{to}\:{reach}\:{back}\:\Rightarrow \\ $$$$\left({e}^{\mathrm{2}} {p}^{\mathrm{2}} \right)_{{min}} =\mathrm{2}\left({g}\mathrm{sin}\:\mathrm{2}\phi\right){R}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\phi\right) \\ $$$${say}\:\:\mathrm{2}\phi=\theta\:,\:{and} \\ $$$${f}\left(\theta\right)=\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$${f}\:'\left(\theta\right)=\mathrm{cos}\:\theta+\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2cos}\:^{\mathrm{2}} \theta+\mathrm{cos}\:\theta−\mathrm{1}\:=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\mathrm{cos}\:\mathrm{2}\phi=\mathrm{cos}\:\theta=−\frac{\mathrm{1}}{\mathrm{4}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{cos}\:\mathrm{2}\phi=\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\phi=\mathrm{30}°\:\:\:\:{Now}\:{from} \\ $$$$\left({e}^{\mathrm{2}} {p}^{\mathrm{2}} \right)_{{min}} =\mathrm{2}\left({g}\mathrm{sin}\:\mathrm{2}\phi\right){R}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\phi\right) \\ $$$$\Rightarrow\:\:{p}_{{min}} ^{\mathrm{2}} =\left(\frac{\mathrm{2}}{{e}^{\mathrm{2}} }\right)\left(\frac{{g}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}{R}}{\mathrm{2}}\right) \\ $$$${p}_{{min}} =\sqrt{\frac{\mathrm{3}\sqrt{\mathrm{3}}{gR}}{\mathrm{2}{e}^{\mathrm{2}} }} \\ $$$${u}_{{y},{min}} ^{\mathrm{2}} ={p}_{{min}} ^{\mathrm{2}} −\mathrm{2}\left({g}\mathrm{sin}\:\mathrm{2}\phi\right){R}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\phi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}\sqrt{\mathrm{3}}{gR}}{\mathrm{2}{e}^{\mathrm{2}} }−\frac{\mathrm{3}\sqrt{\mathrm{3}}{gR}}{\mathrm{2}} \\ $$$${u}_{{y},{min}} =\sqrt{\frac{\mathrm{3}\sqrt{\mathrm{3}}{gR}\left(\mathrm{1}−{e}^{\mathrm{2}} \right)}{\mathrm{2}{e}^{\mathrm{2}} }} \\ $$$${R}\mathrm{sin}\:\mathrm{2}\phi={u}_{{x}} {t}_{\mathrm{1}} −\left(\frac{{g}\mathrm{cos}\:\mathrm{2}\phi}{\mathrm{2}}\right){t}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${u}_{{x}} =\frac{{R}\sqrt{\mathrm{3}}}{\mathrm{2}{t}_{\mathrm{1}} }+\frac{{gt}_{\mathrm{1}} }{\mathrm{4}} \\ $$$${t}_{\mathrm{1}} =\frac{{p}_{{min}} −{u}_{{y},{min}} }{{g}\mathrm{sin}\:\mathrm{2}\phi}=\mathrm{2}\left(\frac{{p}_{{min}} −{u}_{{y},{min}} }{{g}\sqrt{\mathrm{3}}}\right) \\ $$$${t}_{\mathrm{1}} =\frac{\mathrm{2}}{{g}\sqrt{\mathrm{3}}}\left(\sqrt{\frac{\mathrm{3}\sqrt{\mathrm{3}}{gR}}{\mathrm{2}{e}^{\mathrm{2}} }}\:\right)\left(\mathrm{1}\pm\sqrt{\mathrm{1}−{e}^{\mathrm{2}} }\:\right) \\ $$$${u}_{{min}} ^{\mathrm{2}} =\left({u}_{{x}} ^{\mathrm{2}} +{u}_{{y}} ^{\mathrm{2}} \right)\:\:\:\:{with}\:\phi=\mathrm{30}° \\ $$$$\:\:=\left(\frac{{R}\sqrt{\mathrm{3}}}{\mathrm{2}{t}_{\mathrm{1}} }+\frac{{gt}_{\mathrm{1}} }{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{3}\sqrt{\mathrm{3}}{gR}\left(\mathrm{1}−{e}^{\mathrm{2}} \right)}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 15/Feb/21
Please review it Sir, it dont seem to taly with your solution and hence with the original question, but yet i think it is a nice question, and its answer as i have derived, did have a chance (i suspected) to yield the same answer...
