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Question-195870

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Question Number 195870 by mr W last updated on 12/Aug/23
Commented by mr W last updated on 12/Aug/23
find the minimum speed with which the ball is launced from point A such that it can return to this point again after rebounded at point B. assume the coefficient of restitution is e.
$${find}\:{the}\:{minimum}\:{speed}\:{with}\:{which} \\ $$$${the}\:{ball}\:{is}\:{launced}\:{from}\:{point}\:{A}\:{such} \\ $$$${that}\:{it}\:{can}\:{return}\:{to}\:{this}\:{point}\:{again} \\ $$$${after}\:{rebounded}\:{at}\:{point}\:{B}. \\ $$$${assume}\:{the}\:{coefficient}\:{of}\:{restitution} \\ $$$${is}\:{e}.\: \\ $$
Commented by liuxinnan last updated on 12/Aug/23
it as if has not the min v_0 but it has a assured angle between the v_0 and the incline
$${it}\:{as}\:{if}\:\:{has}\:{not}\:{the}\:{min}\:{v}_{\mathrm{0}} \\ $$$${but}\:{it}\:{has}\:{a}\:{assured}\:{angle}\: \\ $$$${between}\:{the}\:{v}_{\mathrm{0}} \:\:{and}\:{the}\:{incline} \\ $$
Commented by mahdipoor last updated on 13/Aug/23
It seems that the question does not haven minimum speed ,instead ,i find relationship between tangential and vertical speed (relative to the surface): (V_t /V_v )=(v/u)=((e^2 +2e+1)/(e+1))tanθ=(e+1)tanθ
$$\mathrm{It}\:\mathrm{seems}\:\mathrm{that}\:\mathrm{the}\:\mathrm{question}\:\mathrm{does}\:\mathrm{not}\:\mathrm{haven} \\ $$$$\mathrm{minimum}\:\mathrm{speed}\:,\mathrm{instead}\:,\mathrm{i}\:\mathrm{find}\:\mathrm{relationship}\: \\ $$$$\mathrm{between}\:\mathrm{tangential}\:\mathrm{and}\:\mathrm{vertical}\:\mathrm{speed}\: \\ $$$$\left(\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{surface}\right): \\ $$$$\frac{{V}_{{t}} }{{V}_{{v}} }=\frac{{v}}{{u}}=\frac{{e}^{\mathrm{2}} +\mathrm{2}{e}+\mathrm{1}}{{e}+\mathrm{1}}{tan}\theta=\left({e}+\mathrm{1}\right){tan}\theta \\ $$
Commented by mr W last updated on 13/Aug/23
it′s correct. there is no minimum for speed u, since the rebound point B is not fixed. but i got tan ϕ=(1/((e+1)tan θ))
$${it}'{s}\:{correct}.\:{there}\:{is}\:{no}\:{minimum}\:{for} \\ $$$${speed}\:{u},\:{since}\:{the}\:{rebound}\:{point}\:{B} \\ $$$${is}\:{not}\:{fixed}. \\ $$$${but}\:{i}\:{got}\:\mathrm{tan}\:\varphi=\frac{\mathrm{1}}{\left({e}+\mathrm{1}\right)\mathrm{tan}\:\theta} \\ $$
Answered by mr W last updated on 13/Aug/23
Commented by mr W last updated on 13/Aug/23
motion from A to B: 0=u sin ϕ t−((g cos θ)/2)t^2 ⇒t=((2u sin ϕ)/(g cos θ)) s=u cos ϕ t−((g sin θ)/2)t^2 s=u cos ϕ×((2u sin ϕ)/(g cos θ))−((g sin θ)/2)×(((2u sin ϕ)/(g cos θ)))^2 s=(u^2 /(g cos θ))(sin 2ϕ−2 sin^2 ϕ tan θ) U_x =u cos ϕ−g sin θ ×((2u sin ϕ)/(g cos θ)) =u(cos ϕ−2 tan θ sin ϕ) U_y =u sin ϕ−g cos θ×((2u sin ϕ)/(g cos θ)) =−u sin ϕ V_x =U_x =u(cos ϕ−2 tan θ sin ϕ) V_y =−eU_y =eu sin ϕ motion from B to A: t=((2eu sin ϕ)/(g cos θ)) −s=u(cos ϕ−2 tan θ sin ϕ)t−((g sin θ)/2)t^2 −s=u(cos ϕ−2 tan θ sin ϕ)×((2eu sin ϕ)/(g cos θ))−((g sin θ)/2)×(((2eu sin ϕ)/(g cos θ)))^2 −(u^2 /(g cos θ))[sin 2ϕ−2 sin^2 ϕ tan θ]=u(cos ϕ−2 tan θ sin ϕ)×((2eu sin ϕ)/(g cos θ))−((g sin θ)/2)×(((2eu sin ϕ)/(g cos θ)))^2 −sin 2ϕ+2 sin^2 ϕ tan θ=2e sin ϕ (cos ϕ−2 tan θ sin ϕ)−2e^2 tan θ sin^2 ϕ (e+1) sin ϕ cos ϕ=(e+1)^2 tan θ sin^2 ϕ ⇒tan ϕ=(1/((e+1) tan θ)) s=(u^2 /(g cos θ))(sin 2ϕ−2 sin^2 ϕ tan θ) ⇒(u^2 /(gs))=[(e+1)^2 +(1/(tan^2 θ))]((sin θ)/(2e))