$${Please}\:{review}\:{it}\:{Sir},\:{it}\:{dont} \\ $$$${seem}\:{to}\:{taly}\:{with}\:{your}\:{solution} \\ $$$${and}\:{hence}\:{with}\:{the}\:{original} \\ $$$${question},\:{but}\:{yet}\:{i}\:{think}\:{it}\:{is} \\ $$$${a}\:{nice}\:{question},\:{and}\:{its}\:{answer} \\ $$$${as}\:{i}\:{have}\:{derived},\:{did}\:{have}\:{a} \\ $$$${chance}\:\left({i}\:{suspected}\right)\:{to}\:{yield}\: \\ $$$${the}\:{same}\:{answer}… \\ $$
Commented by mr W last updated on 15/Feb/21
let me some time to follow your working in detail. i think you interpreted the question differently as i did. the question should ask with which smallest speed one can throw the ball into the bowl so that it can return back to the start point. e.g. if e=0.75 the smallest speed is 1.6082(√(gR)) when the ball follows following track:
$${let}\:{me}\:{some}\:{time}\:{to}\:{follow}\:{your} \\ $$$${working}\:{in}\:{detail}.\:{i}\:{think}\:{you} \\ $$$${interpreted}\:{the}\:{question}\:{differently} \\ $$$${as}\:{i}\:{did}.\:\:{the}\:{question}\:{should}\:{ask} \\ $$$${with}\:{which}\:{smallest}\:{speed}\:{one}\:{can} \\ $$$${throw}\:{the}\:{ball}\:{into}\:{the}\:{bowl}\:{so}\:{that} \\ $$$${it}\:{can}\:{return}\:{back}\:{to}\:{the}\:{start}\:{point}. \\ $$$${e}.{g}.\:\:{if}\:{e}=\mathrm{0}.\mathrm{75}\:{the}\:{smallest}\:{speed}\:{is} \\ $$$$\mathrm{1}.\mathrm{6082}\sqrt{{gR}}\:{when}\:{the}\:{ball}\:{follows} \\ $$$${following}\:{track}: \\ $$
Commented by mr W last updated on 15/Feb/21
Commented by mr W last updated on 15/Feb/21
for any other cases, i think also your result, the ball needs more speed than 1.6082(√(gR)). if e=1, the smallest speed should be 0.9098(√(gR)).
$${for}\:{any}\:{other}\:{cases},\:{i}\:{think}\:{also}\:{your} \\ $$$${result},\:{the}\:{ball}\:{needs}\:{more}\:{speed} \\ $$$${than}\:\mathrm{1}.\mathrm{6082}\sqrt{{gR}}. \\ $$$${if}\:{e}=\mathrm{1},\:{the}\:{smallest}\:{speed}\:{should}\:{be} \\ $$$$\mathrm{0}.\mathrm{9098}\sqrt{{gR}}. \\ $$
Commented by mr W last updated on 15/Feb/21
acc. to your result, for e=1, t_1 =(√((2(√3)R)/g)) u_(min) =(((√3)/2)+((√(2(√3)))/4))(√(gR))≈1.3313(√(gR)) >0.9098(√(gR)) (what i got)
$${acc}.\:{to}\:{your}\:{result},\:{for}\:{e}=\mathrm{1}, \\ $$$${t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}{R}}{{g}}}\: \\ $$$${u}_{{min}} \:=\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}{\mathrm{4}}\right)\sqrt{{gR}}\approx\mathrm{1}.\mathrm{3313}\sqrt{{gR}} \\ $$$$>\mathrm{0}.\mathrm{9098}\sqrt{{gR}}\:\left({what}\:{i}\:{got}\right) \\ $$
Commented by mr W last updated on 15/Feb/21
i think the least energy loss during collision doesn′t mean the minimum total energy the ball needs.
$${i}\:{think}\:{the}\:{least}\:{energy}\:{loss}\:{during} \\ $$$${collision}\:{doesn}'{t}\:{mean}\:{the}\:{minimum} \\ $$$${total}\:{energy}\:{the}\:{ball}\:{needs}. \\ $$
Commented by ajfour last updated on 16/Feb/21
I know you are right Sir.
$${I}\:{know}\:{you}\:{are}\:{right}\:{Sir}. \\ $$
Commented by mr W last updated on 16/Feb/21
i think i have found the logic fault in your solution: with But ball has to reach back ⇒ (e^2 p^2 )_(min) =2(gsin 2φ)R(1+cos 2φ) we only ensure that the ball can reach back its height (position in y− direction), the ball may not come back to its start point (also in x− direction), so it is something like this:
$${i}\:{think}\:{i}\:{have}\:{found}\:{the}\:{logic}\:{fault} \\ $$$${in}\:{your}\:{solution}: \\ $$$${with} \\ $$$${But}\:{ball}\:{has}\:{to}\:{reach}\:{back}\:\Rightarrow \\ $$$$\left({e}^{\mathrm{2}} {p}^{\mathrm{2}} \right)_{{min}} =\mathrm{2}\left({g}\mathrm{sin}\:\mathrm{2}\phi\right){R}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\phi\right) \\ $$$${we}\:{only}\:{ensure}\:{that}\:{the}\:{ball}\:{can}\:{reach} \\ $$$${back}\:{its}\:{height}\:\left({position}\:{in}\:{y}−\right. \\ $$$$\left.{direction}\right),\:{the}\:{ball}\:{may}\:{not}\:{come} \\ $$$${back}\:{to}\:{its}\:{start}\:{point}\:\left({also}\:{in}\:{x}−\right. \\ $$$$\left.{direction}\right),\:{so}\:{it}\:{is}\:{something}\:{like} \\ $$$${this}: \\ $$
Commented by mr W last updated on 16/Feb/21

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