$${motion}\:{from}\:{A}\:{to}\:{B}: \\ $$$$\mathrm{0}={u}\:\mathrm{sin}\:\varphi\:{t}−\frac{{g}\:\mathrm{cos}\:\theta}{\mathrm{2}}{t}^{\mathrm{2}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}{u}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta} \\ $$$${s}={u}\:\mathrm{cos}\:\varphi\:{t}−\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{2}}{t}^{\mathrm{2}} \\ $$$${s}={u}\:\mathrm{cos}\:\varphi×\frac{\mathrm{2}{u}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta}−\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{2}}×\left(\frac{\mathrm{2}{u}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} \\ $$$${s}=\frac{{u}^{\mathrm{2}} }{{g}\:\mathrm{cos}\:\theta}\left(\mathrm{sin}\:\mathrm{2}\varphi−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\varphi\:\mathrm{tan}\:\theta\right) \\ $$$${U}_{{x}} ={u}\:\mathrm{cos}\:\varphi−{g}\:\mathrm{sin}\:\theta\:×\frac{\mathrm{2}{u}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:\:={u}\left(\mathrm{cos}\:\varphi−\mathrm{2}\:\mathrm{tan}\:\theta\:\mathrm{sin}\:\varphi\right)\: \\ $$$${U}_{{y}} ={u}\:\mathrm{sin}\:\varphi−{g}\:\mathrm{cos}\:\theta×\frac{\mathrm{2}{u}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:\:=−{u}\:\mathrm{sin}\:\varphi \\ $$$${V}_{{x}} ={U}_{{x}} ={u}\left(\mathrm{cos}\:\varphi−\mathrm{2}\:\mathrm{tan}\:\theta\:\mathrm{sin}\:\varphi\right)\: \\ $$$${V}_{{y}} =−{eU}_{{y}} ={eu}\:\mathrm{sin}\:\varphi \\ $$$${motion}\:{from}\:{B}\:{to}\:{A}: \\ $$$${t}=\frac{\mathrm{2}{eu}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta} \\ $$$$−{s}={u}\left(\mathrm{cos}\:\varphi−\mathrm{2}\:\mathrm{tan}\:\theta\:\mathrm{sin}\:\varphi\right){t}−\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{2}}{t}^{\mathrm{2}} \\ $$$$−{s}={u}\left(\mathrm{cos}\:\varphi−\mathrm{2}\:\mathrm{tan}\:\theta\:\mathrm{sin}\:\varphi\right)×\frac{\mathrm{2}{eu}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta}−\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{2}}×\left(\frac{\mathrm{2}{eu}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} \\ $$$$−\frac{{u}^{\mathrm{2}} }{{g}\:\mathrm{cos}\:\theta}\left[\mathrm{sin}\:\mathrm{2}\varphi−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\varphi\:\mathrm{tan}\:\theta\right]={u}\left(\mathrm{cos}\:\varphi−\mathrm{2}\:\mathrm{tan}\:\theta\:\mathrm{sin}\:\varphi\right)×\frac{\mathrm{2}{eu}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta}−\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{2}}×\left(\frac{\mathrm{2}{eu}\:\mathrm{sin}\:\varphi}{{g}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} \\ $$$$−\mathrm{sin}\:\mathrm{2}\varphi+\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\varphi\:\mathrm{tan}\:\theta=\mathrm{2}{e}\:\mathrm{sin}\:\varphi\:\left(\mathrm{cos}\:\varphi−\mathrm{2}\:\mathrm{tan}\:\theta\:\mathrm{sin}\:\varphi\right)−\mathrm{2}{e}^{\mathrm{2}} \:\mathrm{tan}\:\theta\:\mathrm{sin}^{\mathrm{2}} \:\varphi \\ $$$$\left({e}+\mathrm{1}\right)\:\mathrm{sin}\:\varphi\:\mathrm{cos}\:\varphi=\left({e}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{tan}\:\theta\:\mathrm{sin}^{\mathrm{2}} \:\varphi \\ $$$$\Rightarrow\mathrm{tan}\:\varphi=\frac{\mathrm{1}}{\left({e}+\mathrm{1}\right)\:\mathrm{tan}\:\theta} \\ $$$${s}=\frac{{u}^{\mathrm{2}} }{{g}\:\mathrm{cos}\:\theta}\left(\mathrm{sin}\:\mathrm{2}\varphi−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\varphi\:\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{{gs}}=\left[\left({e}+\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\theta}\right]\frac{\mathrm{sin}\:\theta}{\mathrm{2}{e}} \\ $$
Commented by mr W last updated on 13/Aug/23
Commented by mr W last updated on 13/Aug/23

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